Math 140. Paul Dawkins

Transcription

1 Math 40 Paul Dawkis

2 Math 40 Table of Cotets Itegrals... Itroductio... Idefiite Itegrals... 5 Computig Idefiite Itegrals... Substitutio Rule for Idefiite Itegrals... More Substitutio Rule... 5 Area Problem The Defiitio of the Defiite Itegral Computig Defiite Itegrals Substitutio Rule for Defiite Itegrals Applicatios of Itegrals... 9 Itroductio... 9 Average Fuctio Value... 9 Area Betwee Curves Volumes of Solids of Revolutio / Method of Rigs Volumes of Solids of Revolutio / Method of Cyliders... 6 More Volume Problems... 4 Extras... 5 Itroductio... 5 Proof of Various Itegral Facts/Formulas/Properties... 6 Area ad Volume Formulas Types of Ifiity... 5 Summatio Notatio Costats of Itegratio Itegratio Techiques... 6 Itroductio... 6 Itegratio by Parts Itegrals Ivolvig Trig Fuctios Trig Substitutios Partial Fractios Itegrals Ivolvig Roots Itegrals Ivolvig Quadratics... 0 Usig Itegral Tables... 9 Itegratio Strategy... Improper Itegrals... 0 Compariso Test for Improper Itegrals... 8 Polar Coordiates Itroductio Polar Coordiates Tagets with Polar Coordiates Sequeces ad Series Itroductio Sequeces... 6 More o Sequeces... 7 Series The Basics Series Covergece/Divergece Series Special Series Vectors... 0 Itroductio... 0 Vectors The Basics Vector Arithmetic Dot Product... 5 Cross Product Paul Dawkis i

3 Math 40 Three Dimesioal Space... 0 Itroductio... 0 The -D Coordiate System... Equatios of Lies... 8 Equatios of Plaes Quadric Surfaces Paul Dawkis ii

4 Math 40 Itegrals Itroductio I this chapter we will be lookig at itegrals. Itegrals are the third ad fial major topic that will be covered i this class. As with derivatives this chapter will be devoted almost exclusively to fidig ad computig itegrals. Applicatios will be give i the followig chapter. There are really two types of itegrals that we ll be lookig at i this chapter : Idefiite Itegrals ad Defiite Itegrals. The first half of this chapter is devoted to idefiite itegrals ad the last half is devoted to defiite itegrals. As we will see i the last half of the chapter if we do t kow idefiite itegrals we will ot be able to do defiite itegrals. Here is a quick listig of the material that is i this chapter. Idefiite Itegrals I this sectio we will start with the defiitio of idefiite itegral. This sectio will be devoted mostly to the defiitio ad properties of idefiite itegrals. Computig Idefiite Itegrals I this sectio we will compute some idefiite itegrals ad take a look at a quick applicatio of idefiite itegrals. Substitutio Rule for Idefiite Itegrals Here we will look at the Substitutio Rule as it applies to idefiite itegrals. May of the itegrals that we ll be doig later o i the course ad i later courses will require use of the substitutio rule. More Substitutio Rule Eve more substitutio rule problems. Area Problem I this sectio we start off with the motivatio for defiite itegrals ad give oe of the iterpretatios of defiite itegrals. Defiitio of the Defiite Itegral We will formally defie the defiite itegral i this sectio ad give may of its properties. We will also take a look at the first part of the Fudametal Theorem of Calculus. Computig Defiite Itegrals We will take a look at the secod part of the Fudametal Theorem of Calculus i this sectio ad start to compute defiite itegrals. Substitutio Rule for Defiite Itegrals I this sectio we will revisit the substitutio rule as it applies to defiite itegrals. 007 Paul Dawkis

5 Math Paul Dawkis 4

6 Math 40 Idefiite Itegrals I the past two chapters we ve bee give a fuctio, f ( x ), ad askig what the derivative of this fuctio was. Startig with this sectio we are ot goig to tur thigs aroud. We ow wat to ask what fuctio we differetiated to get the fuctio f ( x ). Let s take a quick look at a example to get us started. Example What fuctio did we differetiate to get the followig fuctio. f x = x + x- 9 ( ) 4 Solutio Let s actually start by gettig the derivative of this fuctio to help us see how we re goig to have to approach this problem. The derivative of this fuctio is, ( ) f x = 4x + The poit of this was to remid us of how differetiatio works. Whe differetiatig powers of x we multiply the term by the origial expoet ad the drop the expoet by oe. Now, let s go back ad work the problem. I fact let s just start with the first term. We got x 4 by differetiatig a fuctio ad sice we drop the expoet by oe it looks like we must have differetiated x 5. However, if we had differetiated x 5 we would have 5x 4 ad we do t have a 5 i frot our first term, so the 5 eeds to cacel out after we ve differetiated. It looks the like 5 we would have to differetiate 5 x i order to get x4. Likewise for the secod term, i order to get x after differetiatig we would have to differetiate x. Agai, the fractio is there to cacel out the we pick up i the differetiatio. The third term is just a costat ad we kow that if we differetiate x we get. So, it looks like we had to differetiate -9x to get the last term. Puttig all of this together gives the followig fuctio, 5 F( x) = x + x - 9x 5 Our aswer is easy eough to check. Simply differetiate F( x ). 4 F ( x) = x + x- 9= f ( x) 007 Paul Dawkis 5

7 Math 40 So, it looks like we got the correct fuctio. Or did we? We kow that the derivative of a costat is zero ad so ay of the followig will also give f ( x ) upo differetiatig. ( ) 5 F x x x x ( ) 5 F x x x x ( ) = = = F x x x x etc. I fact, ay fuctio of the form, 5 F( x) = x + x - 9 x+ c, c is a costat 5 will give f ( x ) upo differetiatig. There were two poits to this last example. The first poit was to get you thikig about how to do these problems. It is importat iitially to remember that we are really just askig what we differetiated to get the give fuctio. The other poit is to recogize that there are actually a ifiite umber of fuctios that we could use ad they will all differ by a costat. Now that we ve worked a example let s get some of the defiitios ad termiology out of the way. Defiitios Give a fuctio, f ( x ), a ati-derivative of f ( x ) is ay fuctio F( x ) such that F ( x) = f ( x) If F( x ) is ay ati-derivative of f ( x ) the the most geeral ati-derivative of f ( ) a idefiite itegral ad deoted, f ( x) dx= F( x) + c, c is ay costat x is called I this defiitio the f x is called the itegrad, x is called the itegratio variable ad the c is called the costat of itegratio. is called the itegral symbol, ( ) 007 Paul Dawkis 6

8 Math 40 Note that ofte we will just say itegral istead of idefiite itegral (or defiite itegral for that matter whe we get to those). It will be clear from the cotext of the problem that we are talkig about a idefiite itegral (or defiite itegral). The process of fidig the idefiite itegral is called itegratio or itegratig f(x). If we eed to be specific about the itegratio variable we will say that we are itegratig f(x) with respect to x. Let s rework the first problem i light of the ew termiology. Example Evaluate the followig idefiite itegral. 4 x + x-9dx Solutio Sice this is really askig for the most geeral ati-derivative we just eed to reuse the fial aswer from the first example. The idefiite itegral is, x + x- dx= x + x - x+ c A couple of warigs are ow i order. Oe of the more commo mistakes that studets make with itegrals (both idefiite ad defiite) is to drop the dx at the ed of the itegral. This is required! Thik of the itegral sig ad the dx as a set of parethesis. You already kow ad are probably quite comfortable with the idea that every time you ope a parethesis you must close it. With itegrals, thik of the itegral sig as a ope parethesis ad the dx as a close parethesis. If you drop the dx it wo t be clear where the itegrad eds. Cosider the followig variatios of the above example. 4 5 x + x- 9dx= x + x - 9x+ c x + xdx- 9= x + x + c x dx+ x- 9= x + c+ x-9 5 You oly itegrate what is betwee the itegral sig ad the dx. Each of the above itegrals ed i a differet place ad so we get differet aswers because we itegrate a differet umber of terms each time. I the secod itegral the -9 is outside the itegral ad so is left aloe ad ot itegrated. Likewise, i the third itegral the x - 9 is outside the itegral ad so is left aloe. 007 Paul Dawkis 7

9 Math 40 Kowig which terms to itegrate is ot the oly reaso for writig the dx dow. I the Substitutio Rule sectio we will actually be workig with the dx i the problem ad if we are t i the habit of writig it dow it will be easy to forget about it ad the we will get the wrog aswer at that stage. The moral of this is to make sure ad put i the dx! At this stage it may seem like a silly thig to do, but it just eeds to be there, if for o other reaso tha kowig where the itegral stops. O a side ote, the dx otatio should seem a little familiar to you. We saw thigs like this a couple of sectios ago. We called the dx a differetial i that sectio ad yes that is exactly what it is. The dx that eds the itegral is othig more tha a differetial. The ext topic that we should discuss here is the itegratio variable used i the itegral. Actually there is t really a lot to discuss here other tha to ote that the itegratio variable does t really matter. For istace, x + x- dx= x + x - x+ c 4 5 t + t- 9dt = t + t - 9t+ c 4 5 w + w- dw= w + w - w+ c Chagig the itegratio variable i the itegral simply chages the variable i the aswer. It is importat to otice however that whe we chage the itegratio variable i the itegral we also chaged the differetial (dx, dt, or dw) to match the ew variable. This is more importat that we might realize at this poit. Aother use of the differetial at the ed of itegral is to tell us what variable we are itegratig with respect to. At this stage that may seem uimportat sice most of the itegrals that we re goig to be workig with here will oly ivolve a sigle variable. However, if you are o a degree track that will take you ito multi-variable calculus this will be very importat at that stage sice there will be more tha oe variable i the problem. You eed to get ito the habit of writig the correct differetial at the ed of the itegral so whe it becomes importat i those classes you will already be i the habit of writig it dow. To see why this is importat take a look at the followig two itegrals. xdx tdx The first itegral is simple eough. xdx= x + c 007 Paul Dawkis 8

10 Math 40 The secod itegral is also fairly simple, but we eed to be careful. The dx tells us that we are itegratig x s. That meas that we oly itegrate x s that are i the itegrad ad all other variables i the itegrad are cosidered to be costats. The secod itegral is the, tdx= tx+ c So, it may seem silly to always put i the dx, but it is a vital bit of otatio that ca cause us to get the icorrect aswer if we eglect to put it i. Now, there are some importat properties of itegrals that we should take a look at. Properties of the Idefiite Itegral. kf ( x) dx= k f ( x) dx where k is ay umber. So, we ca factor multiplicative costats out of idefiite itegrals. See the Proof of Various Itegral Formulas sectio of the Extras chapter to see the proof of this property. f x dx f x dx. This is really the first property with k =- ad so o proof of this property will be give.. - ( ) =- ( ) f x g x dx f x dx g x dx. I other words, the itegral of a sum or differece of fuctios is the sum or differece of the idividual itegrals. This rule ca be exteded to as may fuctios as we eed.. ( ) ± ( ) = ( ) ± ( ) See the Proof of Various Itegral Formulas sectio of the Extras chapter to see the proof of this property. Notice that whe we worked the first example above we used the first ad third property i the discussio. We itegrated each term idividually, put ay costats back i ad the put everythig back together with the appropriate sig. Not listed i the properties above were itegrals of products ad quotiets. The reaso for this is simple. Just like with derivatives each of the followig will NOT work. ( x) ( ) f f ( x) g( x) dx f ( x) dx g( x) dx Ù dx ı g x f ( ) ( ) x dx g x dx With derivatives we had a product rule ad a quotiet rule to deal with these cases. However, with itegrals there are o such rules. Whe faced with a product ad quotiet i a itegral we will have a variety of ways of dealig with it depedig o just what the itegrad is. 007 Paul Dawkis 9

11 Math 40 There is oe fial topic to be discussed briefly i this sectio. O occasio we will be give f ( x) ad will ask what f ( ) itegral. x was. We ca ow aswer this questio easily with a idefiite ( ) ( ) f x = f x dx 4 Example If f ( x) = x + x- 9 what was ( ) f x? Solutio By this poit i this sectio this is a simple questio to aswer. 4 5 f ( x) = f ( x) dx= x + x- 9dx= x + x - 9x+ c 5 I this sectio we kept evaluatig the same idefiite itegral i all of our examples. The poit of this sectio was ot to do idefiite itegrals, but istead to get us familiar with the otatio ad some of the basic ideas ad properties of idefiite itegrals. The ext couple of sectios are devoted to actually evaluatig idefiite itegrals. 007 Paul Dawkis 0

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13 Math 40 Computig Idefiite Itegrals I the previous sectio we started lookig at idefiite itegrals ad i that sectio we cocetrated almost exclusively o otatio, cocepts ad properties of the idefiite itegral. I this sectio we eed to start thikig about how we actually compute idefiite itegrals. We ll start off with some of the basic idefiite itegrals. The first itegral that we ll look at is the itegral of a power of x. + x x dx= + c, - + The geeral rule whe itegratig a power of x we add oe oto the expoet ad the divide by the ew expoet. It is clear (hopefully) that we will eed to avoid =- i this formula. If we allow =- i this formula we will ed up with divisio by zero. We will take care of this case i a bit. Next is oe of the easier itegrals but always seems to cause problems for people. kdx= kx+ c, c ad k are costats If you remember that all we re askig is what did we differetiate to get the itegrad this is pretty simple, but it does seem to cause problems o occasio. Let s ow take a look at the trig fuctios. sixdx=- cosx+ c cosxdx= si x+ c sec ta sec ta sec xdx= x+ c x xdx= x+ c csc cot csc cot csc xdx=- x+ c x xdx=- x+ c Notice that we oly itegrated two of the six trig fuctios here. The remaiig four itegrals are really itegrals that give the remaiig four trig fuctios. Also, be careful with sigs here. It is easy to get the sigs for derivatives ad itegrals mixed up. Agai, remember that we re askig what fuctio we differetiated to get the itegrad. We will be able to itegrate the remaiig four trig fuctios i a couple of sectios, but they all require the Substitutio Rule. Now, let s take care of expoetial ad logarithm fuctios. Ù ı x x x x x a - dx= + c a dx= + c dx= x dx= l x + c e e l a 007 Paul Dawkis

14 Math 40 Itegratig logarithms requires a topic that is usually taught i Calculus II ad so we wo t be itegratig a logarithm i this class. Also ote the third itegrad ca be writte i a couple of ways ad do t forget the absolute value bars i the x i the aswer to the third itegral. Fially, let s take care of the iverse trig ad hyperbolic fuctios. - - Ù dx= ta x+ c dx si x c Ù = + ı x + ı -x sihxdx= coshx+ c coshxdx= sih x+ c sech tah sech tah sech xdx= x+ c x xdx=- x+ c csch coth csch coth csch xdx=- x+ c x xdx=- x+ c As with logarithms itegratig iverse trig fuctios requires a topic usually taught i Calculus II ad so we wo t be itegratig them i this class. There is also a differet aswer for the secod itegral above. Recallig that sice all we are askig here is what fuctio did we differetiate to get the itegrad the secod itegral could also be, Ù ı -x Traditioally we use the first form of this itegral. cos - dx=- x+ c Okay, ow that we ve got most of the basic itegrals out of the way let s do some idefiite itegrals. I all of these problems remember that we ca always check our aswer by differetiatig ad makig sure that we get the itegrad. Example Evaluate each of the followig idefiite itegrals. -6 (a) 5t - 0t + 4dt [Solutio] 8-8 (b) x + x dx [Solutio] 4 7 (c) Ù x + + dx [Solutio] 5 ı x 6 x (d) dy [Solutio] (e) ( w + w )( 4- w ) dw [Solutio] 0 4 4x - x + 5x (f) Ù dx [Solutio] ı x Solutio Okay, i all of these remember the basic rules of idefiite itegrals. First, to itegrate sums ad differeces all we really do is itegrate the idividual terms ad the put the terms back together 007 Paul Dawkis

15 Math 40 with the appropriate sigs. Next, we ca igore ay coefficiets util we are doe with itegratig that particular term ad the put the coefficiet back i. Also, do ot forget the +c at the ed it is importat ad must be there. So, let s evaluate some itegrals. (a) t - t + dt There s ot really a whole lot to do here other tha use the first two formulas from the begiig of this sectio. Remember that whe itegratig powers (that are t - of course) we just add oe oto the expoet ad the divide by the ew expoet. -6 ʈ 4 Ê ˆ -5 5t - 0t + 4dt = 5Á t - 0Á t + 4t+ c Ë4 Ë = t + t + 4t+ c 4 Be careful whe itegratig egative expoets. Remember to add oe oto the expoet. Oe of the more commo mistakes that studets make whe itegratig egative expoets is to add oe ad ed up with a expoet of -7 istead of the correct expoet of -5. [Retur to Problems] (b) x x dx This is here just to make sure we get the poit about itegratig egative expoets x + x dx= x - x + c 9 7 [Retur to Problems] 4 7 (c) Ù x + + dx 5 ı x 6 x I this case there is t a formula for explicitly dealig with radicals or ratioal expressios. However, just like with derivatives we ca write all these terms so they are i the umerator ad they all have a expoet. This should always be your first step whe faced with this kid of itegral just as it was whe differetiatig x + + dx= Ù x 7x x dx 5 Ù + + ı x 6 x ı Ê ˆ - = x - x + Á x + c 7 4 6Á 4 Ë = x - x + x + c 7 4 Whe dealig with fractioal expoets we usually do t divide by the ew expoet. Doig 007 Paul Dawkis 4

16 Math 40 this is equivalet to multiplyig by the reciprocal of the ew expoet ad so that is what we will usually do. [Retur to Problems] (d) dy Do t make this oe harder tha it is dy = dy = y+ c I this case we are really just itegratig a oe! [Retur to Problems] (e) ( + )( 4 - ) w w w dw We ve got a product here ad as we oted i the previous sectio there is o rule for dealig with products. However, i this case we do t eed a rule. All that we eed to do is multiply thigs out (takig care of the radicals at the same time of course) ad the we will be able to itegrate. ( )( ) 7 w+ w 4- w dw= 4w- w + 4w -wdw = w - w + w - w + c 4 0 [Retur to Problems] 0 4 4x - x + 5x (f) Ù dx ı x As with the previous part it s ot really a problem that we do t have a rule for quotiets for this itegral. I this case all we eed to do is break up the quotiet ad the itegrate the idividual terms x - x + 5x 4x x 5x dx= Ù Ù - + dx ı x ı x x x 7 5 = Ù4x - x+ dx ı x 8 = x - x + 5l x + c Be careful to ot thik of the third term as x to a power for the purposes of itegratio. Usig that rule o the third term will NOT work. The third term is simply a logarithm. Also, do t get excited about the 5. The 5 is just a costat ad so it ca be factored out of the itegral. I other words, here is what we did to itegrate the third term. 5 dx= 5 Ù Ù dx= 5l x + c ı x ı x [Retur to Problems] 007 Paul Dawkis 5

17 Math 40 Always remember that you ca t itegrate products ad quotiets i the same way that we itegrate sums ad differeces. At this poit the oly way to itegrate products ad quotiets is to multiply the product out or break up the quotiet. Evetually we ll see some other products ad quotiets that ca be dealt with i other ways. However, there will ever be a sigle rule that will work for all products ad there will ever be a sigle rule that will work for all quotiets. Every product ad quotiet is differet ad will eed to be worked o a case by case basis. The first set of examples focused almost exclusively o powers of x (or whatever variable we used i each example). It s time to do some examples that ivolve other fuctios. Example Evaluate each of the followig itegrals. x (a) e + 5cosx-0sec xdx [Solutio] (b) Ù secwta w+ dw [Solutio] ı 6w (c) Ù + 6cscycot y+ 9 dy [Solutio] ı y + y (d) Ù 6six 0sih xdx ı -x + + [Solutio] 7-6si q (e) Ù d q [Solutio] ı si q Solutio Most of the problems i this example will simply use the formulas from the begiig of this sectio. More complicated problems ivolvig most of these fuctios will eed to wait util we reach the Substitutio Rule. x (a) e + 5cosx-0sec xdx There is t aythig to this oe other tha usig the formulas. x x e + 5cos x - 0sec xdx = e + 5si x - 0ta x + c [Retur to Problems] (b) Ù secwta w+ dw ı 6w Let s be a little careful with this oe. First break it up ito two itegrals ad ote the rewritte itegrad o the secod itegral. secwtaw+ dw= secwta wdw+ Ù Ù dw ı 6w ı 6 w = secwta wdw+ Ù dw 6 ı w 007 Paul Dawkis 6

18 Math 40 Rewritig the secod itegrad will help a little with the itegratio at this early stage. We ca thik of the 6 i the deomiator as a /6 out i frot of the term ad the sice this is a costat it ca be factored out of the itegral. The aswer is the, Ùsecwtaw+ dw= secw+ l w + c ı 6w 6 Note that we did t factor the out of the first itegral as we factored the /6 out of the secod. I fact, we will geerally ot factor the /6 out either i later problems. It was oly doe here to make sure that you could follow what we were doig. [Retur to Problems] (c) Ù + 6cscycot y+ 9 dy ı y + y I this oe we ll just use the formulas from above ad do t get excited about the coefficiets. They are just multiplicative costats ad so ca be igored while we itegrate each term ad the oce we re doe itegratig a give term we simply put the coefficiets back i. (d) Ù ı Ù ı x -x cscycot y+ dy = ta y- 6cscy+ 9l y + c y + y 6si 0sih xdx [Retur to Problems] Agai, there really is t a whole lot to do with this oe other tha to use the appropriate formula from above. Ù ı -x - + 6si + 0sih = si - 6cos + 0cosh + x xdx x x x c [Retur to Problems] 7-6si q (e) Ù d q ı si q This oe ca be a little tricky if you are t ready for it. As discussed previously, at this poit the oly way we have of dealig with quotiets is to break it up. Ù ı 7-6si 7 q d 6 q = Ù - d q si q ı si q = - 7csc q 6 Notice that upo breakig the itegral up we further simplified the itegrad by recallig the defiitio of cosecat. With this simplificatio we ca do the itegral. 007 Paul Dawkis 7 dq 7-6si q Ù d 7cot 6 c q =- q - q + ı si q [Retur to Problems]

19 Math 40 As show i the last part of this example we ca do some fairly complicated lookig quotiets at this poit if we remember to do simplificatios whe we see them. I fact, this is somethig that you should always keep i mid. I almost ay problem that we re doig here do t forget to simplify where possible. I almost every case this ca oly help the problem ad will rarely complicate the problem. I the ext problem we re goig to take a look at a product ad this time we re ot goig to be able to just multiply the product out. However, if we recall the commet about simplifyig a little this problem becomes fairly simple. Ê t ˆ Ê t ˆ Example Itegrate ÙsiÁ cosá dt ı Ë Ë. Solutio There are several ways to do this itegral ad most of them require the ext sectio. However, there is a way to do this itegral usig oly the material from this sectio. All that is required is to remember the trig formula that we ca use to simplify the itegrad up a little. Recall the followig double agle formula. A small rewrite of this formula gives, ( ) si t = sitcost ( ) sitcost = si t If we ow replace all the t s with t we get, Ê t ˆ Ê t siá cosá ˆ= si Ë Ë 007 Paul Dawkis 8 () t Usig this formula the itegral becomes somethig we ca do. Ê t ˆ Ê t ˆ si cos dt= Ù Á Á Ù si () t dt ı Ë Ë ı =- cos () t + c As oted earlier there is aother method for doig this itegral. I fact there are two alterate methods. To see all three check out the sectio o Costat of Itegratio i the Extras chapter but be aware that the other two do require the material covered i the ext sectio. The formula/simplificatio i the previous problem is a ice trick to remember. It ca be used o occasio to greatly simplify some problems. There is oe more set of examples that we should do before movig out of this sectio.

20 Math 40 Example 4 Give the followig iformatio determie the fuctio f ( x ). x (a) f ( x) = 4x - 9+ six+ 7 e, f ( 0) = 5 [Solutio] = , =-, 4 = 404 [Solutio] 4 (b) f ( x) x x f () f ( ) 5 Solutio I both of these we will eed to remember that ( ) ( ) f x = f x dx Also ote that because we are givig values of the fuctio at specific poits we are also goig to be determiig what the costat of itegratio will be i these problems. = 4-9+ si + 7, 0 = 5 x (a) f ( x) x x e f ( ) The first step here is to itegrate to determie the most geeral possible f ( x ). ( ) x f x = 4x - 9+ six+ 7e dx = e + 4 x x 9x cosx 7 c Now we have a value of the fuctio so let s plug i x = 0 ad determie the value of the costat of itegratio c. So, from this it looks like c = 0 ( ) ( ) ( ) cos 0 7e 4 0 = f = = c = 5+ c. This meas that the fuctio is, ( ) 4 x f x = x -9x- cosx+ 7e + 0 (b) f ( x) x x f () f ( ) 5 c [Retur to Problems] = , =-, 4 = This oe is a little differet form the first oe. I order to get the fuctio we will eed the first derivative ad we have the secod derivative. We ca however, use a itegral to get the first derivative from the secod derivative, just as we used a itegral to get the fuctio from the first derivative. So, let s first get the most geeral possible first derivative. 007 Paul Dawkis 9

21 Math 40 ( ) = ( ) f x f x dx = x + x + dx ʈ 5 4 = 5Á x + x + 6x+ c Ë = 0x + x + 6x+ c 4 Do t forget the costat of itegratio! We ca ow fid the most geeral possible fuctio. 5 4 f ( x) = Ù0x + x + 6x+ cdx ı = 4x + x + x + cx+ d 4 Do ot get excited about itegratig the c. It s just a costat ad we kow how to itegrate costats. Also, there will be o reaso to thik the costats of itegratio from the itegratio i each step will be the same ad so we ll eed to call each costat of itegratio somethig differet. Now, plug i the two values of the fuctio that we ve got = f () = c+ d = + c+ d = f ( 4) = 4 ( ) + ( 04) + 6 ( ) + c( 4) + d = 4+ 4c+ d 4 This gives us a system of two equatios i two ukows that we ca solve = + c+ d c=- 4 4 fi 404= 4+ 4c+ d d =- The fuctio is the, ( ) 5 5 f x = 4x + x + x - x- 4 Do t remember how to solve systems? Check out the Solvig Systems portio of my Algebra/Trig Review. [Retur to Problems] I this sectio we ve started the process of itegratio. We ve see how to do quite a few basic itegrals ad we also saw a quick applicatio of itegrals i the last example. 007 Paul Dawkis 0

22 Math 40 There are may ew formulas i this sectio that we ll ow have to kow. However, if you thik about it, they are t really ew formulas. They are really othig more tha derivative formulas that we should already kow writte i terms of itegrals. If you remember that you should fid it easier to remember the formulas i this sectio. Always remember that itegratio is askig othig more tha what fuctio did we differetiate to get the itegrad. If you ca remember that may of the basic itegrals that we saw i this sectio ad may of the itegrals i the comig sectios are t too bad. 007 Paul Dawkis

23 Math 40 Substitutio Rule for Idefiite Itegrals After the last sectio we ow kow how to do the followig itegrals. 4 y xdx Ù dt cos wdw dy ı t e However, we ca t do the followig itegrals. t + x x dx dt Ù ı ( t + t) Ê ˆ 4y - y ÙÁ- cos( w-lwdw ) ( 8y-) dy Ë w e ı All of these look cosiderably more difficult tha the first set. However, they are t too bad oce you see how to do them. Let s start with the first oe. I this case let s otice that if we let x x dx u = 6x + 5 ad we compute the differetial (you remember how to compute these right?) for this we get, du = 8 xdx Now, let s go back to our itegral ad otice that we ca elimiate every x that exists i the itegral ad write the itegral completely i terms of u usig both the defiitio of u ad its differetial. 4 4 ( ) ( ) 8x 6x + 5dx= Ù 6x + 5 8xdx ı = 4 u du I the process of doig this we ve take a itegral that looked very difficult ad with a quick substitutio we were able to rewrite the itegral ito a very simple itegral that we ca do. Evaluatig the itegral gives, x 6x + 5dx= u du = u + c= 4 ( 6x + 5) + c 5 5 As always we ca check our aswer with a quick derivative if we d like to ad do t forget to back substitute ad get the itegral back ito terms of the origial variable. What we ve doe i the work above is called the Substitutio Rule. Here is the substitutio rule i geeral. 007 Paul Dawkis

24 Math 40 Substitutio Rule ( ( )) ( ) = ( ), where, = ( ) f g x g x dx f u du u g x A atural questio at this stage is how to idetify the correct substitutio. Ufortuately, the aswer is it depeds o the itegral. However, there is a geeral rule of thumb that will work for may of the itegrals that we re goig to be ruig across. Whe faced with a itegral we ll ask ourselves what we kow how to itegrate. With the itegral above we ca quickly recogize that we kow how to itegrate 4 xdx However, we did t have just the root we also had stuff i frot of the root ad (more importatly i this case) stuff uder the root. Sice we ca oly itegrate roots if there is just a x uder the root a good first guess for the substitutio is the to make u be the stuff uder the root. Aother way to thik of this is to ask yourself what portio of the itegrad has a iside fuctio ad ca you do the itegral with that iside fuctio preset. If you ca t the there is a pretty good chace that the iside fuctio will be the substitutio. We will have to be careful however. There are times whe usig this geeral rule ca get us i trouble or overly complicate the problem. We ll evetually see problems where there are more tha oe iside fuctio ad/or itegrals that will look very similar ad yet use completely differet substitutios. The reality is that the oly way to really lear how to do substitutios is to just work lots of problems ad evetually you ll start to get a feel for how these work ad you ll fid it easier ad easier to idetify the proper substitutios. Now, with that out of the way we should ask the followig questio. How, do we kow if we got the correct substitutio? Well, upo computig the differetial ad actually performig the substitutio every x i the itegral (icludig the x i the dx) must disappear i the substitutio process ad the oly letters left should be u s (icludig a du). If there are x s left over the there is a pretty good chace that we chose the wrog substitutio. Ufortuately, however there is at least oe case (we ll be seeig a example of this i the ext sectio) where the correct substitutio will actually leave some x s ad we ll eed to do a little more work to get it to work. Agai, it caot be stressed eough at this poit that the oly way to really lear how to do substitutios is to just work lots of problems. There are lots of differet kids of problems ad after workig may problems you ll start to get a real feel for these problems ad after you work eough of them you ll reach the poit where you ll be able to do simple substitutios i your head without havig to actually write aythig dow. As a fial ote we should poit out that ofte (i fact i almost every case) the differetial will ot appear exactly i the itegrad as it did i the example above ad sometimes we ll eed to do 007 Paul Dawkis

25 Math 40 some maipulatio of the itegrad ad/or the differetial to get all the x s to disappear i the substitutio. Let s work some examples so we ca get a better idea o how the substitutio rule works. Example Evaluate each of the followig itegrals. Ê ˆ (a) ÙÁ- cos( w-l wdw ) [Solutio] ıë w 4y - y (b) 8 ( y-) e dy [Solutio] (c) x ( -0 x ) 4 dx [Solutio] x (d) Ù dx [Solutio] ı - 4x Solutio Ê ˆ (a) ÙÁ- cos( w-l wdw ) ıë w I this case we kow how to itegrate just a cosie so let s make the substitutio the stuff that is iside the cosie. Ê ˆ u = w- lw du = Á- dw Ë w So, as with the first example we worked the stuff i frot of the cosie appears exactly i the differetial. The itegral is the, Ê ˆ Ù Á- cos( w- lwdw ) = cos( u) du Ë w ı = si + ( u) c ( l ) = si w- w + c Do t forget to go back to the origial variable i the problem. [Retur to Problems] 4y - y e dy Agai, we kow how to itegrate a expoetial by itself so it looks like the substitutio for this problem should be, (b) 8 ( y-) ( ) u = 4y - y du = 8y- dy Now, with the exceptio of the the stuff i frot of the expoetial appears exactly i the differetial. Recall however that we ca factor the out of the itegral ad so it wo t cause ay problems. The itegral is the, 007 Paul Dawkis 4

26 Math 40 ( ) 4 y -y u 8y- e dy = e du u = e + c 4 y -y = e + c [Retur to Problems] (c) ( -0 ) 4 x x dx I this case it looks like the followig should be the substitutio. u = - 0x du =- 0x dx Okay, ow we have a small problem. We ve got a x out i frot of the parethesis but we do t have a -0. This is ot really the problem it might appear to be at first. We will simply rewrite the differetial as follows. x dx=- du 0 With this we ca ow substitute the x dx away. I the process we will pick up a costat, but that is t a problem sice it ca always be factored out of the itegral. We ca ow do the itegral. ( - 0 ) = ( -0 ) 4 4 x x dx x x dx 4Ê ˆ = Ù u Á - du ı Ë 0 ʈ 5 =- Á u + c 0Ë5 5 =- ( - 0x ) + c 50 Note that i most problems whe we pick up a costat as we did i this example we will geerally factor it out of the itegral i the same step that we substitute it i. [Retur to Problems] x (d) Ù dx ı - 4x I this example do t forget to brig the root up to the umerator ad chage it ito fractioal expoet form. Upo doig this we ca see that the substitutio is, u = - 4x du =-8xdx fi xdx=- du 8 The itegral is the, 007 Paul Dawkis 5

27 Math 40 Ù ı x -4x - ( 4 ) dx= x - x dx - =- 8u du =- u + c 4 =- ( - 4 x ) + c 4 [Retur to Problems] I the previous set of examples the substitutio was geerally pretty clear. There was exactly oe term that had a iside fuctio that we also could t itegrate. Let s take a look at some more complicated problems to make sure we do t come to expect all substitutios are like those i the previous set of examples. Example Evaluate each of the followig itegrals. (a) si( - x )( -cos( - x )) 4 dx [Solutio] 0 (b) cos( z ) si ( z) dz [Solutio] (c) sec ( 4 t )( -ta( 4 t )) dt [Solutio] Solutio ( ) 4 (a) si( - ) -cos( - ) x x dx I this problem there are two iside fuctios. There is the - x that is iside the two trig fuctios ad there is also the term that is raised to the 4 th power. There are two ways to proceed with this problem. The first idea that may studets have is substitute the - x away. There is othig wrog with doig this but it does t elimiate the problem of the term to the 4 th power. That s still there ad if we used this idea we would the eed to do a secod substitutio to deal with that. The secod (ad much easier) way of doig this problem is to just deal with the stuff raised to the 4 th power ad see what we get. The substitutio i this case would be, ( ) ( ) ( ) u = -cos - x du =-si -x dx fi si - x dx=- du Two thigs to ote here. First, do t forget to correctly deal with the -. A commo mistake at this poit is to lose it. Secodly, otice that the - x turs out to ot really be a problem after all. Because the - x was buried i the substitutio that we actually used it was also take care of at the same time. The itegral is the, 007 Paul Dawkis 6

28 Math 40 ( ) ( ) ( ) 4 4 si -x -cos - x dx=- u du 5 =- u + c 5 5 =- ( - cos ( - x) ) + c 5 As see i this example sometimes there will seem to be two substitutios that will eed to be doe however, if oe of them is buried iside of aother substitutio the we ll oly really eed to do oe. Recogizig this ca save a lot of time i workig some of these problems. [Retur to Problems] 0 (b) cos( z) si ( ) z dz This oe is a little tricky at first. We ca see the correct substitutio by recallig that, ( ) 0 ( z) = ( z) 0 si si Usig this it looks like the correct substitutio is, u = si( z) du = cos( z) dz fi cos( z) dz = du Notice that we agai had two apparet substitutios i this itegral but agai the z is buried i the substitutio we re usig ad so we did t eed to worry about it. Here is the itegral. 0 0 cos( z) si ( z) dz = u du Ê ˆ = Á u + c Ë si = ( z) + c Note that the oe third i frot of the itegral came about from the substitutio o the differetial ad we just factored it out to the frot of the itegral. This is what we will usually do with these costats. [Retur to Problems] ( ) (c) sec ( 4 ) -ta( 4 ) t t dt I this case we ve got a 4t, a secat squared as well as a term cubed. However, it looks like if we use the followig substitutio the first two issues are goig to be take care of for us. ( ) ( ) ( ) u = - ta 4t du =-4sec 4t dt fi sec 4t dt =- du Paul Dawkis 7

29 Math 40 The itegral is ow, sec ( 4t) ( - ta( 4t) ) dt =- 4u du 4 =- u + c 6 4 =- ( - ta ( 4 t) ) + c 6 [Retur to Problems] The most importat thig to remember i substitutio problems is that after the substitutio all the origial variables eed to disappear from the itegral. After the substitutio the oly variables that should be preset i the itegral should be the ew variable from the substitutio (usually u). Note as well that this icludes the variables i the differetial! This ext set of examples, while ot particular difficult, ca cause trouble if we are t payig attetio to what we re doig. Example Evaluate each of the followig itegrals. (a) Ù ı5y+ 4 dy [Solutio] y (b) Ù dy ı5y + 4 [Solutio] y (c) Ù ı 5y + 4 dy [Solutio] ( ) (d) Ù ı5y + 4 dy [Solutio] Solutio (a) Ù ı5y+ 4 dy We have t see a problem quite like this oe yet. Let s otice that if we take the deomiator ad differetiate it we get just a costat ad the oly thig that we have i the umerator is also a costat. This is a pretty good idicatio that we ca use the deomiator for our substitutio so, u = 5y+ 4 du = 5dy fi dy = du 5 The itegral is ow, 007 Paul Dawkis 8

30 Math 40 dy = Ù Ù du ı5y+ 4 5ı u = l u + c 5 = l5 y+ 4 + c 5 Remember that we ca just factor the i the umerator out of the itegral ad that makes the itegral a little clearer i this case. [Retur to Problems] y (b) Ù dy ı5y + 4 The itegral is very similar to the previous oe with a couple of mior differeces but otice that agai if we differetiate the deomiator we get somethig that is differet from the umerator by oly a multiplicative costat. Therefore we ll agai take the deomiator as our substitutio. u = 5y + 4 du = 0ydy fi ydy = du 0 The itegral is, Ù ı y dy = Ù du ı u 5y = l u + c 0 l5 y 4 = + + c 0 [Retur to Problems] (c) Ù ı y ( 5y + 4) dy Now, this oe is almost idetical to the previous part except we added a power oto the deomiator. Notice however that if we igore the power ad differetiate what s left we get the same thig as the previous example so we ll use the same substitutio here. The itegral i this case is, u = 5y + 4 du = 0ydy fi ydy = du Paul Dawkis 9

31 Math 40 Ù ı y 0 ( 5y + 4) - dy = u du 0 - =- u + c - ( 5 4) =- y + + c=- + c ( y + ) Be careful i this case to ot tur this ito a logarithm. After workig problems like the first two i this set a commo error is to tur every ratioal expressio ito a logarithm. If there is a power o the whole deomiator the there is a good chace that it is t a logarithm. The idea that we used i the last three parts to determie the substitutio is ot a bad idea to remember. If we ve got a ratioal expressio try differetiatig the deomiator (igorig ay powers that are o the whole deomiator) ad if the result is the umerator or oly differs from the umerator by a multiplicative costat the we ca usually use that as our substitutio. [Retur to Problems] (d) Ù ı5y + 4 dy Now, this part is completely differet from the first three ad yet seems similar to them as well. I this case if we differetiate the deomiator we get a y that is ot i the umerator ad so we ca t use the deomiator as our substitutio. I fact, because we have y i the deomiator ad o y i the umerator is a idicatio of how to work this problem. This itegral is goig to be a iverse taget whe we are doe. To key to seeig this is to recall the followig formula, - Ù du = ta u+ c ı+ u We clearly do t have exactly this but we do have somethig that is similar. The deomiator has a squared term plus a costat ad the umerator is just a costat. So, with a little work ad the proper substitutio we should be able to get our itegral ito a form that will allow us to use this formula. There is oe part of this formula that is really importat ad that is the + i the deomiator. That must be there ad we ve got a 4+ but it is easy eough to take care of that. We ll just factor a 4 out of the deomiator ad at the same time we ll factor the i the umerator out of the itegral as well. Doig this gives, 007 Paul Dawkis 0

32 Math 40 Ù dy = Ù dy ı5y + 4 Ù Ê5y ˆ ı 4Á + Ë 4 = Ù dy 4Ù 5y ı + 4 = Ù dy 4 Ù Ê 5 yˆ ı Á + Ë Notice that i the last step we rewrote thigs a little i the deomiator. This will help us to see what the substitutio eeds to be. I order to get this itegral ito the formula above we eed to ed up with a u i the deomiator. Our substitutio will the eed to be somethig that upo squarig gives us substitutio. y 5 4. With the rewrite we ca see what that we ll eed to use the followig 5y 5 u = du = dy fi dy = du 5 Do t get excited about the root i the substitutio, these will show up o occasio. Upo pluggig our substitutio i we get, Ù ı Ê ˆ dy = Á Ù Ë 5 ı 5y u + After doig the substitutio, ad factorig ay costats out, we get exactly the itegral that gives a iverse taget ad so we kow that we did the correct substitutio for this itegral. The itegral is the, Ù ı dy = Ù y + 5 ı u + ta Paul Dawkis ( u) du - = + Ê 5 yˆ ta 5 Á Ë du - = + c c [Retur to Problems] I this last set of itegrals we had four itegrals that were similar to each other i may ways ad yet all either yielded differet aswer usig the same substitutio or used a completely differet substitutio tha oe that was similar to it.

33 Math 40 This is a fairly commo occurrece ad so you will eed to be able to deal with these kids of issues. There are may itegrals that o the surface look very similar ad yet will use a completely differet substitutio or will yield a completely differet aswer whe usig the same substitutio. Let s take a look at aother set of examples to give us more practice i recogizig these kids of issues. Note however that we wo t be puttig as much detail ito these as we did with the previous examples. Example 4 Evaluate each of the followig itegrals. t + (a) dt [Solutio] Ù 4 ı t + t ( ) t + (b) Ù dt 4 ı t + t x (c) Ù dx ı - 4x (d) Ù dx ı - 4x Solutio (a) Ù ı + t 4 ( t + t) dt [Solutio] [Solutio] [Solutio] Clearly the derivative of the deomiator, igorig the expoet, differs from the umerator oly by a multiplicative costat ad so the substitutio is, ( ) ( ) ( ) u = t + t du = 4t + dt = t + dt fi t + dt = du 4 After a little maipulatio of the differetial we get the followig itegral. Ù ı t + 4 ( t + t) dt = Ù ı u = - u Ê du du ˆ Á c Ë 4 - =- ( t + t) + c 4 - = - u + [Retur to Problems] 007 Paul Dawkis

34 Math 40 + t (b) Ù dt 4 ı t + t The oly differece betwee this problem ad the previous oe is the deomiator. I the previous problem the whole deomiator is cubed ad i this problem the deomiator has o power o it. The same substitutio will work i this problem but because we o loger have the power the problem will be differet. So, usig the substitutio from the previous example the itegral is, Ù ı t + dt = 4 Ù t + t ı u du = l u + c l 4 = t + t + c So, i this case we get a logarithm from the itegral. [Retur to Problems] x (c) Ù dx ı - 4x Here, if we igore the root we ca agai see that the derivative of the stuff uder the radical differs from the umerator by oly a multiplicative costat ad so we ll use that as the substitutio. u = - 4x du =-8xdx fi xdx=- du 8 The itegral is the, Ù ı x -4x - dx=- 8 u du =- ( ) u + c =- - x + c [Retur to Problems] (d) Ù dx ı - 4x I this case we are missig the x i the umerator ad so the substitutio from the last part will do us o good here. This itegral is aother iverse trig fuctio itegral that is similar to the last part of the previous set of problems. I this case we eed to followig formula. 007 Paul Dawkis

35 Math 40 Ù ı - u - du = si u+ c The itegral i this problem is early this. The oly differece is the presece of the coefficiet of 4 o the x. With the correct substitutio this ca be dealt with however. To see what this substitutio should be let s rewrite the itegral a little. We eed to figure out what we squared to get 4x ad that will be our substitutio. Ù ı dx= Ù - 4x ı - ( x) dx With this rewrite it looks like we ca use the followig substitutio. u = x du = dx fi dx= du The itegral is the, Ù ı dx= Ù -4x ı -u du si u - = + si - c ( x) = + c [Retur to Problems] Sice this documet is also beig preseted o the web we re goig to put the rest of the substitutio rule examples i the ext sectio. With all the examples i oe sectio the sectio was becomig too large for web presetatio. 007 Paul Dawkis 4

36 Math 40 More Substitutio Rule I order to allow these pages to be displayed o the web we ve broke the substitutio rule examples ito two sectios. The previous sectio cotais the itroductio to the substitutio rule ad some fairly basic examples. The examples i this sectio ted towards the slightly more difficult side. Also, we ll ot be puttig quite as much explaatio ito the solutios here as we did i the previous sectio. I the first couple of sets of problems i this sectio the difficulty is ot with the actual itegratio itself, but with the set up for the itegratio. Most of the itegrals are fairly simple ad most of the substitutios are fairly simple. The problems arise i gettig the itegral set up properly for the substitutio(s) to be doe. Oce you see how these are doe it s easy to see what you have to do, but the first time through these ca cause problems if you are t o the lookout for potetial problems. Example Evaluate each of the followig itegrals. t (a) e + sec( t) ta( t) dt [Solutio] (b) si( t )( 4cos ( t ) + 6cos ( t ) -8) dt [Solutio] (c) x Ùxcos( x + ) + dx [Solutio] ı x + Solutio t (a) e + sec( t) ta( t) dt This first itegral has two terms i it ad both will require the same substitutio. This meas that we wo t have to do aythig special to the itegral. Oe of the more commo mistakes here is to break the itegral up ad do a separate substitutio o each part. This is t really mistake but will defiitely icrease the amout of work we ll eed to do. So, sice both terms i the itegral use the same substitutio we ll just do everythig as a sigle itegral usig the followig substitutio. u = t du = dt fi dt = du The itegral is the, t u e + sec( t) ta( t) dt = sec( ) ta ( ) e + u u du u = ( e + sec( u) ) + c t = ( e + sec( t) ) + c Ofte a substitutio ca be used multiple times i a itegral so do t get excited about that if it happes. Also ote that sice there was a i frot of the whole itegral there must also be a i frot of the aswer from the itegral. [Retur to Problems] 007 Paul Dawkis 5

37 Math 40 ( + - ) si t 4cos t 6cos t 8 dt (b) ( ) ( ) ( ) This itegral is similar to the previous oe, but it might ot look like it at first glace. Here is the substitutio for this problem, ( ) ( ) ( ) u = cos t du =-si t dt fi si t dt =- du We ll plug the substitutio ito the problem twice (sice there are two cosies) ad will oly work because there is a sie multiplyig everythig. Without that sie i frot we would ot be able to use this substitutio. The itegral i this case is, ( ) ( ) ( ) ( ) si 4cos 6cos ( u u 8u) c 4 ( cos () cos () 8cos() ) t t + t - dt =- u + u - du = =- t + t - t + c Agai, be careful with the mius sig i frot of the whole itegral. [Retur to Problems] (c) x Ùxcos( x + ) + dx ı x + It should be fairly clear that each term i this itegral will use the same substitutio, but let s rewrite thigs a little to make thigs really clear. x Ê ˆ Ù xcos( x + ) + dx= x cos Ù Á ( x + ) + dx ı x + ı Ë x + Sice each term had a x i it ad we ll eed that for the differetial we factored that out of both terms to get it ito the frot. This itegral is ow very similar to the previous oe. Here s the substitutio. u = x + du = xdx fi xdx= du The itegral is, x xcos( x + ) + dx= Ù Ùcos( u) + du ı x + ı u = ( si ( u) + l u ) + c = ( si ( x + ) + l x + ) + c [Retur to Problems] 007 Paul Dawkis 6

38 Math 40 So, as we ve see i the previous set of examples sometimes we ca use the same substitutio more tha oce i a itegral ad doig so will simplify the work. Example Evaluate each of the followig itegrals. - x (a) x + e dx [Solutio] (b) Ùxcos( x + ) + dx [Solutio] ı x + Solutio - x (a) x + e dx I this itegral the first term does ot eed ay substitutio while the secod term does eed a substitutio. So, to deal with that we ll eed to split the itegral up as follows, x x x + e - dx= x dx+ e - dx The substitutio for the secod itegral is the, u = - x du =-dx fi dx=- du The itegral is, - x u x + e dx= x dx- e du x u = - e + c - x = x - e + c Be careful with this kid of itegral. Oe of the more commo mistakes here is do the followig shortcut. - x u x + e dx=- x + e du I other words, some studets will try do the substitutio just the secod term without breakig up the itegral. There are two issues with this. First, there is a - i frot of the whole itegral that should t be there. It should oly be o the secod term because that is the term gettig the substitutio. Secodly, ad probably more importatly, there are x s i the itegral ad we have a du for the differetial. We ca t mix variables like this. Whe we do itegrals all the variables i the itegrad must match the variable i the differetial. [Retur to Problems] (b) Ùxcos( x + ) + dx ı x + This itegral looks very similar to Example c above, but it is differet. I this itegral we o loger have the x i the umerator of the secod term ad that meas that the substitutio we ll use for the first term will o loger work for the secod term. I fact, 007 Paul Dawkis 7

39 Math 40 the secod term does t eed a substitutio at all sice it is just a iverse taget. The substitutio for the first term is the, u = x + du = xdx fi xdx= du Now let s do the itegral. Remember to first break it up ito two terms before usig the substitutio. xcos( x + ) + dx= xcos ( x + ) dx+ Ù Ù dx ı x + ı x + = cos( u) du+ Ù dx ı x + si ( u) ta - = + ( x) + c si ( x ) ta - = + + ( x) + c [Retur to Problems] I this set of examples we saw that sometimes oe (or potetially more tha oe) term i the itegrad will ot require a substitutio. I these cases we ll eed to break up the itegral ito two itegrals, oe ivolvig the terms that do t eed a substitutio ad aother with the term(s) that do eed a substitutio. Example Evaluate each of the followig itegrals. - z Ê z ˆ (a) Ù + sec Á dz ı e Ë0 [Solutio] (b) Ùsiw - cos w+ dw [Solutio] ı 7w + (c) 0x + Ù dx [Solutio] ı x + 6 Solutio - z Ê z ˆ (a) Ù + sec Á dz ı e Ë0 I this itegral, ulike ay itegrals that we ve yet doe, there are two terms ad each will require a differet substitutio. So, to do this itegral we ll first eed to split up the itegral as follows, z z z z - Ê ˆ sec dz - Ê ˆ Ùe + Á = dz+ sec Á dz Ë0 e Ù ı ı Ë0 Here are the substitutios for each itegral. 007 Paul Dawkis 8

40 Math 40 u =- z du =-dz fi dz =-du z v= dv= dz fi dz = 0dv 0 0 Notice that we used differet letters for each substitutio to avoid cofusio whe we go to plug back i for u ad v. Here is the itegral. Ê z ˆ u + sec Á dz =- e du+ 0 sec ( v) dv Ë0 -z Ùe ı ( v) u =- e + 0ta + c -z Ê z ˆ =- e + 0ta Á + c Ë 0 [Retur to Problems] (b) Ùsiw - cos w+ dw ı 7w + As with the last problem this itegral will require two separate substitutios. Let s first break up the itegral. ( ) si w - cos w + si cos 7 dw = w - w w dw + Ù Ù ı + ı 7w+ dw Here are the substitutios for this itegral. u = - cos( w) du = si( w) dw fi si ( wdw ) = du v= 7w+ dv= 7dw fi dw= dv 7 The itegral is the, siw - cos w+ dw= u du+ Ù Ù dv ı 7w+ 7ı v ʈ l = Á u + v + c Ë 7 = ( - cosw) + l 7w+ + c 7 [Retur to Problems] (c) 0x + Ù dx ı x + 6 The last problem i this set ca be tricky. If there was just a x i the umerator we could do a quick substitutio to get a atural logarithm. Likewise if there was t a x i the umerator we 007 Paul Dawkis 9

41 Math 40 would get a iverse taget after a quick substitutio. To get this itegral ito a form that we ca work with we will first eed to break it up as follows. 0x+ 0x dx= dx+ Ù dx Ù Ù ı x + 6 ı x + 6 ı x + 6 0x = Ù dx + Ù dx ı x Ù x ı + 6 We ow have two itegrals each requirig a differet substitutio. The substitutios for each of the itegrals above are, u = x + 6 du = xdx fi xdx= du x v= dv= dx fi dx= 4dv 4 4 The itegral is the, 0x+ dx= 5 du+ Ù Ù Ù dv ı x + 6 ı u 4ı v + - = 5l u + ta ( v) + c 4 Ê xˆ 5l 6 ta Á 4 Ë4 - = x c [Retur to Problems] We ve ow see a set of itegrals i which we eed to do more tha oe substitutio. I these cases we will eed to break up the itegral ito separate itegrals ad do separate substitutios for each. We ow eed to move oto a differet set of examples that ca be a little tricky. Oce you ve see how to do these they are t too bad, but doig them for the first time ca be difficult if you are t ready for them. 007 Paul Dawkis 40

42 Math 40 Example 4 Evaluate each of the followig itegrals. (a) ta xdx [Solutio] (b) sec ydy (c) (d) (e) [Solutio] cos( x ) Ù dx x ı t+ t e e dt [Solutio] [Solutio] x x + dx [Solutio] Solutio (a) ta xdx The first questio about this problem is probably why is it here? Substitutio rule problems geerally require more tha a sigle fuctio. The key to this problem is to realize that there really are two fuctios here. All we eed to do is remember the defiitio of taget ad we ca write the itegral as, si x ta xdx= Ù dx ı cos x Writte i this way we ca see that the followig substitutio will work for us, u = cosx du =-sixdx fi si xdx=- du The itegral is the, ta xdx=- Ù du ı u =- l u + c =- l cos x + c Now, while this is a perfectly serviceable aswer that mius sig i frot is liable to cause problems if we are t careful. So, let s rewrite thigs a little. Recallig a property of logarithms we ca move the mius sig i frot to a expoet o the cosie ad the do a little simplificatio. taxdx=- l cos x + c - = l cos x + c = l + c cos x = l sec x + c 007 Paul Dawkis 4

43 Math 40 This is the formula that is typically give for the itegral of taget. Note that we could itegrate cotaget i a similar maer. [Retur to Problems] (b) sec ydy This problem also at first appears to ot belog i the substitutio rule problems. This is eve more of a problem upo oticig that we ca t just use the defiitio of the secat fuctio to write this i a form that will allow the use of the substitutio rule. This problem is goig to require a techique that is t used terribly ofte at this level, but is a useful techique to be aware of. Sometimes we ca make a itegral doable by multiplyig the top ad bottom by a commo term. This will ot always work ad eve whe it does it is ot always clear what we should multiply by but whe it works it is very useful. Here is how we ll use this idea for this problem. sec ydy = Ù ı ( secy+ ta y) ( y+ y) sec y sec ta First, we will thik of the secat as a fractio ad the multiply the top ad bottom of the fractio by the same term. It is probably ot clear why oe would wat to do this here but doig this will actually allow us to use the substitutio rule. To see how this will work let s simplify the itegrad somewhat. sec ydy = We ca ow use the followig substitutio. Ù ı sec + ta sec dy y y y dy secy+ ta y ( ) u = secy+ ta y du = secyta y+ sec y dy The itegral is the, sec ydy = Ù du ı u = l u + c = l secy+ ta y + c Sometimes multiplyig the top ad bottom of a fractio by a carefully chose term will allow us to work a problem. It does however take some thought sometimes to determie just what the term should be. We ca use a similar process for itegratig cosecat. [Retur to Problems] 007 Paul Dawkis 4

44 Math 40 (c) cos( x ) Ù dx ı x This ext problem has a subtlety to it that ca get us i trouble if we are t payig attetio. Because of the root i the cosie it makes some sese to use the followig substitutio. - u = x du = x dx This is where we eed to be careful. Upo rewritig the differetial we get, du = dx x The root that is i the deomiator will ot become a u as we might have bee tempted to do. Istead it will get take care of i the differetial. The itegral is, Ù ı ( ) cos x dx = cos( u ) du x ( u) = si + c ( x) = si + c [Retur to Problems] (d) e t+ t e dt With this problem we eed to very carefully pick our substitutio. As the problem is writte we might be tempted to use the followig substitutio, t ( ) t u = t+ e du = + e dt However, this wo t work as you ca probably see. The differetial does t show up aywhere i the itegrad ad we just would t be able to elimiate all the t s with this substitutio. I order to work this problem we will eed to rewrite the itegrad as follows, t+ et t et e dt = ee dt We will ow use the substitutio, t t u = e du = e dt The itegral is, 007 Paul Dawkis 4

45 Math 40 e t+ et dt = e u u = e + et du c = e + c Some substitutios ca be really tricky to see ad it s ot uusual that you ll eed to do some simplificatio ad/or rewritig to get a substitutio to work. [Retur to Problems] (e) + x x dx This last problem i this set is differet from all the other substitutio problems that we ve worked to this poit. Give the fact that we ve got more tha a x uder the root it makes sese that the substitutio pretty much has to be, u = x + du = xdx At first glace it looks like this might ot work for the substitutio because we have a frot of the root. However, if we first rewrite x x ( x) x i = we could the move the x to the ed of the itegral so at least the du will show up explicitly i the itegral. Doig this gives the followig, ( ) x x + dx= x x + x dx = xu du This is a real problem. Our itegrals ca t have two variables i them. Normally this would mea that we chose our substitutio icorrectly. However, i this case we ca rewrite the substitutio as follows, x = u- ad ow, we ca elimiate the remaiig x s from our itegral. Doig this gives, ( ) x x + dx= u- u du ı = u -u du 5 = u - u + c 5 5 = ( x + ) - ( x + ) + c 5 Sometimes, we will eed to use a substitutio more tha oce. This kid of problem does t arise all that ofte ad whe it does there will sometimes be 007 Paul Dawkis 44

46 Math 40 alterate methods of doig the itegral. However, it will ofte work out that the easiest method of doig the itegral is to do what we just did here. [Retur to Problems] This fial set of examples is t too bad oce you see the substitutios ad that is the poit with this set of problems. These all ivolve substitutios that we ve ot see prior to this ad so we eed to see some of these kids of problems. Example 5 Evaluate each of the followig itegrals. (a) Ù ı xl x dx t e (b) Ù dt t ı + e t e (c) Ù dt 4t ı + e - si x (d) Ù dx ı - x [Solutio] [Solutio] [Solutio] [Solutio] Solutio (a) Ù ı xl x dx I this case we kow that we ca t itegrate a logarithm by itself ad so it makes some sese (hopefully) that the logarithm will eed to be i the substitutio. Here is the substitutio for this problem. u = l x du = dx x So the x i the deomiator of the itegrad will get substituted away with the differetial. Here is the itegral for this problem. dx= Ù Ù du ı xl x ı u = l u + c = l l x + c t e (b) Ù dt t ı + e Agai, the substitutio here may seem a little tricky. I this case the substitutio is, [Retur to Problems] 007 Paul Dawkis 45

47 Math 40 u = + e du = e dt fi e dt = du t t t The itegral is the, Ù ı t e + e dt = Ù du ı u t l t = + + e c [Retur to Problems] t e (c) Ù dt 4t ı+ e I this case we ca t use the same type of substitutio that we used i the previous problem. I order to use the substitutio i the previous example the expoetial i the umerator ad the deomiator eed to be the same ad i this case they are t. To see the correct substitutio for this problem ote that, e = e t ( ) 4t Usig this, the itegral ca be writte as follows, Ù ı We ca ow use the followig substitutio. e e dt = + Ù ı + t t 4t e t ( e ) u = e du = e dt fi e dt = du t t t dt The itegral is the, t e dt = Ù Ù ı+ e ı+ u ta ta 4t ( u) du - = + t ( e ) c - = + c [Retur to Problems] - si x (d) Ù dx ı -x This itegral is similar to the first problem i this set. Sice we do t kow how to itegrate iverse sie fuctios it seems likely that this will be our substitutio. If we use this as our 007 Paul Dawkis 46

48 Math 40 substitutio we get, ( ) = = - u si x du dx -x So, the root i the itegral will get take care of i the substitutio process ad this will elimiate all the x s from the itegral. Therefore this was the correct substitutio. The itegral is, Ù ı - si x dx = udu - x = u + c ( si - = x) + c [Retur to Problems] Over the last couple of sectios we ve see a lot of substitutio rule examples. There are a couple of geeral rules that we will eed to remember whe doig these problems. First, whe doig a substitutio remember that whe the substitutio is doe all the x s i the itegral (or whatever variable is beig used for that particular itegral) should all be substituted away. This icludes the x i the dx. After the substitutio oly u s should be left i the itegral. Also, sometimes the correct substitutio is a little tricky to fid ad more ofte tha ot there will eed to be some maipulatio of the differetial or itegrad i order to actually do the substitutio. Also, may itegrals will require us to break them up so we ca do multiple substitutios so be o the lookout for those kids of itegrals/substitutios. 007 Paul Dawkis 47

49 Math 40 Area Problem As oted i the first sectio of this sectio there are two kids of itegrals ad to this poit we ve looked at idefiite itegrals. It is ow time to start thikig about the secod kid of itegral : Defiite Itegrals. However, before we do that we re goig to take a look at the Area Problem. The area problem is to defiite itegrals what the taget ad rate of chage problems are to derivatives. The area problem will give us oe of the iterpretatios of a defiite itegral ad it will lead us to the defiitio of the defiite itegral. To start off we are goig to assume that we ve got a fuctio f ( x ) that is positive o some iterval [a,b]. What we wat to do is determie the area of the regio betwee the fuctio ad the x-axis. It s probably easiest to see how we do this with a example. So let s determie the area betwee ( ) f x = x + o [0,]. I other words, we wat to determie the area of the shaded regio below. Now, at this poit, we ca t do this exactly. However, we ca estimate the area. We will estimate the area by dividig up the iterval ito subitervals each of width, b- a D x = The i each iterval we ca form a rectagle whose height is give by the fuctio value at a specific poit i the iterval. We ca the fid the area of each of these rectagles, add them up ad this will be a estimate of the area. It s probably easier to see this with a sketch of the situatio. So, let s divide up the iterval ito 4 subitervals ad use the fuctio value at the right edpoit of each iterval to defie the height of the rectagle. This gives, 007 Paul Dawkis 48

50 Math 40 Note that by choosig the height as we did each of the rectagles will over estimate the area sice each rectagle takes i more area tha the graph each time. Now let s estimate the area. First, the width of each of the rectagles is. The height of each rectagle is determied by the fuctio value at the right edpoit ad so the height of each rectagle is othig more that the fuctio value at the right edpoit. Here is the estimated area. ʈ ʈ Ar = f Á + f + f Á + f Ë Ë Ê5ˆ ʈ = Á + ( ) + Á + ( 5) Ë4 Ë 4 = 5.75 () ( ) Of course takig the rectagle heights to be the fuctio value at the right edpoit is ot our oly optio. We could have take the rectagle heights to be the fuctio value at the left edpoit. Usig the left edpoits as the heights of the rectagles will give the followig graph ad estimated area. 007 Paul Dawkis 49

51 Math 40 ʈ ʈ Al = f ( 0) + f Á + f () + f Á Ë Ë Ê5ˆ Ê ˆ = () + Á + ( ) + Á Ë4 Ë 4 =.75 I this case we ca see that the estimatio will be a uderestimatio sice each rectagle misses some of the area each time. There is oe more commo poit for gettig the heights of the rectagles that is ofte more accurate. Istead of usig the right or left edpoits of each sub iterval we could take the midpoit of each subiterval as the height of each rectagle. Here is the graph for this case. So, it looks like each rectagle will over ad uder estimate the area. This meas that the approximatio this time should be much better tha the previous two choices of poits. Here is the estimatio for this case. ʈ ʈ Ê5ˆ Ê7ˆ Am = f Á + f Á + f Á + f Á Ë4 Ë4 Ë4 Ë4 Ê7ˆ Ê5ˆ Ê4ˆ Ê65 ˆ = Á + Á + Á + Á Ë6 Ë6 Ë6 Ë6 = 4.65 We ve ow got three estimates. For compariso s sake the exact area is 4 A= = Paul Dawkis 50

52 Math 40 So, both the right ad left edpoit estimatio did ot do all that great of a job at the estimatio. The midpoit estimatio however did quite well. Be careful to ot draw ay coclusio about how choosig each of the poits will affect our estimatio. I this case, because we are workig with a icreasig fuctio choosig the right edpoits will overestimate ad choosig left edpoit will uderestimate. If we were to work with a decreasig fuctio we would get the opposite results. For decreasig fuctios the right edpoits will uderestimate ad the left edpoits will overestimate. Also, if we had a fuctio that both icreased ad decreased i the iterval we would, i all likelihood, ot eve be able to determie if we would get a overestimatio or uderestimatio. Now, let s suppose that we wat a better estimatio, because oe of the estimatios above really did all that great of a job at estimatig the area. We could try to fid a differet poit to use for the height of each rectagle but that would be cumbersome ad there would t be ay guaratee that the estimatio would i fact be better. Also, we would like a method for gettig better approximatios that would work for ay fuctio we would chose to work with ad if we just pick ew poits that may ot work for other fuctios. The easiest way to get a better approximatio is to take more rectagles (i.e. icrease ). Let s double the umber of rectagles that we used ad see what happes. Here are the graphs showig the eight rectagles ad the estimatios for each of the three choices for rectagle heights that we used above. 007 Paul Dawkis 5

53 Math 40 Here are the area estimatios for each of these cases. A = A = A = r l m So, icreasig the umber of rectagles did improve the accuracy of the estimatio as we d guessed that it would. Let s work a slightly more complicated example. Example Estimate the area betwee ( ) f x = x - 5x + 6x+ 5 ad the x-axis usig = 5 subitervals ad all three cases above for the heights of each rectagle. Solutio First, let s get the graph to make sure that the fuctio is positive. So, the graph is positive ad the width of each subiterval will be, 4 D x = = This meas that the edpoits of the subitervals are, 0, 0.8,.6,.4,., 4 Let s first look at usig the right edpoits for the fuctio height. Here is the graph for this case. 007 Paul Dawkis 5

54 Math 40 Notice, that ulike the first area we looked at, the choosig the right edpoits here will both over ad uderestimate the area depedig o where we are o the curve. This will ofte be the case with a more geeral curve that the oe we iitially looked at. The area estimatio usig the right edpoits of each iterval for the rectagle height is, ( ) ( ) ( ) ( ) ( ) A = 0.8f f f f f 4 r = 8.96 Now let s take a look at left edpoits for the fuctio height. Here is the graph. The area estimatio usig the left edpoits of each iterval for the rectagle height is, ( ) ( ) ( ) ( ) ( ) A = 0.8f f f f f. r =.56 Fially, let s take a look at the midpoits for the heights of each rectagle. Here is the graph, 007 Paul Dawkis 5

55 Math 40 The area estimatio usig the midpoit is the, ( ) ( ) ( ) ( ) ( ) A = 0.8f f f + 0.8f f.6 r = 5. For compariso purposes the exact area is, 76 A = = 5. So, agai the midpoit did a better job tha the other two. While this will be the case more ofte tha ot, it wo t always be the case ad so do t expect this to always happe. Now, let s move o to the geeral case. Let s start out with f ( x) 0 o [a,b] ad we ll divide the iterval ito subitervals each of legth, b- a D x = Note that the subitervals do t have to be equal legth, but it will make our work sigificatly easier. The edpoits of each subiterval are, x = a 0 x = a+dx x = a+ Dx M x = a+d i x M i - ( ) x = a+ - Dx x = a+ D x= b 007 Paul Dawkis 54

56 Math 40 Next i each iterval, we choose a poit [ x, x ],[ x, x ], K, [ x, x ], K,[ x, x ] * * * * i 0 i- i - x, x, K, x, K x. These poits will defie the height of the rectagle i each subiterval. Note as well that these poits do ot have to occur at the same poit i each subiterval. Here is a sketch of this situatio. The area uder the curve o the give iterval is the approximately, * * * * ( ) ( ) L ( i ) L ( ) Aª f x D x+ f x D x+ + f x D x+ + f x Dx We will use summatio otatio or sigma otatio at this poit to simplify up our otatio a little. If you eed a refresher o summatio otatio check out the sectio devoted to this i the Extras chapter. Usig summatio otatio the area estimatio is, Â i= * ( i ) Aª f x Dx The summatio i the above equatio is called a Riema Sum. To get a better estimatio we will take larger ad larger. I fact, if we let go out to ifiity we will get the exact area. I other words, Â Æ i = * ( i ) A= lim f x Dx Before leavig this sectio let s address oe more issue. To this poit we ve required the fuctio to be positive i our work. May fuctios are ot positive however. Cosider the case 007 Paul Dawkis 55

57 Math 40 f x = x -4 o [0,]. If we use = 8 ad the midpoits for the rectagle height we get the of ( ) followig graph, I this case let s otice that the fuctio lies completely below the x-axis ad hece is always egative. If we igore the fact that the fuctio is always egative ad use the same ideas above to estimate the area betwee the graph ad the x-axis we get, ʈ ʈ Ê5ˆ Ê7ˆ Ê9ˆ Am = f Á + f Á + f Á + f Á + f Á + 4 Ë8 4 Ë8 4 Ë8 4 Ë8 4 Ë8 ʈ ʈ Ê5 ˆ f Á + f Á + f Á 4 Ë 8 4 Ë 8 4 Ë 8 = Our aswer is egative as we might have expected give that all the fuctio evaluatios are egative. So, usig the techique i this sectio it looks like if the fuctio is above the x-axis we will get a positive area ad if the fuctio is below the x-axis we will get a egative area. Now, what about f x = x - o a fuctio that is both positive ad egative i the iterval? For example, ( ) [0,]. Usig = 8 ad midpoits the graph is, 007 Paul Dawkis 56

58 Math 40 Some of the rectagles are below the x-axis ad so will give egative areas while some are above the x-axis ad will give positive areas. Sice more rectagles are below the x-axis tha above it looks like we should probably get a egative area estimatio for this case. I fact that is correct. Here the area estimatio for this case. ʈ ʈ Ê5ˆ Ê7ˆ Ê9ˆ Am = f Á + f Á + f Á f Á + f Á + 4 Ë8 4 Ë8 4 Ë8 4 Ë8 4 Ë8 ʈ ʈ Ê5 ˆ f Á + f Á + f Á 4 Ë 8 4 Ë 8 4 Ë 8 =-.475 I cases where the fuctio is both above ad below the x-axis the techique give i the sectio will give the et area betwee the fuctio ad the x-axis with areas below the x-axis egative ad areas above the x-axis positive. So, if the et area is egative the there is more area uder the x-axis tha above while a positive et area will mea that more of the area is above the x-axis. 007 Paul Dawkis 57

59 Math 40 The Defiitio of the Defiite Itegral I this sectio we will formally defie the defiite itegral ad give may of the properties of defiite itegrals. Let s start off with the defiitio of a defiite itegral. Defiite Itegral Give a fuctio f ( x ) that is cotiuous o the iterval [a,b] we divide the iterval ito subitervals of equal width, itegral of f(x) from a to b is D x, ad from each iterval choose a poit, b a * ( ) limâ ( i ) f x dx= f x Dx Æ i = * x i. The the defiite The defiite itegral is defied to be exactly the limit ad summatio that we looked at i the last sectio to fid the et area betwee a fuctio ad the x-axis. Also ote that the otatio for the defiite itegral is very similar to the otatio for a idefiite itegral. The reaso for this will be apparet evetually. There is also a little bit of termiology that we should get out of the way here. The umber a that is at the bottom of the itegral sig is called the lower limit of the itegral ad the umber b at the top of the itegral sig is called the upper limit of the itegral. Also, despite the fact that a ad b were give as a iterval the lower limit does ot ecessarily eed to be smaller tha the upper limit. Collectively we ll ofte call a ad b the iterval of itegratio. Let s work a quick example. This example will use may of the properties ad facts from the brief review of summatio otatio i the Extras chapter. Example Usig the defiitio of the defiite itegral compute the followig. x + dx 0 Solutio First, we ca t actually use the defiitio uless we determie which poits i each iterval that well use for * x i. I order to make our life easier we ll use the right edpoits of each iterval. From the previous sectio we kow that for a geeral the width of each subiterval is, - 0 D x = = The subitervals are the, ( i- ) i ( - ) È È 4 È4 6 È È Í 0,,,,,,,,,,, Í Í Í Í K K Î Î Î Î Î As we ca see the right edpoit of the i th subiterval is 007 Paul Dawkis 58

60 Math 40 x i * i = The summatio i the defiitio of the defiite itegral is the,  * ÊiˆÊˆ f ( xi ) D x=  f Á Á Ë Ë i= i= ÊÊiˆ ˆÊˆ =  + ÁÁ Á i= Ë Ë Ë Ê8i ˆ = ÂÁ + i= Ë Now, we are goig to have to take a limit of this. That meas that we are goig to eed to evaluate this summatio. I other words, we are goig to have to use the formulas give i the summatio otatio review to elimiate the actual summatio ad get a formula for this for a geeral. To do this we will eed to recogize that is a costat as far as the summatio otatio is cocered. As we cycle through the itegers from to i the summatio oly i chages ad so aythig that is t a i will be a costat ad ca be factored out of the summatio. I particular ay that is i the summatio ca be factored out if we eed to. Here is the summatio evaluatio. 8i *  f ( xi ) D x=  +  i= i= i= We ca ow compute the defiite itegral. 8 = Âi +  i= i= ( + )( + ) 8 Ê ˆ = Á + Ë 6 4( + )( + ) = = ( ) 007 Paul Dawkis 59

61 Math 40 0 ( i ) * x + dx= lim f x Dx Æ i = = lim Æ 4 = We ve see several methods for dealig with the limit i this problem so I ll leave it to you to verify the results. Wow, that was a lot of work for a fairly simple fuctio. There is a much simpler way of evaluatig these ad we will get to it evetually. The mai purpose to this sectio is to get the mai properties ad facts about the defiite itegral out of the way. We ll discuss how we compute these i practice startig with the ext sectio. So, let s start takig a look at some of the properties of the defiite itegral. Properties b a. ( ) =- ( ) f x dx f x dx a. We ca iterchage the limits o ay defiite itegral, all that b we eed to do is tack a mius sig oto the itegral whe we do. a. f ( x ) dx = 0 a itegral is zero. b b. ( ) = ( ) Â. If the upper ad lower limits are the same the there is o work to do, the cf x dx c f x dx a, where c is ay umber. So, as with limits, derivatives, ad a idefiite itegrals we ca factor out a costat. b b b 4. ( ) ± ( ) = ( ) ± ( ) f x g x dx a f x dx g x dx a. We ca break up defiite itegrals across a a sum or differece. b c b 5. ( ) = ( ) + ( ) f x dx a f x dx f x dx a where c is ay umber. This property is more c importat tha we might realize at first. Oe of the mai uses of this property is to tell us how we ca itegrate a fuctio over the adjacet itervals, [a,c] ad [c,b]. Note however that c does t eed to be betwee a ad b. b b 6. ( ) = () f x dx f t dt a. The poit of this property is to otice that as log as the fuctio a ad limits are the same the variable of itegratio that we use i the defiite itegral wo t affect the aswer. 007 Paul Dawkis 60

62 Math 40 See the Proof of Various Itegral Properties sectio of the Extras chapter for the proof of properties 4. Property 5 is ot easy to prove ad so is ot show there. Property 6 is ot really a property i the full sese of the word. It is oly here to ackowledge that as log as the fuctio ad limits are the same it does t matter what letter we use for the variable. The aswer will be the same. Let s do a couple of examples dealig with these properties. Example Use the results from the first example to evaluate each of the followig. 0 (a) x + dx [Solutio] (b) 0 x dx [Solutio] (c) 0 t + dt [Solutio] Solutio All of the solutios to these problems will rely o the fact we proved i the first example. Namely that, 4 x + dx = 0 (a) 0 x + dx I this case the oly differece betwee the two is that the limits have iterchaged. So, usig the first property gives, (b) 0 x dx 0 x + dx=- x + dx 0 =- 007 Paul Dawkis [Retur to Problems] For this part otice that we ca factor a 0 out of both terms ad the out of the itegral usig the third property. 0 0 ( ) 0x + 0dx= 0 x + dx 0 x 0 = + Ê4 ˆ = 0Á Ë 40 = dx [Retur to Problems]

63 Math 40 (c) 0 t + dt I this case the oly differece is the letter used ad so this is just goig to use property 6. 4 t + dt = x dx 0 + = 0 [Retur to Problems] Here are a couple of examples usig the other properties. Example Evaluate the followig defiite itegral. 0 x - xsi( x) + cos( x) Ù dx ı0 x + Solutio There really is t aythig to do with this itegral oce we otice that the limits are the same. Usig the secod property this is, Ù ı 0 0 ( ) cos( ) x - xsi x + x 0 dx = x Example 4 Give that f ( x) dx = ad ( ) ( )- ( ) g x dx =-9determie the value of f x 0g x dx -0 Solutio We will first eed to use the fourth property to break up the itegral ad the third property to factor out the costats. ( )- ( ) = ( ) - ( ) f x 0g x dx f x dx 0g x dx ( ) ( ) 6 6 = f x dx-0 g x dx -0-0 Now otice that the limits o the first itegral are iterchaged with the limits o the give itegral so switch them usig the first property above (ad addig a mius sig of course). Oce this is doe we ca plug i the kow values of the itegrals. - ( )- ( ) =- ( ) - ( ) f x 0g x dx f x dx 0 g x dx ( ) ( ) = = Paul Dawkis 6

64 Math Example 5 Give that f ( x) dx = 6, f ( x) dx =-, ad f ( ) the value of f ( x ) dx x dx = 4 determie Solutio This example is mostly a example of property 5 although there are a couple of uses of property i the solutio as well. We eed to figure out how to correctly break up the itegral usig property 5 to allow us to use the give pieces of iformatio. First we ll ote that there is a itegral that has a -5 i oe of the limits. It s ot the lower limit, but we ca use property to correct that evetually. The other limit is 00 so this is the umber c that we ll use i property 5. ( ) = ( ) + ( ) 00 f x dx f x dx f x dx We ll be able to get the value of the first itegral, but the secod still is t i the list of kow itegrals. However, we do have secod limit that has a limit of 00 i it. The other limit for this secod itegral is -0 ad this will be c i this applicatio of property 5. - ( ) = ( ) + ( ) + ( ) 00 0 f x dx f x dx f x dx f x dx At this poit all that we eed to do is use the property o the first ad third itegral to get the limits to match up with the kow itegrals. After that we ca plug i for the kow itegrals ( ) =- ( ) + ( ) - ( ) f x dx f x dx f x dx f x dx =-4--6 =- There are also some ice properties that we ca use i comparig the geeral size of defiite itegrals. Here they are. More Properties b 7. cdx = cb ( - a) a, c is ay umber. b 8. If f ( x) 0 for a x b the ( ) 0 f x dx. a b b 9. If f ( x) g( x) for a x bthe ( ) ( ) f x dx g x dx. b 0. If m f ( x) M for a x b the mb ( - a) f ( x) dx M ( b-a) b b. ( ) ( ) a a f x dx f x dx a a a Paul Dawkis 6

65 Math 40 See the Proof of Various Itegral Properties sectio of the Extras chapter for the proof of these properties. Iterpretatios of Defiite Itegral There are a couple of quick iterpretatios of the defiite itegral that we ca give here. First, as we alluded to i the previous sectio oe possible iterpretatio of the defiite itegral is to give the et area betwee the graph of f ( x ) ad the x-axis o the iterval [a,b]. So, the et area betwee the graph of ( ) f x = x + ad the x-axis o [0,] is, 0 x 4 + dx = If you look back i the last sectio this was the exact area that was give for the iitial set of problems that we looked at i this area. Aother iterpretatio is sometimes called the Net Chage Theorem. This iterpretatio says that if f ( x ) is some quatity (so f ( x) is the rate of chage of f ( ) b ( ) ( ) ( ) f x dx= f b - f a a x, the, is the et chage i f ( x ) o the iterval [a,b]. I other words, compute the defiite itegral of a rate of chage ad you ll get the et chage i the quatity. We ca see that the value of the defiite itegral, f ( b) - f ( a), does i fact give use the et chage i f ( ) x ad so there really is t aythig to prove with this statemet. This is really just a ackowledgmet of what the defiite itegral of a rate of chage tells us. So as a quick example, if V ( t ) is the volume of water i a tak the, t t () ( ) ( ) V t dt = V t -V t is the et chage i the volume as we go from time t to time t. Likewise, if s( t ) is the fuctio givig the positio of some object at time t we kow that the velocity of the object at ay time t is : vt ( ) s ( t) time t to time t is, t t =. Therefore the displacemet of the object () = ( )- ( ) v t dt s t s t Note that i this case if vt ( ) is both positive ad egative (i.e. the object moves to both the right ad left) i the time frame this will NOT give the total distace traveled. It will oly give the displacemet, i.e. the differece betwee where the object started ad where it eded up. To get the total distace traveled by a object we d have to compute, 007 Paul Dawkis 64

66 Math 40 t t () vt dt It is importat to ote here that the Net Chage Theorem oly really makes sese if we re itegratig a derivative of a fuctio. Fudametal Theorem of Calculus, Part I As oted by the title above this is oly the first part to the Fudametal Theorem of Calculus. We will give the secod part i the ext sectio as it is the key to easily computig defiite itegrals ad that is the subject of the ext sectio. The first part of the Fudametal Theorem of Calculus tells us how to differetiate certai types of defiite itegrals ad it also tells us about the very close relatioship betwee itegrals ad derivatives. Fudametal Theorem of Calculus, Part I If f ( x ) is cotiuous o [a,b] the, = x ( ) () g x f t dt is cotiuous o [a,b] ad it is differetiable o ( ab, ) ad that, g ( x) = f ( x) A alterate otatio for the derivative portio of this is, d x f t dt f x dx = a a () ( ) To see the proof of this see the Proof of Various Itegral Properties sectio of the Extras chapter. Let s check out a couple of quick examples usig this. Example 6 Differetiate each of the followig. x (a) ( ) t g x = e cos ( -5t) dt [Solutio] (b) Ù ı Solutio 4 x t t -4 + dt + [Solutio] x (a) ( ) t g x = cos ( -5 ) e -4 t dt This oe is othig more tha a quick applicatio of the Fudametal Theorem of Calculus. x g x = e cos -5x ( ) ( ) [Retur to Problems] 007 Paul Dawkis 65

67 Math 40 (b) Ù ı 4 x t t + dt + This oe eeds a little work before we ca use the Fudametal Theorem of Calculus. The first thig to otice is that the FToC requires the lower limit to be a costat ad the upper limit to be the variable. So, usig a property of defiite itegrals we ca iterchage the limits of the itegral we just eed to remember to add i a mius sig after we do that. Doig this gives, 4 x 4 x 4 d t + d Ê t + ˆ d t + dt = - dt =- Ù dt dx x t dxá Ù + t + Ù ı Ë ı dxı t + The ext thig to otice is that the FToC also requires a x i the upper limit of itegratio ad we ve got x. To do this derivative we re goig to eed the followig versio of the chai rule. d d du ( g( u) ) = ( g( u) ) where u = f ( x) dx du dx So, if we let u= x we use the chai rule to get, 4 x 4 d t + d t + dt =- Ù Ù dt dxı t + dxı t + x u 4 d t + du Ù duı t + dx =- dt where u = x 4 u + =- u + 4 u + =-x u + ( x) The fial step is to get everythig back i terms of x. d Ù 4 t + dt =-x ( x ) ( ) dxı x t + x + x =-x x [Retur to Problems] Usig the chai rule as we did i the last part of this example we ca derive some geeral formulas for some more complicated problems. First, d dx ( ) ux a () ( ) ( ( )) f t dt = u x f u x This is simply the chai rule for these kids of problems. 007 Paul Dawkis 66

68 Math 40 Next, we ca get a formula for itegrals i which the upper limit is a costat ad the lower limit is a fuctio of x. All we eed to do here is iterchage the limits o the itegral (addig i a mius sig of course) ad the usig the formula above to get, d dx d ( ) f t dt =- f t dt =-v x f v x dx b vx () () vx ( ) b ( ) ( ( )) Fially, we ca also get a versio for both limits beig fuctios of x. I this case we ll eed to use Property 5 above to break up the itegral as follows, ( ) ( ) () ( ) () () ( ) ux a ux f t dt = f t dt+ f t dt vx vx a We ca use pretty much ay value of a whe we break up the itegral. The oly thig that we eed to avoid is to make sure that f ( a ) exists. So, assumig that ( ) up the itegral we ca the differetiate ad use the two formulas above to get, d dx Let s work a quick example. ( ) () d ( ( ) () ( ) () ) ( ) ( ) ux a u x f t dt = f t dt f t dt vx ( ) dx + vx a =- v x f ( v x ) + u ( x) f ( u( x) ) f a exists after we break Example 7 Differetiate the followig itegral. ( + ) x t si t dt x Solutio This will use the fial formula that we derived above. d dx - ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) x x t si + t dt =- x x si + x + x si + x =- xsi( + x) + 7x si( + 9x ) 007 Paul Dawkis 67

69 Math 40 Computig Defiite Itegrals I this sectio we are goig to cocetrate o how we actually evaluate defiite itegrals i practice. To do this we will eed the Fudametal Theorem of Calculus, Part II. Fudametal Theorem of Calculus, Part II Suppose f ( x ) is a cotiuous fuctio o [a,b] ad also suppose that F( x ) is ay atiderivative for f ( x ). The, b ( ) = b ( ) = ( ) - ( ) f x dx F x F b F a a a To see the proof of this see the Proof of Various Itegral Properties sectio of the Extras chapter. Recall that whe we talk about a ati-derivative for a fuctio we are really talkig about the idefiite itegral for the fuctio. So, to evaluate a defiite itegral the first thig that we re goig to do is evaluate the idefiite itegral for the fuctio. This should explai the similarity i the otatios for the idefiite ad defiite itegrals. Also otice that we require the fuctio to be cotiuous i the iterval of itegratio. This was also a requiremet i the defiitio of the defiite itegral. We did t make a big deal about this i the last sectio. I this sectio however, we will eed to keep this coditio i mid as we do our evaluatios. Next let s address the fact that we ca use ay ati-derivative of f ( x ) i the evaluatio. Let s take a fial look at the followig itegral. 0 x + dx Both of the followig are ati-derivatives of the itegrad. 8 F( x) = x + x ad F( x) = x + x- Usig the FToC to evaluate this itegral with the first ati-derivatives gives, Ê Ë x + dx= x + x 0 Á ˆ 0 Ê ˆ = ( ) + - Á ( 0) + 0 Ë 4 = 007 Paul Dawkis 68

70 Math 40 Much easier tha usig the defiitio was t it? Let s ow use the secod ati-derivative to evaluate this defiite itegral. Ê 8 ˆ + = + - Ë x dx x x 0 Á Ê ˆ = ( ) Á ( 0) + 0- Ë = = The costat that we tacked oto the secod ati-derivative caceled i the evaluatio step. So, whe choosig the ati-derivative to use i the evaluatio process make your life easier ad do t bother with the costat as it will oly ed up cacelig i the log ru. Also, ote that we re goig to have to be very careful with mius sigs ad parethesis with these problems. It s very easy to get i a hurry ad mess them up. Let s start our examples with the followig set desiged to make a couple of quick poits that are very importat. Example Evaluate each of the followig. - (a) y + y dy [Solutio] Solutio (a) (b) (c) + - y y dy y y - + dy [Solutio] y y - + dy - [Solutio] This is the oly idefiite itegral i this sectio ad by ow we should be gettig pretty good with these so we wo t sped a lot of time o this part. This is here oly to make sure that we uderstad the differece betwee a idefiite ad a defiite itegral. The itegral is, - - y + y dy = y - y + c [Retur to Problems] (b) y y - + dy Recall from our first example above that all we really eed here is ay ati-derivative of the itegrad. We just computed the most geeral ati-derivative i the first part so we ca use that if we wat to. However, recall that as we oted above ay costats we tack o will just cacel 007 Paul Dawkis 69

71 Math 40 i the log ru ad so we ll use the aswer from (a) without the +c. Here s the itegral, y + y - dy = Ê Á y - ˆ Ë y Ê ˆ = ( ) - -Á () - Ë 8 = = 6 Remember that the evaluatio is always doe i the order of evaluatio at the upper limit mius evaluatio at the lower limit. Also be very careful with mius sigs ad parethesis. It s very easy to forget them or mishadle them ad get the wrog aswer. Notice as well that, i order to help with the evaluatio, we rewrote the idefiite itegral a little. I particular we got rid of the egative expoet o the secod term. It s geerally easier to evaluate the term with positive expoets. [Retur to Problems] (c) y y - + dy - This itegral is here to make a poit. Recall that i order for us to do a itegral the itegrad must be cotiuous i the rage of the limits. I this case the secod term will have divisio by zero at y = 0 ad sice y = 0 is i the iterval of itegratio, i.e. it is betwee the lower ad upper limit, this itegrad is ot cotiuous i the iterval of itegratio ad so we ca t do this itegral. Note that this problem will ot prevet us from doig the itegral i (b) sice y = 0 is ot i the iterval of itegratio. So what have we leared from this example? [Retur to Problems] First, i order to do a defiite itegral the first thig that we eed to do is the idefiite itegral. So we are t goig to get out of doig idefiite itegrals, they will be i every itegral that we ll be doig i the rest of this course so make sure that you re gettig good at computig them. Secod, we eed to be o the lookout for fuctios that are t cotiuous at ay poit betwee the limits of itegratio. Also, it s importat to ote that this will oly be a problem if the 007 Paul Dawkis 70

72 Math 40 poit(s) of discotiuity occur betwee the limits of itegratio or at the limits themselves. If the poit of discotiuity occurs outside of the limits of itegratio the itegral ca still be evaluated. I the followig sets of examples we wo t make too much of a issue with cotiuity problems, or lack of cotiuity problems, uless it affects the evaluatio of the itegral. Do ot let this covice you that you do t eed to worry about this idea. It arises ofte eough that it ca cause real problems if you are t o the lookout for it. Fially, ote the differece betwee idefiite ad defiite itegrals. Idefiite itegrals are fuctios while defiite itegrals are umbers. Let s work some more examples. Example Evaluate each of the followig. (a) 6x - 5x + dx [Solutio] - 0 (b) t( t- ) dt [Solutio] 4 5 w - w+ (c) Ù dw ı w -0 (d) dr [Solutio] 5 Solutio (a) x x dx [Solutio] There is t a lot to this oe other tha simply doig the work. Ê 5 ˆ x x dx x x x Ë = Á - Ê 5 ˆ Ê 45 ˆ = Á- + -Á Ë Ë = 84 [Retur to Problems] 0 (b) t( t- ) 4 dt Recall that we ca t itegrate products as a product of itegrals ad so we first eed to multiply the itegrad out before itegratig, just as we did i the idefiite itegral case. 007 Paul Dawkis 7

73 Math 40 I the evaluatio process recall that, 0 0 ( ) t t- dt = t -t dt 4 4 Ê 4 = Á t - t Ë5 5 ˆ Ê 4 = 0-Á 4-4 Ë5 =- 5 ( ) ( ) ( ) ( ) ( ) Ê ˆ = Á 4 = = Ë 4 Ê ˆ = Á 4 = = 8 Ë ( ) ( ) ( ) Also, do t get excited about the fact that the lower limit of itegratio is larger tha the upper limit of itegratio. That will happe o occasio ad there is absolutely othig wrog with this. [Retur to Problems] ˆ (c) Ù ı 5 w - w+ dw w First, otice that we will have a divisio by zero issue at w= 0, but sice this is t i the iterval of itegratio we wo t have to worry about it. Next agai recall that we ca t itegrate quotiets as a quotiet of itegrals ad so the first step that we ll eed to do is break up the quotiet so we ca itegrate the fuctio. Ù ı 5 w - w+ - dw= w w dw Ù - + w ı w Ê 4 ˆ = Á w -l w - Ë w Ê ˆ Ê ˆ = Á8-l- -Á -l- Ë Ë = 9-l Do t get excited about aswers that do t come dow to a simple iteger or fractio. Ofte times they wo t. Also do t forget that l( ) = 0. [Retur to Problems] 007 Paul Dawkis 7

74 Math 40 (d) -0 dr 5 This oe is actually pretty easy. Recall that we re just itegratig! dr= R 5 =-0-5 =-5 [Retur to Problems] The last set of examples dealt exclusively with itegratig powers of x. Let s work a couple of examples that ivolve other fuctios. Example Evaluate each of the followig. (a) 4x- 6 x dx [Solutio] 0 (e) Ù ı Solutio (a) p (b) siq - 5cosq dq [Solutio] 0 p 4 (c) 5- sec z ta zdz [Solutio] p 6 - (d) Ù - dz [Solutio] -z ı e z -0 t - t+ dt t x x - dx [Solutio] This oe is here mostly here to cotrast with the ext example. 4x- 6 x dx= 4x-6x dx 0 0 Ê 8 = Áx - x Ë = =- 5 ( ) ˆ 0 [Retur to Problems] p (b) siq - 5cosq dq 0 Be careful with sigs with this oe. Recall from the idefiite itegral sectios that it s easy to mess up the sigs whe itegratig sie ad cosie. 007 Paul Dawkis 7

75 Math 40 p 0 ( ) siq - 5cosq dq = -cosq -5siq p 0 Êp ˆ Êp ˆ =-cosá -5siÁ - -cos0-5si0 Ë Ë 5 = = - ( ) Compare this aswer to the previous aswer, especially the evaluatio at zero. It s very easy to get ito the habit of just writig dow zero whe evaluatig a fuctio at zero. This is especially a problem whe may of the fuctios that we itegrate ivolve oly x s raised to positive itegers ad i these cases evaluate is zero of course. After evaluatig may of these kids of defiite itegrals it s easy to get ito the habit of just writig dow zero whe you evaluate at zero. However, there are may fuctios out there that are t zero whe evaluated at zero so be careful. [Retur to Problems] (c) p 4 p 6 5- sec z ta zdz Not much to do other tha do the itegral. ( ) p 4 p 4 p 6 5- secztazdz = 5z-sec z p 6 Êp ˆ Êp ˆ Ê Êp ˆ Êp ˆˆ = 5Á -secá - 5 sec 4 4 Á Á - Á 6 6 Ë Ë Ë Ë Ë 5p 4 = - + For the evaluatio, recall that sec z = cos z ad so if we ca evaluate cosie at these agles we ca evaluate secat at these agles. [Retur to Problems] (d) Ù ı dz -z e z I order to do this oe will eed to rewrite both of the terms i the itegral a little as follows, - - z - dz = Ù dz -z Ù e - ı-0e z ı-0 z For the first term recall we used the followig fact about expoets. 007 Paul Dawkis 74

76 Math 40 x -a = = x a -a x x a I the secod term, takig the out of the deomiator will just make itegratig that term easier. Now the itegral. Ù ı - - Ê z ˆ - dz = l z -z Á e - -0 e z Ë -0 Ê = e - - -Á e - - Ë - -0 = e - e + l 0 Just leave the aswer like this. It s messy, but it s also exact l l 0 Note that the absolute value bars o the logarithm are required here. Without them we could t have doe the evaluatio. [Retur to Problems] (e) Ù ı t - t+ dt t This itegral ca t be doe. There is divisio by zero i the third term at t = 0 ad t = 0 lies i the iterval of itegratio. The fact that the first two terms ca be itegrated does t matter. If eve oe term i the itegral ca t be itegrated the the whole itegral ca t be doe. [Retur to Problems] So, we ve computed a fair umber of defiite itegrals at this poit. Remember that the vast majority of the work i computig them is first fidig the idefiite itegral. Oce we ve foud that the rest is just some umber cruchig. There are a couple of particularly tricky defiite itegrals that we eed to take a look at ext. Actually they are oly tricky util you see how to do them, so do t get too excited about them. The first oe ivolves itegratig a piecewise fuctio. ˆ Example 4 Give, Evaluate each of the followig itegrals. (a) f ( x ) dx [Solutio] 0 (b) ( ) [Solutio] - f x dx Solutio Let s first start with a graph of this fuctio. f ( x) =Ì Ï6 if x> Óx if x 007 Paul Dawkis 75

77 Math 40 The graph reveals a problem. This fuctio is ot cotiuous at x = ad we re goig to have to watch out for that. (a) f ( x ) dx 0 For this itegral otice that x = is ot i the iterval of itegratio ad so that is somethig that we ll ot eed to worry about i this part. Also ote the limits for the itegral lie etirely i the rage for the first fuctio. What this meas for us is that whe we do the itegral all we eed to do is plug i the first fuctio ito the itegral. Here is the itegral. ( ) f x dx= 6dx 0 0 = 6x 0 = -60 = 7 [Retur to Problems] (b) f ( x ) dx - I this part x = is betwee the limits of itegratio. This meas that the itegrad is o loger cotiuous i the iterval of itegratio ad that is a show stopper as far we re cocered. As oted above we simply ca t itegrate fuctios that are t cotiuous i the iterval of itegratio. Also, eve if the fuctio was cotiuous at x = we would still have the problem that the fuctio is actually two differet equatios depedig where we are i the iterval of itegratio. 007 Paul Dawkis 76

78 Math 40 Let s first address the problem of the fuctio ot begiig cotiuous at x =. As we ll see, i this case, if we ca fid a way aroud this problem the secod problem will also get take care of at the same time. I the previous examples where we had fuctios that were t cotiuous we had divisio by zero ad o matter how hard we try we ca t get rid of that problem. Divisio by zero is a real problem ad we ca t really avoid it. I this case the discotiuity does ot stem from problems with the fuctio ot existig at x =. Istead the fuctio is ot cotiuous because it takes o differet values o either sides of x =. We ca remove this problem by recallig Property 5 from the previous sectio. This property tells us that we ca write the itegral as follows, ( ) = ( ) + ( ) f x dx f x dx f x dx - - O each of these itervals the fuctio is cotiuous. I fact we ca say more. I the first itegral we will have x betwee - ad ad this meas that we ca use the secod equatio for f ( ) x ad likewise for the secod itegral x will be betwee ad ad so we ca use the first fuctio for f ( x ). The itegral i this case is the, ( ) = ( ) + ( ) f x dx f x dx f x dx - - = x dx+ 6dx - 6x - = x + ( ) ( ) = = [Retur to Problems] So, to itegrate a piecewise fuctio, all we eed to do is break up the itegral at the break poit(s) that happe to occur i the iterval of itegratio ad the itegrate each piece. Next we eed to look at is how to itegrate a absolute value fuctio. Example 5 Evaluate the followig itegral. t- 5 dt 0 Solutio Recall that the poit behid idefiite itegratio (which we ll eed to do i this problem) is to determie what fuctio we differetiated to get the itegrad. To this poit we ve ot see ay fuctios that will differetiate to get a absolute value or will we ever see a fuctio that will differetiate to get a absolute value. 007 Paul Dawkis 77

79 Math 40 The oly way that we ca do this problem is to get rid of the absolute value. To do this we eed to recall the defiitio of absolute value. Ïx if x 0 x = Ì Ó - x if x < 0 Oce we remember that we ca defie absolute value as a piecewise fuctio we ca use the work from Example 4 as a guide for doig this itegral. What we eed to do is determie where the quatity o the iside of the absolute value bars is 5 egative ad where it is positive. It looks like if t > the quatity iside the absolute value is 5 positive ad if t < the quatity iside the absolute value is egative. 5 Next, ote that t = is i the iterval of itegratio ad so, if we break up the itegral at this poit we get, t- 5 dt = t- 5 dt+ t-5 dt 5 Now, i the first itegrals we have t < ad so t - 5< 0 i this iterval of itegratio. That meas we ca drop the absolute value bars if we put i a mius sig. Likewise i the secod 5 itegral we have t > which meas that i this iterval of itegratio we have t - 5> 0 ad so we ca just drop the absolute value bars i this itegral. After gettig rid of the absolute value bars i each itegral we ca do each itegral. So, doig the itegratio gives, 5 ( ) t- 5 dt = - t- 5 dt+ t-5dt 5 0 = - t+ 5dt+ t-5dt 5 5 Ê ˆ Ê ˆ = Á- t + 5t + Á t -5t Ë Ë 0 5 Ê5ˆ Ê5ˆ Ê ÊÊ5ˆ Ê5ˆˆˆ =- Á + 5Á - ( 0) + ( ) -5 ( )- Á -5Á Ë Ë Á Á Ë Ë Ë Ë 5 8 = = Paul Dawkis 78

80 Math 40 Itegratig absolute value fuctios is t too bad. It s a little more work tha the stadard defiite itegral, but it s ot really all that much more work. First, determie where the quatity iside the absolute value bars is egative ad where it is positive. Whe we ve determied that poit all we eed to do is break up the itegral so that i each rage of limits the quatity iside the absolute value bars is always positive or always egative. Oce this is doe we ca drop the absolute value bars (addig egative sigs whe the quatity is egative) ad the we ca do the itegral as we ve always doe. 007 Paul Dawkis 79

81 Math 40 Substitutio Rule for Defiite Itegrals We ow eed to go back ad revisit the substitutio rule as it applies to defiite itegrals. At some level there really is t a lot to do i this sectio. Recall that the first step i doig a defiite itegral is to compute the idefiite itegral ad that has t chaged. We will still compute the idefiite itegral first. This meas that we already kow how to do these. We use the substitutio rule to fid the idefiite itegral ad the do the evaluatio. There are however, two ways to deal with the evaluatio step. Oe of the ways of doig the evaluatio is the probably the most obvious at this poit, but also has a poit i the process where we ca get i trouble if we are t payig attetio. Let s work a example illustratig both ways of doig the evaluatio step. Example Evaluate the followig defiite itegral. 0 t 4 - t dt - Solutio Let s start off lookig at the first way of dealig with the evaluatio step. We ll eed to be careful with this method as there is a poit i the process where if we are t payig attetio we ll get the wrog aswer. Solutio : We ll first eed to compute the idefiite itegral usig the substitutio rule. Note however, that we will costatly remid ourselves that this is a defiite itegral by puttig the limits o the itegral at each step. Without the limits it s easy to forget that we had a defiite itegral whe we ve gotte the idefiite itegral computed. I this case the substitutio is, u = - 4t du =-tdt fi tdt =- du Pluggig this ito the itegral gives, t - 4t dt =- u du =- u 9 Notice that we did t do the evaluatio yet. This is where the potetial problem arises with this solutio method. The limits give here are from the origial itegral ad hece are values of t. We have u s i our solutio. We ca t plug values of t i for u. Therefore, we will have to go back to t s before we do the substitutio. This is the stadard step Paul Dawkis 80

82 Math 40 i the substitutio process, but it is ofte forgotte whe doig defiite itegrals. Note as well that i this case, if we do t go back to t s we will have a small problem i that oe of the evaluatios will ed up givig us a complex umber. So, fiishig this problem gives, 0-0 ( ) t - 4t dt =- -4t 9 - Ê =- - Á - 9 Ë 9 = ( - ) 9 ( ) So, that was the first solutio method. Let s take a look at the secod method. Solutio : Note that this solutio method is t really all that differet from the first method. I this method we are goig to remember that whe doig a substitutio we wat to elimiate all the t s i the itegral ad write everythig i terms of u. Whe we say all here we really mea all. I other words, remember that the limits o the itegral are also values of t ad we re goig to covert the limits ito u values. Covertig the limits is pretty simple sice our substitutio will tell us how to relate t ad u so all we eed to do is plug i the origial t limits ito the substitutio ad we ll get the ew u limits. Here is the substitutio (it s the same as the first method) as well as the limit coversios. u = - 4t du =-tdt fi tdt =- du ( ) t =- fi u = -4 - = ( ) t = 0 fi u = - 40 = ˆ The itegral is ow, t - 4t dt =- u du =- u 9 As with the first method let s pause here a momet to remid us what we re doig. I this case, we ve coverted the limits to u s ad we ve also got our itegral i terms of u s ad so here we ca just plug the limits directly ito our itegral. Note that i this case we wo t plug our substitutio back i. Doig this here would cause problems as we would have t s i the itegral 007 Paul Dawkis 8

83 Math 40 ad our limits would be u s. Here s the rest of this problem. 0 - t - 4t dt =- u 9 Ê =- -Á- = - 9 Ë 9 9 ˆ ( ) ( ) We got exactly the same aswer ad this time did t have to worry about goig back to t s i our aswer. So, we ve see two solutio techiques for computig defiite itegrals that require the substitutio rule. Both are valid solutio methods ad each have their uses. We will be usig the secod exclusively however sice it makes the evaluatio step a little easier. Let s work some more examples. Example Evaluate each of the followig. 5 5 (a) ( + w )( w + w ) dw [Solutio] - (b) Ù ı dx + x ( + x) y (c) + cos( ) 0 [Solutio] e py dy [Solutio] Ù ı 0 (d) si 5cos( p ) p Ê z Á ˆ- - z dz Ë [Solutio] Solutio Sice we ve doe quite a few substitutio rule itegrals to this time we are t goig to put a lot of effort ito explaiig the substitutio part of thigs here. 5 5 (a) ( + )( + ) w w w dw - The substitutio ad coverted limits are, ( ) ( ) u = w+ w du = + wdw fi + wdw= du w=- fi u =- w= 5 fi u = 5 Sometimes a limit will remai the same after the substitutio. Do t get excited whe it happes ad do t expect it to happe all the time. Here is the itegral, 007 Paul Dawkis 8

84 Math 40 + w w+ w dw= u du ( )( ) = u = Do t get excited about large umbers for aswers here. Sometime they are. That s life. [Retur to Problems] - (b) Ù ı dx + x ( + x) Here is the substitutio ad coverted limits for this problem, u = + x du = dx fi dx= du x=- fi u =- x=-6 fi u =- The itegral is the, Ù ı dx= 4u du Ù - + x ı - u - ( + x) = ( u 5l u ) - - Ê ˆ Ê ˆ = Á- -5l - Á- -5l Ë Ë = - l+ l 089 [Retur to Problems] e y (c) + cos( ) 0 py dy This itegral eeds to be split ito two itegrals sice the first term does t require a substitutio ad the secod does. y e + cos = e y ( p ) cos( p ) y dy dy y dy Here is the substitutio ad coverted limits for the secod term. u = py du = pdy fi dy = du p p y = 0 fi u = 0 y = fi u = 007 Paul Dawkis 8

85 Math 40 Here is the itegral. Ù ı 0 (d) si 5cos( p ) p y dy dy u du p p y e + cos = y ( p ) e cos( ) Ê z Á ˆ- - z dz Ë y = e + si u 0 p p 0 p = e - e + si - si0 p p = e - + p This itegral will require two substitutios. So first split up the itegral so we ca do a substitutio o each term. 0 0 Ê zˆ Ê zˆ 0 Ù siá -5cos( p - z) dz = Ù siá dz- p5cos( p -z) dz Ë Ë ı ı p p There are the two substitutios for these itegrals. z u = du = dz fi dz = du p p z = fi u = 6 z = 0 fi u = 0 v= p - z dv=-dz fi dz =-dv p p z = fi v= z = 0 fi v= p Here is the itegral for this problem. Ù 0 p Ê z 0 p siá ˆ- 5cos( p - z) dz = 6 psi( u) du + 5 p cos( v) dv Ë ı 6 =- 6cos ( u) p 5si ( v) p p 6 Ê 5 ˆ = - 6+ Á - Ë [Retur to Problems] = - 6 [Retur to Problems] 007 Paul Dawkis 84

86 Math 40 The ext set of examples is desiged to make sure that we do t forget about a very importat poit about defiite itegrals. Example Evaluate each of the followig. 5 (a) 4t Ù dt [Solutio] ı 8t Solutio (a) Ù ı 5 (b) -5 - Ù ı 4t 8t t - 8t dt dt [Solutio] Be careful with this itegral. The deomiator is zero at t =± ad both of these are i the iterval of itegratio. Therefore, this itegrad is ot cotiuous i the iterval ad so the itegral ca t be doe. Be careful with defiite itegrals ad be o the lookout for divisio by zero problems. I the previous sectio they were easy to spot sice all the divisio by zero problems that we had there were at zero. Oce we move ito substitutio problems however they will ot always be so easy to spot so make sure that you first take a quick look at the itegrad ad see if there are ay cotiuity problems with the itegrad ad if they occur i the iterval of itegratio. [Retur to Problems] (b) Ù ı 5 4t - 8t dt Now, i this case the itegral ca be doe because the two poits of discotiuity, t =±, are both outside of the iterval of itegratio. The substitutio ad coverted limits i this case are, u = - 8t du =-6tdt fi dz =- dt 6 t = fi u =- 70 t = 5 fi u =-98 The itegral is the, Ù ı 4t 4 dt =- Ù du -8t 6 ı u =- l 4 u =- l98 - l 70 4 ( ( ) ( )) [Retur to Problems] 007 Paul Dawkis 85

87 Math 40 Let s work aother set of examples. These are a little tougher (at least i appearace) tha the previous sets. Example 4 Evaluate each of the followig. l( -p ) x x (a) e cos( - e ) dx [Solutio] Solutio (b) (c) 0 Ù ı Ù ı 6 e e p 9 [ l t] t 4 p + p dt [Solutio] ( P) ( ) sec( P) sec ta P dp (d) cos( ) cos( si ( )) (e) [Solutio] x x dx [Solutio] Ù ı -p 50 w e dw w ( ) e e l -p x x (a) cos( - ) 0 [Solutio] dx The limits are a little uusual i this case, but that will happe sometimes so do t get too excited about it. Here is the substitutio. x x u = - e du =-e dx x= fi u = - e = - = x ( ) ( -p) ( ) p u e p p l = l - fi = - = - - = The itegral is the, ( -p ) e l x x ( e ) cos - dx=- 0 0 =- p cosudu p si ( u) 0 ( p ) =- si - si0 = 0 [Retur to Problems] 6 e [ ] 4 l t (b) Ù dt ı e t Here is the substitutio ad coverted limits for this problem. 007 Paul Dawkis 86

88 Math 40 u = l t du = dt t t = e fi u = le = 6 t = e fi 6 u = le = 6 The itegral is, (c) Ù ı p 9 p + ( P) ( ) sec( P) sec ta P dp Ù ı 6 e e [ l t] t 4 dt = = u 5 = 6 4 u du [Retur to Problems] Here is the substitutio ad coverted limits ad do t get too excited about the substitutio. It s a little messy i the case, but that ca happe o occasio. u = + sec( P) du = sec( P) ta( P) dp fi sec( P) ta( P) dp= du p Êp ˆ P= fi u = + secá = + Ë 4 p Êp ˆ P= fi u = + secá = 4 9 Ë Here is the itegral, Ù ı p 9 p ( ) ( ) + sec( P) sec ta P P 4 - dp = u du + = u 4 + Ê = - + Á Ë ( ) 4 So, ot oly was the substitutio messy, but we also a messy aswer, but agai that s life o occasio. [Retur to Problems] ˆ (d) cos( ) cos( si ( )) p -p x x dx 007 Paul Dawkis 87

89 Math 40 This problem ot as bad as it looks. Here is the substitutio ad coverted limits. u = six du = cos xdx p p x= fi u = si = x=-p fi u = si( - p) = 0 The cosie i the very frot of the itegrad will get substituted away i the differetial ad so this itegrad actually simplifies dow sigificatly. Here is the itegral. p -p ( ) ( ) ( ) cos x cos si x dx= cosudu 0 = si ( u) 0 () ( ) () = si -si 0 = si Do t get excited about these kids of aswers. O occasio we will ed up with trig fuctio evaluatios like this. [Retur to Problems] (e) Ù ı 50 w e dw w This is also a tricky substitutio (at least util you see it). Here it is, u = du =- dw fi dw=- du w w w w= fi u = w= fi u = Here is the itegral. Ù ı 50 w e w dw=- 00 =- e e u 00 =- - u du 00 ( e e ) [Retur to Problems] I this last set of examples we saw some tricky substitutios ad messy limits, but these are a fact of life with some substitutio problems ad so we eed to be prepared for dealig with them whe the happe. Eve ad Odd Fuctios 007 Paul Dawkis 88

90 Math 40 This is the last topic that we eed to discuss i this chapter. It is probably better suited i the previous sectio, but that sectio has already gotte fairly large so I decided to put it here. First, recall that a eve fuctio is ay fuctio which satisfies, Typical examples of eve fuctios are, A odd fuctio is ay fuctio which satisfies, The typical examples of odd fuctios are, f (- x) = f ( x) ( ) = ( ) = cos( ) f x x f x x (- ) =- ( ) f x f x ( ) = ( ) = si ( ) f x x f x x There are a couple of ice facts about itegratig eve ad odd fuctios over the iterval [-a,a]. If f(x) is a eve fuctio the, Likewise, if f(x) is a odd fuctio the, a -a a ( ) = ( ) f x dx f x dx a 0 ( ) 0 -a f x dx = Note that i order to use these facts the limit of itegratio must be the same umber, but opposite sigs! Example 5 Itegrate each of the followig. 4 (a) 4x - x + dx [Solutio] (b) x + si ( ) x dx [Solutio] -0 Solutio Neither of these are terribly difficult itegrals, but we ca use the facts o them ayway. (a) x x dx I this case the itegrad is eve ad the iterval is correct so, 4 4 4x - x + dx= 4x - x + dx - 0 Ê4 = Á - + Ë 5 5 x x x 748 = 5 So, usig the fact cut the evaluatio i half (i essece sice oe of the ew limits was zero). ˆ Paul Dawkis 89

91 Math 40 [Retur to Problems] 0 5 (b) x + si ( ) -0 x dx The itegrad i this case is odd ad the iterval is i the correct form ad so we do t eve eed to itegrate. Just use the fact. 0-0 x 5 ( ) + si x dx = 0 [Retur to Problems] Note that the limits of itegratio are importat here. Take the last itegral as a example. A small chage to the limits will ot give us zero. 9-0 x si( x) dx = cos( 0) -cos( 9) - = The moral here is to be careful ad ot misuse these facts. 007 Paul Dawkis 90

92 Math 40 Applicatios of Itegrals Itroductio I this last chapter of this course we will be takig a look at a couple of applicatios of itegrals. There are may other applicatios, however may of them require itegratio techiques that are typically taught i Calculus II. We will therefore be focusig o applicatios that ca be doe oly with kowledge taught i this course. Because this chapter is focused o the applicatios of itegrals it is assumed i all the examples that you are capable of doig the itegrals. There will ot be as much detail i the itegratio process i the examples i this chapter as there was i the examples i the previous chapter. Here is a listig of applicatios covered i this chapter. Average Fuctio Value We ca use itegrals to determie the average value of a fuctio. Area Betwee Two Curves I this sectio we ll take a look at determiig the area betwee two curves. Volumes of Solids of Revolutio / Method of Rigs This is the first of two sectios devoted to fid the volume of a solid of revolutio. I this sectio we look at the method of rigs/disks. Volumes of Solids of Revolutio / Method of Cyliders This is the secod sectio devoted to fidig the volume of a solid of revolutio. Here we will look at the method of cyliders. More Volume Problems I this sectio we ll take a look at fidig the volume of some solids that are either ot solids of revolutios or are ot easy to do as a solid of revolutio. 007 Paul Dawkis 9

93 Math 40 Average Fuctio Value The first applicatio of itegrals that we ll take a look at is the average value of a fuctio. The followig fact tells us how to compute this. Average Fuctio Value The average value of a fuctio f ( x ) over the iterval [a,b] is give by, b ( ) f = avg f x dx b a - a To see a justificatio of this formula see the Proof of Various Itegral Properties sectio of the Extras chapter. Let s work a couple of quick examples. Example Determie the average value of each of the followig fuctios o the give iterval. (a) ( ) È 5 f t = t - 5t+ 6cos( p t) o Í -, Î [Solutio] -cos( z) R z si z - pp, [Solutio] (b) ( ) = ( ) e o [ ] Solutio (a) ( ) È 5 f t = t - 5t+ 6cos( p t) o Í -, Î There s really ot a whole lot to do i this problem other tha just use the formula. 5 ( ) 5 - Ê 5 6 si ( p ) ( p ) favg = t - 5t+ 6cos t dt - - = Á t - t + t 7Ë p = - 7p 6 = You caught the substitutio eeded for the third term right? ˆ 5 - So, the average value of this fuctio of the give iterval is [Retur to Problems] -cos( z) (b) R( z) = si( z) e o [- pp, ] Agai, ot much to do here other tha use the formula. Note that the itegral will eed the 007 Paul Dawkis 9

94 Math 40 followig substitutio. ( z) u = - cos Here is the average value of this fuctio, p -cos( z) Ravg = si( z) dz p - -p e -p = e p = 0 ( ) ( z) -cos p -p So, i this case the average fuctio value is zero. Do ot get excited about gettig zero here. It will happe o occasio. I fact, if you look at the graph of the fuctio o this iterval it s ot too hard to see that this is the correct aswer. [Retur to Problems] There is also a theorem that is related to the average fuctio value. The Mea Value Theorem for Itegrals If f ( x ) is a cotiuous fuctio o [a,b] the there is a umber c i [a,b] such that, b ( ) = ( )( - ) a f x dx f c b a Note that this is very similar to the Mea Value Theorem that we saw i the Derivatives Applicatios chapter. See the Proof of Various Itegral Properties sectio of the Extras chapter for the proof. Note that oe way to thik of this theorem is the followig. First rewrite the result as, b ( ) = ( ) b- a f x dx f c a 007 Paul Dawkis 9

95 Math 40 ad from this we ca see that this theorem is tellig us that there is a umber a< c< b such that favg = f ( c). Or, i other words, if f ( ) fuctio will take o its average value. Let s take a quick look at a example usig this theorem. x is a cotiuous fuctio the somewhere i [a,b] the Example Determie the umber c that satisfies the Mea Value Theorem for Itegrals for the fuctio f ( x) = x + x+ o the iterval [,4] Solutio First let s otice that the fuctio is a polyomial ad so is cotiuous o the give iterval. This meas that we ca use the Mea Value Theorem. So, let s do that. 4 x x dx c c ( )( ) + + = ˆ x x x c c Ê Á + + = + + Ë ( ) 99 = c + 9c = c + 9c- This is a quadratic equatio that we ca solve. Usig the quadratic formula we get the followig two solutios, c = = c = =-5.59 Clearly the secod umber is ot i the iterval ad so that is t the oe that we re after. The first however is i the iterval ad so that s the umber we wat. Note that it is possible for both umbers to be i the iterval so do t expect oly oe to be i the iterval. 007 Paul Dawkis 94

96 Math 40 Area Betwee Curves I this sectio we are goig to look at fidig the area betwee two curves. There are actually two cases that we are goig to be lookig at. I the first case we are wat to determie the area betwee y = f ( x) ad y = g( x) o the iterval [a,b]. We are also goig to assume that f ( x) g( x). Take a look at the followig sketch to get a idea of what we re iitially goig to look at. I the Area ad Volume Formulas sectio of the Extras chapter we derived the followig formula for the area i this case. b ( ) ( ) A= f x -g x dx () a The secod case is almost idetical to the first case. Here we are goig to determie the area betwee x f ( y) = ad x= g( y) o the iterval [c,d] with f ( y) g( y). 007 Paul Dawkis 95

97 Math 40 I this case the formula is, c d ( ) ( ) A= f y -g y dy () Now () ad () are perfectly serviceable formulas, however, it is sometimes easy to forget that these always require the first fuctio to be the larger of the two fuctios. So, istead of these formulas we will istead use the followig word formulas to make sure that we remember that the formulas area always the larger fuctio mius the smaller fuctio. I the first case we will use, I the secod case we will use, b Ê upper ˆ Ê lower ˆ A= Ù Á - Á dx, a x b ı Ëfuctio Ëfuctio a d Ê right ˆ Ê left ˆ A= Ù Á - Á dy, c y d ı Ëfuctio Ëfuctio c () (4) Usig these formulas will always force us to thik about what is goig o with each problem ad to make sure that we ve got the correct order of fuctios whe we go to use the formula. Let s work a example. Example Determie the area of the regio eclosed by y = x ad y x =. Solutio First of all, just what do we mea by area eclosed by. This meas that the regio we re iterested i must have oe of the two curves o every boudary of the regio. So, here is a graph of the two fuctios with the eclosed regio shaded. 007 Paul Dawkis 96

98 Math 40 Note that we do t take ay part of the regio to the right of the itersectio poit of these two graphs. I this regio there is o boudary o the right side ad so is ot part of the eclosed area. Remember that oe of the give fuctios must be o the each boudary of the eclosed regio. Also from this graph it s clear that the upper fuctio will be depedet o the rage of x s that we use. Because of this you should always sketch of a graph of the regio. Without a sketch it s ofte easy to mistake which of the two fuctios is the larger. I this case most would probably say that y = x is the upper fuctio ad they would be right for the vast majority of the x s. However, i this case it is the lower of the two fuctios. The limits of itegratio for this will be the itersectio poits of the two curves. I this case it s pretty easy to see that they will itersect at x = 0 ad x = so these are the limits of itegratio. So, the itegral that we ll eed to compute to fid the area is, Ê upper ˆ Ê lower ˆ A= Ù Á -Á dx ı Ëfuctio Ëfuctio 0 b a = x - x dx Ê ˆ = Á x - x Ë = Before movig o to the ext example, there are a couple of importat thigs to ote. First, i almost all of these problems a graph is pretty much required. Ofte the boudig regio, which will give the limits of itegratio, is difficult to determie without a graph. Also, it ca ofte be difficult to determie which of the fuctios is the upper fuctio ad with is the lower fuctio without a graph. This is especially true i cases like the last example where the aswer to that questio actually depeded upo the rage of x s that we were usig. Fially, ulike the area uder a curve that we looked at i the previous chapter the area betwee two curves will always be positive. If we get a egative umber or zero we ca be sure that we ve made a mistake somewhere ad will eed to go back ad fid it. Note as well that sometimes istead of sayig regio eclosed by we will say regio bouded by. They mea the same thig. Let s work some more examples Paul Dawkis 97

99 Math 40 Example Determie the area of the regio bouded by y-axis. y = xe -x, y = x+, x =, ad the Solutio I this case the last two pieces of iformatio, x = ad the y-axis, tell us the right ad left boudaries of the regio. Also, recall that the y-axis is give by the lie x = 0. Here is the graph with the eclosed regio shaded i. Here, ulike the first example, the two curves do t meet. Istead we rely o two vertical lies to boud the left ad right sides of the regio as we oted above Here is the itegral that will give the area. Ê upper ˆ Ê lower ˆ A= Ù Á - dx fuctio Á fuctio ı Ë Ë 0 b a = x + x+ -x = x+ -xe dx Ê Á e Ë -x -4 7 e = + =.509 ˆ 0 Example Determie the area of the regio bouded by y = x + 0 ad y 4x 6 = +. Solutio I this case the itersectio poits (which we ll eed evetually) are ot goig to be easily idetified from the graph so let s go ahead ad get them ow. Note that for most of these problems you ll ot be able to accurately idetify the itersectio poits from the graph ad so you ll eed to be able to determie them by had. I this case we ca get the itersectio poits 007 Paul Dawkis 98

100 Math 40 by settig the two equatios equal. x + 0= 4x+ 6 x -4x- 6= 0 ( x )( x ) + - = 0 So it looks like the two curves will itersect at x=- ad x =. If we eed them we ca get the y values correspodig to each of these by pluggig the values back ito either of the equatios. We ll leave it to you to verify that the coordiates of the two itersectio poits o the graph are (-,) ad (,8). Note as well that if you are t good at graphig kowig the itersectio poits ca help i at least gettig the graph started. Here is a graph of the regio. With the graph we ca ow idetify the upper ad lower fuctio ad so we ca ow fid the eclosed area. Ê upper ˆ Ê lower ˆ A= Ù Á -Á dx ı Ëfuctio Ëfuctio - b a ( ) = 4x+ 6- x + 0 dx = - x + x+ dx Ê = Á Ë = 64 x x x Be careful with parethesis i these problems. Oe of the more commo mistakes studets make with these problems is to eglect parethesis o the secod term. ˆ Paul Dawkis 99

101 Math 40 Example 4 Determie the area of the regio bouded by ad x = 5 y = x + 0, y 4x 6 = +, x =- Solutio So, the fuctios used i this problem are idetical to the fuctios from the first problem. The differece is that we ve exteded the bouded regio out from the itersectio poits. Sice these are the same fuctios we used i the previous example we wo t bother fidig the itersectio poits agai. Here is a graph of this regio. Okay, we have a small problem here. Our formula requires that oe fuctio always be the upper fuctio ad the other fuctio always be the lower fuctio ad we clearly do ot have that here. However, this actually is t the problem that it might at first appear to be. There are three regios i which oe fuctio is always the upper fuctio ad the other is always the lower fuctio. So, all that we eed to do is fid the area of each of the three regios, which we ca do, ad the add them all up. Here is the area ( 4 6) 4 6 ( 0) 0 ( 4 6) A= x + - x+ dx+ x+ - x + dx+ x + - x+ dx = x -4x- 6dx+ - x + 4x+ 6dx+ x -4x-6dx x x 6x x x 6x x x 6x Ê ˆ Ê ˆ Ê ˆ = Á Á Á - - Ë Ë Ë = = Paul Dawkis 00

102 Math 40 Example 5 Determie the area of the regio eclosed by y-axis. Solutio First let s get a graph of the regio. y = si x, y = cos x, p x =, ad the So, we have aother situatio where we will eed to do two itegrals to get the area. The itersectio poit will be where six = cos x p i the iterval. We ll leave it to you to verify that this will be x =. The area is the, 4 p p 4 0 p 4 A= cosx- sixdx+ six-cos xdx ( six cosx 4 ) ( cosx si x) = ( ) = = - = We will eed to be careful with this ext example. p p 0 p 4 Example 6 Determie the area of the regio eclosed by x= y - ad y = x-. Solutio Do t let the first equatio get you upset. We will have to deal with these kids of equatios occasioally so we ll eed to get used to dealig with them. As always, it will help if we have the itersectio poits for the two curves. I this case we ll get 007 Paul Dawkis 0

103 Math 40 the itersectio poits by solvig the secod equatio for x ad the settig them equal. Here is that work, y+ = y - y+ = y -6 = y - y- 0 8 ( y )( y ) 0= So, it looks like the two curves will itersect at y =- ad y = 4 or if we eed the full coordiates they will be : (-,-) ad (5,4). Here is a sketch of the two curves. Now, we will have a serious problem at this poit if we are t careful. To this poit we ve bee usig a upper fuctio ad a lower fuctio. To do that here otice that there are actually two portios of the regio that will have differet lower fuctios. I the rage [-,-] the parabola is actually both the upper ad the lower fuctio. To use the formula that we ve bee usig to this poit we eed to solve the parabola for y. This gives, y =± x+ 6 where the + gives the upper portio of the parabola ad the - gives the lower portio. Here is a sketch of the complete area with each regio shaded that we d eed if we were goig to use the first formula. 007 Paul Dawkis 0

104 Math 40 The itegrals for the area would the be, ( ) ( ) - 5 A= x x+ 6 dx+ x+ 6- x- dx = x+ 6dx+ x+ 6- x+ dx = x+ 6dx+ x+ 6dx+ - x+ dx Ê ˆ = u + u + Á- x + x Ë = While these itegrals are t terribly difficult they are more difficult tha they eed to be. Recall that there is aother formula for determiig the area. It is, d Ê right ˆ Ê left ˆ A= Ù Á - Á dy, c y d ı Ëfuctio Ëfuctio c ad i our case we do have oe fuctio that is always o the left ad the other is always o the right. So, i this case this is defiitely the way to go. Note that we will eed to rewrite the equatio of the lie sice it will eed to be i the form x= f ( y) but that is easy eough to do. Here is the graph for usig this formula Paul Dawkis 0

105 Math 40 The area is, Ê right ˆ Ê left ˆ A= Ù Á -Á dy ı Ëfuctio Ëfuctio c 4 Ê ˆ = Ù ( y+ ) -Á y - dy ı Ë - 4 = Ù ı - y y dy Ê = Á Ë 6 = 8 d y y y ˆ 4 - This is the same that we got usig the first formula ad this was defiitely easier tha the first method. So, i this last example we ve see a case where we could use either formula to fid the area. However, the secod was defiitely easier. Studets ofte come ito a calculus class with the idea that the oly easy way to work with fuctios is to use them i the form y = f ( x). However, as we ve see i this previous example there are defiitely times whe it will be easier to work with fuctios i the form x= f ( y). I fact, there are goig to be occasios whe this will be the oly way i which a problem ca be worked so make sure that you ca deal with fuctios i this form. 007 Paul Dawkis 04

106 Math 40 Let s take a look at oe more example to make sure we ca deal with fuctios i this form. Example 7 Determie the area of the regio bouded by Solutio First, we will eed itersectio poits. y ( y ) = - - y + 0= y - 4y+ 4 x 0= y -4y-6 ( y )( y ) 0= + - =- y + 0 ad x ( y ) The itersectio poits are y =- ad y =. Here is a sketch of the regio. = -. This is defiitely a regio where the secod area formula will be easier. If we used the first formula there would be three differet regios that we d have to look at. The area i this case is, Ê right ˆ Ê left ˆ A= Ù Á - dy fuctio Á fuctio ı Ë Ë c ( ) = - y + 0- y- dy - d = - y + 4y+ 6dy - Ê ˆ 64 = Á- y + y + 6y = Ë Paul Dawkis 05

107 Math 40 Volumes of Solids of Revolutio / Method of Rigs I this sectio we will start lookig at the volume of a solid of revolutio. We should first defie just what a solid of revolutio is. To get a solid of revolutio we start out with a fuctio, y = f ( x), o a iterval [a,b]. We the rotate this curve about a give axis to get the surface of the solid of revolutio. For purposes of this discussio let s rotate the curve about the x-axis, although it could be ay vertical or horizotal axis. Doig this for the curve above gives the followig three dimesioal regio. What we wat to do over the course of the ext two sectios is to determie the volume of this object. I the fial the Area ad Volume Formulas sectio of the Extras chapter we derived the followig formulas for the volume of this solid. 007 Paul Dawkis 06

108 Math 40 b a d ( ) ( ) V = A x dx V = A y dy where, A( x ) ad A( y ) is the cross-sectioal area of the solid. There are may ways to get the cross-sectioal area ad we ll see two (or three depedig o how you look at it) over the ext two sectios. Whether we will use A( x ) or ( ) rotatio used for each problem. c A y will deped upo the method ad the axis of Oe of the easier methods for gettig the cross-sectioal area is to cut the object perpedicular to the axis of rotatio. Doig this the cross sectio will be either a solid disk if the object is solid (as our above example is) or a rig if we ve hollowed out a portio of the solid (we will see this evetually). I the case that we get a solid disk the area is, A = p ( radius) where the radius will deped upo the fuctio ad the axis of rotatio. I the case that we get a rig the area is, A p Ê Êouter ˆ Êier ˆ = Á - ˆ ÁÁ radius Á radius Ë Ë Ë where agai both of the radii will deped o the fuctios give ad the axis of rotatio. Note as well that i the case of a solid disk we ca thik of the ier radius as zero ad we ll arrive at the correct formula for a solid disk ad so this is a much more geeral formula to use. Also, i both cases, whether the area is a fuctio of x or a fuctio of y will deped upo the axis of rotatio as we will see. This method is ofte called the method of disks or the method of rigs. Let s do a example. Example Determie the volume of the solid obtaied by rotatig the regio bouded by y = x - 4x+ 5, x =, x = 4, ad the x-axis about the x-axis. Solutio The first thig to do is get a sketch of the boudig regio ad the solid obtaied by rotatig the regio about the x-axis. Here are both of these sketches. 007 Paul Dawkis 07

109 Math 40 Okay, to get a cross sectio we cut the solid at ay x. Below are a couple of sketches showig a typical cross sectio. The sketch o the right shows a cut away of the object with a typical cross sectio without the caps. The sketch o the left shows just the curve we re rotatig as well as it s mirror image alog the bottom of the solid. I this case the radius is simply the distace from the x-axis to the curve ad this is othig more tha the fuctio value at that particular x as show above. The cross-sectioal area is the, ( ) = p( ) = p( ) A x x x x x x x Next we eed to determie the limits of itegratio. Workig from left to right the first cross sectio will occur at x = ad the last cross sectio will occur at x = 4. These are the limits of itegratio. 007 Paul Dawkis 08

110 Math 40 The volume of this solid is the, V = b a ( ) 4 4 x 8x 6x 40x 5dx Ê 6 = p Á Ë5 78p 5 A x dx = p = 5 4 x x x 0x 5x I the above example the object was a solid object, but the more iterestig objects are those that are ot solid so let s take a look at oe of those. ˆ 4 Example Determie the volume of the solid obtaied by rotatig the portio of the regio x bouded by y = x ad y = that lies i the first quadrat about the y-axis. 4 Solutio First, let s get a graph of the boudig regio ad a graph of the object. Remember that we oly wat the portio of the boudig regio that lies i the first quadrat. There is a portio of the boudig regio that is i the third quadrat as well, but we do't wat that for this problem. There are a couple of thigs to ote with this problem. First, we are oly lookig for the volume of the walls of this solid, ot the complete iterior as we did i the last example. Next, we will get our cross sectio by cuttig the object perpedicular to the axis of rotatio. The cross sectio will be a rig (remember we are oly lookig at the walls) for this example ad it will be horizotal at some y. This meas that the ier ad outer radius for the rig will be x values ad so we will eed to rewrite our fuctios ito the form x= f ( y). Here are the fuctios writte i the correct form for this example. 007 Paul Dawkis 09

111 Math 40 y = x fi x= y x y = fi x= 4y 4 Here are a couple of sketches of the boudaries of the walls of this object as well as a typical rig. The sketch o the left icludes the back portio of the object to give a little cotext to the figure o the right. The ier radius i this case is the distace from the y-axis to the ier curve while the outer radius is the distace from the y-axis to the outer curve. Both of these are the x distaces ad so are give by the equatios of the curves as show above. The cross-sectioal area is the, ( 4 ) p ( 6 6 ) ( ) p ( ) ( ) A y = y - y = y - y Workig from the bottom of the solid to the top we ca see that the first cross-sectio will occur at y = 0 ad the last cross-sectio will occur at y =. These will be the limits of itegratio. The volume is the, V = c d 0 ( ) 6 = p 6y - y dy Ê6 = p Á y - y Ë 7 = 5p A y dy 7 With these two examples out of the way we ca ow make a geeralizatio about this method. If we rotate about a horizotal axis (the x-axis for example) the the cross sectioal area will be a ˆ Paul Dawkis 0

112 Math 40 fuctio of x. Likewise, if we rotate about a vertical axis (the y-axis for example) the the cross sectioal area will be a fuctio of y. The remaiig two examples i this sectio will make sure that we do t get too used to the idea of always rotatig about the x or y-axis. Example Determie the volume of the solid obtaied by rotatig the regio bouded by y = x - x ad y = x about the lie y = 4. Solutio First let s get the boudig regio ad the solid graphed. Agai, we are goig to be lookig for the volume of the walls of this object. Also sice we are rotatig about a horizotal axis we kow that the cross-sectioal area will be a fuctio of x. Here are a couple of sketches of the boudaries of the walls of this object as well as a typical rig. The sketch o the left icludes the back portio of the object to give a little cotext to the figure o the right. 007 Paul Dawkis

113 Math 40 Now, we re goig to have to be careful here i determiig the ier ad outer radius as they are t goig to be quite as simple they were i the previous two examples. Let s start with the ier radius as this oe is a little clearer. First, the ier radius is NOT x. The distace from the x-axis to the ier edge of the rig is x, but we wat the radius ad that is the distace from the axis of rotatio to the ier edge of the rig. So, we kow that the distace from the axis of rotatio to the x-axis is 4 ad the distace from the x-axis to the ier rig is x. The ier radius must the be the differece betwee these two. Or, ier radius= 4- x The outer radius works the same way. The outer radius is, ( ) outer radius= 4- x - x =- x + x+ 4 Note that give the locatio of the typical rig i the sketch above the formula for the outer radius may ot look quite right but it is i fact correct. As sketched the outer edge of the rig is below the x-axis ad at this poit the value of the fuctio will be egative ad so whe we do the subtractio i the formula for the outer radius we ll actually be subtractig off a egative umber which has the et effect of addig this distace oto 4 ad that gives the correct outer radius. Likewise, if the outer edge is above the x-axis, the fuctio value will be positive ad so we ll be doig a hoest subtractio here ad agai we ll get the correct radius i this case. The cross-sectioal area for this case is, ( ) 4 4 p ( ) ( ) p ( ) ( ) A x = - x + x+ - - x = x - x - x + x 007 Paul Dawkis

114 Math 40 The first rig will occur at x = 0 ad the last rig will occur at x = ad so these are our limits of itegratio. The volume is the, V = b a ( ) 4 x 4x 5x 4xdx 0 Ê 5 = p Á x -x - x + x Ë5 5p 5 A x dx = p = 5 4 Example 4 Determie the volume of the solid obtaied by rotatig the regio bouded by y = x-ad y = x- about the lie x =-. Solutio As with the previous examples, let s first graph the bouded regio ad the solid. ˆ 0 Now, let s otice that sice we are rotatig about a vertical axis ad so the cross-sectioal area will be a fuctio of y. This also meas that we are goig to have to rewrite the fuctios to also get them i terms of y. y y = x- fi x= + 4 y = x- fi x= y+ Here are a couple of sketches of the boudaries of the walls of this object as well as a typical rig. The sketch o the left icludes the back portio of the object to give a little cotext to the figure o the right. 007 Paul Dawkis

115 Math 40 The ier ad outer radius for this case is both similar ad differet from the previous example. This example is similar i the sese that the radii are ot just the fuctios. I this example the fuctios are the distaces from the y-axis to the edges of the rigs. The ceter of the rig however is a distace of from the y-axis. This meas that the distace from the ceter to the edges is a distace from the axis of rotatio to the y-axis (a distace of ) ad the from the y-axis to the edge of the rigs. So, the radii are the the fuctios plus ad that is what makes this example differet from the previous example. Here we had to add the distace to the fuctio value whereas i the previous example we eeded to subtract the fuctio from this distace. Note that without sketches the radii o these problems ca be difficult to get. So, i summary, we ve got the followig for the ier ad outer radius for this example. The cross-sectioal area it the, outer radius= y+ + = y+ y y ier radius= + + = Ê 4 Ê y ˆ ˆ Ê y ˆ A( y) = p ( y+ ) - + = p 4y- Á Á Á Ë 4 Ë 6 Ë The first rig will occur at y = 0 ad the fial rig will occur at y = 4 ad so these will be our limits of itegratio. The volume is, 007 Paul Dawkis 4

116 Math 40 V = c d 0 ( ) 4 4 y = p Ù 4y- dy ı 6 96p 5 A y dy Ê 5ˆ = p Áy - y Ë 80 = Paul Dawkis 5

117 Math 40 Volumes of Solids of Revolutio / Method of Cyliders I the previous sectio we started lookig at fidig volumes of solids of revolutio. I that sectio we took cross sectios that were rigs or disks, foud the cross-sectioal area ad the used the followig formulas to fid the volume of the solid. b a d ( ) ( ) V = A x dx V = A y dy I the previous sectio we oly used cross sectios that were i the shape of a disk or a rig. This however does ot always eed to be the case. We ca use ay shape for the cross sectios as log as it ca be expaded or cotracted to completely cover the solid we re lookig at. This is a good thig because as our first example will show us we ca t always use rigs/disks. c Example Determie the volume of the solid obtaied by rotatig the regio bouded by ( )( ) y = x- x- ad the x-axis about the y-axis. Solutio As we did i the previous sectio, let s first graph the bouded regio ad solid. Note that the bouded regio here is the shaded portio show. The curve is exteded out a little past this for the purposes of illustratig what the curve looks like. So, we ve basically got somethig that s roughly doughut shaped. If we were to use rigs o this solid here is what a typical rig would look like. This leads to several problems. First, both the ier ad outer radius are defied by the same fuctio. This, i itself, ca be dealt with o occasio as we saw i a example i the Area 007 Paul Dawkis 6

118 Math 40 Betwee Curves sectio. However, this usually meas more work tha other methods so it s ofte ot the best approach. This leads to the secod problem we got here. I order to use rigs we would eed to put this fuctio ito the form x= f ( y). That is NOT easy i geeral for a cubic polyomial ad i other cases may ot eve be possible to do. Eve whe it is possible to do this the resultig equatio is ofte sigificatly messier tha the origial which ca also cause problems. The last problem with rigs i this case is ot so much a problem as its just added work. If we were to use rigs the limit would be y limits ad this meas that we will eed to kow how high the graph goes. To this poit the limits of itegratio have always bee itersectio poits that were fairly easy to fid. However, i this case the highest poit is ot a itersectio poit, but istead a relative maximum. We spet several sectios i the Applicatios of Derivatives chapter talkig about how to fid maximum values of fuctios. However, fidig them ca, o occasio, take some work. So, we ve see three problems with rigs i this case that will either icrease our work load or outright prevet us from usig rigs. What we eed to do is to fid a differet way to cut the solid that will give us a cross-sectioal area that we ca work with. Oe way to do this is to thik of our solid as a lump of cookie dough ad istead of cuttig it perpedicular to the axis of rotatio we could istead ceter a cylidrical cookie cutter o the axis of rotatio ad push this dow ito the solid. Doig this would give the followig picture, Doig this gives us a cylider or shell i the object ad we ca easily fid its surface area. The surface area of this cylider is, A x = p radius height ( ) ( )( ) ( ) ( x)( x )( x ) = p ( x x x x) = p Notice as well that as we icrease the radius of the cylider we will completely cover the solid ad so we ca use this i our formula to fid the volume of this solid. All we eed are limits of itegratio. The first cylider will cut ito the solid at x = ad as we icrease x to x = we 007 Paul Dawkis 7

119 Math 40 will completely cover both sides of the solid sice expadig the cylider i oe directio will automatically expad it i the other directio as well. The volume of this solid is the, V = b a ( ) 4 p Ê 7 9 = pá x - x + 5x - x Ë5 4 4p 5 A x dx = x - x + x - xdx = 5 4 The method used i the last example is called the method of cyliders or method of shells. The formula for the area i all cases will be, ( )( ) A= p radius height There are a couple of importat differeces betwee this method ad the method of rigs/disks that we should ote before movig o. First, rotatio about a vertical axis will give a area that is a fuctio of x ad rotatio about a horizotal axis will give a area that is a fuctio of y. This is exactly opposite of the method of rigs/disks. Secod, we do t take the complete rage of x or y for the limits of itegratio as we did i the previous sectio. Istead we take a rage of x or y that will cover oe side of the solid. As we oted i the first example if we expad out the radius to cover oe side we will automatically expad i the other directio as well to cover the other side. Let s take a look at aother example. ˆ Example Determie the volume of the solid obtaied by rotatig the regio bouded by y = x, x = 8 ad the x-axis about the x-axis. Solutio First let s get a graph of the bouded regio ad the solid. 007 Paul Dawkis 8

120 Math 40 Okay, we are rotatig about a horizotal axis. This meas that the area will be a fuctio of y ad so our equatio will also eed to be wrote i x= f ( y) form. y = x fi x= y As we did i the rig/disk sectio let s take a couple of looks at a typical cylider. The sketch o the left shows a typical cylider with the back half of the object also i the sketch to give the right sketch some cotext. The sketch o the right cotais a typical cylider ad oly the curves that defie the edge of the solid. I this case the width of the cylider is ot the fuctio value as it was i the previous example. I this case the fuctio value is the distace betwee the edge of the cylider ad the y-axis. The distace from the edge out to the lie is x = 8 ad so the width is the 8 - y. The cross sectioal area i this case is, ( ) = p( radius)( width) A y ( y)( y ) 4 ( y y ) = p 8- = p 8 - The first cylider will cut ito the solid at y = 0 ad the fial cylider will cut i at y = ad so these are our limits of itegratio. The volume of this solid is, 007 Paul Dawkis 9

121 Math 40 V = c d 0 ( ) = p 8y- 96p 5 A y dy 4 y dy Ê 5ˆ = pá4y - y Ë 5 = 0 The remaiig examples i this sectio will have axis of rotatio about axis other tha the x ad y-axis. As with the method of rigs/disks we will eed to be a little careful with these. Example Determie the volume of the solid obtaied by rotatig the regio bouded by y = x- ad y = x- about the lie x= 6. Solutio Here s a graph of the bouded regio ad solid. Here are our sketches of a typical cylider. Agai, the sketch o the left is here to provide some cotext for the sketch o the right. 007 Paul Dawkis 0

122 Math 40 Okay, there is a lot goig o i the sketch to the left. First otice that the radius is ot just a x or y as it was i the previous two cases. I this case x is the distace from the x-axis to the edge of the cylider ad we eed the distace from the axis of rotatio to the edge of the cylider. That meas that the radius of this cylider is 6 - x. Secodly, the height of the cylider is the differece of the two fuctios i this case. The cross sectioal area is the, ( ) = p ( radius)( height) A x ( x)( x x ) ( x x x x x ) = p = p Now the first cylider will cut ito the solid at x = ad the fial cylider will cut ito the solid at x = 5 so there are our limits. Here is the volume. V = b a ( ) A x dx 5 p 7 6 = x - x+ + x- - x x- dx Ê = Á Ë 5 5 p x x 6x 8 x x x Ê6 ˆ = pá Ë 5 7p = 5 ( ) ( ) ( ) 007 Paul Dawkis ˆ 5

123 Math 40 The itegratio of the last term is a little tricky so let s do that here. It will use the substitutio, u = x- du = dx x = u+ ( ) x x- dx= u+ u du ı = u + u du 5 Ê ˆ = Á u + u + c Ë = ( x- ) + ( x- ) + c 5 We saw oe of these kids of substitutios back i the substitutio sectio. Example 4 Determie the volume of the solid obtaied by rotatig the regio bouded by ( y ) x= - ad y = x about the lie y =-. Solutio We should first get the itersectio poits there. ( y ) y = - = y y y = y 5y 4 ( y )( y ) 0= -4 - So, the two curves will itersect at y = ad y = 4. Here is a sketch of the bouded regio ad the solid. Here are our sketches of a typical cylider. The sketch o the left is here to provide some cotext for the sketch o the right. 007 Paul Dawkis

124 Math 40 Here s the cross sectioal area for this cylider. A y = p radius width ( ) ( )( ) ( ) ( y ) y ( y ) = p ( y y y ) = p The first cylider will cut ito the solid at y = ad the fial cylider will cut i at y = 4. The volume is the, V = c d ( ) 4 p 4 4 Ê 4 4 ˆ = p Á- y + y + y -4y Ë 4 6p A y dy = - y + y + y- dy = Paul Dawkis

125 Math 40 More Volume Problems I this sectio we re goig to take a look at some more volume problems. However, the problems we ll be lookig at here will ot be solids of revolutio as we looked at i the previous two sectios. There are may solids out there that caot be geerated as solids of revolutio, or at least ot easily ad so we eed to take a look at how to do some of these problems. Now, havig said that these will ot be solids of revolutios they will still be worked i pretty much the same maer. For each solid we ll eed to determie the cross-sectioal area, either A( x ) or ( ) A y, ad they use the formulas we used i the previous two sectios, d ( ) ( ) b a c V = A x dx V = A y dy The hard part of these problems will be determiig what the cross-sectioal area for each solid is. Each problem will be differet ad so each cross-sectioal area will be determied by differet meas. Also, before we proceed with ay examples we eed to ackowledge that the itegrals i this sectio might look a little tricky at first. There are goig to be very few umbers i these problems. All of the examples i this sectio are goig to be more geeral derivatio of volume formulas for certai solids. As such we ll be workig with thigs like circles of radius r ad we ll ot be givig a specific value of r ad we ll have heights of h istead of specific heights, etc. All the letters i the itegrals are goig to make the itegrals look a little tricky, but all you have to remember is that the r s ad the h s are just letters beig used to represet a fixed quatity for the problem, i.e. it is a costat. So whe we itegrate we oly eed to worry about the letter i the differetial as that is the variable we re actually itegratig with respect to. All other letters i the itegral should be thought of as costats. If you have trouble doig that, just thik about what you d do if the r was a or the h was a for example. Let s start with a simple example that we do t actually eed to do a itegral that will illustrate how these problems work i geeral ad will get us used to seeig multiple letters i itegrals. Example Fid the volume of a cylider of radius r ad height h. Solutio Now, as we metioed before startig this example we really do t eed to use a itegral to fid this volume, but it is a good example to illustrate the method we ll eed to use for these types of problems. We ll start off with the sketch of the cylider below. 007 Paul Dawkis 4

126 Math 40 We ll ceter the cylider o the x-axis ad the cylider will start at x = 0 ad ed at x= h as show. Note that we re oly choosig this particular set up to get a itegral i terms of x ad to make the limits ice to deal with. There are may other orietatios that we could use. What we eed here is to get a formula for the cross-sectioal area at ay x. I this case the crosssectioal area is costat ad will be a disk of radius r. Therefore, for ay x we ll have the followig cross-sectioal area, ( ) = p r A x Next the limits for the itegral will be 0 x h sice that is the rage of x i which the cylider lives. Here is the itegral for the volume, So, we get the expected formula. h h h V = pr dx= pr dx= prx = prh 0 0 Also, recall we are usig r to represet the radius of the cylider. While r ca clearly take differet values it will ever chage oce we start the problem. Cyliders do ot chage their radius i the middle of a problem ad so as we move alog the ceter of the cylider (i.e. the x- axis) r is a fixed umber ad wo t chage. I other words it is a costat that will ot chage as we chage the x. Therefore, because we itegrated with respect to x the r will be a costat as far as the itegral is cocered. The r ca the be pulled out of the itegral as show (although that s ot required, we just did it to make the poit). At this poit we re just itegratig dx ad we kow how to do that. Whe we evaluate the itegral remember that the limits are x values ad so we plug ito the x ad NOT the r. Agai, remember that r is just a letter that is beig used represet the radius of the cylider ad, oce we start the itegratio, is assumed to be a fixed costat. As oted before we started this example if you re havig trouble with the r just thik of what you d do if there was a there istead of a r. I this problem, because we re itegratig with respect to x, both the ad the r will behave i the same maer. Note however that you should NEVER actually replace the r with a as that WILL lead to a wrog aswer. You should just thik of what you would do IF the r was a. 007 Paul Dawkis 5 0

127 Math 40 So, to work these problems we ll first eed to get a sketch of the solid with a set of x ad y axes to help us see what s goig o. At the very least we ll eed the sketch to get the limits of the itegral, but we will ofte eed it to see just what the cross-sectioal area is. Oce we have the sketch we ll eed to determie a formula for the cross-sectioal area ad the do the itegral. Let s work a couple of more complicated examples. I these examples the mai issue is goig to be determiig what the cross-sectioal areas are. Example Fid the volume of a pyramid whose base is a square with sides of legth L ad whose height is h. Solutio Let s start off with a sketch of the pyramid. I this case we ll ceter the pyramid o the y-axis ad to make the equatios easier we are goig to positio the poit of the pyramid at the origi. Now, as show here the cross-sectioal area will be a fuctio of y ad it will also be a square with sides of legth s. The area of the square is easy, but we ll eed to get the legth of the side i terms of y. To determie this cosider the figure o the right above. If we look at the pyramid directly from the frot we ll see that we have two similar triagles ad we kow that the ratio of ay two sides of similar triagles must be equal. I other words, we kow that, s y y L = fi s = L= y L h h h So, the cross-sectioal area is the, L A( y) = s = y h 007 Paul Dawkis 6

128 Math 40 The limit for the itegral will be 0 y h ad the volume will be, h h L L h L Ê ˆ 0 Á 0 h h h Ë 0 V = Ù y dy = y dy = y = Lh ı Agai, do ot get excited about the L ad the h i the itegral. Oce we start the problem if we chage y they will ot chage ad so they are costats as far as the itegral is cocered ad ca get pulled out of the itegral. Also, remember that whe we evaluate will oly plug the limits ito the variable we itegrated with respect to, y i this case. Before we proceed with some more complicated examples we should oce agai remid you to ot get excited about the other letters i the itegrals. If we re itegratig with respect to x or y the all other letters i the formula that represet fixed quatities (i.e. radius, height, legth of a side, etc.) are just costats ad ca be treated as such whe doig the itegral. Now let s do some more examples. Example For a sphere of radius r fid the volume of the cap of height h. Solutio A sketch is probably best to illustrate what we re after here. We are after the top portio of the sphere ad the height of this is portio is h. I this case we ll use a cross-sectioal area that starts at the bottom of the cap, which is at y = r- h, ad moves up towards the top, which is at y = r. So, each cross-sectio will be a disk of radius x. It is a little easier to see that the radius will be x if we look at it from the top as show i the sketch to the right above. The area of this disk is the, p x This is a problem however as we eed the cross-sectioal area i terms of y. So, what we really 007 Paul Dawkis 7

129 Math 40 eed to determie what x will be for ay give y at the cross-sectio. To get this let s look at the sphere from the frot. I particular look at the triagle POR. Because the poit R lies o the sphere itself we ca see that the legth of the hypoteuse of this triagle (the lie OR) is r, the radius of the sphere. The lie PR has a legth of x ad the lie OP has legth y so by the Pythagorea Theorem we kow, x + y = r fi x = r - y So, we ow kow what x will be for ay give y ad so the cross-sectioal area is, ( ) = p ( - ) A y r y As we oted earlier the limits o y will be r- h y r ad so the volume is, r ( ) ( r ) ( ) ( ) V = p r - y dy = p r y - y = p hr - h = p h r - h r-h r-h I the previous example we agai saw a r i the itegral. However, ulike the previous two examples it was ot multiplied times the x or the y ad so could ot be pulled out of the itegral. I this case it was like we were itegratig case we eed to treat the 4 - y ad we kow how itegrate that. So, i this r like the 4 ad so whe we itegrate that we ll pick up a y. 007 Paul Dawkis 8

130 Math 40 Example 4 Fid the volume of a wedge cut out of a cylider of radius r if the agle betwee the top ad bottom of the wedge is 6 p. Solutio We should really start off with a sketch of just what we re lookig for here. O the left we see how the wedge is beig cut out of the cylider. The base of the cylider is the circle give by x + y = r ad the agle betwee this circle ad the top of the wedge is p 6. The sketch i the upper right positio is the actual wedge itself. Give the orietatio of the axes here we get the portio of the circle with positive y ad so we ca write the equatio of the circle as y = r - x sice we oly eed the positive y values. Note as well that this is the reaso for the way we orieted the axes here. We get positive y s ad we ca write the equatio of the circle as a fuctio oly of x s. Now, as we ca see i the two sketches of the wedge the cross-sectioal area will be a right triagle ad the area will be a fuctio of x as we move from the back of the cylider, at x=- r, to the frot of the cylider, at x= r. The right agle of the triagle will be o the circle itself while the poit o the x-axis will have a iterior agle of 6 p. The base of the triagle will have a legth of y ad usig a little right triagle trig we see that the height of the rectagle is, p height = yta = ( ) 6 y So, we ow kow the base ad height of our triagle, i terms of y, ad we have a equatio for y i terms of x ad so we ca see that the area of the triagle, i.e. the cross-sectioal area is, ( ) = ( )( ) ( ) = - - = ( - ) A x y y r x r x r x The limits o x are - r x r ad so the volume is the, ( ) ( ) r r r -r -r V = r - x dx= rx- x = 007 Paul Dawkis 9

131 Math 40 The ext example is very similar to the previous oe except it ca be a little difficult to visualize the solid itself. Example 5 Fid the volume of the solid whose base is a disk of radius r ad whose crosssectios are equilateral triagles. Solutio Let s start off with a couple of sketches of this solid. The sketch o the left is from the frot of the solid ad the sketch o the right is more from the top of the solid. The base of this solid is the disk of radius r ad we move from the back of the disk at x=- r to the frot of the disk at x= r we form equilateral triagles to form the solid. A sample equilateral triagle, which is also the cross-sectioal area, is show above to hopefully make it a little clearer how the solid is formed. Now, let s get a formula for the cross-sectioal area. Let s start with the two sketches below. I the left had sketch we are lookig at the solid from directly above ad otice that we reorieted the sketch a little to put the x ad y-axis i the ormal orietatio. The solid 007 Paul Dawkis 0

132 Math 40 vertical lie i this sketch is the cross-sectioal area. From this we ca see that the cross-sectio occurs at a give x ad the top half will have a legth of y where the value of y will be the y- coordiate of the poit o the circle ad so is, y = r - x Also, because the cross-sectio is a equilateral triagle that is cetered o the x-axis the bottom half will also have a legth of y. Thus the base of the cross-sectio must have a legth of y. The sketch to the right is of oe of the cross-sectios. As oted above the base of the triagle has a legth of y. Also ote that because it is a equilateral triagle the agles are all p. If we divide the cross-sectio i two (as show with the dashed lie) we ow have two right triagles ad usig right triagle trig we ca see that the legth of the dashed lie is, p ( ) dashed lie= yta = y Therefore the height of the cross-sectio is y. Because the cross-sectio is a triagle we kow that that it s area must the be, ( ) = ( )( ) = ( - )( - ) = ( - ) A x y y r x r x r x Note that we used the cross-sectioal area i terms of x because each of the cross-sectios is perpedicular to the x-axis ad this pretty much forces us to itegrate with respect to x. The volume of the solid is the, r 4 -r -r r ( ) ( ) ( ) V x = r - x dx = rx - x = r The fial example we re goig to work here is a little tricky both i seeig how to set it up ad i doig the itegral. 007 Paul Dawkis

133 Math 40 Example 6 Fid the volume of a torus with radii r ad R. Solutio First, just what is a torus? A torus is a dout shaped solid that is geerated by rotatig the circle of radius r ad cetered at (R, 0) about the y-axis. This is show i the sketch to the left below. Oe of the trickiest parts of this problem is seeig what the cross-sectioal area eeds to be. There is a obvious oe. Most people would probably thik of usig the circle of radius r that we re rotatig about the y-axis as the cross-sectio. It is defiitely oe of the more obvious choices, however settig up a itegral usig this is ot so easy. So, what we ll do is use a cross-sectio as show i the sketch to the right above. If we cut the torus perpedicular to the y-axis we ll get a cross-sectio of a rig ad fidig the area of that should t be too bad. To do that let s take a look at the two sketches below. The sketch to the left is a sketch of the full cross-sectio. The sketch to the left is more importat however. This is a sketch of the circle that we are rotatig about the y-axis. Icluded is a lie represetig where the cross-sectioal area would be i the torus. Notice that the ier radius will always be the left portio of the circle ad the outer radius will always be the right portio of the circle. Now, we kow that the equatio of this is, ( ) x- R + y = r 007 Paul Dawkis

134 Math 40 ad so if we solve for x we ca get the equatios for the left ad right sides as show above i the sketch. This however meas that we also ow have equatios for the ier ad outer radii. ier radius: x= R- r - y outer radius: x= R+ r - y The cross-sectioal area is the, ( ) = p ( outer radius) -p( ier radius) A y ( R r y ) ( R r y ) R R r y r y ( R R r y r y ) È = p Í Î = p È ÍÎ = 4pR r - y Next, the lowest cross-sectio will occur at y =- r ad the highest cross-sectio will occur at y = r ad so the limits for the itegral will be - r y r. The itegral givig the volume is the, r r r V = 4p R r - y dy = 4p R r - y dy = 8p R r - y dy -r 0 0 Note that we used the fact that because the itegrad is a eve fuctio ad we re itegratig over [ rr, ] - we could chage the lower limit to zero ad double the value of the itegral. We saw this fact back i the Substitutio Rule for Defiite Itegrals sectio. We ve ow reached the secod really tricky part of this example. With the kowledge that we ve curretly got at this poit this itegral is ot possible to do. It requires somethig called a trig substitutio ad that is a topic for Calculus II. Luckily eough for us, ad this is the tricky part, i this case we ca actually determie what the itegrals value is usig what we kow about itegrals. Just for a secod let s thik about a differet problem. Let s suppose we wated to use a itegral to determie the area uder the portio of the circle of radius r ad cetered at the origi that is i the first quadrat. There are a couple of ways to do this, but to match what we re doig here let s do the followig. We kow that the equatio of the circle is circle i the first (ad fourth for that matter) quadrat is, x + y = r ad if we solve for x the equatio of the x= r - y If we wat a itegral for the area i the first quadrat we ca thik of this area as the regio 007 Paul Dawkis

135 Math 40 betwee the curve x= r - y ad the y-axis for 0 y r ad this is, r A= r - y dy 0 I other words, this itegral represets oe quarter of the area of a circle of radius r ad from basic geometric formulas we ow kow that this itegral must have the value, r A= r - y dy = p r 0 4 So, puttig all this together the volume of the torus is the, r 0 ( 4 ) V = 8Rp r - y dy = 8pR pr = Rp r 007 Paul Dawkis 4

136 Math 40 Extras Itroductio I this chapter material that did t fit ito other sectios for a variety of reasos. Also, i order to ot obscure the mechaics of actually workig problems, most of the proofs of various facts ad formulas are i this chapter as opposed to beig i the sectio with the fact/formula. This chapter cotais those topics. Proof of Various Itegral Facts/Formulas/Properties Here we will give the proofs of some of the facts ad formulas from the Itegral Chapter as well as a couple from the Applicatios of Itegrals chapter. Area ad Volume Formulas Here is the derivatio of the formulas for fidig area betwee two curves ad fidig the volume of a solid of revolutio. Types of Ifiity This is a discussio o the types of ifiity ad how these affect certai limits. Summatio Notatio Here is a quick review of summatio otatio. Costat of Itegratio This is a discussio o a couple of subtleties ivolvig costats of itegratio that may studets do t thik about. 007 Paul Dawkis 5

137 Math 40 Proof of Various Itegral Facts/Formulas/Properties I this sectio we ve got the proof of several of the properties we saw i the Itegrals Chapter as well as a couple from the Applicatios of Itegrals Chapter. kf x dx k f x dx where k is ay umber. Proof of : ( ) = ( ) This is a very simple proof. Suppose that F( x ) is a ati-derivative of f ( ) F ( x) = f ( x). The by the basic properties of derivatives we also have that, ( kf( x) ) = kf ( x) = kf ( x) ad so kf( x ) is a ati-derivative of kf ( x ), i.e. ( ) ( kf x ) kf ( x) ( ) = ( ) + = ( ) kf x dx kf x c k f x dx x, i.e. =. I other words, Proof of : ( ) ± ( ) = ( ) ± ( ) f x g x dx f x dx g x dx This is also a very simple proof Suppose that F( x ) is a ati-derivative of f ( ) G( x ) is a ati-derivative of g( x ). So we have that F ( x) = f ( x) ad G ( x) g( x) Basic properties of derivatives also tell us that ( F( x) ± G( x) ) = F ( x) ± G ( x) = f ( x) ± g( x) ad so F( x) + G( x) is a ati-derivative of f ( x) + g( x) ad F( x) G( x) - is a atiderivative of f ( x) - g( x). I other words, ( ) ± ( ) = ( ) ± ( ) + = ( ) ± ( ) f x g x dx F x G x c f x dx g x dx x ad that =. 007 Paul Dawkis 6

138 Math 40 b a Proof of : ( ) =- ( ) a f x dx f x dx From the defiitio of the defiite itegral we have, b a b b- Æ i = * ( ) limâ ( i ) f x dx= f x Dx D x= a ad we also have, b a a- Æ i = * ( ) limâ ( i ) f x dx= f x Dx D x= b Therefore, b a * ( ) = limâ ( i ) f x dx f x Æ i = = lim Â Æ i = f Æ i = * ( xi ) b- a ( a b) - - Ê * a-bˆ = lim f ( xi ) Æ Á - i Ë = a * a-b =- limâ f ( xi ) =- f ( x) dx b a Proof of : f ( x ) dx = 0 a From the defiitio of the defiite itegral we have, a * a- a f ( x) dx= limâ f ( xi ) Dx D x= = 0 a Æ i = Æ i = * ( xi )( ) = lim f 0 = lim0 Æ = 0  007 Paul Dawkis 7

139 Math 40 b b Proof of : ( ) = ( ) a cf x dx c f x dx From the defiitio of the defiite itegral we have, a b a * ( ) limâ ( i ) cf x dx= cf x Dx Æ i = i= Æ i = ( ) * ( i ) = lim c f x Dx Æ * ( i ) = clim f x Dx b = c f x dx Remember that we ca pull costats out of summatios ad out of limits. a   b b b Proof of : ( ) ± ( ) = ( ) ± ( ) f x g x dx f x dx g x dx a a a First we ll prove the formula for +. From the defiitio of the defiite itegral we have, b a ( ) * * ( ) ( ) limâ ( i ) ( i ) f x + g x dx= f x + g x Dx Æ i = Ê * * ˆ = lim f ( xi ) x g( xi ) x Æ Á D +  D Ë i= i= * * ( i ) limâ ( i ) = lim f x D x+ g x Dx Æ Æ i= i= b a  b ( ) ( ) = f x dx+ g x dx To prove the formula for - we ca either redo the above work with a mius sig istead of a plus sig or we ca use the fact that we ow kow this is true with a plus ad usig the properties proved above as follows. b a b ( )- ( ) = ( ) + - ( ) a b a b a a ( ) f x g x dx f x g x dx ( ) b ( ) ( ) = f x dx+ -g x dx b ( ) ( ) = f x dx- g x dx a a 007 Paul Dawkis 8

140 Math 40 b Proof of : cdx = cb ( - a) a, c is ay umber. If we defie f ( x) = c the from the defiitio of the defiite itegral we have, b a b a ( ) cdx= f x dx * b-a = limâ f ( xi ) Dx D x= Æ i = Ê ˆb-a = lim c Æ ÁÂ Ë i= b-a = lim ( c) Æ = lim c b- Æ ( a) ( a) = c b- b Proof of : If f ( x) 0 for a x b the ( ) 0 From the defiitio of the defiite itegral we have, Now, by assumptio ( ) 0 b a * ( ) limâ ( i ) f x dx. a b- Æ i = f x dx= f x Dx D x= f x ad we also have D x > 0 ad so we kow that  i= * ( i ) f x Dx 0 So, from the basic properties of limits we the have, Â Æ i= * ( i ) lim f x Dx lim0= 0 Æ But the left side is exactly the defiitio of the itegral ad so we have, b a * ( )  ( i ) f x dx= lim f x Dx 0 Æ i = a 007 Paul Dawkis 9

141 Math 40 b b Proof of : If f ( x) g( x) for a x bthe ( ) ( ) f x dx g x dx. Sice we have f ( x) g( x) the we kow that f ( x) g( x) 0 Property 8 proved above we kow that, We also kow from Property 4 that, b a ( ) - ( ) 0 a f x g x dx ( )- ( ) = ( ) - ( ) b b b f x g x dx f x dx g x dx a a a a - o a x b ad so by So, we the have, b a b ( ) ( ) f x dx- g x dx 0 a b a b ( ) ( ) f x dx g x dx a b Proof of : If m f ( x) M for a x b the mb ( - a) f ( x) dx M ( b-a) Give m f ( x) M we ca use Property 9 o each iequality to write, ( ) b b b mdx f x dx Mdx a a a The by Property 7 o the left ad right itegral to get, b ( - ) ( ) ( - ) mb a f x dx M b a a. a b b Proof of : ( ) ( ) a a f x dx f x dx First let s ote that we ca say the followig about the fuctio ad the absolute value, - f ( x) f ( x) f ( x) 007 Paul Dawkis 40

142 Math 40 If we ow use Property 9 o each iequality we get, ( ) ( ) ( ) b b b - f x dx f x dx f x dx a a a We kow that we ca factor the mius sig out of the left itegral to get, b b b ( ) ( ) ( ) - f x dx f x dx f x dx a a a Fially, recall that if p the must have, b the - b p b ad of course this works i reverse as well so we b ( ) ( ) b a a f x dx f x dx Fudametal Theorem of Calculus, Part I If f ( x ) is cotiuous o [a,b] the, = x ( ) () g x f t dt is cotiuous o [a,b] ad it is differetiable o ( ab, ) ad that, g ( x) = f ( x) Proof Suppose that x ad x+ h are i ( ab, ). We the have, x+ h x ( + )- ( ) = () - () a a g x h g x f t dt f t dt Now, usig Property 5 of the Itegral Properties we ca rewrite the first itegral ad the do a little simplificatio as follows. x x h x ( ) () a x a + ( ) ( ) () () g x+ h - g x = f t dt+ f t dt - f t dt Fially assume that h 0 ad we get, = x+ h x () f t ( + )- ( ) g x h g x h dt x+ h 007 Paul Dawkis 4 a () = f t dt h () x

143 Math 40 Let s ow assume that h > 0 ad sice we are still assumig that x+ h are i (, ) that f ( x ) is cotiuous o [ xx, + h] ad so by the Extreme Value Theorem we kow that there are umbers c ad d i [ xx, + h] so that f ( c) = m is the absolute miimum of f ( ) [ xx, + h] ad that f ( d) = M is the absolute maximum of f ( x ) i [ xx, + h]. So, by Property 0 of the Itegral Properties we the kow that we have, x+ h x () mh f t dt Mh ab we kow x i Or, x+ h ( ) () ( ) f c h f t dt f d h x Now divide both sides of this by h to get, ad the use () to get, x h + ( ) () ( ) f c f t dt f d h x ( + ) - ( ) g x h g x f ( c) f ( d) () h Next, if h < 0 we ca go through the same argumet above except we ll be workig o [ x hx, ] + to arrive at exactly the same iequality above. I other words, () is true provided h 0. Now, if we take h Æ 0 we also have cæ x ad d Æ x because both c ad d are betwee x ad x+ h. This meas that we have the followig two limits. ( ) = ( ) = ( ) ( ) = ( ) = ( ) lim f c lim f c f x lim f d lim f d f x hæ0 cæx hæ0 dæx The Squeeze Theorem the tells us that, ( + ) - ( ) h Æ0 g x lim h g x h ( x) = f () but the left side of this is exactly the defiitio of the derivative of g( x ) ad so we get that, g ( x) = f ( x) 007 Paul Dawkis 4

144 Math 40 So, we ve show that g( x ) is differetiable o (, ) ab. Now, the Theorem at the ed of the Defiitio of the Derivative sectio tells us that g( x ) is also cotiuous o ( ab, ). Fially, if we take x = a or x= b we ca go through a similar argumet we used to get () usig oe-sided limits to get the same result ad so the theorem at the ed of the Defiitio of the Derivative sectio will also tell us that g( x ) is cotiuous at x x= b ad so i fact g( x ) is also cotiuous o [ ab, ]. = a or Fudametal Theorem of Calculus, Part II Suppose f ( x ) is a cotiuous fuctio o [a,b] ad also suppose that F( x ) is ay atiderivative for f ( x ). The, Proof x First let ( ) () a b ( ) = b ( ) = ( ) - ( ) f x dx F x F b F a a a g x = f t dt ad the we kow from Part I of the Fudametal Theorem of Calculus that g ( x) = f ( x) ad so g( x ) is a ati-derivative of f ( ) suppose that F( x ) is ay ati-derivative of f ( ) meas that we must have, x o [a,b]. Further x o [a,b] that we wat to choose. So, this ( ) = ( ) g x F x The, by Fact i the Mea Value Theorem sectio we kow that g( x ) ad F( x ) ca differ by o more tha a additive costat o ( ab, ). I other words for a< x< b we have, F( x) = g( x) + c Now because g( x ) ad F( x ) are cotiuous o [a,b], if we take the limit of this as xæ a + ad xæ b - we ca see that this also holds if x= a ad x= b. So, for a x b do the followig. we kow that F( x) = g( x) + c. Let s use this ad the defiitio of ( ) g x to 007 Paul Dawkis 4

145 Math 40 ( ) ( ( ) ) ( ) ( ) ( ) = g( b) -g( a) F b - F a = g b + c - g a + c b a b a () () = f t dt+ f t dt () = f t dt + 0 = a b a f ( ) x dx Note that i the last step we used the fact that the variable used i the itegral does ot matter ad so we could chage the t s to x s. a Average Fuctio Value The average value of a fuctio f ( x ) over the iterval [a,b] is give by, b ( ) f = avg f x dx b a - a Proof We kow that the average value of umbers is simply the sum of all the umbers divided by so let s start off with that. Let s take the iterval [a,b] ad divide it ito subitervals each of legth, b- a D x = * * * Now from each of these itervals choose the poits x, x, K, x ad ote that it does t really matter how we choose each of these umbers as log as they come from the appropriate iterval. We ca the compute the average of the fuctio values f ( x * ) ( * ) ( *, f x,, f x ) computig, K by * * * ( ) + ( ) + L+ ( ) f x f x f x (4) Now, from our defiitio of D x we ca get the followig formula for. b- a = D x 007 Paul Dawkis 44

146 Math 40 ad we ca plug this ito (4) to get, * * * * * * ( ) ( ) L ( ) È ( ) ( ) L ( ) f x + f x + + f x f x + f x + + f x Dx = Î b-a b-a Dx * * * = Èf ( x) x f ( x) x f ( x) x b a Î D + D + L + D - * =  f ( xi ) Dx b-a i = Let s ow icrease. Doig this will mea that we re takig the average of more ad more fuctio values i the iterval ad so the larger we chose the better this will approximate the average value of the fuctio. If we the take the limit as goes to ifiity we should get the average fuctio value. Or, * * favg = lim  f ( xi ) D x= limâ f ( xi ) Dx Æ b-a b-a Æ i= i= We ca factor the b- a out of the limit as we ve doe ad ow the limit of the sum should look familiar as that is the defiitio of the defiite itegral. So, puttig i defiite itegral we get the formula that we were after. b ( ) f = avg f x dx b a - a The Mea Value Theorem for Itegrals If f ( x ) is a cotiuous fuctio o [a,b] the there is a umber c i [a,b] such that, b ( ) = ( )( - ) a f x dx f c b a Proof Let s start off by defiig, x ( ) () F x = a f t dt Sice f ( x ) is cotiuous we kow from the Fudametal Theorem of Calculus, Part I that F( x ) is cotiuous o [a,b], differetiable o (a,b) ad that F ( x) = f ( x). 007 Paul Dawkis 45

147 Math 40 Now, from the Mea Value Theorem we kow that there is a umber c such that a< c< b ad that, F b - F a = F c b- a However we kow that F ( c) = f ( c) ad, ( ) ( ) ( )( ) b b a ( ) = () = ( ) ( ) = () = 0 F b f t dt f x dx F a f t dt a a a So, we the have, b ( ) = ( )( - ) a f x dx f c b a Work The work doe by the force F( x ) (assumig that F( x ) is cotiuous) over the rage a x b is, W = b a ( ) F x dx Proof Let s start off by dividig the rage a x b ito subitervals of width D x ad from each of these itervals choose the poits * * * x, x, K, x. Now, if is large ad because F( x ) is cotiuous we ca assume that F( x ) wo t vary by much over each iterval ad so i the i th iterval we ca assume that the force is approximately * costat with a value of F( x) F( x i ) ª. The work o each iterval is the approximately, i * ( i ) W ª F x D x The total work over a x b is the approximately, Â Â i i= i= 0 * ( i ) W ª W = F x Dx Fially, if we take the limit of this as goes to ifiity we ll get the exact work doe. So, Â Æ i = 0 * ( i ) W = lim F x Dx 007 Paul Dawkis 46

148 Math 40 This is, however, othig more tha the defiitio of the defiite itegral ad so the work doe by the force F( x ) over a x b is, b a ( ) W = F x dx 007 Paul Dawkis 47

149 Math 40 Area ad Volume Formulas I this sectio we will derive the formulas used to get the area betwee two curves ad the volume of a solid of revolutio. Area Betwee Two Curves We will start with the formula for determiig the area betwee y = f ( x) ad y g( x) iterval [a,b]. We will also assume that f ( x) g( x) o [a,b]. = o the We will ow proceed much as we did whe we looked that the Area Problem i the Itegrals Chapter. We will first divide up the iterval ito equal subitervals each with legth, b- a D x = Next, pick a poit i each subiterval, follows. x, ad we ca the use rectagles o each iterval as * i The height of each of these rectagles is give by, ad the area of each rectagle is the, * * ( i ) - ( i ) f x g x ( ( * ) ( * )) i i f x -g x D x So, the area betwee the two curves is the approximated by, Â ( ( * ) ( * i i )) Aª f x -g x Dx i= The exact area is, Â ( ( * ) ( * i i )) A= lim f x -g x Dx Æ i = 007 Paul Dawkis 48

150 Math 40 Now, recallig the defiitio of the defiite itegral this is othig more tha, b a ( ) ( ) A= f x -g x dx The formula above will work provided the two fuctios are i the form y = f ( x) ad ( ) y = g x. However, ot all fuctios are i that form. Sometimes we will be forced to work with fuctios i the form betwee x= f ( y) ad x g( y) y values ). = o the iterval [c,d] (a iterval of Whe this happes the derivatio is idetical. First we will start by assumig that f ( y) g( y) o [c,d]. We ca the divide up the iterval ito equal subitervals ad build rectagles o each of these itervals. Here is a sketch of this situatio. Followig the work from above, we will arrive at the followig for the area, c d ( ) ( ) A= f y -g y dy So, regardless of the form that the fuctios are i we use basically the same formula. 007 Paul Dawkis 49

151 Math 40 Volumes for Solid of Revolutio Before derivig the formula for this we should probably first defie just what a solid of revolutio is. To get a solid of revolutio we start out with a fuctio, y = f ( x), o a iterval [a,b]. We the rotate this curve about a give axis to get the surface of the solid of revolutio. For purposes of this derivatio let s rotate the curve about the x-axis. Doig this gives the followig three dimesioal regio. We wat to determie the volume of the iterior of this object. To do this we will proceed much as we did for the area betwee two curves case. We will first divide up the iterval ito subitervals of width, b- a D x = We will the choose a poit from each subiterval, Now, i the area betwee two curves case we approximated the area usig rectagles o each subiterval. For volumes we will use disks o each subiterval to approximate the area. The area * of the face of each disk is give by ( i ) Here is a sketch of this, x. * i A x ad the volume of each disk is i * ( i ) V = A x D x 007 Paul Dawkis 50

152 Math 40 The volume of the regio ca the be approximated by,  i= * ( i ) V ª A x Dx The exact volume is the, Æ i = ( ) * ( i ) V = lim A x Dx = b a  A x dx So, i this case the volume will be the itegral of the cross-sectioal area at ay x, A( x ). Note as well that, i this case, the cross-sectioal area is a circle ad we could go farther ad get a formula for that as well. However, the formula above is more geeral ad will work for ay way of gettig a cross sectio so we will leave it like it is. I the sectios where we actually use this formula we will also see that there are ways of geeratig the cross sectio that will actually give a cross-sectioal area that is a fuctio of y istead of x. I these cases the formula will be, c d ( ), V = A y dy c y d I this case we looked at rotatig a curve about the x-axis, however, we could have just as easily rotated the curve about the y-axis. I fact we could rotate the curve about ay vertical or horizotal axis ad i all of these, case we ca use oe or both of the followig formulas. b a d ( ) ( ) V = A x dx V = A y dy c 007 Paul Dawkis 5

153 Math 40 Types of Ifiity Most studets have ru across ifiity at some poit i time prior to a calculus class. However, whe they have dealt with it, it was just a symbol used to represet a really, really large positive or really, really large egative umber ad that was the extet of it. Oce they get ito a calculus class studets are asked to do some basic algebra with ifiity ad this is where they get ito trouble. Ifiity is NOT a umber ad for the most part does t behave like a umber. However, despite that we ll thik of ifiity i this sectio as a really, really, really large umber that is so large there is t aother umber larger tha it. This is ot correct of course, but may help with the discussio i this sectio. Note as well that everythig that we ll be discussig i this sectio applies oly to real umbers. If you move ito complex umbers for istace thigs ca ad do chage. So, let s start thikig about additio with ifiity. Whe you add two o-zero umbers you get a ew umber. For example, 4+ 7=. With ifiity this is ot true. With ifiity you have the followig. + a = where a - + = I other words, a really, really large positive umber ( ) plus ay positive umber, regardless of the size, is still a really, really large positive umber. Likewise, you ca add a egative umber (i.e. a < 0 ) to a really, really large positive umber ad stay really, really large ad positive. So, additio ivolvig ifiity ca be dealt with i a ituitive way if you re careful. Note as well that the a must NOT be egative ifiity. If it is, there are some serious issues that we eed to deal with as we ll see i a bit. Subtractio with egative ifiity ca also be dealt with i a ituitive way i most cases as well. A really, really large egative umber mius ay positive umber, regardless of its size, is still a really, really large egative umber. Subtractig a egative umber (i.e. a < 0 ) from a really, really large egative umber will still be a really, really large egative umber. Or, - - a=- where a =- Agai, a must ot be egative ifiity to avoid some potetially serious difficulties. Multiplicatio ca be dealt with fairly ituitively as well. A really, really large umber (positive, or egative) times ay umber, regardless of size, is still a really, really large umber we ll just eed to be careful with sigs. I the case of multiplicatio we have ( a)( ) = if a > 0 ( a)( ) =- if a< 0 ( )( ) ( )( ) ( )( ) = - - = - =- What you kow about products of positive ad egative umbers is still true here. Some forms of divisio ca be dealt with ituitively as well. A really, really large umber divided by a umber that is t too large is still a really, really large umber. 007 Paul Dawkis 5

154 Math 40 = if a > 0, a =- a a if a< 0, a =- if a> 0, a = a a if a< 0, a - Divisio of a umber by ifiity is somewhat ituitive, but there are a couple of subtleties that you eed to be aware of. Whe we talk about divisio by ifiity we are really talkig about a limitig process i which the deomiator is goig towards ifiity. So, a umber that is t too large divided a icreasigly large umber is a icreasigly small umber. I other words i the limit we have, a a = 0 = 0 - So, we ve dealt with almost every basic algebraic operatio ivolvig ifiity. There are two cases that that we have t dealt with yet. These are ± - =? =? ± The problem with these two cases is that ituitio does t really help here. A really, really large umber mius a really, really large umber ca be aythig ( -, a costat, or ). Likewise, a really, really large umber divided by a really, really large umber ca also be aythig ( ± this depeds o sig issues, 0, or a o-zero costat). What we ve got to remember here is that there are really, really large umbers ad the there are really, really, really large umbers. I other words, some ifiities are larger tha other ifiities. With additio, multiplicatio ad the first sets of divisio we worked this was t a issue. The geeral size of the ifiity just does t affect the aswer i those cases. However, with the subtractio ad divisio cases listed above, it does matter as we will see. Here is oe way to thik of this idea that some ifiities are larger tha others. This is a fairly dry ad techical way to thik of this ad your calculus problems will probably ever use this stuff, but this it is a ice way of lookig at this. Also, please ote that I m ot tryig to give a precise proof of aythig here. I m just tryig to give you a little isight ito the problems with ifiity ad how some ifiities ca be thought of as larger tha others. For a much better (ad defiitely more precise) discussio see, Let s start by lookig at how may itegers there are. Clearly, I hope, there are a ifiite umber of them, but let s try to get a better grasp o the size of this ifiity. So, pick ay two itegers completely at radom. Start at the smaller of the two ad list, i icreasig order, all the itegers that come after that. Evetually we will reach the larger of the two itegers that you picked. 007 Paul Dawkis 5

155 Math 40 Depedig o the relative size of the two itegers it might take a very, very log time to list all the itegers betwee them ad there is t really a purpose to doig it. But, it could be doe if we wated to ad that s the importat part. Because we could list all these itegers betwee two radomly chose itegers we say that the itegers are coutably ifiite. Agai, there is o real reaso to actually do this, it is simply somethig that ca be doe if we should chose to do so. I geeral a set of umbers is called coutably ifiite if we ca fid a way to list them all out. I a more precise mathematical settig this is geerally doe with a special kid of fuctio called a bijectio that associates each umber i the set with exactly oe of the positive itegers. To see some more details of this see the pdf give above. It ca also be show that the set of all fractios are also coutably ifiite, although this is a little harder to show ad is ot really the purpose of this discussio. To see a proof of this see the pdf give above. It has a very ice proof of this fact. Let s cotrast this by tryig to figure out how may umbers there are i the iterval (0,). By umbers, I mea all possible fractios that lie betwee zero ad oe as well as all possible decimals (that are t fractios) that lie betwee zero ad oe. The followig is similar to the proof give i the pdf above, but was ice eough ad easy eough (I hope) that I wated to iclude it here. To start let s assume that all the umbers i the iterval (0,) are coutably ifiite. This meas that there should be a way to list all of them out. We could have somethig like the followig, x = L Now, select the i th x x x M 4 = 0.704L = L = L decimal out of x i as show below x = L x x x 4 M = 0.704L = L = L M M ad form a ew umber with these digits. So, for our example we would have the umber x = 0.679L I this ew decimal replace all the s with a ad replace every other umbers with a. I the case of our example this would yield the ew umber 007 Paul Dawkis 54

156 Math 40 x = 0.L Notice that this umber is i the iterval (0,) ad also otice that give how we choose the digits of the umber this umber will ot be equal to the first umber i our list, x, because the first digit of each is guarateed to ot be the same. Likewise, this ew umber will ot get the same umber as the secod i our list, x, because the secod digit of each is guarateed to ot be the same. Cotiuig i this maer we ca see that this ew umber we costructed, x, is guarateed to ot be i our listig. But this cotradicts the iitial assumptio that we could list out all the umbers i the iterval (0,). Hece, it must ot be possible to list out all the umbers i the iterval (0,). Sets of umbers, such as all the umbers i (0,), that we ca t write dow i a list are called ucoutably ifiite. The reaso for goig over this is the followig. A ifiity that is ucoutably ifiite is sigificatly larger tha a ifiity that is oly coutably ifiite. So, if we take the differece of two ifiities we have a couple of possibilities. ( ucoutable) ( coutable) ( coutable) ( ucoutable) - = - =- Notice that we did t put dow a differece of two ifiities of the same type. Depedig upo the cotext there might still have some ambiguity about just what the aswer would be i this case, but that is a whole differet topic. We could also do somethig similar for quotiets of ifiities. ( coutable) = 0 ( ucoutable) ( ucoutable) = ( coutable) Agai, we avoided a quotiet of two ifiities of the same type sice, agai depedig upo the cotext, there might still be ambiguities about its value. So, that s it ad hopefully you ve leared somethig from this discussio. Ifiity simply is t a umber ad because there are differet kids of ifiity it geerally does t behave as a umber does. Be careful whe dealig with ifiity. 007 Paul Dawkis 55

157 Math 40 Summatio Notatio I this sectio we eed to do a brief review of summatio otatio or sigma otatio. We ll start out with two itegers, ad m, with < m ad a list of umbers deoted as follows, a, a+, a+, K, am-, am-, am We wat to add them up, i other words we wat, a + a + a + K + a + a + a + + m- m- m For large lists this ca be a fairly cumbersome otatio so we itroduce summatio otatio to deote these kids of sums. The case above is deoted as follows. m  a = a + a + a + K + a + a + a i= i + + m- m- m The i is called the idex of summatio. This otatio tells us to add all the a i s up for all itegers startig at ad edig at m. For istace, 4  i= 0 6  i= 4 4  i= i = = =.7666 i i x = x + x + x = 6x + x + 64x i * * * * * ( i ) = ( ) + ( ) + ( ) + ( 4) f x f x f x f x f x Properties Here are a couple of formulas for summatio otatio..   ca = c a where c is ay umber. So, we ca factor costats out of a summatio. i i= i0 i= i0. ( ) i i i i i= i0 i= i0 i= i0 i  a ± b = Âa ±  b So we ca break up a summatio across a sum or differece. Note that we started the series at i 0 to deote the fact that they ca start at ay value of i that we eed them to. Also ote that while we ca break up sums ad differeces as we did i above we ca t do the same thig for products ad quotiets. I other words, Ê ˆÊ ˆ i i= i0 Â( ab i i) ai bi Á Á  i= i0 i= i0 i= i0 i= i b 0 i Ë Ë a   i= i0 a b i i 007 Paul Dawkis 56

158 Math 40 Formulas Here are a couple of ice formulas that we will fid useful i a couple of sectios. Note that these formulas are oly true if startig at i =. You ca, of course, derive other formulas from these for differet startig poits if you eed to  i=  i=  i=  i= c= c ( + ) i = + + i = 6 i ( )( ) ( + ) È = Í Î Here is a quick example o how to use these properties to quickly evaluate a sum that would ot be easy to do by had. Example Usig the formulas ad properties from above determie the value of the followig summatio. 00  i= ( - i) Solutio The first thig that we eed to do is square out the stuff beig summed ad the break up the summatio usig the properties as follows, ( )   i= i= - i = 9- i+ 4i Now, usig the formulas, this is easy to compute, 00  i= ( i) ( )    = 9- i+ 4i i= i= i=    = 9- i+ 4 i= i= i= ( ) ( )( ) Ê00 0 ˆ Ê ˆ - = Á + 4Á Ë Ë 6 = 9700 Doig this by had would defiitely take some time ad there s a good chace that we might have made a mior mistake somewhere alog the lie. i 007 Paul Dawkis 57

159 Math 40 Costats of Itegratio I this sectio we eed to address a couple of topics about the costat of itegratio. Throughout most calculus classes we play pretty fast ad loose with it ad because of that may studets do t really uderstad it or how it ca be importat. First, let s address how we play fast ad loose with it. Recall that techically whe we itegrate a sum or differece we are actually doig multiple itegrals. For istace, x - x dx= x dx- x dx Upo evaluatig each of these itegrals we should get a costat of itegratio for each itegral sice we really are doig two itegrals. 5-9 = x x dx x dx x dx = x c x k = x x c k Sice there is o reaso to thik that the costats of itegratio will be the same from each itegral we use differet costats for each itegral. Now, both c ad k are ukow costats ad so the sum of two ukow costats is just a ukow costat ad we ackowledge that by simply writig the sum as a c. So, the itegral is the, x - x dx= x + x + c We also ted to play fast ad loose with costats of itegratio i some substitutio rule problems. Cosider the followig problem, cos( + x) + si( + x) dx= cosu+ siudu u = + x Techically whe we itegrate we should get, cos si si cos ( + x) + ( + x) dx= ( u- u+ c) Sice the whole itegral is multiplied by, the whole aswer, icludig the costat of itegratio, should be multiplied by. Upo multiplyig the through the aswer we get, c cos( + x) + si( + x) dx= siu- cosu+ 007 Paul Dawkis 58

160 Math 40 However, sice the costat of itegratio is a ukow costat dividig it by is t goig to chage that fact so we ted to just write the fractio as a c. cos( + x) + si( + x) dx= siu- cosu+ c I geeral, we do t really eed to worry about how we ve played fast ad loose with the costat of itegratio i either of the two examples above. The real problem however is that because we play fast ad loose with these costats of itegratio most studets do t really have a good grasp of them ad do t uderstad that there are times where the costats of itegratio are importat ad that we eed to be careful with them. To see how a lack of uderstadig about the costat of itegratio ca cause problems cosider the followig itegral. Ù ı x dx This is a really simple itegral. However, there are two ways (both simple) to itegrate it ad that is where the problem arises. The first itegratio method is to just break up the fractio ad do the itegral. l dx = Ù Ù x x dx = x + ı ı c The secod way is to use the followig substitutio. u = x du = dx fi dx= du l l dx = Ù Ù du = x u u + c = x + ı ı c Ca you see the problem? We itegrated the same fuctio ad got very differet aswers. This does t make ay sese. Itegratig the same fuctio should give us the same aswer. We oly used differet methods to do the itegral ad both are perfectly legitimate itegratio methods. So, how ca usig differet methods produce differet aswer? The first thig that we should otice is that because we used a differet method for each there is o reaso to thik that the costat of itegratio will i fact be the same umber ad so we really should use differet letters for each. More appropriate aswers would be, l l dx = x + c Ù Ù x x dx = x + ı ı k 007 Paul Dawkis 59

161 Math 40 Now, let s take aother look at the secod aswer. Usig a property of logarithms we ca write the aswer to the secod itegral as follows, Ù dx= l x + k ıx = ( l + l x ) + k = l x + l+ k Upo doig this we ca see that the aswers really are t that differet after all. I fact they oly differ by a costat ad we ca eve fid a relatioship betwee c ad k. It looks like, c= l + k So, without a proper uderstadig of the costat of itegratio, i particular usig differet itegratio techiques o the same itegral will likely produce a differet costat of itegratio, we might ever figure out why we got differet aswers for the itegral. Note as well that gettig aswers that differ by a costat does t violate ay priciples of calculus. I fact, we ve actually see a fact that suggested that this might happe. We saw a fact f x g x f x = g x + c. I i the Mea Value Theorem sectio that said that if ( ) = ( ) the ( ) ( ) other words, if two fuctios have the same derivative the they ca differ by o more tha a costat. This is exactly what we ve got here. The two fuctios, f ( x) = l x g( x) = l x have exactly the same derivative, x ad as we ve show they really oly differ by a costat. There is aother itegral that also exhibits this behavior. Cosider, si ( x) cos( ) x dx There are actually three differet methods for doig this itegral. Method : This method uses a trig formula, ( x) = ( x) ( x) si si cos Usig this formula (ad a quick substitutio) the itegral becomes, 007 Paul Dawkis 60

162 Math 40 si cos si cos 4 ( x) ( x) dx= ( x) dx=- ( x) + c Method : This method uses the substitutio, ( ) si ( ) u = cos x du =- x dx si( x) cos( x) dx=- udu =- u + c =- cos ( x) + c Method : Here is aother substitutio that could be doe here as well. ( ) cos( ) u = si x du = x dx si( x) cos( x) dx= udu = u + c = si ( x) + c So, we ve got three differet aswers each with a differet costat of itegratio. However, accordig to the fact above these three aswers should oly differ by a costat sice they all have the same derivative. I fact they do oly differ by a costat. We ll eed the followig trig formulas to prove this. ( x) ( x) ( x) ( x) ( x) cos = cos - si cos + si = Start with the aswer from the first method ad use the double agle formula above. - ( cos ( x) - si ( x) ) + c 4 Now, from the secod idetity above we have, so, plug this i, ( x) = - ( x) si cos ( ( ) ( ( ))) ( ) - cos x - - cos x + c =- ( cos x - ) + c 4 4 =- cos ( x) + + c 4 This is the aswer we got from the secod method with a slightly differet costat. I other words, c = + c 4 We ca do a similar maipulatio to get the aswer from the third method. Agai, startig with the aswer from the first method use the double agle formula ad the substitute i for the cosie istead of the sie usig, 007 Paul Dawkis 6

163 Math 40 Doig this gives, ( x) = - ( x) cos si (( si ( )) si ( )) si ( ) - - x - x + c =- ( - x ) + c 4 4 = si ( x) - + c 4 which is the aswer from the third method with a differet costat ad agai we ca relate the two costats by, c =- + c 4 So, what have we leared here? Hopefully we ve see that costats of itegratio are importat ad we ca t forget about them. We ofte do t work with them i a Calculus I course, yet without a good uderstadig of them we would be hard pressed to uderstad how differet itegratio methods ad apparetly produce differet aswers. 007 Paul Dawkis 6

164 Math 40 Itegratio Techiques Itroductio I this chapter we are goig to be lookig at various itegratio techiques. There are a fair umber of them ad some will be easier tha others. The poit of the chapter is to teach you these ew techiques ad so this chapter assumes that you ve got a fairly good workig kowledge of basic itegratio as well as substitutios with itegrals. I fact, most itegrals ivolvig simple substitutios will ot have ay of the substitutio work show. It is goig to be assumed that you ca verify the substitutio portio of the itegratio yourself. Also, most of the itegrals doe i this chapter will be idefiite itegrals. It is also assumed that oce you ca do the idefiite itegrals you ca also do the defiite itegrals ad so to coserve space we cocetrate mostly o idefiite itegrals. There is oe exceptio to this ad that is the Trig Substitutio sectio ad i this case there are some subtleties ivolved with defiite itegrals that we re goig to have to watch out for. Outside of that however, most sectios will have at most oe defiite itegral example ad some sectios will ot have ay defiite itegral examples. Here is a list of topics that are covered i this chapter. Itegratio by Parts Of all the itegratio techiques covered i this chapter this is probably the oe that studets are most likely to ru ito dow the road i other classes. Itegrals Ivolvig Trig Fuctios I this sectio we look at itegratig certai products ad quotiets of trig fuctios. Trig Substitutios Here we will look usig substitutios ivolvig trig fuctios ad how they ca be used to simplify certai itegrals. Partial Fractios We will use partial fractios to allow us to do itegrals ivolvig some ratioal fuctios. Itegrals Ivolvig Roots We will take a look at a substitutio that ca, o occasio, be used with itegrals ivolvig roots. Itegrals Ivolvig Quadratics I this sectio we are goig to look at some itegrals that ivolve quadratics. 007 Paul Dawkis 6

165 Math 40 Usig Itegral Tables Here we look at usig Itegral Tables as well as relatig ew itegrals back to itegrals that we already kow how to do. Itegratio Strategy We give a geeral set of guidelies for determiig how to evaluate a itegral. Improper Itegrals We will look at itegrals with ifiite itervals of itegratio ad itegrals with discotiuous itegrads i this sectio. Compariso Test for Improper Itegrals Here we will use the Compariso Test to determie if improper itegrals coverge or diverge. Approximatig Defiite Itegrals There are may ways to approximate the value of a defiite itegral. We will look at three of them i this sectio. 007 Paul Dawkis 64

166 Math 40 Itegratio by Parts Let s start off with this sectio with a couple of itegrals that we should already be able to do to get us started. First let s take a look at the followig. x x e dx= e + c So, that was simple eough. Now, let s take a look at, x e To do this itegral we ll use the followig substitutio. x u = x du = xdx fi xdx= du x u u x xe dx= du = + c= + c e e e dx Agai, simple eough to do provided you remember how to do substitutios. By the way make sure that you ca do these kids of substitutios quickly ad easily. From this poit o we are goig to be doig these kids of substitutios i our head. If you have to stop ad write these out with every problem you will fid that it will take you sigificatly loger to do these problems. Now, let s look at the itegral that we really wat to do. 6x x e dx 6x If we just had a x by itself or e by itself we could do the itegral easily eough. But, we do t have them by themselves, they are istead multiplied together. There is o substitutio that we ca use o this itegral that will allow us to do the itegral. So, at this poit we do t have the kowledge to do this itegral. To do this itegral we will eed to use itegratio by parts so let s derive the itegratio by parts formula. We ll start with the product rule. Now, itegrate both sides of this. fg = f g+ fg ( ) fg dx= f g+ fgdx ( ) The left side is easy eough to itegrate ad we ll split up the right side of the itegral. fg = f gdx+ fgdx 007 Paul Dawkis 65

167 Math 40 Note that techically we should have had a costat of itegratio show up o the left side after doig the itegratio. We ca drop it at this poit sice other costats of itegratio will be showig up dow the road ad they would just ed up absorbig this oe. Fially, rewrite the formula as follows ad we arrive at the itegratio by parts formula. fgdx = fg- f gdx This is ot the easiest formula to use however. So, let s do a couple of substitutios. ( ) ( ) ( ) ( ) u = f x v= g x du = f x dx dv= g x dx Both of these are just the stadard Calc I substitutios that hopefully you are used to by ow. Do t get excited by the fact that we are usig two substitutios here. They will work the same way. Usig these substitutios gives us the formula that most people thik of as the itegratio by parts formula. udv= uv- vdu To use this formula we will eed to idetify u ad dv, compute du ad v ad the use the formula. Note as well that computig v is very easy. All we eed to do is itegrate dv. v= dv So, let s take a look at the itegral above that we metioed we wated to do. Example Evaluate the followig itegral. 6 x x e dx Solutio So, o some level, the problem here is the x that is i frot of the expoetial. If that was t there we could do the itegral. Notice as well that i doig itegratio by parts aythig that we choose for u will be differetiated. So, it seems that choosig u = x will be a good choice sice upo differetiatig the x will drop out. Now that we ve chose u we kow that dv will be everythig else that remais. So, here are the choices for u ad dv as well as du ad v. 6 x u = x dv= e dx du = dx v= e dx = e 6 6x 6x 007 Paul Dawkis 66

168 Math 40 The itegral is the, x e 6 e Ù ı 6 e x 6 6 6x 6x 6x x dx= - dx 6x 6x = e - e + c Oce we have doe the last itegral i the problem we will add i the costat of itegratio to get our fial aswer. Next, let s take a look at itegratio by parts for defiite itegrals. The itegratio by parts formula for defiite itegrals is, Itegratio by Parts, Defiite Itegrals b b b = - a a a udv uv vdu b Note that the uv i the first term is just the stadard itegral evaluatio otatio that you should a be familiar with at this poit. All we do is evaluate the term, uv i this case, at b the subtract off the evaluatio of the term at a. At some level we do t really eed a formula here because we kow that whe doig defiite itegrals all we eed to do is do the idefiite itegral ad the do the evaluatio. Let s take a quick look at a defiite itegral usig itegratio by parts. Example Evaluate the followig itegral. 6x xe dx - Solutio This is the same itegral that we looked at i the first example so we ll use the same u ad dv to get, 6x x 6x 6x xe dx= e - e dx x 6x x = e - e = e + e Sice we eed to be able to do the idefiite itegral i order to do the defiite itegral ad doig the defiite itegral amouts to othig more tha evaluatig the idefiite itegral at a couple of poits we will cocetrate o doig idefiite itegrals i the rest of this sectio. I fact, 007 Paul Dawkis 67

169 Math 40 throughout most of this chapter this will be the case. We will be doig far more idefiite itegrals tha defiite itegrals. Let s take a look at some more examples. Example Evaluate the followig itegral. Ê t ˆ Ù ( t+ 5cos ) Á dt ı Ë 4 Solutio There are two ways to proceed with this example. For may, the first thig that they try is multiplyig the cosie through the parethesis, splittig up the itegral ad the doig itegratio by parts o the first itegral. While that is a perfectly acceptable way of doig the problem it s more work tha we really eed to do. Istead of splittig the itegral up let s istead use the followig choices for u ad dv. The itegral is the, u = t+ 5 Ê t ˆ dv= cosá dt Ë 4 du = dt Ê t ˆ v= 4si Á Ë 4 Ê t ˆ Ê t ˆ Ê t ˆ Ù( t+ 5cos ) Á dt = 4 ( t+ 5si ) Á -Ùsi Á dt ı Ë4 Ë4 ı Ë4 Ê t ˆ Ê t ˆ = 4 ( t+ 5si ) Á + 48cosÁ + c Ë4 Ë4 Notice that we pulled ay costats out of the itegral whe we used the itegratio by parts formula. We will usually do this i order to simplify the itegral a little. Example 4 Evaluate the followig itegral. w si 0w dw ( ) Solutio For this example we ll use the followig choices for u ad dv. The itegral is the, ( ) u = w dv= si 0wdw du = wdw v=- cos 0w 0 w ( ) ( ) =- ( ) + ( ) w si 0wdw cos 0w wcos 0w dw 0 5 I this example, ulike the previous examples, the ew itegral will also require itegratio by parts. For this secod itegral we will use the followig choices. 007 Paul Dawkis 68

170 Math 40 So, the itegral becomes, ( ) u = w dv= cos 0wdw du = dw v= si 0 w 0 ( ) w w Á 0 5Ë0 0 Ê w si 0wdw cos 0w si 0w si 0wdw ( ) =- ( ) + ( )- ( ) w Ê w ˆ =- cos( 0w) + Á si( 0w) + cos( 0w) + c 0 5Ë0 00 w w =- cos( 0w) + si( 0w) + cos( 0w) + c Be careful with the coefficiet o the itegral for the secod applicatio of itegratio by parts. Sice the itegral is multiplied by 5 we eed to make sure that the results of actually doig the itegral are also multiplied by 5. Forgettig to do this is oe of the more commo mistakes with itegratio by parts problems. As this last example has show us, we will sometimes eed more tha oe applicatio of itegratio by parts to completely evaluate a itegral. This is somethig that will happe so do t get excited about it whe it does. I this ext example we eed to ackowledge a importat poit about itegratio techiques. Some itegrals ca be doe i usig several differet techiques. That is the case with the itegral i the ext example. ˆ Example 5 Evaluate the followig itegral x x+ dx (a) Usig Itegratio by Parts. [Solutio] (b) Usig a stadard Calculus I substitutio. [Solutio] Solutio (a) Evaluate usig Itegratio by Parts. First otice that there are o trig fuctios or expoetials i this itegral. While a good may itegratio by parts itegrals will ivolve trig fuctios ad/or expoetials ot all of them will so do t get too locked ito the idea of expectig them to show up. I this case we ll use the followig choices for u ad dv. The itegral is the, u = x dv= x+ dx du = dx v= x+ ( ) 007 Paul Dawkis 69

171 Math 40 x x+ dx= x( x+ ) - ( ) x+ dx 5 4 = x( x+ ) - ( x+ ) + c 5 (b) Evaluate Usig a stadard Calculus I substitutio. [Retur to Problems] Now let s do the itegral with a substitutio. We ca use the followig substitutio. u = x+ x = u- du = dx Notice that we ll actually use the substitutio twice, oce for the quatity uder the square root ad oce for the x i frot of the square root. The itegral is the, ( ) x x+ dx= u- udu = u -u du 5 = u - u + c 5 5 = ( x+ ) - ( x+ ) + c 5 [Retur to Problems] So, we used two differet itegratio techiques i this example ad we got two differet aswers. The obvious questio the should be : Did we do somethig wrog? Actually, we did t do aythig wrog. We eed to remember the followig fact from Calculus I. If ( ) ( ) the ( ) ( ) f x = g x f x = g x + c I other words, if two fuctios have the same derivative the they will differ by o more tha a costat. So, how does this apply to the above problem? First defie the followig, ( ) = ( ) = + f x g x x x The we ca compute f ( x) ad g( x ) by itegratig as follows, ( ) = ( ) ( ) = ( ) f x f x dx g x g x dx We ll use itegratio by parts for the first itegral ad the substitutio for the secod itegral. The accordig to the fact f ( x ) ad ( ) g x should differ by o more tha a costat. Let s verify this ad see if this is the case. We ca verify that they differ by o more tha a costat if 007 Paul Dawkis 70

172 Math 40 we take a look at the differece of the two ad do a little algebraic maipulatio ad simplificatio. 5 5 Ê 4 ˆ Ê ˆ ( ) ( ) ( ) ( ) Á x x+ - x+ - Á x+ - x+ Ë 5 Ë5 Ê 4 ˆ = ( x+ ) Á x- ( x+ ) - ( x+ ) + Ë 5 5 ( x ) ( 0) = + = 0 So, i this case it turs out the two fuctios are exactly the same fuctio sice the differece is zero. Note that this wo t always happe. Sometimes the differece will yield a ozero costat. For a example of this check out the Costat of Itegratio sectio i my Calculus I otes. So just what have we leared? First, there will, o occasio, be more tha oe method for evaluatig a itegral. Secodly, we saw that differet methods will ofte lead to differet aswers. Last, eve though the aswers are differet it ca be show, sometimes with a lot of work, that they differ by o more tha a costat. Whe we are faced with a itegral the first thig that we ll eed to decide is if there is more tha oe way to do the itegral. If there is more tha oe way we ll the eed to determie which method we should use. The geeral rule of thumb that I use i my classes is that you should use the method that you fid easiest. This may ot be the method that others fid easiest, but that does t make it the wrog method. Oe of the more commo mistakes with itegratio by parts is for people to get too locked ito perceived patters. For istace, all of the previous examples used the basic patter of takig u to be the polyomial that sat i frot of aother fuctio ad the lettig dv be the other fuctio. This will ot always happe so we eed to be careful ad ot get locked ito ay patters that we thik we see. Let s take a look at some itegrals that do t fit ito the above patter. Example 6 Evaluate the followig itegral. l xdx Solutio So, ulike ay of the other itegral we ve doe to this poit there is oly a sigle fuctio i the itegral ad o polyomial sittig i frot of the logarithm. The first choice of may people here is to try ad fit this ito the patter from above ad make the followig choices for u ad dv. 007 Paul Dawkis 7

173 Math 40 u = dv= l xdx This leads to a real problem however sice that meas v must be, v= l xdx I other words, we would eed to kow the aswer ahead of time i order to actually do the problem. So, this choice simply wo t work. Also otice that with this choice we d get that du = 0 which also causes problems ad is aother reaso why this choice will ot work. Therefore, if the logarithm does t belog i the dv it must belog istead i the u. So, let s use the followig choices istead u = l x dv= dx The itegral is the, Example 7 Evaluate the followig itegral. du = dx v= x x lxdx= xl x- Ù xdx ı x = xl x- dx = xl x- x+ c 5 x x + dx Solutio So, if we agai try to use the patter from the first few examples for this itegral our choices for u ad dv would probably be the followig. 5 u = x dv= x + dx However, as with the previous example this wo t work sice we ca t easily compute v. v= x + dx This is ot a easy itegral to do. However, otice that if we had a x i the itegral alog with the root we could very easily do the itegral with a substitutio. Also otice that we do have a lot of x s floatig aroud i the origial itegral. So istead of puttig all the x s (outside of the root) i the u let s split them up as follows. u x dv x x dx = = + du = xdx v= x + 9 ( ) We ca ow easily compute v ad after usig itegratio by parts we get, 007 Paul Dawkis 7

174 Math 40 5 x x + dx= x ( x + ) - Ù x ( x + ) dx 9 ı 5 4 = x ( x + ) - ( x + ) + c 9 45 So, i the previous two examples we saw cases that did t quite fit ito ay perceived patter that we might have gotte from the first couple of examples. This is always somethig that we eed to be o the lookout for with itegratio by parts. Let s take a look at aother example that also illustrates aother itegratio techique that sometimes arises out of itegratio by parts problems. Example 8 Evaluate the followig itegral. q e cosq dq Solutio Okay, to this poit we ve always picked u i such a way that upo differetiatig it would make that portio go away or at the very least put it the itegral ito a form that would make it easier to deal with. I this case o matter which part we make u it will ever go away i the differetiatio process. It does t much matter which we choose to be u so we ll choose i the followig way. Note however that we could choose the other way as well ad we ll get the same result i the ed. q u = cosq dv= e dq du =- siq dq v= e q The itegral is the, q q q e cosq dq = e cosq + e siq dq So, it looks like we ll do itegratio by parts agai. Here are our choices this time. q u = siq dv= e dq du = cosq dq v= e q The itegral is ow, q q q q e cosq dq = e cosq + e siq - e cosq dq Now, at this poit it looks like we re just ruig i circles. However, otice that we ow have the same itegral o both sides ad o the right side it s got a mius sig i frot of it. This meas that we ca add the itegral to both sides to get, e q cosq dq = e q cosq + e q siq 007 Paul Dawkis 7

175 Math 40 All we eed to do ow is divide by ad we re doe. The itegral is, q q q e cosq dq = ( cosq + siq) + c e e Notice that after dividig by the two we add i the costat of itegratio at that poit. This idea of itegratig util you get the same itegral o both sides of the equal sig ad the simply solvig for the itegral is kid of ice to remember. It does t show up all that ofte, but whe it does it may be the oly way to actually do the itegral. We ve got oe more example to do. As we will see some problems could require us to do itegratio by parts umerous times ad there is a short had method that will allow us to do multiple applicatios of itegratio by parts quickly ad easily. Example 9 Evaluate the followig itegral. x e dx Solutio We start off by choosig u ad dv as we always would. However, istead of computig du ad v we put these ito the followig table. We the differetiate dow the colum correspodig to u util we hit zero. I the colum correspodig to dv we itegrate oce for each etry i the first colum. There is also a third colum which we will explai i a bit ad it always starts with a + ad the alterates sigs as show. 4 x Now, multiply alog the diagoals show i the table. I frot of each product put the sig i the third colum that correspods to the u term for that product. I this case this would give, x x x x x x 4 4 Ê ˆ Ê ˆ Ê ˆ Ê ˆ Ê ˆ x e dx= ( x ) Áe - ( 4x ) Á4e + ( x ) Á8e - ( 4x) Á6e + ( 4) Áe Ë Ë Ë Ë Ë x x x x x 4 = x e - 6x e + 96x e - 84xe + 768e + c 007 Paul Dawkis 74

176 Math 40 We ve got the itegral. This is much easier tha writig dow all the various u s ad dv s that we d have to do otherwise. So, i this sectio we ve see how to do itegratio by parts. I your later math classes this is liable to be oe of the more frequet itegratio techiques that you ll ecouter. It is importat to ot get too locked ito patters that you may thik you ve see. I most cases ay patter that you thik you ve see ca (ad will be) violated at some poit i time. Be careful! Also, do t forget the shorthad method for multiple applicatios of itegratio by parts problems. It ca save you a fair amout of work o occasio. 007 Paul Dawkis 75

177 Math 40 Itegrals Ivolvig Trig Fuctios I this sectio we are goig to look at quite a few itegrals ivolvig trig fuctios ad some of the techiques we ca use to help us evaluate them. Let s start off with a itegral that we should already be able to do. 5 5 cos si = usig the substitutio = si x xdx u du u x si 6 6 = x+ c This itegral is easy to do with a substitutio because the presece of the cosie, however, what about the followig itegral. Example Evaluate the followig itegral. 5 si xdx Solutio This itegral o loger has the cosie i it that would allow us to use the substitutio that we used above. Therefore, that substitutio wo t work ad we are goig to have to fid aother way of doig this itegral. Let s first otice that we could write the itegral as follows, Now recall the trig idetity, 5 4 = = ( ) si xdx si xsixdx si x si xdx cos x+ si x= fi si x= - cos x With this idetity the itegral ca be writte as, 5 = ( - ) si xdx cos x si xdx ad we ca ow use the substitutio u = cos x. Doig this gives us, ( ) u 5 si xdx=- -u du 4 =- - + u du Ê 5ˆ =-Áu- u + u + c Ë 5 =- cos + cos - cos x x x c So, with a little rewritig o the itegrad we were able to reduce this to a fairly simple 007 Paul Dawkis 76

178 Math 40 substitutio. Notice that we were able to do the rewrite that we did i the previous example because the expoet o the sie was odd. I these cases all that we eed to do is strip out oe of the sies. The expoet o the remaiig sies will the be eve ad we ca easily covert the remaiig sies to cosies usig the idetity, cos x si x + = () If the expoet o the sies had bee eve this would have bee difficult to do. We could strip out a sie, but the remaiig sies would the have a odd expoet ad while we could covert them to cosies the resultig itegral would ofte be eve more difficult tha the origial itegral i most cases. Let s take a look at aother example. Example Evaluate the followig itegral. si xcos 6 Solutio So, i this case we ve got both sies ad cosies i the problem ad i this case the expoet o the sie is eve while the expoet o the cosie is odd. So, we ca use a similar techique i this itegral. This time we ll strip out a cosie ad covert the rest to sies. 6 6 si cos = si cos cos 6 ( ) ( ) 6 u u du xdx x xdx x x xdx = si x - si x cosxdx u = si x = = u -u du x x c = si - si + At this poit let s pause for a secod to summarize what we ve leared so far about itegratig powers of sie ad cosie. m si xcos xdx () I this itegral if the expoet o the sies () is odd we ca strip out oe sie, covert the rest to cosies usig () ad the use the substitutio u = cos x. Likewise, if the expoet o the cosies (m) is odd we ca strip out oe cosie ad covert the rest to sies ad the use the substitutio u = si x. Of course, if both expoets are odd the we ca use either method. However, i these cases it s usually easier to covert the term with the smaller expoet. 007 Paul Dawkis 77

179 Math 40 The oe case we have t looked at is what happes if both of the expoets are eve? I this case the techique we used i the first couple of examples simply wo t work ad i fact there really is t ay oe set method for doig these itegrals. Each itegral is differet ad i some cases there will be more tha oe way to do the itegral. With that beig said most, if ot all, of itegrals ivolvig products of sies ad cosies i which both expoets are eve ca be doe usig oe or more of the followig formulas to rewrite the itegrad. cos x= ( + cos( x) ) si x= ( -cos( x) ) sixcosx= si( x) The first two formulas are the stadard half agle formula from a trig class writte i a form that will be more coveiet for us to use. The last is the stadard double agle formula for sie, agai with a small rewrite. Let s take a look at a example. Example Evaluate the followig itegral. si xcos Solutio As oted above there are ofte more tha oe way to do itegrals i which both of the expoets are eve. This itegral is a example of that. There are at least two solutio techiques for this problem. We will do both solutios startig with what is probably the harder of the two, but it s also the oe that may people see first. Solutio I this solutio we will use the two half agle formulas above ad just substitute them ito the itegral. ʈ si xcos xdx= Ù ( - cos( x) ) Á ( + cos( x) ) dx ı Ë cos = - ( x) dx 4 So, we still have a itegral that ca t be completely doe, however otice that we have maaged to reduce the itegral dow to just oe term causig problems (a cosie with a eve power) rather tha two terms causig problems. I fact to elimiate the remaiig problem term all that we eed to do is reuse the first half agle xdx 007 Paul Dawkis 78

180 Math 40 formula give above. si xcos xdx= Ù- ( + cos( 4x) ) dx 4ı = Ù - cos ( 4 x) dx 4ı Ê ˆ = Á x- si ( 4 x) + c 4Ë 8 = x- si ( 4 x) + c 8 So, this solutio required a total of three trig idetities to complete. Solutio I this solutio we will use the half agle formula to help simplify the itegral as follows. ( ) si cos = si cos x xdx x x dx Ê ˆ = Ù Á si ( x) ıë si ( ) = x dx 4 Now, we use the double agle formula for sie to reduce to a itegral that we ca do. ( ) si xcos xdx= -cos 4x dx 8 = x- si ( 4 x) + c 8 This method required oly two trig idetities to complete. Notice that the differece betwee these two methods is more oe of messiess. The secod method is ot appreciably easier (other tha eedig oe less trig idetity) it is just ot as messy ad that will ofte traslate ito a easier process. I the previous example we saw two differet solutio methods that gave the same aswer. Note that this will ot always happe. I fact, more ofte tha ot we will get differet aswers. However, as we discussed i the Itegratio by Parts sectio, the two aswers will differ by o more tha a costat. I geeral whe we have products of sies ad cosies i which both expoets are eve we will eed to use a series of half agle ad/or double agle formulas to reduce the itegral ito a form that we ca itegrate. Also, the larger the expoets the more we ll eed to use these formulas ad hece the messier the problem. dx 007 Paul Dawkis 79

181 Math 40 Sometimes i the process of reducig itegrals i which both expoets are eve we will ru across products of sie ad cosie i which the argumets are differet. These will require oe of the followig formulas to reduce the products to itegrals that we ca do. siacosb = Èsi( ) si ( ) Î a - b + a + b siasib = Ècos( a b) cos( a b) Î cosacosb = Ècos( a - b) + cos( a + b) Î Let s take a look at a example of oe of these kids of itegrals. Example 4 Evaluate the followig itegral. cos 5x cos 4x dx ( ) ( ) Solutio This itegral requires the last formula listed above. cos( 5x) cos( 4x) dx= cos( ) cos( 9 ) x + x dx Ê ˆ = Á si( x) + si( 9x) + c Ë 9 Okay, at this poit we ve covered pretty much all the possible cases ivolvig products of sies ad cosies. It s ow time to look at itegrals that ivolve products of secats ad tagets. This time, let s do a little aalysis of the possibilities before we just jump ito examples. The geeral itegral will be, m sec xta xdx () The first thig to otice is that we ca easily covert eve powers of secats to tagets ad eve powers of tagets to secats by usig a formula similar to (). I fact, the formula ca be derived from () so let s do that. si x+ cos x= si cos x x + = cos x cos x cos x 007 Paul Dawkis 80

182 Math 40 ta x+ = sec x (4) Now, we re goig to wat to deal with () similarly to how we dealt with (). We ll wat to evetually use oe of the followig substitutios. u tax du sec xdx = = u = secx du = secxta xdx So, if we use the substitutio u = ta x we will eed two secats left for the substitutio to work. This meas that if the expoet o the secat () is eve we ca strip two out ad the covert the remaiig secats to tagets usig (4). Next, if we wat to use the substitutio u = sec xwe will eed oe secat ad oe taget left over i order to use the substitutio. This meas that if the expoet o the taget (m) is odd ad we have at least oe secat i the itegrad we ca strip out oe of the tagets alog with oe of the secats of course. The taget will the have a eve expoet ad so we ca use (4) to covert the rest of the tagets to secats. Note that this method does require that we have at least oe secat i the itegral as well. If there are t ay secats the we ll eed to do somethig differet. If the expoet o the secat is eve ad the expoet o the taget is odd the we ca use either case. Agai, it will be easier to covert the term with the smallest expoet. Let s take a look at a couple of examples. Example 5 Evaluate the followig itegral. sec xta 9 5 Solutio First ote that sice the expoet o the secat is t eve we ca t use the substitutio u = ta x. However, the expoet o the taget is odd ad we ve got a secat i the itegral ad so we will be able to use the substitutio u = sec x. This meas stripig out a sigle taget (alog with a secat) ad covertig the remaiig tagets to secats usig (4). Here s the work for this itegral sec ta sec ta ta sec 8 ( ) ( ) 8 u u du 0 8 u u u du xdx x xdx= x x x xdx = sec x sec x- taxsecxdx u = sec x = - = - + sec x sec x sec x c 9 9 = Paul Dawkis 8

183 Math 40 Example 6 Evaluate the followig itegral. sec xta 4 6 Solutio So, i this example the expoet o the taget is eve so the substitutio u = sec x wo t work. The expoet o the secat is eve ad so we ca use the substitutio u = ta x for this itegral. That meas that we eed to strip out two secats ad covert the rest to tagets. Here is the work for this itegral. ( ) 6 ( u ) 8 6 u u du xdx sec xta xdx= sec xta xsec xdx 6 = ta x+ ta xsec xdx u = ta x = + = + u du x x c = ta + ta + Both of the previous examples fit very icely ito the patters discussed above ad so were ot all that difficult to work. However, there are a couple of exceptios to the patters above ad i these cases there is o sigle method that will work for every problem. Each itegral will be differet ad may require differet solutio methods i order to evaluate the itegral. Let s first take a look at a couple of itegrals that have odd expoets o the tagets, but o secats. I these cases we ca t use the substitutio u = sec xsice it requires there to be at least oe secat i the itegral. Example 7 Evaluate the followig itegral. ta xdx Solutio To do this itegral all we eed to do is recall the defiitio of taget i terms of sie ad cosie ad the this itegral is othig more tha a Calculus I substitutio. si x taxdx= Ù dx u = cos x ı cos x =- Ù du ı u =- l cosx + c rlx= l x - = l cos x + c l sec x + c r Example 8 Evaluate the followig itegral. ta xdx 007 Paul Dawkis 8

184 Math 40 Solutio The trick to this oe is do the followig maipulatio of the itegrad. ta xdx= taxta xdx ( ) = tax sec x- dx ta sec ta = x xdx- xdx We ca ow use the substitutio u = ta x o the first itegral ad the results from the previous example to o the secod itegral. The itegral is the, xdx= x- x + c ta ta l sec Note that all odd powers of taget (with the exceptio of the first power) ca be itegrated usig the same method we used i the previous example. For istace, = ( - ) = - 5 ta xdx ta x sec x dx ta xsec xdx ta xdx So, a quick substitutio ( u = ta x) will give us the first itegral ad the secod itegral will always be the previous odd power. Now let s take a look at a couple of examples i which the expoet o the secat is odd ad the expoet o the taget is eve. I these cases the substitutios used above wo t work. It should also be oted that both of the followig two itegrals are itegrals that we ll be seeig o occasio i later sectios of this chapter ad i later chapters. Because of this it would t be a bad idea to make a ote of these results so you ll have them ready whe you eed them later. Example 9 Evaluate the followig itegral. sec xdx Solutio This oe is t too bad oce you see what you ve got to do. By itself the itegral ca t be doe. However, if we maipulate the itegrad as follows we ca do it. sec xdx= = Ù ı Ù ı ( + ) secx secx ta x secx+ ta x sec + ta sec dx x x x dx secx+ ta x I this form we ca do the itegral usig the substitutio u = secx+ ta x. Doig this gives, secxdx= l secx+ ta x + c 007 Paul Dawkis 8

185 Math 40 The idea used i the above example is a ice idea to keep i mid. Multiplyig the umerator ad deomiator of a term by the same term above ca, o occasio, put the itegral ito a form that ca be itegrated. Note that this method wo t always work ad eve whe it does it wo t always be clear what you eed to multiply the umerator ad deomiator by. However, whe it does work ad you ca figure out what term you eed it ca greatly simplify the itegral. Here s the ext example. Example 0 Evaluate the followig itegral. sec xdx Solutio This oe is differet from ay of the other itegrals that we ve doe i this sectio. The first step to doig this itegral is to perform itegratio by parts usig the followig choices for u ad dv. u = secx dv= sec xdx du = secxtaxdx v= ta x Note that usig itegratio by parts o this problem is ot a obvious choice, but it does work very icely here. After doig itegratio by parts we have, sec xdx= secxtax- secxta xdx Now the ew itegral also has a odd expoet o the secat ad a eve expoet o the taget ad so the previous examples of products of secats ad tagets still wo t do us ay good. To do this itegral we ll first write the tagets i the itegral i terms of secats. Agai, this is ot ecessarily a obvious choice but it s what we eed to do i this case. ( ) sec = sec ta - sec sec - xdx x x x x dx = secxtax- sec xdx+ sec xdx Now, we ca use the results from the previous example to do the secod itegral ad otice that the first itegral is exactly the itegral we re beig asked to evaluate with a mius sig i frot. So, add it to both sides to get, Fially divide by a two ad we re doe. sec = sec ta + l sec + ta 007 Paul Dawkis 84 xdx x x x x sec xdx= secxtax+ l secx+ ta x + c ( ) Agai, ote that we ve agai used the idea of itegratig the right side util the origial itegral shows up ad the movig this to the left side ad dividig by its coefficiet to complete the evaluatio. We first saw this i the Itegratio by Parts sectio ad oted at the time that this was a ice techique to remember. Here is aother example of this techique.

186 Math 40 Now that we ve looked at products of secats ad tagets let s also ackowledge that because we ca relate cosecats ad cotagets by + cot x= csc x all of the work that we did for products of secats ad tagets will also work for products of cosecats ad cotagets. We ll leave it to you to verify that. There is oe fial topic to be discussed i this sectio before movig o. To this poit we ve looked oly at products of sies ad cosies ad products of secats ad tagets. However, the methods used to do these itegrals ca also be used o some quotiets ivolvig sies ad cosies ad quotiets ivolvig secats ad tagets (ad hece quotiets ivolvig cosecats ad cotagets). Let s take a quick look at a example of this. Example Evaluate the followig itegral. 7 si x Ù 4 dx ı cos x Solutio If this were a product of sies ad cosies we would kow what to do. We would strip out a sie (sice the expoet o the sie is odd) ad covert the rest of the sies to cosies. The same idea will work i this case. We ll strip out a sie from the umerator ad covert the rest to cosies as follows, si Ù ı cos x si dx= Ù x ı cos = = Ù ı Ù ı ( si x) cos x si xdx x ( -cos x) 007 Paul Dawkis cos x 4 x si xdx si xdx At this poit all we eed to do is use the substitutio u = cos xad we re doe. 7 si x Ù dx =-Ù ı cos x ı ( -u ) u 4 4 du = u u u du Ê ˆ =-Á- + + u- u c + Ë u u = - - cos + cos + x x c cos x cosx

187 Math 40 So, uder the right circumstaces, we ca use the ideas developed to help us deal with products of trig fuctios to deal with quotiets of trig fuctios. The atural questio the, is just what are the right circumstaces? First otice that if the quotiet had bee reversed as i this itegral, cos Ù ı si we would t have bee able to strip out a sie. Ù ı 4 7 x dx x 4 4 cos x cos x dx= 7 Ù 6 si si si x ı x x I this case the stripped out sie remais i the deomiator ad it wo t do us ay good for the substitutio u = cos xsice this substitutio requires a sie i the umerator of the quotiet. Also ote that, while we could covert the sies to cosies, the resultig itegral would still be a fairly difficult itegral. So, we ca use the methods we applied to products of trig fuctios to quotiets of trig fuctios provided the term that eeds parts stripped out i is the umerator of the quotiet. dx 007 Paul Dawkis 86

188 Math 40 Trig Substitutios As we have doe i the last couple of sectios, let s start off with a couple of itegrals that we should already be able to do with a stadard substitutio. ( ) x x 5x - 4dx= 5x c Ù dx= 5x - 4+ c 75 ı 5x -4 5 Both of these used the substitutio u = 5x - 4 ad at this poit should be pretty easy for you to do. However, let s take a look at the followig itegral. Example Evaluate the followig itegral. Solutio Ù ı 5x 4 I this case the substitutio u = 5x - 4 will ot work ad so we re goig to have to do somethig differet for this itegral. x It would be ice if we could get rid of the square root somehow. The followig substitutio will do that for us. x= sec q 5 Do ot worry about where this came from at this poit. As we work the problem you will see that it works ad that if we have a similar type of square root i the problem we ca always use a similar substitutio. Before we actually do the substitutio however let s verify the claim that this will allow us to get rid of the square root. Ê 4 ˆ Á Ë5 - dx ( ) 5x - 4 = 5 sec q - 4 = 4sec q - = sec q - To get rid of the square root all we eed to do is recall the relatioship, ta q + = sec q fi sec q - = ta q Usig this fact the square root becomes, 5-4 = ta = ta x q q Note the presece of the absolute value bars there. These are importat. Recall that x = x There should always be absolute value bars at this stage. If we kew that taq was always 007 Paul Dawkis 87

189 Math 40 positive or always egative we could elimiate the absolute value bars usig, Ïx if x 0 x = Ì Ó - x if x < 0 Without limits we wo t be able to determie if taq is positive or egative, however, we will eed to elimiate them i order to do the itegral. Therefore, sice we are doig a idefiite itegral we will assume that taq will be positive ad so we ca drop the absolute value bars. This gives, 5x - 4 = ta So, we were able to elimiate the square root usig this substitutio. Let s ow do the substitutio ad see what we get. I doig the substitutio do t forget that we ll also eed to substitute for the dx. This is easy eough to get from the substitutio. x= secq fi dx= secq taq dq 5 5 Usig this substitutio the itegral becomes, Ù ı - q Ê ˆ = Á q q dq 5x 4 ta dx Ù sec ta x ı 5 secq Ë5 = ta q dq With this substitutio we were able to reduce the give itegral to a itegral ivolvig trig fuctios ad we saw how to do these problems i the previous sectio. Let s fiish the itegral. 5x -4 Ù dx= - ı x q sec q ( q q) dq = ta - + c So, we ve got a aswer for the itegral. Ufortuately the aswer is t give i x s as it should be. So, we eed to write our aswer i terms of x. We ca do this with some right triagle trig. From our origial substitutio we have, 5x hypoteuse secq = = adjacet This gives the followig right triagle. 007 Paul Dawkis 88

190 Math 40 From this we ca see that, taq = 5x - 4 We ca deal with the q i oe of ay variety of ways. From our substitutio we ca see that, - Ê5x ˆ q = sec Á Ë While this is a perfectly acceptable method of dealig with the q we ca use ay of the possible six iverse trig fuctios ad sice sie ad cosie are the two trig fuctios most people are familiar with we will usually use the iverse sie or iverse cosie. I this case we ll use the iverse cosie. So, with all of this the itegral becomes, - Ê ˆ q = cos Á Ë5x 5x -4 Ê 5x -4 -Ê ˆˆ Ù dx= - cos Á + c ı x Á 5x Ë Ë -Ê ˆ = 5x -4- cos Á + c Ë5x We ow have the aswer back i terms of x. Wow! That was a lot of work. Most of these wo t take as log to work however. This first oe eeded lot s of explaatio sice it was the first oe. The remaiig examples wo t eed quite as much explaatio ad so wo t take as log to work. However, before we move oto more problems let s first address the issue of defiite itegrals ad how the process differs i these cases. Example Evaluate the followig itegral. Ù ı x - 4 x dx 007 Paul Dawkis 89

191 Math 40 Solutio The limits here wo t chage the substitutio so that will remai the same. x = sec q 5 Usig this substitutio the square root still reduces dow to, 5x - 4 = ta However, ulike the previous example we ca t just drop the absolute value bars. I this case we ve got limits o the itegral ad so we ca use the limits as well as the substitutio to determie the rage of q that we re i. Oce we ve got that we ca determie how to drop the absolute value bars. Here s the limits of q. x = fi = sec q fi q = p x = fi = sec q fi q = So, if we are i the rage 4 p 5 x 5 the q is i the rage of 0 q ad i this rage of q s taget is positive ad so we ca just drop the absolute value bars. Let s do the substitutio. Note that the work is idetical to the previous example ad so most of it is left out. We ll pick up at the fial itegral ad the do the substitutio. Ù ı x -4 x q p sec q 0 dx= - ( q q) = ta - p = - Note that because of the limits we did t eed to resort to a right triagle to complete the problem. Let s take a look at a differet set of limits for this itegral. p 0 dq Example Evaluate the followig itegral. Ù ı x - 4 Solutio Agai, the substitutio ad square root are the same as the first two examples. x dx 007 Paul Dawkis 90

192 Math 40 sec 5 q x 4 ta x= - = 5 q Let s ext see the limits q for this problem. x =- fi - = sec q fi q = p 4 4 p x =- fi - = secq fi q = Note that i determiig the value of q we used the smallest positive value. Now for this rage of x s we have p q p ad i this rage of q taget is egative ad so i this case we ca drop the absolute value bars, but will eed to add i a mius sig upo doig so. I other words, 5x - 4 =- ta So, the oly chage this will make i the itegratio process is to put a mius sig i frot of the itegral. The itegral is the, Ù ı x -4 x q p p sec q dx=- - ( q q) =- ta - p = - I the last two examples we saw that we have to be very careful with defiite itegrals. We eed to make sure that we determie the limits o q ad whether or ot this will mea that we ca drop the absolute value bars or if we eed to add i a mius sig whe we drop them. Before movig o to the ext example let s get the geeral form for the substitutio that we used i the previous set of examples. p dq p a - fi = secq b bx a x Let s work a ew ad differet type of example. Example 4 Evaluate the followig itegral. Ù dx 4 ı x 9-x Solutio Now, the square root i this problem looks to be (almost) the same as the previous oes so let s try the same type of substitutio ad see if it will work here as well. 007 Paul Dawkis 9

193 Math 40 Usig this substitutio the square root becomes, x = secq 9- = 9-9sec = - sec = - ta x q q q So usig this substitutio we will ed up with a egative quatity (the taget squared is always positive of course) uder the square root ad this will be trouble. Usig this substitutio will give complex values ad we do t wat that. So, usig secat for the substitutio wo t work. However, the followig substitutio (ad differetial) will work. x= siq dx= cosq dq With this substitutio the square root is, 9- = - si = cos = cos = cos x q q q q We were able to drop the absolute value bars because we are doig a idefiite itegral ad so we ll assume that everythig is positive. The itegral is ow, Ù ı x dx= Ù 4 9 -x ı8si q = Ù dq 4 8ı si q csc 4 = q dq 8 4 ( cosq) cosq dq I the previous sectio we saw how to deal with itegrals i which the expoet o the secat was eve ad sice cosecats behave a awful lot like secats we should be able to do somethig similar with this. Here is the itegral. Ù ı x dx= csc csc 8 q q dq 9 -x = ( cot csc ) cot 8 q + q dq u = q =- u du 8 + Ê cot ˆ =- Á q + cot q + c 8Ë 4 Now we eed to go back to x s usig a right triagle. Here is the right triagle for this problem 007 Paul Dawkis 9

194 Math 40 ad trig fuctios for this problem. x siq = cotq = 9 - x x The itegral is the, Ù ı x Ê Ê ˆ 9-x ˆ 9-x dx=- Á + + c 9- x 8ÁÁ x x Ë Ë 4 ( x ) 9-9- x =- - + c x x 4 8 Here s the geeral form for this type of square root. a a -bx fi x= siq b There is oe fial case that we eed to look at. The ext itegral will also cotai somethig that we eed to make sure we ca deal with. Example 5 Evaluate the followig itegral. 6 5 x Ù ı 0 6 ( x + ) Solutio First, otice that there really is a square root i this problem eve though it is t explicitly writte out. To see the root let s rewrite thigs a little. dx Ê ˆ ( 6x + ) = Á( 6x + ) = ( 6x + ) Ë This square root is ot i the form we saw i the previous examples. Here we will use the substitutio for this root. x= taq dx= sec q dq Paul Dawkis 9

195 Math 40 With this substitutio the deomiator becomes, ( ) ( ) ( ) 6x + = ta q + = sec q = secq Now, because we have limits we ll eed to covert them to q so we ca determie how to drop the absolute value bars. x = 0 fi 0= taq fi q = 0 6 p x = fi = ta q fi q = I this rage of q secat is positive ad so we ca drop the absolute value bars. Here is the itegral, Ù ı p x 7776 ta q Ê ˆ dx= Ù Á sec q ı 0 sec q Ë6 0 ( 6x + ) p 5 4 ta q = Ù d q 46656ı secq 0 dq There are several ways to proceed from this poit. Normally with a odd expoet o the taget we would strip oe of them out ad covert to secats. However, that would require that we also have a secat i the umerator which we do t have. Therefore, it seems like the best way to do this oe would be to covert the itegrad to sies ad cosies. Ù ı 6 5 p x 4 ı 0 ( 6x + ) si q dx= Ù d q cos q p 4 0 ( - cos q ) = Ù ı cos q siqdq We ca ow use the substitutio u = cosq ad we might as well covert the limits as well. q = 0 u = cos0= The itegral is the, p p q = u = cos = Paul Dawkis 94

196 Math 40 Ù ı 6 5 x -4 - dx=- u u du ( 6x + ) The geeral form for this fial type of square root is Ê ˆ =- Á- + + u 46656Ë u u = a + fi = taq b a bx x We have a couple of fial examples to work i this sectio. Not all trig substitutios will just jump right out at us. Sometimes we eed to do a little work o the itegrad first to get it ito the correct form ad that is the poit of the remaiig examples. Example 6 Evaluate the followig itegral. x Ù dx ı x -4x-7 Solutio I this case the quatity uder the root does t obviously fit ito ay of the cases we looked at above ad i fact is t i the ay of the forms we saw i the previous examples. Note however that if we complete the square o the quadratic we ca make it look somewhat like the above itegrals. Remember that completig the square requires a coefficiet of oe i frot of the x. Oce we have that we take half the coefficiet of the x, square it, ad the add ad subtract it to the quatity. Here is the completig the square for this problem. Ê 7ˆ Ê 7ˆ Ê 9ˆ Áx -x- = Áx - x+ -- = Á( x-) - = ( x-) -9 Ë Ë Ë So, the root becomes, ( ) x -4x- 7 = x- - 9 This looks like a secat substitutio except we do t just have a x that is squared. That is okay, it will work the same way. x- = secq x= + secq dx= secq taq dq 007 Paul Dawkis 95

197 Math 40 Usig this substitutio the root reduces to, ( ) -4-7 = = 9sec - 9 = ta = ta = ta x x x q q q q Note we could drop the absolute value bars sice we are doig a idefiite itegral. Here is the itegral. x + secq Ê ˆ Ù dx= sec ta Ù q q dq x 4x 7 taq Á ı - - ı Ë = Ù secq + sec q dq ı = l secq + taq + taq + c Ad here is the right triagle for this problem. ( ) 4 7 sec x - ta x - q q x - = = The itegral is the, Ù ı 4 7 x ( x ) x - x -4x-7 x -4x-7 dx= l c - x- Example 7 Evaluate the followig itegral. 4x e + e x dx Solutio This does t look to be aythig like the other problems i this sectio. However it is. To see this we first eed to otice that, x ( ) x e = e With this we ca use the followig substitutio. 007 Paul Dawkis 96

198 Math 40 e x = taq e x dx = sec q d q Remember that to compute the differetial all we do is differetiate both sides ad the tack o dx or dq oto the appropriate side. With this substitutio the square root becomes, x ( ) x + e = + e = + ta = sec = sec = sec q q q q Agai, we ca drop the absolute value bars because we are doig a idefiite itegral. Here s the itegral. dx x ( e ) x x e ( e ) dx ta ( sec )( sec ) ( ) 4 u u du e + e dx= e e + e 4x x x x x = + = q q q dq = sec q - sec qsecq taq dq u = secq = - sec sec 5 5 = q - q + Here is the right triagle for this itegral. x x e + e taq = secq = = + e c x The itegral is the, 5 4x x x x e + e dx= ( + ) - ( + ) + c 5 e e So, as we ve see i the fial two examples i this sectio some itegrals that look othig like the first few examples ca i fact be tured ito a trig substitutio problem with a little work. Before leavig this sectio let s summarize all three cases i oe place. 007 Paul Dawkis 97

199 Math 40 a - fi = siq b a - fi = secq b a + fi = taq b a bx x bx a x a bx x 007 Paul Dawkis 98

200 Math 40 Partial Fractios I this sectio we are goig to take a look at itegrals of ratioal expressios of polyomials ad oce agai let s start this sectio out with a itegral that we ca already do so we ca cotrast it with the itegrals that we ll be doig i this sectio. x- dx= Ù du usig u x x 6 ad du Ù = - - = ( x-) dx ı x -x-6 ı u = l x x c So, if the umerator is the derivative of the deomiator (or a costat multiple of the derivative of the deomiator) doig this kid of itegral is fairly simple. However, ofte the umerator is t the derivative of the deomiator (or a costat multiple). For example, cosider the followig itegral. x+ Ù dx ı x -x-6 I this case the umerator is defiitely ot the derivative of the deomiator or is it a costat multiple of the derivative of the deomiator. Therefore, the simple substitutio that we used above wo t work. However, if we otice that the itegrad ca be broke up as follows, x+ 4 = - x -x-6 x- x+ the the itegral is actually quite simple. x+ 4 dx= Ù Ù - dx ı x -x-6 ı x- x+ = 4l x- - l x+ + c This process of takig a ratioal expressio ad decomposig it ito simpler ratioal expressios that we ca add or subtract to get the origial ratioal expressio is called partial fractio decompositio. May itegrals ivolvig ratioal expressios ca be doe if we first do partial fractios o the itegrad. So, let s do a quick review of partial fractios. We ll start with a ratioal expressio i the form, P( x) f ( x) = Q x where both P(x) ad Q(x) are polyomials ad the degree of P(x) is smaller tha the degree of Q(x). Recall that the degree of a polyomial is the largest expoet i the polyomial. Partial fractios ca oly be doe if the degree of the umerator is strictly less tha the degree of the deomiator. That is importat to remember. So, oce we ve determied that partial fractios ca be doe we factor the deomiator as completely as possible. The for each factor i the deomiator we ca use the followig table to determie the term(s) we pick up i the partial fractio decompositio. 007 Paul Dawkis 99 ( )

201 Math 40 Factor i deomiator ax+ b ( ax+ b) k ( ) Term i partial fractio decompositio A ax+ b A A A ax+ b ax+ b ax+ b k L, k =,,, k ( ) ( ) Ax+ B ax + bx+ c ax + bx+ c Ax k + B Ax + B Ax k + Bk ax + bx+ c + + L +, k =,,, K k ax + bx+ c ax + bx+ c ax + bx+ c ( ) ( ) Notice that the first ad third cases are really special cases of the secod ad fourth cases respectively. There are several methods for determiig the coefficiets for each term ad we will go over each of those i the followig examples. Let s start the examples by doig the itegral above. K Example Evaluate the followig itegral. x + Ù dx ı x -x-6 Solutio The first step is to factor the deomiator as much as possible ad get the form of the partial fractio decompositio. Doig this gives, x+ A B = + x- x+ x- x+ ( )( ) The ext step is to actually add the right side back up. x+ A x+ + B x- = x- x+ x- x+ ( )( ) ( ) ( ) ( )( ) Now, we eed to choose A ad B so that the umerators of these two are equal for every x. To do this we ll eed to set the umerators equal. ( ) ( ) x+ = A x+ + B x- 007 Paul Dawkis 00

202 Math 40 Note that i most problems we will go straight from the geeral form of the decompositio to this step ad ot bother with actually addig the terms back up. The oly poit to addig the terms is to get the umerator ad we ca get that without actually writig dow the results of the additio. At this poit we have oe of two ways to proceed. Oe way will always work, but is ofte more work. The other, while it wo t always work, is ofte quicker whe it does work. I this case both will work ad so we ll use the quicker way for this example. We ll take a look at the other method i a later example. What we re goig to do here is to otice that the umerators must be equal for ay x that we would choose to use. I particular the umerators must be equal for x =- ad x =. So, let s plug these i ad see what we get. ( ) ( ) ( ) ( ) x=- 5= A 0 + B -5 fi B=- x= 0= A 5 + B 0 fi A= 4 So, by carefully pickig the x s we got the ukow costats to quickly drop out. Note that these are the values we claimed they would be above. At this poit there really is t a whole lot to do other tha the itegral. x+ 4 dx= Ù Ù - dx ı x -x-6 ı x- x+ 4 = dx- Ù Ù dx ı x- ı x+ = 4l x- - l x+ + c Recall that to do this itegral we first split it up ito two itegrals ad the used the substitutios, u = x- v = x+ o the itegrals to get the fial aswer. Before movig oto the ext example a couple of quick otes are i order here. First, may of the itegrals i partial fractios problems come dow to the type of itegral see above. Make sure that you ca do those itegrals. There is also aother itegral that ofte shows up i these kids of problems so we may as well give the formula for it here sice we are already o the subject. Ù ı x = ta + c + Ë dx - Ê Á ˆ x a a a It will be a example or two before we use this so do t forget about it. Now, let s work some more examples. 007 Paul Dawkis 0

203 Math 40 Example Evaluate the followig itegral. x + 4 Ù dx ıx + 4x -4x Solutio We wo t be puttig as much detail ito this solutio as we did i the previous example. The first thig is to factor the deomiator ad get the form of the partial fractio decompositio. x 4 A B C + = + + x x+ x- x x+ x- ( )( ) The ext step is to set umerators equal. If you eed to actually add the right side together to get the umerator for that side the you should do so, however, it will defiitely make the problem quicker if you ca do the additio i your head to get, ( )( ) ( ) ( ) x A x x Bx x Cx x + 4= As with the previous example it looks like we ca just pick a few values of x ad fid the costats so let s do that. x= 0 4= A - fi A=- ( )( ) x=- 8= B( -)(-8) fi B= 40 ʈÊ8ˆ 40 5 x= = CÁ Á fi C = = 9 Ë Ë 6 Note that ulike the first example most of the coefficiets here are fractios. That is ot uusual so do t get excited about it whe it happes. Now, let s do the itegral. 5 x + 4 dx= Ù Ù dx ı x + 4x - 4x ı x x+ x- 5 =- l x + l x+ + l x- + c 6 Agai, as oted above, itegrals that geerate atural logarithms are very commo i these problems so make sure you ca do them. Example Evaluate the followig itegral. x - 9x+ 5 Ù ı x- 4 x + Solutio ( ) ( ) dx 007 Paul Dawkis 0

204 Math 40 This time the deomiator is already factored so let s just jump right to the partial fractio decompositio. Settig umerators gives, 9 5 x - x+ A B Cx+ D = x + ( x- 4) ( x + ) x ( x-4) ( )( ) ( ) ( )( ) x x A x x B x Cx D x = I this case we are t goig to be able to just pick values of x that will give us all the costats. Therefore, we will eed to work this the secod (ad ofte loger) way. The first step is to multiply out the right side ad collect all the like terms together. Doig this gives, ( ) ( ) ( ) x - 9x+ 5= A+ C x + - 4A+ B- 8C+ D x + A+ 6C-8D x- A+ B+ 6D Now we eed to choose A, B, C, ad D so that these two are equal. I other words we will eed to set the coefficiets of like powers of x equal. This will give a system of equatios that ca be solved. : + = 0 Ô : = Ô : =-9Ô : = 5 Ô x A C x A B C D x A C D 0 x A B D fi A=, B=- 5, C =-, D= Note that we used x 0 to represet the costats. Also ote that these systems ca ofte be quite large ad have a fair amout of work ivolved i solvig them. The best way to deal with these is to use some form of computer aided solvig techiques. Now, let s take a look at the itegral. Ù ı dx= dx Ù - + ı - 4 x + x x x ( x- 4) ( x + ) x ( x-4) 5 x = Ù ı x- x + x + ( x-4) 4 5 Ê x ˆ = l l + + ta + x-4 Á Ë - x x c I order to take care of the third term we eeded to split it up ito two separate terms. Oce we ve doe this we ca do all the itegrals i the problem. The first two use the substitutio u = x- 4, the third uses the substitutio v= x + ad the fourth term uses the formula give above for iverse tagets. Example 4 Evaluate the followig itegral. 007 Paul Dawkis 0 dx

205 Math 40 Ù ı dx x x x ( x- )( x + 4) Solutio Let s first get the geeral form of the partial fractio decompositio. x + 0x + x+ 6 A Bx+ C Dx+ E = + + ( x- )( x + 4) x- x + 4 ( x + 4) Now, set umerators equal, expad the right side ad collect like terms. ( ) ( )( )( ) ( )( ) x x x A x Bx C x x Dx E x = ( ) ( ) ( 8 4 ) ( ) Settig coefficiet equal gives the followig system. 4 x A B x C B 4 = A+ B x + C- B x + A+ B- C+ D x + - 4B+ 4C- D+ E x+ 6A-4C-E : + = 0 Ô : - = ÔÔ : = 0 fi =, =-, =-, =, = 0 : = Ô Ô : = 6Ô x A B C D A B C D E x B C D E 0 x A C E Do t get excited if some of the coefficiets ed up beig zero. It happes o occasio. Here s the itegral. Ù ı dx= Ù + + x x x x x ( x- )( x + 4) ı x- x + 4 ( x + 4) x x = Ù ı x- x + x + x + ( ) Ê xˆ = l -- l ta Á - + Ë x + 4 dx - x x c To this poit we ve oly looked at ratioal expressios where the degree of the umerator was strictly less that the degree of the deomiator. Of course ot all ratioal expressios will fit ito this form ad so we eed to take a look at a couple of examples where this is t the case. dx Example 5 Evaluate the followig itegral. 007 Paul Dawkis 04

206 Math 40 4 x - 5x + 6x -8 Ù dx ı x -x Solutio So, i this case the degree of the umerator is 4 ad the degree of the deomiator is. Therefore, partial fractios ca t be doe o this ratioal expressio. To fix this up we ll eed to do log divisio o this to get it ito a form that we ca deal with. Here is the work for that. x- So, from the log divisio we see that, ad the itegral becomes, 4 x -x x - 5x + 6x -8 4 ( x x ) x 6x 8 ( x 6x ) x - 5x + 6x -8 8 = x-- x -x x -x 4 x - 5x + 6x -8 8 dx= Ù Ù x- - dx ı x -x ı x -x 8 = x- dx- Ù dx ı x -x The first itegral we ca do easily eough ad the secod itegral is ow i a form that allows us to do partial fractios. So, let s get the geeral form of the partial fractios for the secod itegrad. 8 = A + B + C x x- x x x- Settig umerators equal gives us, ( ) ( ) ( ) 8= Ax x- + B x- + Cx Now, there is a variatio of the method we used i the first couple of examples that will work here. There are a couple of values of x that will allow us to quickly get two of the three costats, but there is o value of x that will just had us the third. What we ll do i this example is pick x s to get the two costats that we ca easily get ad the we ll just pick aother value of x that will be easy to work with (i.e. it wo t give large/messy umbers aywhere) ad the we ll use the fact that we also kow the other two costats to fid 007 Paul Dawkis 05

207 Math 40 the third. ( ) ( ) ( ) ( ) x= 0 8= B - fi B=-6 x= 8= C 9 fi C = x= 8= A - + B - + C =- A+ 4 fi A=- The itegral is the, 4 x - 5x + 6x -8 6 dx= x-dx- Ù Ù dx ı x -x ı x x x- 6 = - + l - -l - + x x x x x c I the previous example there were actually two differet ways of dealig with the x i the deomiator. Oe is to treat is as a quadratic which would give the followig term i the decompositio Ax+ B x ad the other is to treat it as a liear term i the followig way, x ( x 0) = - which gives the followig two terms i the decompositio, A B + x x We used the secod way of thikig about it i our example. Notice however that the two will give idetical partial fractio decompositios. So, why talk about this? Simple. This will work for x, but what about x or x 4? I these cases we really will eed to use the secod way of thikig about these kids of terms. x Let s take a look at oe more example. A B C A B C D fi + + x fi x x x x x x x 4 4 Example 6 Evaluate the followig itegral. x Ù dx ı x - Solutio I this case the umerator ad deomiator have the same degree. As with the last example we ll eed to do log divisio to get this ito the correct form. I ll leave the details of that to you to check. x Ù dx= Ù + dx= dx+ dx Ù ı x - ı x - ı x Paul Dawkis 06

208 Math 40 So, we ll eed to partial fractio the secod itegral. Here s the decompositio. A B = + x- x+ x- x+ Settig umerator equal gives, ( )( ) ( ) B( x ) = A x+ + - Pickig value of x gives us the followig coefficiets. x=- = B( -) fi B=- x= = A( ) fi A= The itegral is the, x dx= dx+ Ù Ù - dx ı x - ı x- x+ = x+ l x-- l x+ + c 007 Paul Dawkis 07

209 Math 40 Itegrals Ivolvig Roots I this sectio we re goig to look at a itegratio techique that ca be useful for some itegrals with roots i them. We ve already see some itegrals with roots i them. Some ca be doe quickly with a simple Calculus I substitutio ad some ca be doe with trig substitutios. However, ot all itegrals with roots will allow us to use oe of these methods. Let s look at a couple of examples to see aother techique that ca be used o occasio to help with these itegrals. Example Evaluate the followig itegral. x + Ù dx ı x - Solutio Sometimes whe faced with a itegral that cotais a root we ca use the followig substitutio to simplify the itegral ito a form that ca be easily worked with. u = x- So, istead of lettig u be the stuff uder the radical as we ofte did i Calculus I we let u be the whole radical. Now, there will be a little more work here sice we will also eed to kow what x is so we ca substitute i for that i the umerator ad so we ca compute the differetial, dx. This is easy eough to get however. Just solve the substitutio for x as follows, Usig this substitutio the itegral is ow, Ù ı x= u + dx= u du ( u ) = + 5 u udu u udu 5 5 = u + u + c = ( x- ) + ( x- ) + c 5 So, sometimes, whe a itegral cotais the root g( x ) the substitutio, u = ( ) g x ca be used to simplify the itegral ito a form that we ca deal with. Let s take a look at aother example real quick. Example Evaluate the followig itegral. 007 Paul Dawkis 08

210 Math 40 Ù ı x- x+ 0 dx Solutio We ll do the same thig we did i the previous example. Here s the substitutio ad the extra work we ll eed to do to get x i terms of u. 0 0 u = x+ x= u - dx= udu With this substitutio the itegral is, 4u Ù dx= Ù ( u) du = Ù du ı x- x+ 0 ıu -0-u ı u -u-0 This itegral ca ow be doe with partial fractios. 4 u = A + B u- 5 u+ u- 5 u+ ( )( ) Settig umerators equal gives, ( ) ( ) 4u = Au+ + B u- 5 Pickig value of u gives the coefficiets. 8 u =- - 8= B( - 7) B= 7 0 u = 5 0= A( 7) A= 7 The itegral is the, dx= Ù Ù + du ı x- x+ 0 ı u- 5 u+ 0 8 = l u- 5 + l u+ + c = l x l x c 7 7 So, we ve see a ice method to elimiate roots from the itegral ad put ito a form that we ca deal with. Note however, that this wo t always work ad sometimes the ew itegral will be just as difficult to do. 007 Paul Dawkis 09

211 Math 40 Itegrals Ivolvig Quadratics To this poit we ve see quite a few itegrals that ivolve quadratics. A couple of examples are, x -Ê xˆ Ù dx= l x ± a + c dx ta Ù = Á ı x ± a ı x + a a Ëa We also saw that itegrals ivolvig with a trig substitutio. bx - a, a - bx ad a + bx could be doe Notice however that all of these itegrals were missig a x term. They all cosist of a quadratic term ad a costat. Some itegrals ivolvig geeral quadratics are easy eough to do. For istace, the followig itegral ca be doe with a quick substitutio. x + dx= Ù ( 4 4 Ù du u = x + x- du = ( x+ ) dx) ı 4x + x- 4ıu l4 = x + x- + c 4 Some itegrals with quadratics ca be doe with partial fractios. For istace, 0x- 6 4 dx= Ù 4l 5 l Ù - dx= x+ - x+ + c ıx + 6x+ 5 ı x+ 5 x+ Ufortuately, these methods wo t work o a lot of itegrals. A simple substitutio will oly work if the umerator is a costat multiple of the derivative of the deomiator ad partial fractios will oly work if the deomiator ca be factored. This sectio is how to deal with itegrals ivolvig quadratics whe the techiques that we ve looked at to this poit simply wo t work. Back i the Trig Substitutio sectio we saw how to deal with square roots that had a geeral quadratic i them. Let s take a quick look at aother oe like that sice the idea ivolved i doig that kid of itegral is exactly what we are goig to eed for the other itegrals i this sectio. Example Evaluate the followig itegral. x + 4x+ 5dx Solutio Recall from the Trig Substitutio sectio that i order to do a trig substitutio here we first eeded to complete the square o the quadratic. This gives, x x x x x ( ) = = Paul Dawkis 0

212 Math 40 After completig the square the itegral becomes, ( ) x + 4x+ 5dx= x+ + dx Upo doig this we ca idetify the trig substitutio that we eed. Here it is, x+ = taq x= taq - dx= sec q dq ( ) x+ + = ta q + = sec q = secq = secq Recall that sice we are doig a idefiite itegral we ca drop the absolute value bars. Usig this substitutio the itegral becomes, x + 4x+ 5dx= sec q dq = ( sec q ta q + l sec q + ta q ) + c We ca fiish the itegral out with the followig right triagle. 4 5 ta x + sec x + x + q = q = = x + 4x+ 5 (( ) ) x + 4x+ 5dx= x+ x + 4x+ 5+ l x+ + x + 4x+ 5 + c So, by completig the square we were able to take a itegral that had a geeral quadratic i it ad covert it ito a form that allowed use a kow itegratio techique. Let s do a quick review of completig the square before proceedig. Here is the geeral completig the square formula that we ll use. Êbˆ Êbˆ Ê bˆ b x + bx+ c= x + bx+ - + c= x+ + c- Á Á Á Ë Ë Ë 4 This will always take a geeral quadratic ad write it i terms of a squared term ad a costat term. Recall as well that i order to do this we must have a coefficiet of oe i frot of the x. If ot we ll eed to factor out the coefficiet before completig the square. I other words, 007 Paul Dawkis

213 Math 40 Ê ˆ Á b c + + = Á + + Á a a Á 44 complete the Ë square o this! ax bx c a x x Now, let s see how completig the square ca be used to do itegrals that we are t able to do at this poit. Example Evaluate the followig itegral. Ù ı x - x+ dx Solutio Okay, this does t factor so partial fractios just wo t work o this. Likewise, sice the umerator is just we ca t use the substitutio u = x - x+ 8. So, let s see what happes if we complete the square o the deomiator. x - x+ = Ê ˆ Áx - x+ Ë = Ê 9 9 Áx - x+ - + Ë 6 6 ÊÊ ˆ 7 ˆ x- + ÁÁ Ë 4 6 Ë ˆ With this the itegral is, Ù dx= ı x - x+ Ù x- + ( ) 7 ı 4 6 dx Now this may ot seem like all that great of a chage. However, otice that we ca ow use the followig substitutio. u = x- du = dx 4 ad the itegral is ow, Ù dx= Ù du 7 ı x - x+ ı u + We ca ow see that this is a iverse taget! So, usig the formula from above we get, Ù ı Ê 4 ˆ Ê 4u ˆ - + Ë Ë - dx= ta x x Á Á Paul Dawkis 6 Ê4x- ˆ Á 7 Ë 7 - = ta + c c

214 Math 40 Example Evaluate the followig itegral. x - Ù dx ı x + 0x+ 8 Solutio This example is a little differet from the previous oe. I this case we do have a x i the umerator however the umerator still is t a multiple of the derivative of the deomiator ad so a simple Calculus I substitutio wo t work. So, let s agai complete the square o the deomiator ad see what we get, x x x x x ( ) = = Upo completig the square the itegral becomes, x- x- Ù dx= dx ı x + 0x+ 8 Ù ı x+ 5 + ( ) At this poit we ca use the same type of substitutio that we did i the previous example. The oly real differece is that we ll eed to make sure that we plug the substitutio back ito the umerator as well. u = x+ 5 x= u- 5 dx= du ( u ) x Ù dx= Ù du ı x + 0x+ 8 ı u + u 6 = Ù - du ı u + u + 6 Ê u ˆ l u ta Á Ë - = So, i geeral whe dealig with a itegral i the form, 6 -Ê x+ 5ˆ = l ( x+ 5) + - ta c Á + Ë c + B Ù Ax dx ı ax + bx+ c () Here we are goig to assume that the deomiator does t factor ad the umerator is t a costat multiple of the derivative of the deomiator. I these cases we complete the square o the deomiator ad the do a substitutio that will yield a iverse taget ad/or a logarithm depedig o the exact form of the umerator. 007 Paul Dawkis

215 Math 40 Let s ow take a look at a couple of itegrals that are i the same geeral form as () except the deomiator will also be raised to a power. I other words, let s look at itegrals i the form, Ù ı Ax+ B ( ax + bx+ c) Example 4 Evaluate the followig itegral. x Ù ı x - 6x+ ( ) Solutio For the most part this itegral will work the same as the previous two with oe exceptio that will occur dow the road. So, let s start by completig the square o the quadratic i the deomiator. dx dx x x x x x ( ) = = - + () The itegral is the, Ù ı x x dx= Ù Î ( x x Ù ) ı È( x ) dx Now, we will use the same substitutio that we ve used to this poit i the previous two examples. u = x- x = u+ dx = du Ù ı x dx= Ù u+ ( x - 6x+ ) ı ( u + ) du u = Ù du+ Ù ı ( u + ) ı ( u + ) Now, here is where the differeces start croppig up. The first itegral ca be doe with the substitutio v= u + ad is t too difficult. The secod itegral however, ca t be doe with the substitutio used o the first itegral ad it is t a iverse taget. It turs out that a trig substitutio will work icely o the secod itegral ad it will be the same as we did whe we had square roots i the problem. u = du = d With these two substitutios the itegrals become, taq sec q q du 007 Paul Dawkis 4

216 Math 40 Ù ı dx dv dq x ( ) ( ) ( = sec ) Ù + Ù q x - 6x+ ı v ı ta q + sec q =- + Ù ı ( q + ) 4 v 8ta Ù ( u + ) ı ( sec q) (( x- ) + ) (( x- ) + ) =- + Ù dq ı sec q =- + cos 4 8 dq sec q = dq q dq Okay, at this poit we ve got two optios for the remaiig itegral. We ca either use the ideas we leared i the sectio about itegrals ivolvig trig itegrals or we could use the followig formula. m cos q m m m d q si cos cos d m q q - = m q q Let s use this formula to do the itegral. 4 cos q dq = siqcos q + cos 4 4 q dq Ê 0 ˆ 0 = siqcos q + Á siqcosq + cos d cos! 4 4 q q q = Ë = siqcos q + siqcosq + q Next, let s use the followig right triagle to get this back to x s. u x- x- taq = = siq = cosq = ( x ) ( x ) Paul Dawkis 5

217 Math 40 The cosie itegral is the, ( ) (( x- ) + ) ( ) ( x ) 4 x- x- -Ê x-ˆ cos q dq = + + ta Á - + Ë All told the the origial itegral is, Ù ı ( x x ) 4 ( x ) x- x- Ê x-ˆ Ë - = + + ta Á (( x- ) + ) ( x ) x dx = ( ) Ê ˆ Á x- x- -Ê x-ˆ ta (( x ) 8 ) ( x ) 8 Á - + Á - + Ë Ë x- 9 x- 9 Ê x-ˆ 8 64 Á - + Ë - = + + ta + (( x- ) + ) ( x ) It s a log ad messy aswer, but there it is. Example 5 Evaluate the followig itegral. x- Ù ı 4-x-x ( ) Solutio As with the other problems we ll first complete the square o the deomiator. 007 Paul Dawkis 6 dx ( ) ( ) ( ) ( ) ( ) 4-x- x =- x + x- 4 =- x + x =- x+ - 5 = 5- x+ The itegral is, x- x- Ù dx= dx ( 4-x-x Ù ı ) ı È5- ( x+ ) Î Now, let s do the substitutio. u = x+ x = u- dx = du ad the itegral is ow, Ù ı x- u-4 dx= Ù ( 4-x-x ) ı( 5-u ) du u = Ù du-ù ı 4 ( 5-u ) ı ( 5-u ) du c

218 Math 40 I the first itegral we ll use the substitutio v= 5 - u ad i the secod itegral we ll use the followig trig substitutio u = 5siq du = 5cosq dq Usig these substitutios the itegral becomes, Ù ı dx dv dq x - 4 =- ( ) ( ) ( 5cos ) Ù - Ù q 4-x-x ı v ı 5-5si q 4 5 cosq = - Ù v 5 ı si ( - q ) 4 5 cosq = - Ù dq 4 v 5 ı cos q 4 5 = - sec v 5 5 = - ( sec q ta q + l sec q + ta q ) + c v 5 We ll eed the followig right triagle to fiish this itegral out. q dq u x+ 5 x+ siq = = secq = taq = dq ( x ) ( x ) So, goig back to x s the itegral becomes, Ù ı Ê ˆ x- 5 5( x+ ) 5 x+ dx= - Á + l + + c x-x - u Á5- ( x + ) 5- ( x+ ) 5- ( x+ ) Ë ( ) 4x- 5 x+ + 5 = + l + c ( x ) ( x ) 007 Paul Dawkis 7

219 Math 40 Ofte the followig formula is eeded whe usig the trig substitutio that we used i the previous example. m sec q m m m d q ta sec sec d m q q - = m- q q Note that we ll oly eed the two trig substitutios that we used here. The third trig substitutio that we used will ot be eeded here. 007 Paul Dawkis 8

220 Math 40 Usig Itegral Tables Note : Of all the otes that I ve writte up for dowload, this is the oe sectio that is tied to the book that we are curretly usig here at Lamar Uiversity. I this sectio we discuss usig tables of itegrals to help us with some itegrals. However, I have t had the time to costruct a table of my ow ad so I will be usig the tables give i Stewart s Calculus, Early Trascedetals (6 th editio). As soo as I get aroud to writig my ow table I ll post it olie ad make ay appropriate chages to this sectio. So, with that out of the way let s get o with this sectio. This sectio is etitled Usig Itegral Tables ad we will be usig itegral tables. However, at some level, this is t really the poit of this sectio. To a certai extet the real subject of this sectio is how to take advatage of kow itegrals to do itegrals that may ot look like aythig the oes that we do kow how to do or are give i a table of itegrals. For the most part we ll be doig this by usig substitutio to put itegrals ito a form that we ca deal with. However, ot all of the itegrals will require a substitutio. For some itegrals all that we eed to do is a little rewritig of the itegrad to get ito a form that we ca deal with. We ve already related a ew itegral to oe we could deal with least oce. I the last example i the Trig Substitutio sectio we looked at the followig itegral. 4x e + e x dx At first glace this looks othig like a trig substitutio problem. However, with the substitutio x u =e we could tur the itegral ito, u + u du which defiitely is a trig substitutio problem ( u = taq ). We actually did this process i a x sigle step by usig e = taq, but the poit is that with a substitutio we were able to covert a itegral ito a form that we could deal with. So, let s work a couple examples usig substitutios ad tables. Example Evaluate the followig itegral. Ù ı 7+ 9x dx x Solutio So, the first thig we should do is go to the tables ad see if there is aythig i the tables that is close to this. I the tables i Stewart we fid the followig itegral, 007 Paul Dawkis 9

221 Math 40 Ù ı ( ) a + u a + u du =- + l u+ a + u + c u u This is early what we ve got i our itegral. The oly real differece is that we ve got a coefficiet i frot of the x ad the formula does t. This is easily eough dealt with. All we eed to do is the followig maipulatio o the itegrad. ( ) x x 9 + x 9 + x Ù dx = dx dx dx Ù = = x x Ù x Ù ı ı ı ı x So, we ca ow use the formula with Ù ı 7 a = x Ê 9 + x Ê 7 ˆˆ l dx = - + x + + x + c x Á x Á 9 Ë Ë Example Evaluate the followig itegral. cos x Ù dx ısix 9six-4 Solutio Goig through our tables we are t goig to fid aythig that looks like this i them. However, otice that with the substitutio u = si x we ca rewrite the itegral as, ad this is i the tables. cosx Ù dx= Ù du ısix 9six-4 ı u 9u-4 a+ bu - a Ù du = l + c if a> 0 ıu a+ bu a a+ bu + a Ê ˆ = + < -a Á -a Ë ta - a+ bu c if a 0 Notice that this is a formula that will deped upo the value of a. This will happe o occasio. I our case we have a =- 4 ad b= 9 so we ll use the secod formula. Ù ı cosx Ê 9u-4 ˆ dx six 9six-4 -(-4) Á -(-4) Ë - = ta + 9six-4 ˆ - = ta + This fial example uses a type of formula kow as a reductio formula. 007 Paul Dawkis 0 Ê Á Ë c c

222 Math 40 Example Evaluate the followig itegral. 4 Ê x ˆ Ù cot Á dx ı Ë Solutio x We ll first eed to use the substitutio u = sice oe of the formulas i our tables have that i them. Doig this gives, 4Ê x ˆ 4 Ùcot Á dx = cot udu Ë ı To help us with this itegral we ll use the followig formula cot udu = cot u- cot udu - Formulas like this are called reductio formulas. Reductio formulas geerally do t explicitly give the itegral. Istead they reduce the itegral to a easier oe. I fact they ofte reduce the itegral to a differet versio of itself! For our itegral we ll use = 4. 4Ê cot x ˆ Ê ˆ Ù Á dx = Á- cot u - cot udu Ë Ë ı At this stage we ca either reuse the reductio formula with = or use the formula cot udu =-cot u- u+ c We ll reuse the reductio formula with = so we ca address somethig that happes o occasio. 007 Paul Dawkis

223 Math 40 Ù ı Ê xˆ Ê Ê ˆˆ Á dx u u udu u Á Á Ë Ë Ë cot = - cot - - cot - cot cot =! cot cot =- u+ u+ du cot cot =- u+ u+ u+ c cot Ê xˆ Ê xˆ =- Á + cot Á + x+ c Ë Ë Do t forget that 0 a =. Ofte people forget that ad the get stuck o the fial itegral! There really was t a lot to this sectio. Just do t forget that sometimes a simple substitutio or rewrite of a itegral ca take it from udoable to doable. 007 Paul Dawkis

224 Math 40 Itegratio Strategy We ve ow see a fair umber of differet itegratio techiques ad so we should probably pause at this poit ad talk a little bit about a strategy to use for determiig the correct techique to use whe faced with a itegral. There are a couple of poits that eed to be made about this strategy. First, it is t a hard ad fast set of rules for determiig the method that should be used. It is really othig more tha a geeral set of guidelies that will help us to idetify techiques that may work. Some itegrals ca be doe i more tha oe way ad so depedig o the path you take through the strategy you may ed up with a differet techique tha somebody else who also wet through this strategy. Secod, while the strategy is preseted as a way to idetify the techique that could be used o a itegral also keep i mid that, for may itegrals, it ca also automatically exclude certai techiques as well. Whe goig through the strategy keep two lists i mid. The first list is itegratio techiques that simply wo t work ad the secod list is techiques that look like they might work. After goig through the strategy ad the secod list has oly oe etry the that is the techique to use. If, o the other had, there are more tha oe possible techique to use we will the have to decide o which is liable to be the best for us to use. Ufortuately there is o way to teach which techique is the best as that usually depeds upo the perso ad which techique they fid to be the easiest. Third, do t forget that may itegrals ca be evaluated i multiple ways ad so more tha oe techique may be used o it. This has already bee metioed i each of the previous poits, but is importat eough to warrat a separate metio. Sometimes oe techique will be sigificatly easier tha the others ad so do t just stop at the first techique that appears to work. Always idetify all possible techiques ad the go back ad determie which you feel will be the easiest for you to use. Next, it s etirely possible that you will eed to use more tha oe method to completely do a itegral. For istace a substitutio may lead to usig itegratio by parts or partial fractios itegral. Fially, i my class I will accept ay valid itegratio techique as a solutio. As already oted there is ofte more tha oe way to do a itegral ad just because I fid oe techique to be the easiest does t mea that you will as well. So, i my class, there is o oe right way of doig a itegral. You may use ay itegratio techique that I ve taught you i this class or you leared i Calculus I to evaluate itegrals i this class. I other words, always take the approach that you fid to be the easiest. Note that this fial poit is more geared towards my class ad it s completely possible that your istructor may ot agree with this ad so be careful i applyig this poit if you are t i my class. 007 Paul Dawkis

225 Math 40 Okay, let s get o with the strategy.. Simplify the itegrad, if possible. This step is very importat i the itegratio process. May itegrals ca be take from impossible or very difficult to very easy with a little simplificatio or maipulatio. Do t forget basic trig ad algebraic idetities as these ca ofte be used to simplify the itegral. We used this idea whe we were lookig at itegrals ivolvig trig fuctios. For example cosider the followig itegral. cos xdx This itegral ca t be doe as is however, simply by recallig the idetity, x= ( + ( x) ) cos cos the itegral becomes very easy to do. Note that this example also shows that simplificatio does ot ecessarily mea that we ll write the itegrad i a simpler form. It oly meas that we ll write the itegrad ito a form that we ca deal with ad this is ofte loger ad/or messier tha the origial itegral.. See if a simple substitutio will work. Look to see if a simple substitutio ca be used istead of the ofte more complicated methods from Calculus II. For example cosider both if the followig itegrals. x Ù dx x x -dx ı x - The first itegral ca be doe with partial fractios ad the secod could be doe with a trig substitutio. However, both could also be evaluated usig the substitutio u = x - ad the work ivolved i the substitutio would be sigificatly less tha the work ivolved i either partial fractios or trig substitutio. So, always look for quick, simple substitutios before movig o to the more complicated Calculus II techiques.. Idetify the type of itegral. Note that ay itegral may fall ito more tha oe of these types. Because of this fact it s usually best to go all the way through the list ad idetify 007 Paul Dawkis 4

226 Math 40 all possible types sice oe may be easier tha the other ad it s etirely possible that the easier type is listed lower i the list. a. Is the itegrad a ratioal expressio (i.e is the itegrad a polyomial divided by a polyomial)? If so, the partial fractios may work o the itegral. b. Is the itegrad a polyomial times a trig fuctio, expoetial, or logarithm? If so, the itegratio by parts may work. c. Is the itegrad a product of sies ad cosies, secat ad tagets, or cosecats ad cotagets? If so, the the topics from the secod sectio may work. Likewise, do t forget that some quotiets ivolvig these fuctios ca also be doe usig these techiques. d. Does the itegrad ivolve bx + a, bx - a, or a - bx? If so, the a trig substitutio might work icely. e. Does the itegrad have roots other tha those listed above i it? If so, the the substitutio u g( x) = might work. f. Does the itegrad have a quadratic i it? If so, the completig the square o the quadratic might put it ito a form that we ca deal with. 4. Ca we relate the itegral to a itegral we already kow how to do? I other words, ca we use a substitutio or maipulatio to write the itegrad ito a form that does fit ito the forms we ve looked at previously i this chapter. A typical example here is the followig itegral. cosx + si xdx This itegral does t obviously fit ito ay of the forms we looked at i this chapter. However, with the substitutio u = si x we ca reduce the itegral to the form, which is a trig substitutio problem. + u du 5. Do we eed to use multiple techiques? I this step we eed to ask ourselves if it is possible that we ll eed to use multiple techiques. The example i the previous part is a good example. Usig a substitutio did t allow us to actually do the itegral. All it did was put the itegral ad put it ito a form that we could use a differet techique o. Do t ever get locked ito the idea that a itegral will oly require oe step to completely evaluate it. May will require more tha oe step. 6. Try agai. If everythig that you ve tried to this poit does t work the go back through the process ad try agai. This time try a techique that that you did t use the first time aroud. As oted above this strategy is ot a hard ad fast set of rules. It is oly iteded to guide you through the process of best determiig how to do ay give itegral. Note as well that the oly place Calculus II actually arises is i the third step. Steps, ad 4 ivolve othig more tha 007 Paul Dawkis 5

227 Math 40 maipulatio of the itegrad either through direct maipulatio of the itegrad or by usig a substitutio. The last two steps are simply ideas to thik about i goig through this strategy. May studets go through this process ad cocetrate almost exclusively o Step (after all this is Calculus II, so it s easy to see why they might do that.) to the exclusio of the other steps. Oe very large cosequece of that exclusio is that ofte a simple maipulatio or substitutio is overlooked that could make the itegral very easy to do. Before movig o to the ext sectio we should work a couple of quick problems illustratig a couple of ot so obvious simplificatios/maipulatios ad a ot so obvious substitutio. Example Evaluate the followig itegral. ta x Ù 4 dx ı sec x Solutio This itegral almost falls ito the form give i c. It is a quotiet of taget ad secat ad we kow that sometimes we ca use the same methods for products of tagets ad secats o quotiets. The process from that sectio tells us that if we have eve powers of secat to strip two of them off ad covert the rest to tagets. That wo t work here. We ca split two secats off, but they would be i the deomiator ad they wo t do us ay good there. Remember that the poit of splittig them off is so they would be there for the substitutio u = ta x. That requires them to be i the umerator. So, that wo t work ad so we ll have to fid aother solutio method. There are i fact two solutio methods to this itegral depedig o how you wat to go about it. We ll take a look at both. Solutio I this solutio method we could just covert everythig to sies ad cosies ad see if that gives us a itegral we ca deal with. tax si x 4 dx= Ù 4 Ù cos xdx ısec x ı cos x = sixcos xdx u = cos x =- u du cos 4 =- x+ c 4 Note that just covertig to sies ad cosies wo t always work ad if it does it wo t always work this icely. Ofte there will be a lot more work that would eed to be doe to complete the itegral. 007 Paul Dawkis 6

228 Math 40 Solutio This solutio method goes back to dealig with secats ad tagets. Let s otice that if we had a secat i the umerator we could just use u = sec x as a substitutio ad it would be a fairly quick ad simple substitutio to use. We do t have a secat i the umerator. However we could very easily get a secat i the umerator simply by multiplyig the umerator ad deomiator by secat. tax taxsec x dx= Ù sec 4 Ù dx u = x 5 ısec x ı sec x = Ù du 5 ı u =- + c 4 4sec x cos 4 =- x+ c 4 I the previous example we saw two simplificatios that allowed us to do the itegral. The first was usig idetities to rewrite the itegral ito terms we could deal with ad the secod ivolved multiplyig the umerator ad the deomiator by somethig to agai put the itegral ito terms we could deal with. Usig idetities to rewrite a itegral is a importat simplificatio ad we should ot forget about it. Itegrals ca ofte be greatly simplified or at least put ito a form that ca be dealt with by usig a idetity. The secod simplificatio is ot used as ofte, but does show up o occasio so agai, it s best to ot forget about it. I fact, let s take aother look at a example i which multiplyig the umerator ad deomiator by somethig will allow us to do a itegral. Example Evaluate the followig itegral. Ù ı + si x dx Solutio This is a itegral i which if we just cocetrate o the third step we wo t get aywhere. This itegral does t appear to be ay of the kids of itegrals that we worked i this chapter. We ca do the itegral however, if we do the followig, - si x dx= Ù Ù dx ı+ six ı+ six-si x -si x = Ù dx ı-si x 007 Paul Dawkis 7

229 Math 40 This does ot appear to have doe aythig for us. However, if we ow remember the first simplificatio we looked at above we will otice that we ca use a idetity to rewrite the deomiator. Oce we do that we ca further reduce the itegral ito somethig we ca deal with. - si x dx= Ù Ù dx ı+ six ı cos x six = Ù - dx ı cos x cosx cos x = sec x-taxsec xdx = tax- sec x+ c So, we ve see oce agai that multiplyig the umerator ad deomiator by somethig ca put the itegral ito a form that we ca itegrate. Notice as well that this example also showed that simplificatios do ot ecessarily put a itegral ito a simpler form. They oly put the itegral ito a form that is easier to itegrate. Let s ow take a quick look at a example of a substitutio that is ot so obvious. Example Evaluate the followig itegral. cos ( ) x dx Solutio We itroduced this example sayig that the substitutio was ot so obvious. However, this is really a itegral that falls ito the form give by e i our strategy above. However, may people miss that form ad so do t thik about it. So, let s try the followig substitutio. With this substitutio the itegral becomes, u = x x= u dx= udu ( ) cos x dx= ucosudu This is ow a itegratio by parts itegral. Remember that ofte we will eed to use more tha oe techique to completely do the itegral. This is a fairly simple itegratio by parts problem so I ll leave the remaider of the details to you to check. ( ) ( ) ( ) ( ) cos x dx= cos x + x si x + c Before leavig this sectio we should also poit out that there are itegrals out there i the world that just ca t be doe i terms of fuctios that we kow. Some examples of these are. -x si ( x) x dx cos( x ) dx e Ùı dx cos( ) dx x e 007 Paul Dawkis 8

230 Math 40 That does t mea that these itegrals ca t be doe at some level. If you go to a computer algebra system such as Maple or Mathematica ad have it do these itegrals here is what it will retur the followig. Ù ı - x p e dx= erf ( x) p Ê ˆ cos( x ) dx= FreselC x Á p Ë si ( x) dx = Si( x ) x cos x x ( e ) dx = Ci( e ) So it appears that these itegrals ca i fact be doe. However this is a little misleadig. Here are the defiitios of each of the fuctios give above. Error Fuctio The Sie Itegral The Fresel Cosie Itegral The Cosie Itegral erf = p e x ( x) 0 Si FreselC ( x) = Ù ı 0 x -t si t dt t dt Êp ˆ x = Ù Á t dt ı Ë ( ) cos x cost - Ci( x) = g + l ( x) + Ù dt ı 0 t Where g is the Euler-Mascheroi costat. 0 x Note that the first three are simply defied i terms of themselves ad so whe we say we ca itegrate them all we are really doig is reamig the itegral. The fourth oe is a little differet ad yet it is still defied i terms of a itegral that ca t be doe i practice. It will be possible to itegrate every itegral give i this class, but it is importat to ote that there are itegrals that just ca t be doe. We should also ote that after we look at Series we will be able to write dow series represetatios of each of the itegrals above. 007 Paul Dawkis 9

231 Math 40 Improper Itegrals I this sectio we eed to take a look at a couple of differet kids of itegrals. Both of these are examples of itegrals that are called Improper Itegrals. Let s start with the first kid of improper itegrals that we re goig to take a look at. Ifiite Iterval I this kid of itegrals we are goig to take a look at itegrals that i which oe or both of the limits of itegratio are ifiity. I these cases the iterval of itegratio is said to be over a ifiite iterval. Let s take a look at a example that will also show us how we are goig to deal with these itegrals. Example Evaluate the followig itegral. Ù dx ı x Solutio This is a iocet eough lookig itegral. However, because ifiity is ot a real umber we ca t just itegrate as ormal ad the plug i the ifiity to get a aswer. To see how we re goig to do this itegral let s thik of this as a area problem. So istead of askig what the itegral is, let s istead ask what the area uder f ( x) = o the iterval x [, ) is. We still are t able to do this, however, let s step back a little ad istead ask what the area uder f ( x ) is o the iterval [ ], t were t > ad t is fiite. This is a problem that we ca do. At t = Ù dx =- = - ı x x t Now, we ca get the area uder f ( x ) o [, ) simply by takig the limit of A t as t goes to ifiity. Ê ˆ A= limat = limá- = tæ tæ Ë t This is the how we will do the itegral itself. t 007 Paul Dawkis 0

232 Math 40 Ù ı t dx= lim Ù dx t x Æ ı x Ê ˆ = lim tæ Á - Ë x Ê ˆ = limá- = tæ Ë t t So, this is how we will deal with these kids of itegrals i geeral. We will replace the ifiity with a variable (usually t), do the itegral ad the take the limit of the result as t goes to ifiity. O a side ote, otice that the area uder a curve o a ifiite iterval was ot ifiity as we might have suspected it to be. I fact, it was a surprisigly small umber. Of course this wo t always be the case, but it is importat eough to poit out that ot all areas o a ifiite iterval will yield ifiite areas. Let s ow get some defiitios out of the way. We will call these itegrals coverget if the associated limit exists ad is a fiite umber (i.e. it s ot plus or mius ifiity) ad diverget if the associated limits either does t exist or is (plus or mius) ifiity. Let s ow formalize up the method for dealig with ifiite itervals. There are essetially three cases that we ll eed to look at. t. If ( ) x dx exists for every t > athe, a f ( ) = lim t ( ) a tæ a provided the limit exists ad is fiite. b. If f ( ) t f x dx f x dx x dx exists for every t < b the, b b ( ) = lim ( ) f x dx f x dx - tæ- provided the limits exists ad is fiite. c. If f ( x) dx ad f ( ) - c x dx are both coverget the, c ( ) = ( ) + ( ) f x dx f x dx f x dx - - Where c is ay umber. Note as well that this requires BOTH of the itegrals to be coverget i order for this itegral to also be coverget. If either of the two itegrals is diverget the so is this itegral. Let s take a look at a couple more examples. 007 Paul Dawkis t c

233 Math 40 Example Determie if the follow itegral is coverget or diverget ad if it s coverget fid its value. Ù dx ı x Solutio So, the first thig we do is covert the itegral to a limit. Now, do the itegral ad the limit. Ù ı Ù ı t dx= lim Ù dx t x Æ ı x dx= liml t x Æ ( x) ( () t l) = lim l - tæ = So, the limit is ifiite ad so the itegral is diverget. t If we go back to thikig i terms of area otice that the area uder g( x ) = x o the iterval [, ) is ifiite. This is i cotrast to the area uder ( ) f x = which was quite small. There really is t all that much differece betwee these two fuctios ad yet there is a large differece i the area uder them. We ca actually exted this out to the followig fact. Fact If a > 0 the Ù ı is coverget if p > ad diverget if p. dx p a x x Oe thig to ote about this fact is that it s i essece sayig that if a itegrad goes to zero fast eough the the itegral will coverge. How fast is fast eough? If we use this fact as a guide it looks like itegrads that go to zero faster tha x goes to zero will probably coverge. Let s take a look at a couple more examples. Example Determie if the followig itegral is coverget or diverget. If it is coverget fid its value. 0 Ù dx ı- - x 007 Paul Dawkis

234 Math 40 Solutio There really is t much to do with these problems oce you kow how to do them. We ll covert the itegral to a limit/itegral pair, evaluate the itegral ad the the limit. Ù ı dx= lim Ù dx -x tæ- ı -x = lim - -x tæ- ( t) = lim tæ- =- + = So, the limit is ifiite ad so this itegral is diverget. t 0 t Example 4 Determie if the followig itegral is coverget or diverget. If it is coverget fid its value. -x xe dx - Solutio I this case we ve got ifiities i both limits ad so we ll eed to split the itegral up ito two separate itegrals. We ca split the itegral up at ay poit, so let s choose a = 0 sice this will be a coveiet poit for the evaluate process. The itegral is the, 0 -x -x -x - xe dx= xe dx+ xe dx We ve ow got to look at each of the idividual limits x -x - xe dx= lim xe dx tæ- Ê = lim tæ- Á - e Ë t 0 -x ˆ Ê = lim Á- + e tæ- Ë =- So, the first itegral is coverget. Note that this does NOT mea that the secod itegral will also be coverget. So, let s take a look at that oe. 0 t -t ˆ 007 Paul Dawkis

235 Math 40 t -x -x xe dx= lim xe dx 0 tæ 0 Ê = lim tæ Á - e Ë -x ˆ 0 -t Ê ˆ = limá- e + tæ Ë = t This itegral is coverget ad so sice they are both coverget the itegral we were actually asked to deal with is also coverget ad its value is, 0 -x -x x x dx x dx - e = x dx 0 - e + =- + = - e 0 Example 5 Determie if the followig itegral is coverget or diverget. If it is coverget fid its value. si xdx Solutio First covert to a limit. - t sixdx= lim si xdx - tæ - = lim - tæ ( cos x) - ( cost) = lim cos- tæ This limit does t exist ad so the itegral is diverget. I most examples i a Calculus II class that are worked over ifiite itervals the limit either exists or is ifiite. However, there are limits that do t exist, as the previous example showed, so do t forget about those. Discotiuous Itegrad We ow eed to look at the secod type of improper itegrals that we ll be lookig at i this sectio. These are itegrals that have discotiuous itegrads. The process here is basically the same with oe o subtle differece. Here are the geeral cases that we ll look at for these itegrals.. If f ( x ) is cotiuous o the iterval [, ) a b t ab ad ot cotiuous at x t ( ) = lim ( ) f x dx f x dx - tæb a = b the, provided the limit exists ad is fiite. Note as well that we do eed to use a left had limit here sice the iterval of itegratio is etirely o the left side of the upper limit. 007 Paul Dawkis 4

236 Math 40. If f ( x ) is cotiuous o the iterval (, ] a b ab ad ot cotiuous at x b ( ) = lim ( ) f x dx f x dx + tæa t = a the, provided the limit exists ad is fiite. I this case we eed to use a right had limit here sice the iterval of itegratio is etirely o the right side of the lower limit. c b. If f ( x ) is ot cotiuous at x= c where a< c< b ad ( ) ad f ( ) are both coverget the, ( ) = ( ) + ( ) b c b f x dx f x dx f x dx a a c a f x dx c x dx As with the ifiite iterval case this requires BOTH of the itegrals to be coverget i order for this itegral to also be coverget. If either of the two itegrals is diverget the so is this itegral. c b 4. If f ( x ) is ot cotiuous at x= aad x= bad if ( ) ad f ( ) both coverget the, ( ) = ( ) + ( ) b c b a a c a f x dx f x dx f x dx f x dx x dx are Where c is ay umber. Agai, this requires BOTH of the itegrals to be coverget i order for this itegral to also be coverget. Note that the limits i these cases really do eed to be right or left haded limits. Sice we will be workig iside the iterval of itegratio we will eed to make sure that we stay iside that iterval. This meas that we ll use oe-sided limits to make sure we stay iside the iterval. Let s do a couple of examples of these kids of itegrals. c Example 6 Determie if the followig itegral is coverget or diverget. If it is coverget fid its value. Ù dx ı0 - x Solutio The problem poit is the upper limit so we are i the first case above. Ù ı dx= lim Ù dx -x ı -x - tæ 0 0 t ( x) ( t ) = lim tæ 0 = lim tæ = 007 Paul Dawkis 5 t

237 Math 40 The limit exists ad is fiite ad so the itegral coverges ad the itegrals value is. Example 7 Determie if the followig itegral is coverget or diverget. If it is coverget fid its value. Ù dx ı - x Solutio This itegrad is ot cotiuous at x = 0 ad so we ll eed to split the itegral up at that poit. 0 dx= dx+ Ù Ù Ù dx ı x ı x ı x Now we eed to look at each of these itegrals ad see if they are coverget. Ù ı 0 t dx= lim Ù dx x ı x - - tæ0 - Ê = - Ë lim- tæ0 Á Ê ˆ = lim-á- + tæ0 Ë t 8 =- At this poit we re doe. Oe of the itegrals is diverget that meas the itegral that we were asked to look at is diverget. We do t eve eed to bother with the secod itegral. Before leavig this sectio lets ote that we ca also have itegrals that ivolve both of these cases. Cosider the followig itegral. x ˆ t - Example 8 Determie if the followig itegral is coverget or diverget. If it is coverget fid its value. Ù dx ı 0 x Solutio This is a itegral over a ifiite iterval that also cotais a discotiuous itegrad. To do this itegral we ll eed to split it up ito two itegrals. We ca split it up aywhere, but pick a value that will be coveiet for evaluatio purposes. dx= dx+ Ù Ù Ù dx ı x ı x ı x 0 0 I order for the itegral i the example to be coverget we will eed BOTH of these to be coverget. If oe or both are diverget the the whole itegral will also be diverget. We kow that the secod itegral is coverget by the fact give i the ifiite iterval portio 007 Paul Dawkis 6

238 Math 40 above. So, all we eed to do is check the first itegral. dx= lim Ù Ù dx ı x ı x 0 + tæ0 t Ê ˆ = lim+ tæ0 Á - Ë x Ê ˆ = lim + Á- + tæ0 Ë t = So, the first itegral is diverget ad so the whole itegral is diverget. t 007 Paul Dawkis 7

239 Math 40 Compariso Test for Improper Itegrals Now that we ve see how to actually compute improper itegrals we eed to address oe more topic about them. Ofte we are t cocered with the actual value of these itegrals. Istead we might oly be iterested i whether the itegral is coverget or diverget. Also, there will be some itegrals that we simply wo t be able to itegrate ad yet we would still like to kow if they coverge or diverge. To deal with this we ve got a test for covergece or divergece that we ca use to help us aswer the questio of covergece for a improper itegral. We will give this test oly for a sub-case of the ifiite iterval itegral, however versios of the test exist for the other sub-cases of the ifiite iterval itegrals as well as itegrals with discotiuous itegrads. Compariso Test If f ( x) g( x) 0 o the iterval [, ) a the,. If f ( x) dx coverges the so does ( ) a g x dx.. If g ( x ) dx diverges the so does f ( ) a a a x dx. Note that if you thik i terms of area the Compariso Test makes a lot of sese. If f ( x ) is larger tha g( x ) the the area uder f ( x ) must also be larger tha the area uder g( x ). So, if the area uder the larger fuctio is fiite (i.e. f ( ) the smaller fuctio must also be fiite (i.e. ( ) the smaller fuctio is ifiite (i.e. ( ) must also be ifiite (i.e. f ( ) a a a x dx coverges) the the area uder a g x dx coverges). Likewise, if the area uder g x dx diverges) the the area uder the larger fuctio x dx diverges). Be careful ot to misuse this test. If the smaller fuctio coverges there is o reaso to believe that the larger will also coverge (after all ifiity is larger tha a fiite umber ) ad if the larger fuctio diverges there is o reaso to believe that the smaller fuctio will also diverge. Let s work a couple of example usig the compariso test. Note that all we ll be able to do is determie the covergece of the itegral. We wo t be able to determie the value of the itegrals ad so wo t eve bother with that. 007 Paul Dawkis 8

240 Math 40 Example Determie if the followig itegral is coverget or diverget. cos x Ù dx ı x Solutio Let s take a secod ad thik about how the Compariso Test works. If this itegral is coverget the we ll eed to fid a larger fuctio that also coverges o the same iterval. Likewise, if this itegral is diverget the we ll eed to fid a smaller fuctio that also diverges. So, it seems like it would be ice to have some idea as to whether the itegral coverges or diverges ahead of time so we will kow whether we will eed to look for a larger (ad coverget) fuctio or a smaller (ad diverget) fuctio. To get the guess for this fuctio let s otice that the umerator is ice ad bouded ad simply wo t get too large. Therefore, it seems likely that the deomiator will determie the covergece/divergece of this itegral ad we kow that Ù dx ı x coverges sice p = > by the fact i the previous sectio. So let s guess that this itegral will coverge. So we ow kow that we eed to fid a fuctio that is larger tha cos x x ad also coverges. Makig a fractio larger is actually a fairly simple process. We ca either make the umerator larger or we ca make the deomiator smaller. I this case ca t do a lot about the deomiator. However we ca use the fact that 0 cos x to make the umerator larger (i.e. we ll replace the cosie with somethig we kow to be larger, amely ). So, Now, as we ve already oted cos x x x Ù dx ı x coverges ad so by the Compariso Test we kow that must also coverge. Ù ı cos x dx x 007 Paul Dawkis 9

241 Math 40 Example Determie if the followig itegral is coverget or diverget. Ù dx x ı x + e Solutio Let s first take a guess about the covergece of this itegral. As oted after the fact i the last sectio about Ù ı dx p a x if the itegrad goes to zero faster tha x the the itegral will probably coverge. Now, we ve got a expoetial i the deomiator which is approachig ifiity much faster tha the x ad so it looks like this itegral should probably coverge. So, we eed a larger fuctio that will also coverge. I this case we ca t really make the umerator larger ad so we ll eed to make the deomiator smaller i order to make the fuctio larger as a whole. We will eed to be careful however. There are two ways to do this ad oly oe, i this case oly oe, of them will work for us. First, otice that sice the lower limit of itegratio is we ca say that x > 0 ad we kow that expoetials are always positive. So, the deomiator is the sum of two positive terms ad if we were to drop oe of them the deomiator would get smaller. This would i tur make the fuctio larger. The questio the is which oe to drop? Let s first drop the expoetial. Doig this gives, < x x+ e x This is a problem however, sice Ù dx ı x diverges by the fact. We ve got a larger fuctio that is diverget. This does t say aythig about the smaller fuctio. Therefore, we chose the wrog oe to drop. Let s try it agai ad this time let s drop the x. Also, -x < = e x x x + e e t - x - e dx= lim tæ - x e dx -t - ( e e ) = lim - + tæ = e 007 Paul Dawkis 40

242 Math 40 So, - x dx e is coverget. Therefore, by the Compariso test is also coverget. Ù ı dx x x + e Example Determie if the followig itegral is coverget or diverget. Ù dx -x ı x - e Solutio This is very similar to the previous example with a couple of very importat differeces. First, otice that the expoetial ow goes to zero as x icreases istead of growig larger as it did i the previous example (because of the egative i the expoet). Also ote that the expoetial is ow subtracted off the x istead of added oto it. The fact that the expoetial goes to zero meas that this time the x i the deomiator will probably domiate the term ad that meas that the itegral probably diverges. We will therefore eed to fid a smaller fuctio that also diverges. Makig fractios smaller is pretty much the same as makig fractios larger. I this case we ll eed to either make the umerator smaller or the deomiator larger. This is where the secod chage will come ito play. As before we kow that both x ad the expoetial are positive. However, this time sice we are subtractig the expoetial from the x if we were to drop the expoetial the deomiator will become larger ad so the fractio will become smaller. I other words, > - x x- e x ad we kow that Ù dx ı x diverges ad so by the Compariso Test we kow that must also diverge. Ù ı dx -x x - e Example 4 Determie if the followig itegral is coverget or diverget. 007 Paul Dawkis 4

243 Math 40 ( ) 4 si x dx + Ù ı Solutio First otice that as with the first example, the umerator i this fuctio is goig to be bouded sice the sie is ever larger tha. Therefore, sice the expoet o the deomiator is less tha we ca guess that the itegral will probably diverge. We will eed a smaller fuctio that also diverges. 4 We kow that ( x) x 0 si. I particular, this term is positive ad so if we drop it from the umerator the umerator will get smaller. This gives, ad diverges so by the Compariso Test also diverges. ( x) 4 + si Ù ı x > dx x ( ) x 4 si x dx + Ù ı Okay, we ve see a few examples of the Compariso Test ow. However, most of them worked pretty much the same way. All the fuctios were ratioal ad all we did for most of them was add or subtract somethig from the umerator or deomiator to get what we wat. Let s take a look at a example that works a little differetly so we do t get too locked ito these ideas. x Example 5 Determie if the followig itegral is coverget or diverget. -x e Ù dx ı x Solutio Normally, the presece of just a x i the deomiator would lead us to guess diverget for this itegral. However, the expoetial i the umerator will approach zero so fast that istead we ll eed to guess that this itegral coverges. To get a larger fuctio we ll use the fact that we kow from the limits of itegratio that x >. This meas that if we just replace the x i the deomiator with (which is always smaller tha x) we will make the deomiator smaller ad so the fuctio will get larger. e x x e < = e -x Paul Dawkis 4 -x

244 Math 40 ad we ca show that - x dx e coverges. I fact, we ve already doe this for a lower limit of ad chagig that to a wo t chage the covergece of the itegral. Therefore, by the Compariso Test also coverges. Ù ı - x e x dx We should also really work a example that does t ivolve a ratioal fuctio sice there is o reaso to assume that we ll always be workig with ratioal fuctios. Example 6 Determie if the followig itegral is coverget or diverget. - e x dx Solutio We kow that expoetials with egative expoets die dow to zero very fast so it makes sese to guess that this itegral will be coverget. We eed a larger fuctio, but this time we do t have a fractio to work with so we ll eed to do somethig differet. We ll take advatage of the fact that - x e is a decreasig fuctio. This meas that x > x fi e < e I other words, plug i a larger umber ad the fuctio gets smaller. -x -x From the limits of itegratio we kow that x > ad this meas that if we square x we will get larger. Or, x x x > provided > Note that we ca oly say this sice x >. This wo t be true if x! We ca ow use the fact that So, - x e is a decreasig fuctio to get, - x e is a larger fuctio tha -x -x e < e -x e ad we kow that - x dx e coverges so by the Compariso Test we also kow that is coverget. - x e dx The last two examples made use of the fact that x >. Let s take a look at a example to see how do we would have to go about these if the lower limit had bee smaller tha. 007 Paul Dawkis 4

245 Math 40 Example 7 Determie if the followig itegral is coverget or diverget. -x e Ù dx ı x Solutio I this case we ca t just replace x with i the deomiator ad get a larger fuctio for all x i the iterval of itegratio as we did i Example 5 above. Remember that we eed a fuctio (that is also coverget) that is always larger tha the give fuctio.. To see why we ca t just replace x with plug i x = 4 ito the deomiator ad compare this to what we would have if we plug i x =. -x -x e 4 -x -x e = e < e = 4 So, for x s i the rage x < we wo t get a larger fuctio as required for use i the Compariso Test. However, this is t the problem it might at first appear to be. We ca always write the itegral as follows, -x -x -x e dx= e dx+ e Ù Ù Ù dx ı x ı x ı x = Ù ı -x We used Maple to get the value of the first itegral. Now, if the secod itegral coverges it will have a fiite value ad so the sum of two fiite values will also be fiite ad so the origial itegral will coverge. Likewise, if the secod itegral diverges it will either be ifiite or ot have a value at all ad addig a fiite umber oto this will ot all of a sudde make it fiite or exist ad so the origial itegral will diverge. Therefore, this itegral will coverge or diverge depedig oly o the covergece of the secod itegral. As we saw i Example 5 the secod itegral does coverge ad so the whole itegral must also coverge. As we saw i this example, if we eed to, we ca split the itegral up ito oe that does t ivolve ay problems ad ca be computed ad oe that may cotai problem what we ca use the Compariso Test o to determie its coverge. e x dx 007 Paul Dawkis 44

246 Math 40 Polar Coordiates Itroductio I this sectio we will be lookig at polar coordiates. Here is a list of topics that we ll be coverig i this chapter. Polar Coordiates We ll itroduce polar coordiates i this sectio. We ll look at covertig betwee polar coordiates ad Cartesia coordiates as well as some basic graphs i polar coordiates. Tagets with Polar Coordiates Fidig taget lies of polar curves. 007 Paul Dawkis 45

247 Math 40 Polar Coordiates Up to this poit we ve dealt exclusively with the Cartesia (or Rectagular, or x-y) coordiate system. However, as we will see, this is ot always the easiest coordiate system to work i. So, i this sectio we will start lookig at the polar coordiate system. Coordiate systems are really othig more tha a way to defie a poit i space. For istace i the Cartesia coordiate system at poit is give the coordiates (x,y) ad we use this to defie the poit by startig at the origi ad the movig x uits horizotally followed by y uits vertically. This is show i the sketch below. This is ot, however, the oly way to defie a poit i two dimesioal space. Istead of movig vertically ad horizotally from the origi to get to the poit we could istead go straight out of the origi util we hit the poit ad the determie the agle this lie makes with the positive x- axis. We could the use the distace of the poit from the origi ad the amout we eeded to rotate from the positive x-axis as the coordiates of the poit. This is show i the sketch below. Coordiates i this form are called polar coordiates. The above discussio may lead oe to thik that r must be a positive umber. However, we also -, p. allow r to be egative. Below is a sketch of the two poits ( ) 6, p ad ( ) Paul Dawkis 46

248 Math 40 From this sketch we ca see that if r is positive the poit will be i the same quadrat as q. O the other had if r is egative the poit will ed up i the quadrat exactly opposite q. Notice as, p 7, p do. The well that the coordiates (- 6 ) describe the same poit as the coordiates ( 6 ) 7 coordiates ( ), p 6 tells us to rotate a agle of 7 6 p from the positive x-axis, this would put us o the dashed lie i the sketch above, ad the move out a distace of. This leads to a importat differece betwee Cartesia coordiates ad polar coordiates. I Cartesia coordiates there is exactly oe set of coordiates for ay give poit. With polar coordiates this is t true. I polar coordiates there is literally a ifiite umber of coordiates for a give poit. For istace, the followig four poits are all coordiates for the same poit. Ê 5 4 5, p ˆ Ê 5, p ˆ Ê 5, p ˆ Ê 5, p ˆ Á = Á - = Á- = Á- - Ë Ë Ë Ë Here is a sketch of the agles used i these four sets of coordiates. 007 Paul Dawkis 47

249 Math 40 I the secod coordiate pair we rotated i a clock-wise directio to get to the poit. We should t forget about rotatig i the clock-wise directio. Sometimes it s what we have to do. The last two coordiate pairs use the fact that if we ed up i the opposite quadrat from the poit we ca use a egative r to get back to the poit ad of course there is both a couter clock-wise ad a clock-wise rotatio to get to the agle. These four poits oly represet the coordiates of the poit without rotatig aroud the system more tha oce. If we allow the agle to make as may complete rotatio about the axis system as we wat the there are a ifiite umber of coordiates for the same poit. I fact the poit ( r, q ) ca be represeted by ay of the followig coordiate pairs. ( q + p ) - q + ( + ) p ( ) r, r,, where is ay iteger. Next we should talk about the origi of the coordiate system. I polar coordiates the origi is ofte called the pole. Because we are t actually movig away from the origi/pole we kow that r = 0. However, we ca still rotate aroud the system by ay agle we wat ad so the coordiates of the origi/pole are ( 0,q ). Now that we ve got a grasp o polar coordiates we eed to thik about covertig betwee the two coordiate systems. Well start out with the followig sketch remidig us how both coordiate systems work. Note that we ve got a right triagle above ad with that we ca get the followig equatios that will covert polar coordiates ito Cartesia coordiates. Polar to Cartesia Coversio Formulas x= rcosq y = rsiq Covertig from Cartesia is almost as easy. Let s first otice the followig. ( cosq) ( siq) x y r r + = + = r cos q + r ( cos q si q) 007 Paul Dawkis 48 si q = r + = r

250 Math 40 This is a very useful formula that we should remember, however we are after a equatio for r so let s take the square root of both sides. This gives, r = x + y Note that techically we should have a plus or mius i frot of the root sice we kow that r ca be either positive or egative. We will ru with the covetio of positive r here. Gettig a equatio for q is almost as simple. We ll start with, y rsiq = = taq x rcosq Takig the iverse taget of both sides gives, - Ê y ˆ q = ta Á Ë x We will eed to be careful with this because iverse tagets oly retur values i the rage - p < q < p. Recall that there is a secod possible agle ad that the secod agle is give by q + p. Summarizig the gives the followig formulas for covertig from Cartesia coordiates to polar coordiates. Cartesia to Polar Coversio Formulas Let s work a quick example. r = x + y r = x + y q ta Ê y ˆ Ë x - = Á Example Covert each of the followig poits ito the give coordiate system. Ê p ˆ (a) Á-4, ito Cartesia coordiates. [Solutio] Ë (b) (-,-) ito polar coordiates. [Solutio] Solutio (a) Covert Ê p ˆ Á-4, Ë ito Cartesia coordiates. This coversio is easy eough. All we eed to do is plug the poits ito the formulas. 007 Paul Dawkis 49

251 Math 40 Êp ˆ Ê ˆ x=- 4cosÁ =-4Á- = Ë Ë Êp ˆ Ê ˆ y =- 4siÁ =- 4 =- Ë Á Ë So, i Cartesia coordiates this poit is (, - ). [Retur to Problems] (b) Covert (-,-) ito polar coordiates. Let s first get r. Now, let s get q. ( ) ( ) r = = -Ê-ˆ - p q = ta Á = ta () = Ë- 4 This is ot the correct agle however. This value of q is i the first quadrat ad the poit we ve bee give is i the third quadrat. As oted above we ca get the correct agle by addig p oto this. Therefore, the actual agle is, p 5p q = + p = So, i polar coordiates the poit is ( 4 ), p. Note as well that we could have used the first q that we got by usig a egative r. I this case the poit could also be writte i polar coordiates as (, p 4) -. [Retur to Problems] We ca also use the above formulas to covert equatios from oe coordiate system to the other. Example Covert each of the followig ito a equatio i the give coordiate system. (a) Covert x- 5x = + xy ito polar coordiates. [Solutio] (b) Covert r =- 8cosq ito Cartesia coordiates. [Solutio] Solutio (a) Covert x- 5x = + xy ito polar coordiates. I this case there really is t much to do other tha pluggig i the formulas for x ad y (i.e. the Cartesia coordiates) i terms of r ad q (i.e. the polar coordiates). ( r q) - ( r q) = + ( r q)( r q) cos 5 cos cos si rcos - 5r cos = + r cos si q q q q 007 Paul Dawkis 50

252 Math 40 (b) Covert r =- 8cosq ito Cartesia coordiates. [Retur to Problems] This oe is a little trickier, but ot by much. First otice that we could substitute straight for the r. However, there is o straight substitutio for the cosie that will give us oly Cartesia coordiates. If we had a r o the right alog with the cosie the we could do a direct substitutio. So, if a r o the right side would be coveiet let s put oe there, just do t forget to put o the left side as well. r =- 8rcosq We ca ow make some substitutios that will covert this ito Cartesia coordiates. x + y =- 8x [Retur to Problems] Before movig o to the ext subject let s do a little more work o the secod part of the previous example. The equatio give i the secod part is actually a fairly well kow graph; it just is t i a form that most people will quickly recogize. To idetify it let s take the Cartesia coordiate equatio ad do a little rearragig. x + 8x+ y = 0 Now, complete the square o the x portio of the equatio. x + 8x+ 6+ y = 6 ( x ) So, this was a circle of radius 4 ad ceter (-4,0). This leads us ito the fial topic of this sectio y = 6 Commo Polar Coordiate Graphs Let s idetify a few of the more commo graphs i polar coordiates. We ll also take a look at a couple of special polar graphs. Lies Some lies have fairly simple equatios i polar coordiates.. q b =. We ca see that this is a lie by covertig to Cartesia coordiates as follows 007 Paul Dawkis 5

253 Math 40 ta - q = b Ê y Á ˆ= b Ë x y x y = = ta b ( ta b ) This is a lie that goes through the origi ad makes a agle of b with the positive x- axis. Or, i other words it is a lie through the origi with slope of ta b.. rcosq = a This is easy eough to covert to Cartesia coordiates to x= a. So, this is a vertical lie.. rsiq = b Likewise, this coverts to y = b ad so is a horizotal lie. x p Example Graph q =, r cosq = 4 ad r siq =- o the same axis system. 4 Solutio There really is t too much to this oe other tha doig the graph so here it is. Circles Let s take a look at the equatios of circles i polar coordiates.. r = a. This equatio is sayig that o matter what agle we ve got the distace from the origi must be a. If you thik about it that is exactly the defiitio of a circle of radius a cetered at the origi. So, this is a circle of radius a cetered at the origi. This is also oe of the reasos why 007 Paul Dawkis 5

254 Math 40 we might wat to work i polar coordiates. The equatio of a circle cetered at the origi has a very ice equatio, ulike the correspodig equatio i Cartesia coordiates.. r = acosq. We looked at a specific example of oe of these whe we were covertig equatios to Cartesia coordiates. This is a circle of radius a ad ceter ( a,0). Note that a might be egative (as it was i our example above) ad so the absolute value bars are required o the radius. They should ot be used however o the ceter.. r = bsiq. 0,b. This is similar to the previous oe. It is a circle of radius b ad ceter ( ) 4. r = acosq + bsiq. This is a combiatio of the previous two ad by completig the square twice it ca be show that this is a circle of radius a + b ad ceter ( ab, ). I other words, this is the geeral equatio of a circle that is t cetered at the origi. Example 4 Graph r = 7, r = 4cosq, ad r =- 7siq o the same axis system. Solutio The first oe is a circle of radius 7 cetered at the origi. The secod is a circle of radius cetered at (,0). The third is a circle of radius 7 cetered at Ê 7 ˆ Á0, - Here is the graph of the Ë. three equatios. 007 Paul Dawkis 5

255 Math 40 Note that it takes a rage of 0 q p for a complete graph of r = a ad it oly takes a rage of 0 q p to graph the other circles give here. Cardioids ad Limacos These ca be broke up ito the followig three cases.. Cardioids : r = a± acosq ad r = a± asiq. These have a graph that is vaguely heart shaped ad always cotai the origi.. Limacos with a ier loop : r = a± bcosq ad r = a± bsiq with a< b. These will have a ier loop ad will always cotai the origi.. Limacos without a ier loop : r = a± bcosq ad r = a± bsiq with a > b. These do ot have a ier loop ad do ot cotai the origi. Example 5 Graph r = 5-5siq, r = 7-6cosq, ad r = + 4cosq. Solutio These will all graph out oce i the rage 0 q p. Here is a table of values for each followed by graphs of each. q r = 5-5siq r = 7-6cosq r = + 4cosq p 0 7 p 5 - p 0 7 p Paul Dawkis 54

256 Math Paul Dawkis 55

257 Math 40 There is oe fial thig that we eed to do i this sectio. I the third graph i the previous example we had a ier loop. We will, o occasio, eed to kow the value of q for which the graph will pass through the origi. To fid these all we eed to do is set the equatio equal to zero ad solve as follows, p 4p 0= + 4cosq fi cos q =- fi q =, 007 Paul Dawkis 56

258 Math 40 Tagets with Polar Coordiates We ow eed to discuss some calculus topics i terms of polar coordiates. We will start with fidig taget lies to polar curves. I this case we are goig to assume that the equatio is i the form r f ( q ) =. With the equatio i this form we ca actually use the equatio for the derivative dy we derived whe we looked at taget lies with parametric dx equatios. To do this however requires us to come up with a set of parametric equatios to represet the curve. This is actually pretty easy to do. From our work i the previous sectio we have the followig set of coversio equatios for goig from polar coordiates to Cartesia coordiates. x= rcosq y = rsiq Now, we ll use the fact that we re assumig that the equatio is i the form r f ( q ) =. Substitutig this ito these equatios gives the followig set of parametric equatios (with q as the parameter) for the curve. ( ) cos ( ) x= f q q y = f q siq Now, we will eed the followig derivatives. dx dy = f ( q) cosq - f ( q) siq = f ( q) siq + f ( q) cosq dq dq dr dr = cosq - rsiq = siq + rcosq dq dq The derivative dy dx is the, Derivative with Polar Coordiates dy = dx dr siq + r cosq dq dr cosq - r siq dq Note that rather tha tryig to remember this formula it would probably be easier to remember how we derived it ad just remember the formula for parametric equatios. Let s work a quick example with this. 007 Paul Dawkis 57

259 Math 40 Example Determie the equatio of the taget lie to r = + 8siq at Solutio We ll first eed the followig derivative. dr 8cosq dq = The formula for the derivative dy dx becomes, dy dx The slope of the taget lie is, ( ) ( ) p q =. 6 8cosqsiq + + 8siq cosq 6cosqsiq + cosq = = 8cos q - + 8siq siq 8cos q -siq -8si q 4 + dy m = = = dx p q = Now, at q = p 6 we have r = 7. We ll eed to get the correspodig x-y coordiates so we ca get the taget lie. Êp ˆ 7 Êp ˆ 7 x= 7cosÁ = y = 7si Á = Ë 6 Ë 6 The taget lie is the, 7 Ê 7 ˆ y = + x- 5 Á Ë For the sake of completeess here is a graph of the curve ad the taget lie. 007 Paul Dawkis 58

260 Math Paul Dawkis 59

261 Math 40 Sequeces ad Series Itroductio I this chapter we ll be takig a look at sequeces ad (ifiite) series. Actually, this chapter will deal almost exclusively with series. However, we also eed to uderstad some of the basics of sequeces i order to properly deal with series. We will therefore, sped a little time o sequeces as well. Series is oe of those topics that may studets do t fid all that useful. To be hoest, may studets will ever see series outside of their calculus class. However, series do play a importat role i the field of ordiary differetial equatios ad without series large portios of the field of partial differetial equatios would ot be possible. I other words, series is a importat topic eve if you wo t ever see ay of the applicatios. Most of the applicatios are beyod the scope of most Calculus courses ad ted to occur i classes that may studets do t take. So, as you go through this material keep i mid that these do have applicatios eve if we wo t really be coverig may of them i this class. Here is a list of topics i this chapter. Sequeces We will start the chapter off with a brief discussio of sequeces. This sectio will focus o the basic termiology ad covergece of sequeces More o Sequeces Here we will take a quick look about mootoic ad bouded sequeces. Series The Basics I this sectio we will discuss some of the basics of ifiite series. Series Covergece/Divergece Most of this chapter will be about the covergece/divergece of a series so we will give the basic ideas ad defiitios i this sectio. Series Special Series We will look at the Geometric Series, Telescopig Series, ad Harmoic Series i this sectio. 007 Paul Dawkis 60

262 Math 40 Sequeces Let s start off this sectio with a discussio of just what a sequece is. A sequece is othig more tha a list of umbers writte i a specific order. The list may or may ot have a ifiite umber of terms i them although we will be dealig exclusively with ifiite sequeces i this class. Geeral sequece terms are deoted as follows, a -first term a a a -secod term - + th M term ( ) st - + term M Because we will be dealig with ifiite sequeces each term i the sequece will be followed by aother term as oted above. I the otatio above we eed to be very careful with the subscripts. The subscript of + deotes the ext term i the sequece ad NOT oe plus the th term! I other words, a a + + so be very careful whe writig subscripts to make sure that the + does t migrate out of the subscript! This is a easy mistake to make whe you first start dealig with this kid of thig. There is a variety of ways of deotig a sequece. Each of the followig are equivalet ways of deotig a sequece. {,, K,,, K } { } { } a a a a a a + = I the secod ad third otatios above a is usually give by a formula. A couple of otes are ow i order about these otatios. First, ote the differece betwee the secod ad third otatios above. If the startig poit is ot importat or is implied i some way by the problem it is ofte ot writte dow as we did i the third otatio. Next, we used a startig poit of = i the third otatio oly so we could write oe dow. There is absolutely o reaso to believe that a sequece will start at =. A sequece will start where ever it eeds to start. Let s take a look at a couple of sequeces. 007 Paul Dawkis 6

263 Math 40 Example Write dow the first few terms of each of the followig sequeces. Solutio Ï + (a) Ì Ó (b) ÏÔ Ì ÔÓ (-) b = (c) { } Ï + (a) Ì Ó = = + Ô Ô = 0 [Solutio] [Solutio] th, where b digit of p = [Solutio] To get the first few sequece terms here all we eed to do is plug i values of ito the formula give ad we ll get the sequece terms. Ï Ï+ Ô Ô Ì = Ì, {,,,, K Ó = Ô= { 4 { 9 6 { { 5 Ô Ó = = = 4 = 5 Note the iclusio of the at the ed! This is a importat piece of otatio as it is the oly thig that tells us that the sequece cotiues o ad does t termiate at the last term. [Retur to Problems] + ÏÔ( -) Ô (b) Ì ÔÓ Ô = 0 This oe is similar to the first oe. The mai differece is that this sequece does t start at =. + ÏÔ( -) Ô Ï Ì,,,,, = Ì- - - K ÔÓ Ô Ó = 0 Note that the terms i this sequece alterate i sigs. Sequeces of this kid are sometimes called alteratig sequeces. [Retur to Problems] b = (c) { } th, where b = digit of p This sequece is differet from the first two i the sese that it does t have a specific formula for 007 Paul Dawkis 6

264 Math 40 each term. However, it does tell us what each term should be. Each term should be the th digit of p. So we kow that p = K The sequece is the, {,,4,,5,9,,6,5,,5,K } [Retur to Problems] I the first two parts of the previous example ote that we were really treatig the formulas as fuctios that ca oly have itegers plugged ito them. Or, + = = ( ) g( ) f (-) This is a importat idea i the study of sequeces (ad series). Treatig the sequece terms as fuctio evaluatios will allow us to do may thigs with sequeces that could t do otherwise. Before delvig further ito this idea however we eed to get a couple more ideas out of the way. First we wat to thik about graphig a sequece. To graph the sequece { a } we plot the poits ( a, ) as rages over all possible values o a graph. For istace, let s graph the Ï + sequece Ì Ó =. The first few poits o the graph are, ( ) + Ê ˆ Ê 4ˆ Ê 5 ˆ Ê 6 ˆ,, Á,, Á,, Á4,, Á5,, Ë 4 Ë 9 Ë 6 Ë 5 K The graph, for the first 0 terms of the sequece, is the, This graph leads us to a importat idea about sequeces. Notice that as icreases the sequece terms from our sequece terms, i this case, get closer ad closer to zero. We the say that zero is the limit (or sometimes the limitig value) of the sequece ad write, 007 Paul Dawkis 6

265 Math 40 + lima = lim = 0 Æ Æ This otatio should look familiar to you. It is the same otatio we used whe we talked about the limit of a fuctio. I fact, if you recall, we said earlier that we could thik of sequeces as fuctios i some way ad so this otatio should t be too surprisig. Usig the ideas that we developed for limits of fuctios we ca write dow the followig workig defiitio for limits of sequeces. Workig Defiitio of Limit. We say that lim a = L Æ if we ca make a as close to L as we wat for all sufficietly large. I other words, the value of the a s approach L as approaches ifiity.. We say that lim a = Æ if we ca make a as large as we wat for all sufficietly large. Agai, i other words, the value of the a s get larger ad larger without boud as approaches ifiity.. We say that lim a =- Æ if we ca make a as large ad egative as we wat for all sufficietly large. Agai, i other words, the value of the a s are egative ad get larger ad larger without boud as approaches ifiity. The workig defiitios of the various sequece limits are ice i that they help us to visualize what the limit actually is. Just like with limits of fuctios however, there is also a precise defiitio for each of these limits. Let s give those before proceedig Precise Defiitio of Limit. We say that lim a = L if for every umber e > 0 there is a iteger N such that Æ a - L < e wheever > N. We say that lim a Æ = if for every umber M > 0 there is a iteger N such that a > M wheever > N. We say that lim a Æ =- if for every umber M < 0 there is a iteger N such that 007 Paul Dawkis 64

266 Math 40 a < M wheever > N We wo t be usig the precise defiitio ofte, but it will show up occasioally. Note that both defiitios tell us that i order for a limit to exist ad have a fiite value all the sequece terms must be gettig closer ad closer to that fiite value as icreases. Now that we have the defiitios of the limit of sequeces out of the way we have a bit of termiology that we eed to look at. If lim a exists ad is fiite we say that the sequece is Æ Æ coverget. If lim a does t exist or is ifiite we say the sequece diverges. Note that sometimes we will say the sequece diverges to if lim a = ad if lim a =- we will sometimes say that the sequece diverges to -. Get used to the terms coverget ad diverget as we ll be seeig them quite a bit throughout this chapter. So just how do we fid the limits of sequeces? Most limits of most sequeces ca be foud usig oe of the followig theorems. Theorem Give the sequece { a } if we have a fuctio f ( x ) such that f ( ) = a ad lim ( ) the lim a = L Æ Æ Æ f x = L x Æ This theorem is basically tellig us that we take the limits of sequeces much like we take the limit of fuctios. I fact, i most cases we ll ot eve really use this theorem by explicitly writig dow a fuctio. We will more ofte just treat the limit as if it were a limit of a fuctio ad take the limit as we always did back i Calculus I whe we were takig the limits of fuctios. So, ow that we kow that takig the limit of a sequece is early idetical to takig the limit of a fuctio we also kow that all the properties from the limits of fuctios will also hold. Properties lim a ± b = lima ± limb. ( ) Æ Æ Æ. limca = clim a Æ Æ. lim( ab ) = ( lima)( lim b) Æ Æ Æ 007 Paul Dawkis 65

267 Math a lim a Æ lim =, provided limb 0 Æ b lim b Æ Æ p 5. lima = Èlim a provided a 0 Î Æ Æ p These properties ca be proved usig Theorem above ad the fuctio limit properties we saw i Calculus I or we ca prove them directly usig the precise defiitio of a limit usig early idetical proofs of the fuctio limit properties. Next, just as we had a Squeeze Theorem for fuctio limits we also have oe for sequeces ad it is pretty much idetical to the fuctio limit versio. Squeeze Theorem for Sequeces If a c b for all > N for some N ad lima = lim b = L the lim c Æ Æ Æ = L. Note that i this theorem the for all > N for some N is really just tellig us that we eed to have a c b for all sufficietly large, but if it is t true for the first few that wo t ivalidate the theorem. As we ll see ot all sequeces ca be writte as fuctios that we ca actually take the limit of. This will be especially true for sequeces that alterate i sigs. While we ca always write these sequece terms as a fuctio we simply do t kow how to take the limit of a fuctio like that. The followig theorem will help with some of these sequeces. Theorem If lim a = 0 the lima = 0. Æ Æ Note that i order for this theorem to hold the limit MUST be zero ad it wo t work for a sequece whose limit is ot zero. This theorem is easy eough to prove so let s do that. Proof of Theorem The mai thig to this proof is to ote that, - a a a The ote that, ( a ) lim - =- lim a = 0 Æ Æ We the have ( a ) lim - = lim a = 0 ad so by the Squeeze Theorem we must also have, Æ Æ 007 Paul Dawkis 66

268 Math 40 lima = 0 Æ The ext theorem is a useful theorem givig the covergece/divergece ad value (for whe it s coverget) of a sequece that arises o occasio. Theorem The sequece { } = 0 r coverges if - < r ad diverges for all other value of r. Also, lim r Æ Ï0 if - < r < = Ì Ó if r = Here is a quick (well ot so quick, but defiitely simple) partial proof of this theorem. Partial Proof of Theorem We ll do this by a series of cases although the last case will ot be completely prove. Case : r > We kow from Calculus I that lim r x that lim r Æ xæ = ad so the sequece diverges if r >. = if r > ad so by Theorem above we also kow Case : r = I this case we have, limr = lim = lim= Æ Æ Æ So, the sequece coverges for r = ad i this case its limit is. Case : 0< r < x We kow from Calculus I that limr = 0 if 0< r < ad so by Theorem above we also kow xæ that limr = 0 ad so the sequece coverges if 0< r < ad i this case its limit is zero. Æ Case 4 : r = 0 I this case we have, limr = lim0 = lim0= 0 Æ Æ Æ So, the sequece coverges for r = 0 ad i this case its limit is zero. 007 Paul Dawkis 67

269 Math 40 Case 5 : - < r < 0 First let s ote that if - < r < 0 the 0< r < the by Case above we have, lim r = lim r = 0 Æ Theorem above ow tells us that we must also have, limr = 0 ad so if - < r < 0 the sequece coverges ad has a limit of 0. Case 6 : r =- I this case the sequece is, { r } ( ) Æ Æ { } {,,,,,,,, } = 0 = - = K = 0 = 0 ad hopefully it is clear that lim( ) Æ - does t exist. Recall that i order of this limit to exist the terms must be approachig a sigle value as icreases. I this case however the terms just alterate betwee ad - ad so the limit does ot exist. So, the sequece diverges for r =-. Case 7 : r <- I this case we re ot goig to go through a complete proof. Let s just see what happes if we let r =- for istace. If we do that the sequece becomes, { r } ( ) { } {,,4, 8,6,, } = 0 = - = K = 0 = 0 So, if r =- we get a sequece of terms whose values alterate i sig ad get larger ad larger ad so lim( ) Æ - does t exist. It does ot settle dow to a sigle value as icreases or do the terms ALL approach ifiity. So, the sequece diverges for r =-. We could do somethig similar for ay value of r such that r <- ad so the sequece diverges for r <-. Let s take a look at a couple of examples of limits of sequeces. Example Determie if the followig sequeces coverge or diverge. If the sequece coverges determie its limit. (a) Ï Ì Ó = [Solutio] 007 Paul Dawkis 68

270 Math 40 Ïe (b) Ì Ó (c) ÏÔ Ì ÔÓ (- ) (d) ( ) = Ô Ô = { } = 0 [Solutio] [Solutio] - [Solutio] Solutio Ï - (a) Ì Ó0+ 5 = I this case all we eed to do is recall the method that was developed i Calculus I to deal with the limits of ratioal fuctios. See the Limits At Ifiity, Part I sectio of my Calculus I otes for a review of this if you eed to. To do a limit i this form all we eed to do is factor from the umerator ad deomiator the largest power of, cacel ad the take the limit. Ê ˆ - - = = = Á Ë - Á lim lim Ë lim Æ + Æ Ê 0 ˆ Æ 0 So the sequece coverges ad its limit is 5. [Retur to Problems] (b) Ïe Ì Ó = We will eed to be careful with this oe. We will eed to use L Hospital s Rule o this sequece. The problem is that L Hospital s Rule oly works o fuctios ad ot o sequeces. Normally this would be a problem, but we ve got Theorem from above to help us out. Let s defie ad ote that, f ( x) ( ) f 007 Paul Dawkis 69 x = e x = e

271 Math 40 Theorem says that all we eed to do is take the limit of the fuctio. x x e e e lim = lim = lim = Æ xæ x xæ So, the sequece i this part diverges (to ). More ofte tha ot we just do L Hospital s Rule o the sequece terms without first covertig to x s sice the work will be idetical regardless of whether we use x or. However, we really should remember that techically we ca t do the derivatives while dealig with sequece terms. [Retur to Problems] (c) ÏÔ Ì ÔÓ (- ) Ô Ô = We will also eed to be careful with this sequece. We might be tempted to just say that the limit of the sequece terms is zero (ad we d be correct). However, techically we ca t take the limit of sequeces whose terms alterate i sig, because we do t kow how to do limits of fuctios that exhibit that same behavior. Also, we wat to be very careful to ot rely too much o ituitio with these problems. As we will see i the ext sectio, ad i later sectios, our ituitio ca lead us astray i these problem if we are t careful. So, let s work this oe by the book. We will eed to use Theorem o this problem. To this we ll first eed to compute, (- ) lim = lim = 0 Æ Æ Therefore, sice the limit of the sequece terms with absolute value bars o them goes to zero we kow by Theorem that, (-) lim = 0 Æ which also meas that the sequece coverges to a value of zero. { } = 0 (d) (- ) [Retur to Problems] For this theorem ote that all we eed to do is realize that this is the sequece i Theorem above usig r =-. So, by Theorem this sequece diverges. [Retur to Problems] 007 Paul Dawkis 70

272 Math 40 We ow eed to give a warig about misusig Theorem. Theorem oly works if the limit is zero. If the limit of the absolute value of the sequece terms is ot zero the the theorem will ot hold. The last part of the previous example is a good example of this (ad i fact this warig the whole reaso that part is there). Notice that ad yet, lim( ) Æ ( ) lim - = lim= Æ Æ - does t eve exist let aloe equal. So, be careful usig this Theorem. You must always remember that it oly works if the limit is zero. Before movig oto the ext sectio we eed to give oe more theorem that we ll eed for a proof dow the road. Theorem 4 For the sequece { } lim a = L. Æ a if both lim a Æ = L ad lim a + Æ = L the { a } is coverget ad Proof of Theorem 4 Let e > 0. The sice lim a Æ = L there is a N > 0 such that if > N we kow that, a - L < e Likewise, because lim a + = L there is a N > 0 such that if > N we kow that, Æ Now, let N max{ N,N } a a + a - + L < e = + ad let > N. The either a = ak for some k > N or = k for some k N > ad so i either case we have that, a - L < e Therefore, lim a Æ = L ad so { a } is coverget. 007 Paul Dawkis 7

273 Math 40 More o Sequeces I the previous sectio we itroduced the cocept of a sequece ad talked about limits of sequeces ad the idea of covergece ad divergece for a sequece. I this sectio we wat to take a quick look at some ideas ivolvig sequeces. Let s start off with some termiology ad defiitios. Give ay sequece { a } we have the followig.. We call the sequece icreasig if a < a + for every.. We call the sequece decreasig if a > a + for every.. If { a } is a icreasig sequece or { } a is a decreasig sequece we call it mootoic. 4. If there exists a umber m such that m a for every we say the sequece is bouded below. The umber m is sometimes called a lower boud for the sequece. 5. If there exists a umber M such that a M for every we say the sequece is bouded above. The umber M is sometimes called a upper boud for the sequece. 6. If the sequece is both bouded below ad bouded above we call the sequece bouded. Note that i order for a sequece to be icreasig or decreasig it must be icreasig/decreasig for every. I other words, a sequece that icreases for three terms ad the decreases for the rest of the terms is NOT a decreasig sequece! Also ote that a mootoic sequece must always icrease or it must always decrease. Before movig o we should make a quick poit about the bouds for a sequece that is bouded above ad/or below. We ll make the poit about lower bouds, but we could just as easily make it about upper bouds. A sequece is bouded below if we ca fid ay umber m such that m a for every. Note however that if we fid oe umber m to use for a lower boud the ay umber smaller tha m will also be a lower boud. Also, just because we fid oe lower boud that does t mea there wo t be a better lower boud for the sequece tha the oe we foud. I other words, there are a ifiite umber of lower bouds for a sequece that is bouded below, some will be better tha others. I my class all that I m after will be a lower boud. I do t ecessarily eed the best lower boud, just a umber that will be a lower boud for the sequece. Let s take a look at a couple of examples. 007 Paul Dawkis 7

274 Math 40 Example Determie if the followig sequeces are mootoic ad/or bouded. Solutio (a) {- } = 0 { + } = - [Solutio] (a) { } (b) ( ) - [Solutio] Ï (c) Ì Ó = 5 = 0 [Solutio] This sequece is a decreasig sequece (ad hece mootoic) because, for every. ( ) - >- + Also, sice the sequece terms will be either zero or egative this sequece is bouded above. We ca use ay positive umber or zero as the boud, M, however, it s stadard to choose the smallest possible boud if we ca ad it s a ice umber. So, we ll choose M = 0 sice, - 0 for every This sequece is ot bouded below however sice we ca always get below ay potetial boud by takig large eough. Therefore, while the sequece is bouded above it is ot bouded. As a side ote we ca also ote that this sequece diverges (to - if we wat to be specific). [Retur to Problems] { + } = (b) (- ) The sequece terms i this sequece alterate betwee ad - ad so the sequece is either a icreasig sequece or a decreasig sequece. Sice the sequece is either a icreasig or decreasig sequece it is ot a mootoic sequece. The sequece is bouded however sice it is bouded above by ad bouded below by -. Agai, we ca ote that this sequece is also diverget. [Retur to Problems] 007 Paul Dawkis 7

275 Math 40 Ï (c) Ì Ó = 5 This sequece is a decreasig sequece (ad hece mootoic) sice, > + ( ) The terms i this sequece are all positive ad so it is bouded below by zero. Also, sice the sequece is a decreasig sequece the first sequece term will be the largest ad so we ca see that the sequece will also be bouded above by 5. Therefore, this sequece is bouded. We ca also take a quick limit ad ote that this sequece coverges ad its limit is zero. [Retur to Problems] Now, let s work a couple more examples that are desiged to make sure that we do t get too used to relyig o our ituitio with these problems. As we oted i the previous sectio our ituitio ca ofte lead us astray with some of the cocepts we ll be lookig at i this chapter. Example Determie if the followig sequeces are mootoic ad/or bouded. Solutio Ï (a) Ì Ó + (a) (b) Ï Ì Ó + Ï Ì Ó = = [Solutio] = 0 [Solutio] We ll start with the bouded part of this example first ad the come back ad deal with the icreasig/decreasig questio sice that is where studets ofte make mistakes with this type of sequece. First, is positive ad so the sequece terms are all positive. The sequece is therefore bouded below by zero. Likewise each sequece term is the quotiet of a umber divided by a larger umber ad so is guarateed to be less that oe. The sequece is the bouded above by oe. So, this sequece is bouded. Now let s thik about the mootoic questio. First, studets will ofte make the mistake of assumig that because the deomiator is larger the quotiet must be decreasig. This will ot always be the case ad i this case we would be wrog. This sequece is icreasig as we ll see. 007 Paul Dawkis 74

276 Math 40 To determie the icreasig/decreasig ature of this sequece we will eed to resort to Calculus I techiques. First cosider the followig fuctio ad its derivative. x f ( x) = f ( x) = x+ x+ ( ) We ca see that the first derivative is always positive ad so from Calculus I we kow that the fuctio must the be a icreasig fuctio. So, how does this help us? Notice that, f ( ) = = a + Therefore because < + ad f ( ) x is icreasig we ca also say that, + a = = f < f + = = a fi a < a ( ) ( ) + + I other words, the sequece must be icreasig. Note that ow that we kow the sequece is a icreasig sequece we ca get a better lower boud for the sequece. Sice the sequece is icreasig the first term i the sequece must be the smallest term ad so sice we are startig at = we could also use a lower boud of for this sequece. It is importat to remember that ay umber that is always less tha or equal to all the sequece terms ca be a lower boud. Some are better tha others however. A quick limit will also tell us that this sequece coverges with a limit of. Before movig o to the ext part there is a atural questio that may studets will have at this poit. Why did we use Calculus to determie the icreasig/decreasig ature of the sequece whe we could have just plugged i a couple of s ad quickly determied the same thig? The aswer to this questio is the ext part of this example! [Retur to Problems] (b) Ï Ì Ó = 0 This is a messy lookig sequece, but it eeds to be i order to make the poit of this part. First, otice that, as with the previous part, the sequece terms are all positive ad will all be less tha oe (sice the umerator is guarateed to be less tha the deomiator) ad so the sequece is bouded. 007 Paul Dawkis 75

277 Math 40 Now, let s move o to the icreasig/decreasig questio. As with the last problem, may studets will look at the expoets i the umerator ad deomiator ad determie based o that that sequece terms must decrease. This however, is t a decreasig sequece. Let s take a look at the first few terms to see this. a = ª a = ª a = ª a4 = ª a5 = ª a6 = ª a7 = ª a8 = ª a9 = ª a0 = = The first 0 terms of this sequece are all icreasig ad so clearly the sequece ca t be a decreasig sequece. Recall that a sequece ca oly be decreasig if ALL the terms are decreasig. Now, we ca t make aother commo mistake ad assume that because the first few terms icrease the whole sequece must also icrease. If we did that we would also be mistake as this is also ot a icreasig sequece. This sequece is either decreasig or icreasig. The oly sure way to see this is to do the Calculus I approach to icreasig/decreasig fuctios. I this case we ll eed the followig fuctio ad its derivative. 4 x -x x f ( x) = f 4 ( x) = 4 x x This fuctio will have the followig three critical poits, x= x= ª x=- ª- ( ) ( ) 4 4 0, , Why critical poits? Remember these are the oly places where the fuctio may chage sig! Our sequece starts at = 0 ad so we ca igore the third oe sice it lies outside the values of that we re cosiderig. By pluggig i some test values of x we ca quickly determie that the 4 derivative is positive for 0< x< 0000 ª.6 ad so the fuctio is icreasig i this rage. Likewise, we ca see that the derivative is egative for will be decreasig i this rage. x> ª.6 ad so the fuctio 007 Paul Dawkis 76

278 Math 40 So, our sequece will be icreasig for 0 ad decreasig for. Therefore the fuctio is ot mootoic. Fially, ote that this sequece will also coverge ad has a limit of zero. [Retur to Problems] So, as the last example has show we eed to be careful i makig assumptios about sequeces. Our ituitio will ofte ot be sufficiet to get the correct aswer ad we ca NEVER make assumptios about a sequece based o the value of the first few terms. As the last part has show there are sequeces which will icrease or decrease for a few terms ad the chage directio after that. Note as well that we said first few terms here, but it is completely possible for a sequece to decrease for the first 0,000 terms ad the start icreasig for the remaiig terms. I other words, there is o magical value of for which all we have to do is check up to that poit ad the we ll kow what the whole sequece will do. The oly time that we ll be able to avoid usig Calculus I techiques to determie the icreasig/decreasig ature of a sequece is i sequeces like part (c) of Example. I this case icreasig oly chaged (i fact icreased) the deomiator ad so we were able to determie the behavior of the sequece based o that. I Example however, icreasig icreased both the deomiator ad the umerator. I cases like this there is o way to determie which icrease will wi out ad cause the sequece terms to icrease or decrease ad so we eed to resort to Calculus I techiques to aswer the questio. We ll close out this sectio with a ice theorem that we ll use i some of the proofs later i this chapter. Theorem If { a } is bouded ad mootoic the { } a is coverget. Be careful to ot misuse this theorem. It does ot say that if a sequece is ot bouded ad/or ot mootoic that it is diverget. Example b is a good case i poit. The sequece i that example was ot mootoic but it does coverge. Note as well that we ca make several variats of this theorem. If { a } is bouded above ad icreasig the it coverges ad likewise if { a } is bouded below ad decreasig the it coverges. 007 Paul Dawkis 77

279 Math 40 Series The Basics I this sectio we will itroduce the topic that we will be discussig for the rest of this chapter. That topic is ifiite series. So just what is a ifiite series? Well, let s start with a sequece { a } = The (ote the = is for coveiece, it ca be aythig) ad defie the followig, s = a s = a + a s = a + a + a s = a + a + a + a 4 4 M s = a + a + a + a + L+ a =Âa 4 i i= s are called partial sums ad otice that they will form a sequece, { s } =. Also recall that the S is used to represet this summatio ad called a variety of ames. The most commo ames are : series otatio, summatio otatio, ad sigma otatio. You should have see this otatio, at least briefly, back whe you saw the defiitio of a defiite itegral i Calculus I. If you eed a quick refresher o summatio otatio see the review of summatio otatio i my Calculus I otes. s. = Now back to series. We wat to take a look at the limit of the sequece of partial sums, { } Notatioally we ll defie, lims = limâai = ai Æ Æ i =  i = We will call  ai a ifiite series ad ote that the series starts at i = because that is i= where our origial sequece, { } a =, started. Had our origial sequece started at the our ifiite series would also have started at. The ifiite series will start at the same value that the sequece of terms (as opposed to the sequece of partial sums) starts. s If the sequece of partial sums, { } = ifiite series, Âai i= series is also called diverget., is coverget ad its limit is fiite the we also call the coverget ad if the sequece of partial sums is diveret the the ifiite Note that sometimes it is coveiet to write the ifiite series as, 007 Paul Dawkis 78

280 Math 40 Âai = a+ a + a+ L+ a + L i= We do have to be careful with this however. This implies that a ifiite series is just a ifiite sum of terms ad as well see i the ext sectio this is ot really true. I the ext sectio we re goig to be discussig i greater detail the value of a ifiite series, provided it has oe of course as well as the ideas of covergece ad divergece. This sectio is goig to be devoted mostly to otatioal issues as well as makig sure we ca do some basic maipulatios with ifiite series so we are ready for them whe we eed to be able to deal with them i later sectios. First, we should ote that i most of this chapter we will refer to ifiite series as simply series. If we ever eed to work with both ifiite ad fiite series we ll be more careful with termiology, but i most sectios we ll be dealig exclusively with ifiite series ad so we ll just call them series. Now, i  ai the i is called the idex of summatio or just idex for short ad ote that the i= letter we use to represet the idex does ot matter. So for example the followig series are all the same. The oly differece is the letter we ve used for the idex. = = i + k + +    i= 0 k= 0 = 0 etc. It is importat to agai ote that the idex will start at whatever value the sequece of series terms starts at ad this ca literally be aythig. So far we ve used = 0 ad = but the idex could have started aywhere. I fact, we will usually use  a to represet a ifiite series i which the startig poit for the idex is ot importat. Whe we drop the iitial value of the idex we ll also drop the ifiity from the top so do t forget that it is still techically there. We will be droppig the iitial value of the idex i quite a few facts ad theorems that we ll be seeig throughout this chapter. I these facts/theorems the startig poit of the series will ot affect the result ad so to simplify the otatio ad to avoid givig the impressio that the startig poit is importat we will drop the idex from the otatio. Do ot forget however, that there is a startig poit ad that this will be a ifiite series. Note however, that if we do put a iitial value of the idex o a series i a fact/theorem it is there because it really does eed to be there. 007 Paul Dawkis 79

281 Math 40 Now that some of the otatioal issues are out of the way we eed to start thikig about various ways that we ca maipulate series. We ll start this off with basic arithmetic with ifiite series as we ll eed to be able to do that o occasio. We have the followig properties. Properties If  a ad  b are both coverget series the, 6.  ca, where c is ay umber, is also coverget ad  =  ca c a 7. Âa ± b = k = k  is also coverget ad, a b a b  ±  =  ( ± ). = k = k = k The first property is simply tellig us that we ca always factor a multiplicative costat out of a ifiite series ad agai recall that if we do t put i a iitial value of the idex that the series ca start at ay value. Also recall that i these cases we wo t put a ifiity at the top either. The secod property says that if we add/subtract series all we really eed to do is add/subtract the series terms. Note as well that i order to add/subtract series we eed to make sure that both have the same iitial value of the idex ad the ew series will also start at this value. Before we move o to a differet topic let s discuss multiplicatio of series briefly. We ll start both series at = 0 for a later formula ad the ote that, Ê ˆÊ ÁÂa ÁÂb ˆ  ab Ë Ë ( ) = 0 = 0 = 0 To covice yourself that this is t true cosider the followig product of two fiite sums. ( )( ) + x - 5x+ x = 6-7x- x + x Yeah, it was just the multiplicatio of two polyomials. Each is a fiite sum ad so it makes the poit. I doig the multiplicatio we did t just multiply the costat terms, the the x terms, etc. Istead we had to distribute the through the secod polyomial, the distribute the x through the secod polyomial ad fially combie like terms. 007 Paul Dawkis 80

282 Math 40 Multiplyig ifiite series (eve though we said we ca t thik of a ifiite series as a ifiite sum) eeds to be doe i the same maer. With multiplicatio we re really askig us to do the followig, Ê ˆÊ Á Á ˆ= Ë Ë Â a  b a0 a a a L b0 b b b L = 0 = 0 ( )( ) To do this multiplicatio we would have to distribute the a 0 through the secod term, distribute the a through, etc the combie like terms. This is pretty much impossible sice both series have a ifiite set of terms i them, however the followig formula ca be used to determie the product of two series. Ê ˆÊ Á a Á b ˆ= c Ë Ë Â Â Â where = 0 = 0 = 0 = i i i= 0 c ab - We also ca t say a lot about the covergece of the product. Eve if both of the origial series are coverget it is possible for the product to be diverget. The reality is that multiplicatio of series is a somewhat difficult process ad i geeral is avoided if possible. We will take a brief look at it towards the ed of the chapter whe we ve got more work uder our belt ad we ru across a situatio where it might actually be what we wat to do. Util the, do t worry about multiplyig series. The ext topic that we eed to discuss i this sectio is that of idex shift. To be hoest this is ot a topic that we ll see all that ofte i this course. I fact, we ll use it oce i the ext sectio ad the ot use it agai i all likelihood. Despite the fact that we wo t use it much i this course does t mea however that it is t used ofte i other classes where you might ru across series. So, we will cover it briefly here so that you ca say you ve see it. The basic idea behid idex shifts is to start a series at a differet value for whatever the reaso (ad yes, there are legitimate reasos for doig that). Cosider the followig series, + 5  = Suppose that for some reaso we wated to start this series at = 0, but we did t wat to chage the value of the series. This meas that we ca t just chage the = to = 0 as this would add i two ew terms to the series ad thus chagig its value. Performig a idex shift is a fairly simple process to do. We ll start by defiig a ew idex, say i, as follows, i = Paul Dawkis 8

283 Math 40 Now, whe =, we will get i = 0. Notice as well that if = the i = - =, so oly the lower limit will chage here. Next, we ca solve this for to get, = i+ We ca ow completely rewrite the series i terms of the idex i istead of the idex simply by pluggig i our equatio for i terms of i. + 5 ( i+ ) + 5 i+ 7  =  = i+  i+ = i= 0 i= 0 To fiish the problem out we ll recall that the letter we used for the idex does t matter ad so we ll chage the fial i back ito a to get, = +   = = 0 To covice yourselves that these really are the same summatio let s write out the first couple of terms for each of them,  = L =  + = L = 0 So, sure eough the two series do have exactly the same terms. There is actually a easier way to do a idex shift. The method give above is the techically correct way of doig a idex shift. However, otice i the above example we decreased the iitial value of the idex by ad all the s i the series terms icreased by as well. This will always work i this maer. If we decrease the iitial value of the idex by a set amout the all the other s i the series term will icrease by the same amout. Likewise, if we icrease the iitial value of the idex by a set amout, the all the s i the series term will decrease by the same amout. Let s do a couple of examples usig this shorthad method for doig idex shifts. Example Perform the followig idex shifts. - (a) Write  ar as a series that starts at = 0. = (b) Write  as a series that starts at =. + = - Solutio (a) I this case we eed to decrease the iitial value by ad so the s (okay the sigle ) i the 007 Paul Dawkis 8

284 Math 40 term must icrease by as well. - ( + ) - Â = Â = Â ar ar ar = = 0 = 0 (b) For this problem we wat to icrease the iitial value by ad so all the s i the series term must decrease by. ( ) ( ) - - = = ( ) - - Â Â Â = = - = The fial topic i this sectio is agai a topic that we ll ot be seeig all that ofte i this class, although we will be seeig it more ofte tha the idex shifts. This fial topic is really more about alterate ways to write series whe the situatio requires it. Let s start with the followig series ad ote that the = startig poit is oly for coveiece sice we eed to start the series somewhere. Âa = a+ a + a+ a4 + a5 + L = Notice that if we igore the first term the remaiig terms will also be a series that will start at = istead of = So, we ca rewrite the origial series as follows, Â Â a = a + a = = I this example we say that we ve stripped out the first term. We could have stripped out more terms if we wated to. I the followig series we ve stripped out the first two terms ad the first four terms respectively. Â Â a = a + a + a = = Â a = a + a + a + a + a 4 = = 5 Beig able to strip out terms will, o occasio, simplify our work or allow us to reuse a prior result so it s a importat idea to remember. Notice that i the secod example above we could have also deoted the four terms that we stripped out as a fiite series as follows, Â 4 Â Â Â Â a = a + a + a + a + a = a + a 4 = = 5 = = Paul Dawkis 8

285 Math 40 This is a coveiet otatio whe we are strippig out a large umber of terms or if we eed to strip out a udetermied umber of terms. I geeral, we ca write a series as follows, N Â Â Â a = a + a = = = N+ We ll leave this sectio with a importat warig about termiology. Do t get sequeces ad series cofused! A sequece is a list of umbers writte i a specific order while a ifiite series is a limit of a sequece of fiite series ad hece, if it exists will be a sigle value. So, oce agai, a sequece is a list of umbers while a series is a sigle umber, provided it makes sese to eve compute the series. Studets will ofte cofuse the two ad try to use facts pertaiig to oe o the other. However, sice they are differet beasts this just wo t work. There will be problems where we are usig both sequeces ad series so we ll always have to remember that they are differet. 007 Paul Dawkis 84

286 Math 40 Series Covergece/Divergece I the previous sectio we spet some time gettig familiar with series ad we briefly defied covergece ad divergece. Before worryig about covergece ad divergece of a series we wated to make sure that we ve started to get comfortable with the otatio ivolved i series ad some of the various maipulatios of series that we will, o occasio, eed to be able to do. As oted i the previous sectio most of what we were doig there wo t be doe much i this chapter. So, it is ow time to start talkig about the covergece ad divergece of a series as this will be a topic that we ll be dealig with to oe extet of aother i almost all of the remaiig sectios of this chapter. So, let s recap just what a ifiite series is ad what it meas for a series to be coverget or a = diverget. We ll start with a sequece { } = oly for the sake of coveiece ad it ca, i fact, be aythig. Next we defie the partial sums of the series as, s = a s = a + a s = a + a + a s = a + a + a + a ad these form a ew sequece, { } 4 4 M ad agai ote that we re startig the sequece at s = a + a + a + a + L+ a =Âa 4 i i= s. = A ifiite series, or just series here sice almost every series that we ll be lookig at will be a ifiite series, is the the limit of the partial sums. Or, Â ai = lim s Æ i= If the sequece of partial sums is a coverget sequece (i.e. its limit exists ad is fiite) the the series is also called coverget ad i this case if lim s = s the, Æ Â ai i= = s. Likewise, if the sequece of partial sums is a diverget sequece (i.e. its limit does t exist or is plus or mius ifiity) the the series is also called diverget. Let s take a look at some series ad see if we ca determie if they are coverget or diverget ad see if we ca determie the value of ay coverget series we fid. 007 Paul Dawkis 85

287 Math 40 Example Determie if the followig series is coverget or diverget. If it coverges determie its value. Solutio To determie if the series is coverget we first eed to get our hads o a formula for the geeral term i the sequece of partial sums. s  = =  i i= This is a kow series ad its value ca be show to be, + s =  i = i= ( ) Do t worry if you did t kow this formula (I d be surprised if ayoe kew it ) as you wo t be required to kow it i my course. So, to determie if the series is coverget we will first eed to see if the sequece of partial sums, ( + ) Ï Ì Ó = is coverget or diverget. That s ot terribly difficult i this case. The limit of the sequece terms is, ( + ) lim Æ Therefore, the sequece of partial sums diverges to ad so the series also diverges. So, as we saw i this example we had to kow a fairly obscure formula i order to determie the covergece of this series. I geeral fidig a formula for the geeral term i the sequece of partial sums is a very difficult process. I fact after the ext sectio we ll ot be doig much with the partial sums of series due to the extreme difficulty faced i fidig the geeral formula. This also meas that we ll ot be doig much work with the value of series sice i order to get the value we ll also eed to kow the geeral formula for the partial sums. We will cotiue with a few more examples however, sice this is techically how we determie covergece ad the value of a series. Also, the remaiig examples we ll be lookig at i this sectio will lead us to a very importat fact about the covergece of series. = 007 Paul Dawkis 86

288 Math 40 So, let s take a look at a couple more examples. Example Determie if the followig series coverges or diverges. If it coverges determie its sum. Â = - Solutio This is actually oe of the few series i which we are able to determie a formula for the geeral term i the sequece of partial fractios. However, i this sectio we are more iterested i the geeral idea of covergece ad divergece ad so we ll put off discussig the process for fidig the formula util the ext sectio. The geeral formula for the partial sums is, s = Â = - - i ad i this case we have, i= ( ) Ê ˆ lims = lim Æ Æ Á - - = 4 ( ) Ë + 4 The sequece of partial sums coverges ad so the series coverges also ad its value is, Â = - 4 = Example Determie if the followig series coverges or diverges. If it coverges determie its sum. Â = 0 (- ) Solutio I this case we really do t eed a geeral formula for the partial sums to determie the covergece of this series. Let s just write dow the first few partial sums. s 0 etc. = s = - = 0 s = - + = s = = Paul Dawkis 87

289 Math 40 So, it looks like the sequece of partial sums is, { } { } s =,0,,0,,0,,0,, K = 0 ad this sequece diverges sice lim s does t exist. Therefore, the series also diverges. Æ Example 4 Determie if the followig series coverges or diverges. If it coverges determie its sum. Â - = Solutio Here is the geeral formula for the partial sums for this series. Ê ˆ Ë s = Â = i- Á - i= Agai, do ot worry about kowig this formula. This is ot somethig that you ll ever be asked to kow i my class. I this case the limit of the sequece of partial sums is, Ê ˆ lims = lim Á- Æ Æ = Ë The sequece of partial sums is coverget ad so the series will also be coverget. The value of the series is, Â = - = As we already oted, do ot get excited about determiig the geeral formula for the sequece of partial sums. There is oly goig to be oe type of series where you will eed to determie this formula ad the process i that case is t too bad. I fact, you already kow how to do most of the work i the process as you ll see i the ext sectio. So, we ve determied the covergece of four series ow. Two of the series coverged ad two diverged. Let s go back ad examie the series terms for each of these. For each of the series let s take the limit as goes to ifiity of the series terms (ot the partial sums!!). 007 Paul Dawkis 88

290 Math 40 lim= Æ (- ) this series diverged lim = 0 this series coverged Æ - lim does't exist this series diverged Æ lim = 0 this series coverged Æ - Notice that for the two series that coverged the series term itself was zero i the limit. This will always be true for coverget series ad leads to the followig theorem. Theorem If Âa coverges the lima = 0. Æ Proof First let s suppose that the series starts at =. If it does t the we ca modify thigs as appropriate below. The the partial sums are, -  s = a = a + a + a + a + L+ a s = a = a + a + a + a + L + a + a - i 4 - i 4 - i= i= Next, we ca use these two partial sums to write, a = s - s -  Now because we kow that  a is coverget we also kow that the sequece { s } = is also coverget ad that lim s = s for some fiite value s. However, sice - Æ as Æ we Æ also have lim s - = s. Æ We ow have, ( ) lima = lim s - s = lims - lims = s- s = Æ Æ Æ Æ Be careful to ot misuse this theorem! This theorem gives us a requiremet for covergece but ot a guaratee of covergece. I other words, the coverse is NOT true. If lima = 0 the series may actually diverge! Cosider the followig two series.   = = Æ 007 Paul Dawkis 89

291 Math 40 I both cases the series terms are zero i the limit as goes to ifiity, yet oly the secod series coverges. The first series diverges. It will be a couple of sectios before we ca prove this, so at this poit please believe this ad kow that you ll be able to prove the covergece of these two series i a couple of sectios. Agai, as oted above, all this theorem does is give us a requiremet for a series to coverge. I order for a series to coverge the series terms must go to zero i the limit. If the series terms do ot go to zero i the limit the there is o way the series ca coverge sice this would violate the theorem. This leads us to the first of may tests for the covergece/divergece of a series that we ll be seeig i this chapter. Divergece Test If lima 0 the a will diverge. Æ Â Agai, do NOT misuse this test. This test oly says that a series is guarateed to diverge if the series terms do t go to zero i the limit. If the series terms do happe to go to zero the series may or may ot coverge! Agai, recall the followig two series, Â diverges = Â = coverges Oe of the more commo mistakes that studets make whe they first get ito series is to assume Æ Â that if lima = 0 the a will coverge. There is just o way to guaratee this so be careful! Let s take a quick look at a example of how this test ca be used. Example 5 Determie if the followig series is coverget or diverget. 4 - Â = 0 0+ Solutio With almost every series we ll be lookig at i this chapter the first thig that we should do is take a look at the series terms ad see if they go to zero of ot. If it s clear that the terms do t go to zero use the Divergece Test ad be doe with the problem. That s what we ll do here. 4 - Æ 0+ lim = Paul Dawkis 90

292 Math 40 The limit of the series terms is t zero ad so by the Divergece Test the series diverges. The divergece test is the first test of may tests that we will be lookig at over the course of the ext several sectios. You will eed to keep track of all these tests, the coditios uder which they ca be used ad their coclusios all i oe place so you ca quickly refer back to them as you eed to. Next we should talk briefly revisit arithmetic of series ad covergece/divergece. As we saw i the previous sectio if  a ad  b are both coverget series the so are  ca ad  ( a ± b) = k. Furthermore, these series will have the followig sums or values.  =  Â( ± ) =  ±  ca c a a b a b = k = k = k We ll see a example of this i the ext sectio after we get a few more examples uder our belt. At this poit just remember that a sum of coverget sequeces is coverget ad multiplyig a coverget sequece by a umber will ot chage its covergece. We eed to be a little careful with these facts whe it comes to diverget series. I the first case if  a is diverget the  ca will also be diverget (provided c is t zero of course) sice multiplyig a series that is ifiite i value or does t have a value by a fiite value (i.e. c) wo t chage the fact that the series has a ifiite or o value. However, it is possible to have both  a ad  b be diverget series ad yet have  ( a ± b) be a coverget series. Now, sice the mai topic of this sectio is the covergece of a series we should metio a stroger type of covergece. A series  a is said to coverge absolutely if  a also coverges. Absolute covergece is stroger tha covergece i the sese that a series that is absolutely coverget will also be coverget, but a series that is coverget may or may ot be absolutely coverget. I fact if  a coverges ad  a diverges the series  a is called coditioally coverget. At this poit we do t really have the tools at had to properly ivestigate this topic i detail or do we have the tools i had to determie if a series is absolutely coverget or ot. So we ll ot say aythig more about this subject for a while. Whe we fially have the tools i had to discuss this topic i more detail we will revisit it. Util the do t worry about it. The idea is = k 007 Paul Dawkis 9

293 Math 40 metioed here oly because we were already discussig covergece i this sectio ad it ties ito the last topic that we wat to discuss i this sectio. I the previous sectio after we d itroduced the idea of a ifiite series we commeted o the fact that we should t thik of a ifiite series as a ifiite sum despite the fact that the otatio we use for ifiite series seems to imply that it is a ifiite sum. It s ow time to briefly discuss this. First, we eed to itroduce the idea of a rearragemet. A rearragemet of a series is exactly what it might soud like, it is the same series with the terms rearraged ito a differet order. For example, cosider the followig the ifiite series. A rearragemet of this series is, Âa = a+ a + a+ a4 + a5 + a6 + a7 + L = Âa = a + a+ a + a4 + a5 + a9 + a4 + L = The issue we eed to discuss here is that for some series each of these arragemets of terms ca have a differet values despite the fact that they are usig exactly the same terms. Here is a example of this. It ca be show that, + (-) = L = l () Â = Sice this series coverges we kow that if we multiply it by a costat c its value will also be multiplied by c. So, let s multiply this by to get, L = l () Now, let s add i a zero betwee each term as follows L = l () Note that this wo t chage the value of the series because the partial sums for this series will be the partial sums for the () except that each term will be repeated. Repeatig terms i a series will ot affect its limit however ad so both () ad () will be the same. We kow that if two series coverge we ca add them by addig term by term ad so add () ad () to get, 007 Paul Dawkis 9

294 Math L = l (4) Now, otice that the terms of (4) are simply the terms of () rearraged so that each egative term comes after two positive terms. The values however are defiitely differet despite the fact that the terms are the same. Note as well that this is ot oe of those tricks that you see occasioally where you get a cotradictory result because of a hard to spot math/logic error. This is a very real result ad we ve ot made ay logic mistakes/errors. Here is a ice set of facts that gover this idea of whe a rearragemet will lead to a differet value of a series. Facts Give the seriesâ a,. If  a is absolutely coverget ad its value is s the ay rearragemet of  a will also have a value of s. Â. If a is coditioally coverget ad r is ay real umber the there is a rearragemet of  a whose value will be r. Agai, we do ot have the tools i had yet to determie if a series is absolutely coverget ad so do t worry about this at this poit. This is here just to make sure that you uderstad that we have to be very careful i thikig of a ifiite series as a ifiite sum. There are times whe we ca (i.e. the series is absolutely coverget) ad there are times whe we ca t (i.e. the series is coditioally coverget). As a fial ote, the fact above tells us that the series,  = (-) + must be coditioally coverget sice two rearragemets gave two separate values of this series. Evetually it will be very simple to show that this series coditioally coverget. 007 Paul Dawkis 9

295 Math 40 Series Special Series I this sectio we are goig to take a brief look at three special series. Actually, special may ot be the correct term. All three have bee amed which makes them special i some way, however the mai reaso that we re goig to look at two of them i this sectio is that they are the oly types of series that we ll be lookig at for which we will be able to get actual values for the series. The third type is diverget ad so wo t have a value to worry about. I geeral, determiig the value of a series is very difficult ad outside of these two kids of series that we ll look at i this sectio we will ot be determiig the value of series i this chapter. So, let s get started. Geometric Series A geometric series is ay series that ca be writte i the form, -  = or, with a idex shift the geometric series will ofte be writte as, = 0 ar These are idetical series ad will have idetical values, provided they coverge of course. If we start with the first form it ca be show that the partial sums are, s  ( - ) ar a r a ar = = - -r -r -r The series will coverge provided the partial sums form a coverget sequece, so let s take the limit of the partial sums. Ê a ar ˆ lims = lim Á - Æ Æ Ë -r -r a ar = lim -lim Æ -r Æ -r a a = - lim r -r -r Æ Now, from Theorem from the Sequeces sectio we kow that the limit above will exist ad be fiite provided - < r. However, ote that we ca t let r = sice this will give divisio by zero. Therefore, this will exist ad be fiite provided - < r < ad i this case the limit is zero ad so we get, 007 Paul Dawkis 94

296 Math 40 lim s Æ a = - r Therefore, a geometric series will coverge if - < r <, which is usually writte r <, its value is, a = = - r - Âar  ar = = 0 Note that i usig this formula we ll eed to make sure that we are i the correct form. I other words, if the series starts at = 0 the the expoet o the r must be. Likewise if the series starts at = the the expoet o the r must be -. Example Determie if the followig series coverge or diverge. If they coverge give the value of the series. Solutio (a) (a) (b)  9 4 [Solutio] =  = 9 4 (-4)  [Solutio] - = 0 5 This series does t really look like a geometric series. However, otice that both parts of the series term are umbers raised to a power. This meas that it ca be put ito the form of a geometric series. We will just eed to decide which form is the correct form. Sice the series starts at = we will wat the expoets o the umbers to be -. It will be fairly easy to get this ito the correct form. Let s first rewrite thigs slightly. Oe of the s i the expoet has a egative i frot of it ad that ca t be there i the geometric form. So, let s first get rid of that ( -) + Â9 4 = Â9 4 =  = = = Now let s get the correct expoet o each of the umbers. This ca be doe usig simple expoet properties. Now, rewrite the term a little Â9 4 =  = -  - - = = 9 = Paul Dawkis 95

297 Math Ê ˆ 9 4 = 6 9 = 44Á 9 Ë Â Â ( ) - Â = = = 4 So, this is a geometric series with a = 44 ad r = <. Therefore, sice r < we kow the 9 series will coverge ad its value will be, Â 9 4 = = ( 44) = 4 = [Retur to Problems] (b) ( 4) - Â = Agai, this does t look like a geometric series, but it ca be put ito the correct form. I this case the series starts at = 0 so we ll eed the expoets to be o the terms. Note that this meas we re goig to eed to rewrite the expoet o the umerator a little ( ) (-4) ( ) ( ) Ê-64 ˆ = = 5 = 5Á Ë 5 Â Â Â Â - - = 0 = 0 = 0 = 0 64 So, we ve got it ito the correct form ad we ca see that a = 5 ad r =-. Also ote that r ad so this series diverges. 5 [Retur to Problems] Back i the Series Basics sectio we talked about strippig out terms from a series, but did t really provide ay examples of how this idea could be used i practice. We ca ow do some examples. Example Use the results from the previous example to determie the value of the followig series. Solutio (a) (a) (b) Â 9 4 [Solutio] = Â 9 4 [Solutio] = Â = I this case we could just ackowledge that this is a geometric series that starts at = 0 ad so 007 Paul Dawkis 96

298 Math 40 we could put it ito the correct form ad be doe with it. However, this does provide us with a ice example of how to use the idea of strippig out terms to our advatage. Let s otice that if we strip out the first term from this series we arrive at,    9 4 = = = 0 = = From the previous example we kow the value of the ew series that arises here ad so the value of the series i this example is,  9 4 = 4 + = = [Retur to Problems] (b)  = 9 4 I this case we ca t strip out terms from the give series to arrive at the series used i the previous example. However, we ca start with the series used i the previous example ad strip terms out of it to get the series i this example. So, let s do that. We will strip out the first two terms from the series we looked at i the previous example    9 4 = = = = = We ca ow use the value of the series from the previous example to get the value of this series Â9 4 =  = - 08 = = = 5 5 [Retur to Problems] Notice that we did t discuss the covergece of either of the series i the above example. Here s why. Cosider the followig series writte i two separate ways (i.e. we stripped out a couple of terms from it). Let s suppose that we kow  a =   a = a + a + a + a 0 = 0 = is a coverget series. This meas that it s got a fiite value ad addig three fiite terms oto this will ot chage that fact. So the value of fiite ad so is coverget.  a = 0 is also 007 Paul Dawkis 97

299 Math 40 Likewise, suppose that  a = 0 this value we will remai fiite ad arrive at the value of so this series will also be coverget. is coverget. I this case if we subtract three fiite values from  a =. This is ow a fiite value ad I other words, if we have two series ad they differ oly by the presece, or absece, of a fiite umber of fiite terms they will either both be coverget or they will both be diverget. The differece of a few terms oe way or the other will ot chage the covergece of a series. This is a importat idea ad we will use it several times i the followig sectios to simplify some of the tests that we ll be lookig at. Telescopig Series It s ow time to look at the secod of the three series i this sectio. I this portio we are goig to look at a series that is called a telescopig series. The ame i this case comes from what happes with the partial sums ad is best show i a example. Example Determie if the followig series coverges or diverges. If it coverges fid its value.  = Solutio We first eed the partial sums for this series. s =  i + i+ Now, let s otice that we ca use partial fractios o the series term to get, = = - i + i+ i+ i+ i+ i+ i= 0 ( )( ) I ll leave the details of the partial fractios to you. By ow you should be fairly adept at this sice we spet a fair amout of time doig partial fractios back i the Itegratio Techiques chapter. If you eed a refresher you should go back ad review that sectio. So, what does this do for us? Well, let s start writig out the terms of the geeral partial sum for this series usig the partial fractio form. 007 Paul Dawkis 98

300 Math 40 s Ê ˆ = ÂÁ - i= 0 Ëi+ i+ Ê Ê Ê Ê = Á - ˆ+ Á - ˆ+ Á - ˆ+ L + Ë Ë Ë 4 - Ê ˆ Á ˆ+ Á - Ë + Ë + + = - + Notice that every term except the first ad last term caceled out. This is the origi of the ame telescopig series. This also meas that we ca determie the covergece of this series by takig the limit of the partial sums. Ê ˆ lims = limá- = Æ Æ Ë + The sequece of partial sums is coverget ad so the series is coverget ad has a value of  = = I telescopig series be careful to ot assume that successive terms will be the oes that cacel. Cosider the followig example. Example 4 Determie if the followig series coverges or diverges. If it coverges fid its value.  = + 4+ Solutio As with the last example we ll leave the partial fractios details to you to verify. The partial sums are, s Ê ˆ Ê ˆ = ÂÁ - i= i i = ÂÁ - Ë + + i= Ëi+ i+ ÈÊ Ê Ê Ê = ÍÁ - ˆ+ Á - ˆ+ Á - ˆ+ L + Á ÍÎË 4 Ë 5 Ë 4 6 Ë È = Í Î + + ˆ Ê ˆ - + Á - + Ë + + I this case istead of successive terms cacelig a term will cacel with a term that is farther dow the list. The ed result this time is two iitial ad two fial terms are left. Notice as well that i order to help with the work a little we factored the out of the series. 007 Paul Dawkis 99

301 Math 40 The limit of the partial sums is, Ê5 ˆ 5 lims = lim Á - - = Æ Æ Ë So, this series is coverget (because the partial sums form a coverget sequece) ad its value is, 5 Â = = + 4+ Note that it s ot always obvious if a series is telescopig or ot util you try to get the partial sums ad the see if they are i fact telescopig. There is o test that will tell us that we ve got a telescopig series right off the bat. Also ote that just because you ca do partial fractios o a series term does ot mea that the series will be a telescopig series. The followig series, for example, is ot a telescopig series despite the fact that we ca partial fractio the series terms. + Ê ˆ = Á Ë+ + Â Â = = I order for a series to be a telescopig we must get terms to cacel ad all of these terms are positive ad so oe will cacel. Next, we eed to go back ad address a issue that was first raised i the previous sectio. I that sectio we stated that the sum or differece of coverget series was also coverget ad that the presece of a multiplicative costat would ot affect the covergece of a series. Now that we have a few more series i had let s work a quick example showig that. Example 5 Determie the value of the followig series. Ê ˆ Â Á = Ë + 4+ Solutio To get the value of this series all we eed to do is rewrite it ad the use the previous results. Ê ˆ ÂÁ = Â - Â = Ë + 4+ = + 4+ = = 4Â Â + 4+ = = Ê 5 ˆ 96 = 4Á - Ë 5 86 =- 5 We did t discuss the covergece of this series because it was the sum of two coverget series 007 Paul Dawkis 00

302 Math 40 ad that guarateed that the origial series would also be coverget. Harmoic Series This is the third ad fial series that we re goig to look at i this sectio. Here is the harmoic series. Â = The harmoic series is diverget ad we ll eed to wait util the ext sectio to show that. This series is here because it s got a ame ad so I wated to put it here with the other two amed series that we looked at i this sectio. We re also goig to use the harmoic series to illustrate a couple of ideas about diverget series that we ve already discussed for coverget series. We ll do that with the followig example. Example 6 Show that each of the followig series are diverget. 5 (a) Â = (b) Â = 4 Solutio 5 (a) Â = To see that this series is diverget all we eed to do is use the fact that we ca factor a costat out of a series as follows, 5 Â = 5Â = = Now, Â is diverget ad so five times this will still ot be a fiite umber ad so the series = has to be diverget. I other words, if we multiply a diverget series by a costat it will still be diverget. (b) Â = 4 I this case we ll start with the harmoic series ad strip out the first three terms. Ê ˆ = fi = Á - Ë 6 Â Â Â Â = = 4 = 4 = I this case we are subtractig a fiite umber from a diverget series. This subtractio will ot chage the divergece of the series. We will either have ifiity mius a fiite umber, which is still ifiity, or a series with o value mius a fiite umber, which will still have o value. 007 Paul Dawkis 0

303 Math 40 Therefore, this series is diverget. Just like with coverget series, addig/subtractig a fiite umber from a diverget series is ot goig to chage the fact the covergece of the series. So, some geeral rules about the covergece/divergece of a series are ow i order. Multiplyig a series by a costat will ot chage the covergece/divergece of the series ad addig or subtractig a costat from a series will ot chage the covergece/divergece of the series. These are ice ideas to keep i mid. 007 Paul Dawkis 0

304 Math 40 Vectors Itroductio This is a fairly short chapter. We will be takig a brief look at vectors ad some of their properties. We will eed some of this material i the ext chapter ad those of you headig o towards Calculus III will use a fair amout of this there as well. Here is a list of topics i this chapter. Vectors The Basics I this sectio we will itroduce some of the basic cocepts about vectors. Vector Arithmetic Here we will give the basic arithmetic operatios for vectors. Dot Product We will discuss the dot product i this sectio as well as a applicatio or two. Cross Product I this sectio we ll discuss the cross product ad see a quick applicatio. 007 Paul Dawkis 0

305 Math 40 Vectors The Basics Let s start this sectio off with a quick discussio o what vectors are used for. Vectors are used to represet quatities that have both a magitude ad a directio. Good examples of quatities that ca be represeted by vectors are force ad velocity. Both of these have a directio ad a magitude. Let s cosider force for a secod. A force of say 5 Newtos that is applied i a particular directio ca be applied at ay poit i space. I other words, the poit where we apply the force does ot chage the force itself. Forces are idepedet of the poit of applicatio. To defie a force all we eed to kow is the magitude of the force ad the directio that the force is applied i. The same idea holds more geerally with vectors. Vectors oly impart magitude ad directio. They do t impart ay iformatio about where the quatity is applied. This is a importat idea to always remember i the study of vectors. I a graphical sese vectors are represeted by directed lie segmets. The legth of the lie segmet is the magitude of the vector ad the directio of the lie segmet is the directio of the vector. However, because vectors do t impart ay iformatio about where the quatity is applied ay directed lie segmet with the same legth ad directio will represet the same vector. Cosider the sketch below. Each of the directed lie segmets i the sketch represets the same vector. I each case the vector starts at a specific poit the moves uits to the left ad 5 uits up. The otatio that we ll use for this vector is, r v = -,5 ad each of the directed lie segmets i the sketch are called represetatios of the vector. 007 Paul Dawkis 04

306 Math 40 Be careful to distiguish vector otatio, -,5, from the otatio we use to represet coordiates of poits, (-,5). The vector deotes a magitude ad a directio of a quatity while the poit deotes a locatio i space. So do t mix the otatios up! r A represetatio of the vector v = a, a uuur AB, from the poit A= ( xy, ) to the poit B ( x a, y a ) r the vector v = a, a, a i two dimesioal space is ay directed lie segmet, = + +. Likewise a represetatio of uuur i three dimesioal space is ay directed lie segmet, AB, from the poit A= ( xyz,, ) to the poit B ( x a, y a, z a ) = Note that there is very little differece betwee the two dimesioal ad three dimesioal formulas above. To get from the three dimesioal formula to the two dimesioal formula all we did is take out the third compoet/coordiate. Because of this most of the formulas here are give oly i their three dimesioal versio. If we eed them i their two dimesioal form we ca easily modify the three dimesioal form. There is oe represetatio of a vector that is special i some way. The represetatio of the r vector v = a, a, a that starts at the poit ( 0,0,0) B= a, a, a is called the positio vector of the poit (,, ) A = ad eds at the poit ( ) we are specifyig the iitial ad fial poit of the vector. a a a. So, whe we talk about positio vectors Positio vectors are useful if we ever eed to represet a poit as a vector. As we ll see there are times i which we defiitely are goig to wat to represet poits as vectors. I fact, we re goig to ru ito topics that ca oly be doe if we represet poits as vectors. Next we eed to discuss briefly how to geerate a vector give the iitial ad fial poits of the represetatio. Give the two poits A= ( a, a, a ) ad B ( b, b, b ) uuur represetatio AB is, r v = b -a, b -a, b -a = the vector with the Note that we have to be very careful with directio here. The vector above is the vector that starts uuur at A ad eds at B. The vector that starts at B ad eds at A, i.e. with represetatio BA is, r w= a -b, a -b, a -b These two vectors are differet ad so we do eed to always pay attetio to what poit is the startig poit ad what poit is the edig poit. Whe determiig the vector betwee two poits we always subtract the iitial poit from the termial poit. 007 Paul Dawkis 05

307 Math 40 Example Give the vector for each of the followig. (a) The vector from (, - 7,0) to (, -,- 5). (b) The vector from (, -,- 5) to(, - 7,0). (c) The positio vector for (- 90,4) Solutio (a) Remember that to costruct this vector we subtract coordiates of the startig poit from the edig poit. (b) Same thig here. -, --(-7, ) -5-0 = -,4,- 5 -, -7-(-,0 ) -(- 5) =, - 4,5 Notice that the oly differece betwee the first two is the sigs are all opposite. This differece is importat as it is this differece that tells us that the two vectors poit i opposite directios. (c) Not much to this oe other tha ackowledgig that the positio vector of a poit is othig more tha a vector with the poits coordiates as its compoets. - 90,4 We ow eed to start discussig some of the basic cocepts that we will ru ito o occasio. Magitude r The magitude, or legth, of the vector v = a, a, a is give by, v = a + a + a Example Determie the magitude of each of the followig vectors. r (a) a =, -5,0 r (b) u =, r (c) w = 0,0 r (d) i =,0,0 Solutio There is t too much to these other tha plug ito the formula. r (a) a = = 4 (b) r 4 u = + = = 5 5 r (c) w = 0+ 0 = 0 r (d) i = = 007 Paul Dawkis 06

308 Math 40 We also have the followig fact about the magitude. r r If a = 0 the a = 0 r This should make sese. Because we square all the compoets the oly way we ca get zero out of the formula was for the compoets to be zero i the first place. Uit Vector r Ay vector with magitude of, i.e. u =, is called a uit vector. Example Which of the vectors from Example are uit vectors? Solutio Both the secod ad fourth vectors had a legth of ad so they are the oly uit vectors from the first example. Zero Vector r The vector w = 0,0 that we saw i the first example is called a zero vector sice its compoets are all zero. Zero vectors are ofte deoted by 0 r. Be careful to distiguish 0 (the umber) from 0 r (the vector). The umber 0 deote the origi i space, while the vector 0 r deotes a vector that has o magitude or directio. Stadard Basis Vectors r The fourth vector from the secod example, i =,0,0 three dimesioal space there are three stadard basis vectors, r r r i =,0,0 j = 0,,0 k = 0,0, I two dimesioal space there are two stadard basis vectors, r r i =,0 j = 0, Note that stadard basis vectors are also uit vectors., is called a stadard basis vector. I Warig We are pretty much doe with this sectio however, before proceedig to the ext sectio we should poit out that vectors are ot restricted to two dimesioal or three dimesioal space. Vectors ca exist i geeral -dimesioal space. The geeral otatio for a -dimesioal vector is, 007 Paul Dawkis 07

309 Math 40 r v = a, a, a, K, a ad each of the a i s are called compoets of the vector. Because we will be workig almost exclusively with two ad three dimesioal vectors i this course most of the formulas will be give for the two ad/or three dimesioal cases. However, most of the cocepts/formulas will work with geeral vectors ad the formulas are easily (ad aturally) modified for geeral -dimesioal vectors. Also, because it is easier to visualize thigs i two dimesios most of the figures related to vectors will be two dimesioal figures. So, we eed to be careful to ot get too locked ito the two or three dimesioal cases from our discussios i this chapter. We will be workig i these dimesios either because it s easier to visualize the situatio or because physical restrictios of the problems will eforce a dimesio upo us. 007 Paul Dawkis 08

310 Math 40 Vector Arithmetic I this sectio we eed to have a brief discussio of vector arithmetic. r We ll start with additio of two vectors. So, give the vectors a = a, a, a ad r b = b, b, b the additio of the two vectors is give by the followig formula. r r a+ b = a + b, a + b, a + b The followig figure gives the geometric iterpretatio of the additio of two vectors. This is sometimes called the parallelogram law or triagle law. r Computatioally, subtractio is very similar. Give the vectors a = a, a, a r b = b, b, b the differece of the two vectors is give by, ad r r a- b = a -b, a -b, a -b Here is the geometric iterpretatio of the differece of two vectors. 007 Paul Dawkis 09

311 Math 40 It is a little harder to see this geometric iterpretatio. To help see this let s istead thik of subtractio as the additio of a r ad -b r. First, as we ll see i a bit -b r is the same vector as b r with opposite sigs o all the compoets. I other words, a r + - b r. b r. Here is the vector set up for ( ) -b r goes i the opposite directio as r + - r As we ca see from this figure we ca move the vector represetig a ( b) we ve got i the first figure showig the differece of the two vectors. to the positio Note that we ca t add or subtract two vectors uless they a have the same umber of compoets. If they do t have the same umber of compoets the additio ad subtractio ca t be doe. The ext arithmetic operatio that we wat to look at is scalar multiplicatio. Give the vector r a = a, a, a ad ay umber c the scalar multiplicatio is, r ca = ca, ca, ca 007 Paul Dawkis 0

312 Math 40 So, we multiply all the compoets by the costat c. To see the geometric iterpretatio of scalar multiplicatio let s take a look at a example. r Example For the vector a =,4 the same axis system. compute a r, ar ad -a r. Graph all four vectors o Solutio Here are the three scalar multiplicatios. a r = 6, a r =, - a r = -4,-8 Here is the graph for each of these vectors. I the previous example we ca see that if c is positive all scalar multiplicatio will do is stretch (if c > ) or shrik (if c < ) the origial vector, but it wo t chage the directio. Likewise, if c is egative scalar multiplicatio will switch the directio so that the vector will poit i exactly the opposite directio ad it will agai stretch or shrik the magitude of the vector depedig upo the size of c. There are several ice applicatios of scalar multiplicatio that we should ow take a look at. The first is parallel vectors. This is a cocept that we will see quite a bit over the ext couple of sectios. Two vectors are parallel if they have the same directio or are i exactly opposite directios. Now, recall agai the geometric iterpretatio of scalar multiplicatio. Whe we performed scalar multiplicatio we geerated ew vectors that were parallel to the origial vectors (ad each other for that matter). So, let s suppose that a r ad b r are parallel vectors. If they are parallel the there must be a umber c so that, r a = cb r So, two vectors are parallel if oe is a scalar multiple of the other. 007 Paul Dawkis

313 Math 40 Example Determie if the sets of vectors are parallel or ot. r r (a) a =, - 4,, b = -6,,- r r (b) a = 4,0, b =, -9 Solutio r r (a) These two vectors are parallel sice b =-a (b) These two vectors are t parallel. This ca be see by oticig that ( ) ( ) 0 = 5-9. I other words we ca t make a r be a scalar multiple of b r. 4 = ad yet The ext applicatio is best see i a example. r Example Fid a uit vector that poits i the same directio as w = -5,,. Solutio Okay, what we re askig for is a ew parallel vector (poits i the same directio) that happes to be a uit vector. We ca do this with a scalar multiplicatio sice all scalar multiplicatio does is chage the legth of the origial vector (alog with possibly flippig the directio to the opposite directio). Here s what we ll do. First let s determie the magitude of w r. r w = = 0 Now, let s form the followig ew vector, r r 5 u = r w= - 5,, = -,, w The claim is that this is a uit vector. That s easy eough to check r u = + + = = This vector also poits i the same directio as w r sice it is oly a scalar multiple of w r ad we used a positive multiple. So, i geeral, give a vector w r, as w r. r u = r w r will be a uit vector that poits i the same directio w 007 Paul Dawkis

314 Math 40 Stadard Basis Vectors Revisited I the previous sectio we itroduced the idea of stadard basis vectors without really discussig why they were importat. We ca ow do that. Let s start with the vector r a = a, a, a We ca use the additio of vectors to break this up as follows, r a = a, a, a = a,0,0 + 0, a,0 + 0,0, a Usig scalar multiplicatio we ca further rewrite the vector as, r a = a,0,0 + 0, a,0 + 0,0, a = a,0,0 + a 0,,0 + a 0,0, Fially, otice that these three ew vectors are simply the three stadard basis vectors for three dimesioal space. r r r a, a, a = ai + a j + ak So, we ca take ay vector ad write it i terms of the stadard basis vectors. From this poit o we will use the two otatios iterchageably so make sure that you ca deal with both otatios. r Example 4 If a =, -9, r ad w=- i r + 8k r compute a r -w r. Solutio I order to do the problem we ll covert to oe otatio ad the perform the idicated operatios. r r a- w=, -9, - -,0,8 = 6, -8, - -,0,4 = 9, -8,- We will leave this sectio with some basic properties of vector arithmetic. Properties If v r, w r ad u r are vectors (each with the same umber of compoets) ad a ad b are two umbers the we have the followig properties. r r r r r r r r r r v+ w= w+ v u+ ( v+ w) = ( u+ v) + w r r r r r v+ 0= v v = v r r r r r r r a v+ w = av+ aw a+ b v = av+ bv ( ) ( ) 007 Paul Dawkis

315 Math 40 The proofs of these are pretty much just computatio proofs so we ll prove oe of them ad leave the others to you to prove. r r r r Proof of a( v+ w) = av+ aw r r We ll start with the two vectors, v = v, v, K, v ad w= w, w, K, w ad yes we did mea for these to each have compoets. The theorem works for geeral vectors so we may as well do the proof for geeral vectors. Now, as oted above this is pretty much just a computatioal proof. What that meas is that we ll compute the left side ad the do some basic arithmetic o the result to show that we ca make the left side look like the right side. Here is the work. r r a v+ w = a v, v, K, v + w, w, K, w ( ) ( ) = a v + w, v + w, K, v + w ( ), ( ), K, ( ) = a v + w a v + w a v + w = av + aw, av + aw, K, av + aw = av, av, K, av + aw, aw, K, aw r r = a v, v, K, v + a w, w, K, w = av+ aw 007 Paul Dawkis 4

316 Math 40 Dot Product The ext topic for discussio is that of the dot product. Let s jump right ito the defiitio of the r r dot product. Give the two vectors a = a, a, a ad b = b, b, b the dot product is, ab r r g = ab + ab + ab () Sometimes the dot product is called the scalar product. The dot product is also a example of a ier product ad so o occasio you may hear it called a ier product. Example Compute the dot product for each of the followig. r r r r r r (a) v = 5i - 8 j, w= i + j r r (b) a = 0,, - 7, b =,, Solutio Not much to do with these other tha use the formula. r r (a) vw= g 5-6=- r r (b) ab= g = Here are some properties of the dot product. Properties r r r rr r r r r r r r r ug v w uv g uw g cv gw vg cw c vw g r r rr r r vw g = wv g vg0= 0 rr r rr r r vv g = v If vv g = 0 the v = 0 ( + ) = + ( ) = ( ) = ( ) The proofs of these properties are mostly computatioal proofs ad so we re oly goig to do a couple of them ad leave the rest to you to prove. Proof of u r g( v r + w r ) = uv rr g + uw r g r r r We ll start with the three vectors, u = u, u, K, u, v = v, v, K, v ad r w= w, w, K, w ad yes we did mea for these to each have compoets. The theorem works for geeral vectors so we may as well do the proof for geeral vectors. Now, as oted above this is pretty much just a computatioal proof. What that meas is that we ll compute the left side ad the do some basic arithmetic o the result to show that we ca make the left side look like the right side. Here is the work. 007 Paul Dawkis 5

317 Math 40 r r r ug v w u u K u g v v K v w w K w ( + ) =,,, (,,, +,,, ) = u, u, K, u g v + w, v + w, K, v + w ( ), ( ), K, ( ) = u v + w u v + w u v + w = uv + uw, uv + uw, K, uv + uw = uv, uv, K, uv + uw, uw, K, uw = u, u, K, u g v, v, K, v + u, u, K, u g w, w, K, w rr r r = uv g + uw g rr r r Proof of : If vv= g 0 the v = 0 r This is a pretty simple proof. Let s start with v = v, v, K, v ad compute the dot product. rr vv g = v, v, K, v g v, v, K, v = 0 = v + v + L + v Now, sice we kow vi 0 for all i the the oly way for this sum to be zero is to i fact have r v = 0. This i tur however meas that we must have v = 0 ad so we must have had v = 0 r. i i There is also a ice geometric iterpretatio to the dot product. First suppose that q is the agle betwee a r ad b r such that 0 q p as show i the image below. We ca the have the followig theorem. 007 Paul Dawkis 6

318 Math 40 Theorem r r r r ab g = a b cosq () Proof Let s give a modified versio of the sketch above. The three vectors above form the triagle AOB ad ote that the legth of each side is othig more tha the magitude of the vector formig that side. The Law of Cosies tells us that, r r r r r r a- b = a + b - a b cosq Also usig the properties of dot products we ca write the left side as, r r r r r r a- b = ( a-b) g( a-b) rr r r r r rr = aa g -ab g - ba g + bb g r r r r = a - ab g + b Our origial equatio is the, 007 Paul Dawkis 7

319 Math 40 r r a- b = r r r a + b - a r b cosq r a r r r - ab g + b = r r r a + b - a r b cosq r r r - ab g =- a r b cosq r r ab g = r a r b cosq The formula from this theorem is ofte used ot to compute a dot product but istead to fid the agle betwee two vectors. Note as well that while the sketch of the two vectors i the proof is for two dimesioal vectors the theorem is valid for vectors of ay dimesio (as log as they have the same dimesio of course). Let s see a example of this. r Example Determie the agle betwee a =, -4,- r ad b = 0,5,. Solutio We will eed the dot product as well as the magitudes of each vector. r r r r ab g =- a = 6 b = 9 The agle is the, r ab g - cosq = rr r = = a b 6 9 ( ) q = - = - cos radias=4.4 degrees The dot product gives us a very ice method for determiig if two vectors are perpedicular ad it will give aother method for determiig whe two vectors are parallel. Note as well that ofte we will use the term orthogoal i place of perpedicular. Now, if two vectors are orthogoal the we kow that the agle betwee them is 90 degrees. From () this tells us that if two vectors are orthogoal the, r r ab=0 g 007 Paul Dawkis 8

320 Math 40 Likewise, if two vectors are parallel the the agle betwee them is either 0 degrees (poitig i the same directio) or 80 degrees (poitig i the opposite directio). Oce agai usig () this would mea that oe of the followig would have to be true. r r r r r r r r ab g = a b = ab g =- a b = ( q 0 ) OR ( q 80 ) Example Determie if the followig vectors are parallel, orthogoal, or either. r r (a) a = 6, -,-, b =,5, r r r r r r (b) u = i - j, v =- i + j 4 Solutio (a) First get the dot product to see if they are orthogoal. r r ab= g -0- = 0 The two vectors are orthogoal. (b) Agai, let s get the dot product first. rr 5 uv=- g - =- 4 4 So, they are t orthogoal. Let s get the magitudes ad see if they are parallel. Now, otice that, r u r 5 5 = 5 v = = 6 4 rr 5 5 uv 5 Ê ˆ r r g =- =- =- u v 4 Á 4 Ë So, the two vectors are parallel. There are several ice applicatios of the dot product as well that we should look at. Projectios The best way to uderstad projectios is to see a couple of sketches. So, give two vectors a r ad b r we wat to determie the projectio of b r oto a r r. The projectio is deoted by projr a b. Here are a couple of sketches illustratig the projectio. 007 Paul Dawkis 9

321 Math 40 So, to get the projectio of b r oto a r we drop straight dow from the ed of b r util we hit (ad form a right agle) with the lie that is parallel to a r. The projectio is the the vector that is parallel to a r, starts at the same poit both of the origial vectors started at ad eds where the dashed lie hits the lie parallel to a r. There is a ice formula for fidig the projectio of b r oto a r. Here it is, r r r g r projr a b = a ab r a Note that we also eed to be very careful with otatio here. The projectio of a r oto b r is give by r r r g r projr b a = b ab r b We ca see that this will be a totally differet vector. This vector is parallel to b r r, while projr a b is parallel to a r. So, be careful with otatio ad make sure you are fidig the correct projectio. Here s a example. r Example 4 Determie the projectio of b =,, - r oto a =,0, -. Solutio We eed the dot product ad the magitude of a r. r r r ab g = 4 a = 5 The projectio is the, 007 Paul Dawkis 0

322 Math 40 proj 4 =,0, =,0,- 5 5 For compariso purposes let s do it the other way aroud as well. r a r r r ab g r b = r a a r Example 5 Determie the projectio of a =,0, - r otob =,, -. Solutio We eed the dot product ad the magitude of b r. r r r ab g = 4 b = 6 The projectio is the, proj r b r a = r r ab g r r b b 4 =,, =,,- As we ca see from the previous two examples the two projectios are differet so be careful. Directio Cosies This applicatio of the dot product requires that we be i three dimesioal space ulike all the other applicatios we ve looked at to this poit. Let s start with a vector, a r, i three dimesioal space. This vector will form agles with the x- axis (a ), the y-axis (b ), ad the z-axis (g ). These agles are called directio agles ad the cosies of these agles are called directio cosies. Here is a sketch of a vector ad the directio agles. 007 Paul Dawkis

323 Math 40 The formulas for the directio cosies are, r r r r r r ai g a agj a ak g a a = r = r b = r = r g = r = r a a a a a a cos cos cos where i r, j r ad k r are the stadard basis vectors. Let s verify the first dot product above. We ll leave the rest to you to verify. r r ai g = a, a, a g,0,0 = a Here are a couple of ice facts about the directio cosies. r. The vector u = cos a,cos b,cosg cos a + cos b + cos g =. r r. a = a cos a,cos b,cosg is a uit vector. Let s do a quick example ivolvig directio cosies. r Example 6 Determie the directio cosies ad directio agles for a =,, -4. Solutio We will eed the magitude of the vector. r a = = 007 Paul Dawkis

324 Math 40 The directio cosies ad agles are the, cosa = a =.9 radias= 64. degrees cosb = b =.5 radias= degrees cosg = -4 g =.6 radias= degrees 007 Paul Dawkis

325 Math 40 Cross Product I this fial sectio of this chapter we will look at the cross product of two vectors. We should ote that the cross product requires both of the vectors to be three dimesioal vectors. Also, before gettig ito how to compute these we should poit out a major differece betwee dot products ad cross products. The result of a dot product is a umber ad the result of a cross product is a vector! Be careful ot to cofuse the two. r So, let s start with the two vectors a = a, a, a give by the formula, r b = b, b, b ad the the cross product is r r a b = ab -ab, ab -ab, ab -ab This is ot a easy formula to remember. There are two ways to derive this formula. Both of them use the fact that the cross product is really the determiat of a x matrix. If you do t kow what this is that is do t worry about it. You do t eed to kow aythig about matrices or determiats to use either of the methods. The otatio for the determiat is as follows, r r r i j k r r a b = a a a b b b The first row is the stadard basis vectors ad must appear i the order give here. The secod row is the compoets of a r ad the third row is the compoets of b r. Now, let s take a look at the differet methods for gettig the formula. The first method uses the Method of Cofactors. If you do t kow the method of cofactors that is fie, the result is all that we eed. Here is the formula. where, r r a a r a a r a a a b = i - j + k b b b b b b a b ad bc c d = - r This formula is ot as difficult to remember as it might at first appear to be. First, the terms alterate i sig ad otice that the x is missig the colum below the stadard basis vector that multiplies it as well as the row of stadard basis vectors. 007 Paul Dawkis 4

326 Math 40 The secod method is slightly easier; however, may textbooks do t cover this method as it will oly work o x determiats. This method says to take the determiat as listed above ad the copy the first two colums oto the ed as show below. r r r r r i j k i j r r a b = a a a a a b b b b b We ow have three diagoals that move from left to right ad three diagoals that move from right to left. We multiply alog each diagoal ad add those that move from left to right ad subtract those that move from right to left. This is best see i a example. We ll also use this example to illustrate a fact about cross products. r r Example If a =,, - ad b = -,4, r (a) a b r r r (b) b a Solutio (a) Here is the computatio for this oe. r r r r r i j k i j r r a b = - compute each of the followig r r r r r r = i + j k - j -i - -k - r r r = 5i + j + k (b) Ad here is the computatio for this oe. r r r r r i j k i j r r b a =- 4-4 ()() ( )( ) ( )( 4) ( )() ( )( 4) ()( ) - r r r r r r = i - + j + k - - j - - -i -k r r r =-5i - j -k ( 4)( ) ()( ) ( )() ( )( ) ()() ( 4)( ) Notice that switchig the order of the vectors i the cross product simply chaged all the sigs i the result. Note as well that this meas that the two cross products will poit i exactly opposite directios sice they oly differ by a sig. We ll formalize up this fact shortly whe we list several facts. There is also a geometric iterpretatio of the cross product. First we will let q be the agle betwee the two vectors a r ad b r ad assume that 0 q p, the we have the followig fact, 007 Paul Dawkis 5

327 Math 40 ad the followig figure. r r r r a b = a b siq () There should be a atural questio at this poit. How did we kow that the cross product poited i the directio that we ve give it here? First, as this figure, implies the cross product is orthogoal to both of the origial vectors. This will always be the case with oe exceptio that we ll get to i a secod. Secod, we kew that it poited i the upward directio (i this case) by the right had rule. This says that if we take our right had, start at a r ad rotate our figers towards b r our thumb r r will poit i the directio of the cross product. Therefore, if we d sketched i b a above we would have gotte a vector i the dowward directio. Example A plae is defied by ay three poits that are i the plae. If a plae cotais the,0,0,, R =, -, fid a vector that is orthogoal to the plae. poits P = ( ), Q = ( ) ad ( ) Solutio The oe way that we kow to get a orthogoal vector is to take a cross product. So, if we could fid two vectors that we kew were i the plae ad took the cross product of these two vectors we kow that the cross product would be orthogoal to both the vectors. However, sice both the vectors are i the plae the cross product would the also be orthogoal to the plae. So, we eed two vectors that are i the plae. This is where the poits come ito the problem. Sice all three poits lie i the plae ay vector betwee them must also be i the plae. There are may ways to get two vectors betwee these poits. We will use the followig two, uuur PQ = -,-0,- 0 = 0,, uuur PR = -, --0,- 0 =, -, The cross product of these two vectors will be orthogoal to the plae. So, let s fid the cross 007 Paul Dawkis 6

328 Math 40 product. uuur uuur PQ PR = r i r j r k r i r j r r r = 4i + j -k So, the vector 4i r + r j -k r will be orthogoal to the plae cotaiig the three poits. Now, let s address the oe time where the cross product will ot be orthogoal to the origial vectors. If the two vectors, a r ad b r, are parallel the the agle betwee them is either 0 or 80 degrees. From () this implies that, r r a b = 0 From a fact about the magitude we saw i the first sectio we kow that this implies r r r a b = 0 I other words, it wo t be orthogoal to the origial vectors sice we have the zero vector. This does give us aother test for parallel vectors however. Fact r r r If a b = 0 the a r ad b r will be parallel vectors. Let s also formalize up the fact about the cross product beig orthogoal to the origial vectors. Fact r r r Provided a b 0 r the a b r is orthogoal to both a r ad b r. Here are some ice properties about the cross product. Properties If u r, v r ad w r are vectors ad c is a umber the, r r r r r r r r r r u v =- v u cu v = u cv = cu v r r r r r r r r r r r r r u v+ w = u v+ u w ug v w = u v gw ( ) ( ) ( ) ( ) ( ) ( ) u u u r r r ug v w = v v v ( ) w w w The determiat i the last fact is computed i the same way that the cross product is computed. We will see a example of this computatio shortly. 007 Paul Dawkis 7

329 Math 40 There are a couple of geometric applicatios to the cross product as well. Suppose we have three vectors a r, b r ad c r ad we form the three dimesioal figure show below. The area of the parallelogram (two dimesioal frot of this object) is give by, r Area = a b r ad the volume of the parallelpiped (the whole three dimesioal object) is give by, r r Volume = ag b c r ( ) Note that the absolute value bars are required sice the quatity could be egative ad volume is t egative. We ca use this volume fact to determie if three vectors lie i the same plae or ot. If three vectors lie i the same plae the the volume of the parallelpiped will be zero. r Example Determie if the three vectors a =,4, -7 i the same plae or ot. r, b =, -,4 r ad c = 0, -9,8 lie Solutio So, as we oted prior to this example all we eed to do is compute the volume of the parallelpiped formed by these three vectors. If the volume is zero they lie i the same plae ad if the volume is t zero they do t lie i the same plae. 007 Paul Dawkis 8

330 Math r r ag b = r ( c) ()( )( 8) ( 4)( 4)( 0) ( 7)( )( 9) ( 4)( )( 8) -()( 4)( -9)-(-7)(-)( 0) = = = 0 So, the volume is zero ad so they lie i the same plae. 007 Paul Dawkis 9

331 Math 40 Three Dimesioal Space Itroductio I this chapter we will start takig a more detailed look at three dimesioal space (-D space or ). This is a very importat topic for Calculus III sice a good portio of Calculus III is doe i three (or higher) dimesioal space. We will be lookig at the equatios of graphs i -D space as well as vector valued fuctios ad how we do calculus with them. We will also be takig a look at a couple of ew coordiate systems for -D space. This is the oly chapter that exists i two places i my otes. Whe I origially wrote these otes all of these topics were covered i Calculus II however, we have sice moved several of them ito Calculus III. So, rather tha split the chapter up I have kept it i the Calculus II otes ad also put a copy i the Calculus III otes. May of the sectios ot covered i Calculus III will be used o occasio there ayway ad so they serve as a quick referece for whe we eed them. Here is a list of topics i this chapter. The -D Coordiate System We will itroduce the cocepts ad otatio for the three dimesioal coordiate system i this sectio. Equatios of Lies I this sectio we will develop the various forms for the equatio of lies i three dimesioal space. Equatios of Plaes Here we will develop the equatio of a plae. Quadric Surfaces I this sectio we will be lookig at some examples of quadric surfaces. 007 Paul Dawkis 0

332 Math 40 The -D Coordiate System We ll start the chapter off with a fairly short discussio itroducig the -D coordiate system ad the covetios that we ll be usig. We will also take a brief look at how the differet coordiate systems ca chage the graph of a equatio. Let s first get some basic otatio out of the way. The -D coordiate system is ofte deoted by. Likewise the -D coordiate system is ofte deoted by ad the -D coordiate system is deoted by. Also, as you might have guessed the a geeral dimesioal coordiate system is ofte deoted by. Next, let s take a quick look at the basic coordiate system. This is the stadard placemet of the axes i this class. It is assumed that oly the positive directios are show by the axes. If we eed the egative axis for ay reaso we will put them i as eeded. Also ote the various poits o this sketch. The poit P is the geeral poit sittig out i -D space. If we start at P ad drop straight dow util we reach a z-coordiate of zero we arrive at the poit Q. We say that Q sits i the xy-plae. The xy-plae correspods to all the poits which have a zero z-coordiate. We ca also start at P ad move i the other two directios as show to get poits i the xz-plae (this is S with a y-coordiate of zero) ad the yz-plae (this is R with a x-coordiate of zero). Collectively, the xy, xz, ad yz-plaes are sometimes called the coordiate plaes. I the remaider of this class you will eed to be able to deal with the various coordiate plaes so make sure that you ca. Also, the poit Q is ofte referred to as the projectio of P i the xy-plae. Likewise, R is the projectio of P i the yz-plae ad S is the projectio of P i the xz-plae. 007 Paul Dawkis

333 Math 40 May of the formulas that you are used to workig with i For istace the distace betwee two poits i While the distace betwee ay two poits i is give by, (, ) = ( - ) + ( - ) have atural extesios i d P P x x y y is give by, (, ) = ( - ) + ( - ) + ( - ) d P P x x y y z z Likewise, the geeral equatio for a circle with ceter ( hk, ) ad radius r is give by, ( x- h) + ( y- k) = r ad the geeral equatio for a sphere with ceter ( hkl,, ) ad radius r is give by, ( ) ( ) ( ) x- h + y- k + z- l = r With that said we do eed to be careful about just traslatig everythig we kow about. ito ad assumig that it will work the same way. A good example of this is i graphig to some extet. Cosider the followig example. Example Graph x = i, ad. Solutio I we have a sigle coordiate system ad so x = is a poit i a -D coordiate system. I the equatio x = tells us to graph all the poits that are i the form (, y ). This is a vertical lie i a -D coordiate system. I the equatio x = tells us to graph all the poits that are i the form (,, ) yz. If you go back ad look at the coordiate plae poits this is very similar to the coordiates for the yz-plae except this time we have x = istead of x = 0. So, i a -D coordiate system this is a plae that will be parallel to the yz-plae ad pass through the x-axis at x =. Here is the graph of x = i. Here is the graph of x = i. 007 Paul Dawkis

334 Math 40 Fially, here is the graph of x = i. Note that we ve preseted this graph i two differet styles. O the left we ve got the traditioal axis system ad we re used to seeig ad o the right we ve put the graph i a box. Both views ca be coveiet o occasio to help with perspective ad so we ll ofte do this with D graphs ad sketches. Note that at this poit we ca ow write dow the equatios for each of the coordiate plaes as well usig this idea. z = 0 xy -plae y = 0 xz -plae x= 0 yz -plae Let s take a look at a slightly more geeral example. 007 Paul Dawkis

335 Math 40 Example Graph y = x- i ad. Solutio Of course we had to throw out for this example sice there are two variables which meas that we ca t be i a -D space. I this is a lie with slope ad a y itercept of -. However, i this is ot ecessarily a lie. Because we have ot specified a value of z we are forced to let z take ay value. This meas that at ay particular value of z we will get a copy of this lie. So, the graph is the a vertical plae that lies over the lie give by y = x- i the xy-plae. Here is the graph i. here is the graph i. 007 Paul Dawkis 4

336 Math 40 Notice that if we look to where the plae itersects the xy-plae we will get the graph of the lie i as oted i the above graph by the red lie through the plae. Let s take a look at oe more example of the differece betwee graphs i the differet coordiate systems. Example Graph x + y = 4 i ad. Solutio As with the previous example this wo t have a -D graph sice there are two variables. I this is a circle cetered at the origi with radius. I however, as with the previous example, this may or may ot be a circle. Sice we have ot specified z i ay way we must assume that z ca take o ay value. I other words, at ay value of z this equatio must be satisfied ad so at ay value z we have a circle of radius cetered o the z-axis. This meas that we have a cylider of radius cetered o the z-axis. Here are the graphs for this example. 007 Paul Dawkis 5

337 Math 40 Notice that agai, if we look to where the cylider itersects the xy-plae we will agai get the circle from. We eed to be careful with the last two examples. It would be temptig to take the results of these ad say that we ca t graph lies or circles i There is o reaso for there to ot be graphs lies or circles i of the circle. To graph a circle i ad yet that does t really make sese. we would eed to do somethig like. Let s thik about the example x + y = 4 at z = 5. This would be a circle of radius cetered o the z-axis at the level of z = 5. So, as log as we specify a z we will get a circle ad ot a cylider. We will see a easier way to specify circles i a later sectio. We could do the same thig with the lie from the secod example. However, we will be lookig at lie i more geerality i the ext sectio ad so we ll see a better way to deal with lies i there. The poit of the examples i this sectio is to make sure that we are beig careful with graphig equatios ad makig sure that we always remember which coordiate system that we are i. 007 Paul Dawkis 6

338 Math 40 Aother quick poit to make here is that, as we ve see i the above examples, may graphs of equatios i will graph curves i are surfaces. That does t mea that we ca t graph curves i as well as we ll see later i this chapter.. We ca ad 007 Paul Dawkis 7

339 Math 40 Equatios of Lies I this sectio we eed to take a look at the equatio of a lie i sectio the equatio y = mx+ b does ot describe a lie i. As we saw i the previous, istead it describes a plae. This does t mea however that we ca t write dow a equatio for a lie i -D space. We re just goig to eed a ew way of writig dow the equatio of a curve. So, before we get ito the equatios of lies we first eed to briefly look at vector fuctios. We re goig to take a more i depth look at vector fuctios later. At this poit all that we eed to worry about is otatioal issues ad how they ca be used to give the equatio of a curve. The best way to get a idea of what a vector fuctio is ad what its graph looks like is to look at a example. So, cosider the followig vector fuctio. r t ( ) = t, A vector fuctio is a fuctio that takes oe or more variables, oe i this case, ad returs a vector. Note as well that a vector fuctio ca be a fuctio of two or more variables. However, i those cases the graph may o loger be a curve i space. The vector that the fuctio gives ca be a vector i whatever dimesio we eed it to be. I the example above it returs a vector i. Whe we get to the real subject of this sectio, equatios of lies, we ll be usig a vector fuctio that returs a vector i Now, we wat to determie the graph of the vector fuctio above. I order to fid the graph of our fuctio we ll thik of the vector that the vector fuctio returs as a positio vector for r poits o the graph. Recall that a positio vector, say v = ab,, is a vector that starts at the origi ad eds at the poit ( ab, ). So, to get the graph of a vector fuctio all we eed to do is plug i some values of the variable ad the plot the poit that correspods to each positio vector we get out of the fuctio ad play coect the dots. Here are some evaluatios for our example. r - = -, r - = -, r =, r 5 = 5, ( ) ( ) ( ) ( ) So, each of these are positio vectors represetig poits o the graph of our vector fuctio. The poits, (-,) (-,) (,) ( 5,) are all poits that lie o the graph of our vector fuctio. If we do some more evaluatios ad plot all the poits we get the followig sketch. 007 Paul Dawkis 8

340 Math 40 I this sketch we ve icluded the positio vector (i gray ad dashed) for several evaluatios as well as the t (above each poit) we used for each evaluatio. It looks like, i this case the graph of the vector equatio is i fact the lie y =. r t t t Here s aother quick example. Here is the graph of ( ) = 6cos,si. I this case we get a ellipse. It is importat to ot come away from this sectio with the idea that vector fuctios oly graph out lies. We ll be lookig at lies i this sectio, but the graphs of vector fuctio do ot have to be lies as the example above shows. We ll leave this brief discussio of vector fuctio with aother way to thik of the graph of a vector fuctio. Imagie that a pecil/pe is attached to the ed of the positio vector ad as we icrease the variable the resultig positio vector moves ad as it moves the pecil/pe o the ed sketches out the curve for the vector fuctio. 007 Paul Dawkis 9

341 Math 40 Okay, we ow eed to move ito the actual topic of this sectio. We wat to write dow the equatio of a lie i ad as suggested by the work above we will eed a vector fuctio to do this. To see how we re goig to do this let s thik about what we eed to write dow the equatio of a lie i. I two dimesios we eed the slope (m) ad a poit that was o the lie i order to write dow the equatio. I that is still all that we eed except i this case the slope wo t be a simple umber as it was i two dimesios. I this case we will eed to ackowledge that a lie ca have a three dimesioal slope. So, we eed somethig that will allow us to describe a directio that is potetially i three dimesios. We already have a quatity that will do this for us. Vectors give directios ad ca be three dimesioal objects. So, let s start with the followig iformatio. Suppose that we kow a poit that is o the lie, r P = x, y, z, ad that v = abc,, is some vector that is parallel to the lie. Note, i all ( ) likelihood, v r will ot be o the lie itself. We oly eed v r to be parallel to the lie. Fially, let ( xyz,, ) P= be ay poit o the lie. Now, sice our slope is a vector let s also represet the two poits o the lie as vectors. We ll do this with positio vectors. So, let r ur 0 ad r be the positio vectors for P 0 ad P respectively. Also, for o apparet reaso, let s defie a r uuur to be the vector with represetatio PP 0. We ow have the followig sketch with all these poits ad vectors o it. Now, we ve show the parallel vector, v r, as a positio vector but it does t eed to be a positio vector. It ca be aywhere, a positio vector, o the lie or off the lie, it just eeds to be parallel to the lie. 007 Paul Dawkis 40

342 Math 40 Next, otice that we ca write r as follows, r ur r = r0 + a If you re ot sure about this go back ad check out the sketch for vector additio i the vector arithmetic sectio. Now, otice that the vectors a r ad v r are parallel. Therefore there is a umber, t, such that r r a = tv We ow have, r ur r = r + tv = x, y, z + t abc,, This is called the vector form of the equatio of a lie. The oly part of this equatio that is ot kow is the t. Notice that tv r will be a vector that lies alog the lie ad it tells us how far from the origial poit that we should move. If t is positive we move away from the origial poit i the directio of v r (right i our sketch) ad if t is egative we move away from the origial poit i the opposite directio of v r (left i our sketch). As t varies over all possible values we will completely cover the lie. The followig sketch shows this depedece o t of our sketch. There are several other forms of the equatio of a lie. To get the first alterate form let s start with the vector form ad do a slight rewrite. r = x, y, z + t abc,, xyz,, = x + ta, y + tb, z + tc Paul Dawkis 4

343 Math 40 The oly way for two vectors to be equal is for the compoets to be equal. I other words, x= x + ta 0 y = y + tb 0 0 z = z + tc This set of equatios is called the parametric form of the equatio of a lie. Notice as well that this is really othig more tha a extesio of the parametric equatios we ve see previously. The oly differece is that we are ow workig i three dimesios istead of two dimesios. To get a poit o the lie all we do is pick a t ad plug ito either form of the lie. I the vector form of the lie we get a positio vector for the poit ad i the parametric form we get the actual coordiates of the poit. There is oe more form of the lie that we wat to look at. If we assume that a, b, ad c are all o-zero umbers we ca solve each of the equatios i the parametric form of the lie for t. We ca the set all of them equal to each other sice t will be the same umber i each. Doig this gives the followig, x-x y- y z-z = = a b c This is called the symmetric equatios of the lie. If oe of a, b, or c does happe to be zero we ca still write dow the symmetric equatios. To see this let s suppose that b = 0. I this case t will ot exist i the parametric equatio for y ad so we will oly solve the parametric equatios for x ad z for t. We the set those equal ad ackowledge the parametric equatio for y as follows, x-x0 z-z0 = y = y0 a c Let s take a look at a example. Example Write dow the equatio of the lie that passes through the poits (, -,) ad (,4, - ). Write dow all three forms of the equatio of the lie. Solutio To do this we eed the vector v r that will be parallel to the lie. This ca be ay vector as log as it s parallel to the lie. I geeral, v r wo t lie o the lie itself. However, i this case it will. All we eed to do is let v r be the vector that starts at the secod poit ad eds at the first poit. Sice these two poits are o the lie the vector betwee them will also lie o the lie ad will 007 Paul Dawkis 4

344 Math 40 hece be parallel to the lie. So, r v =, -5,6 Note that the order of the poits was chose to reduce the umber of mius sigs i the vector. We could just have easily goe the other way. Oce we ve got v r there really is t aythig else to do. To use the vector form we ll eed a poit o the lie. We ve got two ad so we ca use either oe. We ll use the first poit. Here is the vector form of the lie. r =, -, + t, - 5,6 = + t, -- 5, t + 6t Oce we have this equatio the other two forms follow. Here are the parametric equatios of the lie. x= + t y =--5t z = + 6t Here is the symmetric form. x - y + z - = = -5 6 Example Determie if the lie that passes through the poit ( 0, -,8) ad is parallel to the lie give by x = 0+ t, y = t ad z =-- t passes through the xz-plae. If it does give the coordiates of that poit. Solutio To aswer this we will first eed to write dow the equatio of the lie. We kow a poit o the lie ad just eed a parallel vector. We kow that the ew lie must be parallel to the lie give by the parametric equatios i the problem statemet. That meas that ay vector that is parallel to the give lie must also be parallel to the ew lie. Now recall that i the parametric form of the lie the umbers multiplied by t are the compoets of the vector that is parallel to the lie. Therefore, the vector, r v =,, - is parallel to the give lie ad so must also be parallel to the ew lie. The equatio of ew lie is the, r = 0, -,8 + t,, - =, t - + t,8-t If this lie passes through the xz-plae the we kow that the y-coordiate of that poit must be zero. So, let s set the y compoet of the equatio equal to zero ad see if we ca solve for t. If 007 Paul Dawkis 4

345 Math 40 we ca, this will give the value of t for which the poit will pass through the xz-plae. - + t = 0 fi t = 4 So, the lie does pass through the xz-plae. To get the complete coordiates of the poit all we eed to do is plug t = ito ay of the equatios. We ll use the vector form. 4 r ʈ ʈ = Á, - + Á,8 - =,0, Ë4 Ë Recall that this vector is the positio vector for the poit o the lie ad so the coordiates of the Ê ˆ poit where the lie will pass through the xz-plae are Á,0, Ë Paul Dawkis 44

346 Math 40 Equatios of Plaes I the first sectio of this chapter we saw a couple of equatios of plaes. However, oe of those equatios had three variables i them ad were really extesios of graphs that we could look at i two dimesios. We would like a more geeral equatio for plaes. So, let s start by assumig that we kow a poit that is o the plae, P ( x, y, z ) 0 = Let s also r suppose that we have a vector that is orthogoal (perpedicular) to the plae, = abc,,. This vector is called the ormal vector. Now, assume that P ( xyz,, ) = is ay poit i the plae. Fially, sice we are goig to be workig with vectors iitially we ll let r ur 0 ad r be the positio vectors for P 0 ad P respectively. Here is a sketch of all these vectors. Notice that we added i the vector r r ur - 0 which will lie completely i the plae. Also otice that we put the ormal vector o the plae, but there is actually o reaso to expect this to be the case. We put it here to illustrate the poit. It is completely possible that the ormal vector does ot touch the plae i ay way. Now, because r is orthogoal to the plae, it s also orthogoal to ay vector that lies i the plae. I particular it s orthogoal to r r ur - 0. Recall from the Dot Product sectio that two orthogoal vectors will have a dot product of zero. I other words, r r ur rr r ur g - r = 0 fi r g = r g ( ) Paul Dawkis 45

347 Math 40 This is called the vector equatio of the plae. A slightly more useful form of the equatios is as follows. Start with the first form of the vector equatio ad write dow a vector for the differece. Now, actually compute the dot product to get, ( ) abc,, g xyz,, - x, y, z = 0 abc,, g x-x, y- y, z- z = ( ) ( ) ( ) a x- x + b y- y + c z- z = This is called the scalar equatio of plae. Ofte this will be writte as, ax+ by+ cz = d where d = ax0 + by0 + cz0. This secod form is ofte how we are give equatios of plaes. Notice that if we are give the equatio of a plae i this form we ca quickly get a ormal vector for the plae. A ormal vector is, r = abc,, Let s work a couple of examples. Example Determie the equatio of the plae that cotais the poits P = (, -,0), Q = (,,4) ad R = ( 0, -,). Solutio I order to write dow the equatio of plae we eed a poit (we ve got three so we re cool there) ad a ormal vector. We eed to fid a ormal vector. Recall however, that we saw how to do this i the Cross Product sectio. We ca form the followig two vectors from the give poits. uuur uuur PQ=,,4 PR = -,, These two vectors will lie completely i the plae sice we formed them from poits that were i the plae. Notice as well that there are may possible vectors to use here, we just chose two of the possibilities. Now, we kow that the cross product of two vectors will be orthogoal to both of these vectors. Sice both of these are i the plae ay vector that is orthogoal to both of these will also be orthogoal to the plae. Therefore, we ca use the cross product as the ormal vector. 007 Paul Dawkis 46

348 Math 40 r r r r r i j k i j r uuur uuur r r r = PQ PR= 4 = i - 8j + 5k - - The equatio of the plae is the, x- - 8 y+ + 5 z- 0 = 0 ( ) ( ) ( ) x- 8y+ 5z = 8 We used P for the poit, but could have used ay of the three poits. Example Determie if the plae give by - x+ z = 0 ad the lie give by r = 5, - t,0+ 4t are orthogoal, parallel or either. Solutio This is ot as difficult a problem as it may at first appear to be. We ca pick off a vector that is r ormal to the plae. This is = -,0,. We ca also get a vector that is parallel to the lie. This is v = 0, -,4. Now, if these two vectors are parallel the the lie ad the plae will be orthogoal. If you thik about it this makes some sese. If r ad v r are parallel, the v r is orthogoal to the plae, but v r is also parallel to the lie. So, if the two vectors are parallel the lie ad plae will be orthogoal. Let s check this. r r r r r i j k i j r r r r r r v =- 0-0 = i + 4j + k So, the vectors are t parallel ad so the plae ad the lie are ot orthogoal. Now, let s check to see if the plae ad lie are parallel. If the lie is parallel to the plae the ay vector parallel to the lie will be orthogoal to the ormal vector of the plae. I other words, if r ad v r are orthogoal the the lie ad the plae will be parallel. Let s check this. rr v= g = 8 0 The two vectors are t orthogoal ad so the lie ad plae are t parallel. So, the lie ad the plae are either orthogoal or parallel. 007 Paul Dawkis 47

349 Math 40 Quadric Surfaces I the previous two sectios we ve looked at lies ad plaes i three dimesios (or ) ad while these are used quite heavily at times i a Calculus class there are may other surfaces that are also used fairly regularly ad so we eed to take a look at those. I this sectio we are goig to be lookig at quadric surfaces. Quadric surfaces are the graphs of ay equatio that ca be put ito the geeral form Ax + By + Cz + Dxy+ Exz+ Fyz+ Gx+ Hy+ Iz+ J = 0 where A,, J are costats. There is o way that we ca possibly list all of them, but there are some stadard equatios so here is a list of some of the more commo quadric surfaces. Ellipsoid Here is the geeral equatio of a ellipsoid. Here is a sketch of a typical ellipsoid. x y z a b c + + = If a = b= c the we will have a sphere. Notice that we oly gave the equatio for the ellipsoid that has bee cetered o the origi. Clearly ellipsoids do t have to be cetered o the origi. However, i order to make the discussio i this sectio a little easier we have chose to cocetrate o surfaces that are cetered o the origi i oe way or aother. Coe Here is the geeral equatio of a coe. x y z + = a b c 007 Paul Dawkis 48

350 Math 40 Here is a sketch of a typical coe. Note that this is the equatio of a coe that will ope alog the z-axis. To get the equatio of a coe that opes alog oe of the other axes all we eed to do is make a slight modificatio of the equatio. This will be the case for the rest of the surfaces that we ll be lookig at i this sectio as well. I the case of a coe the variable that sits by itself o oe side of the equal sig will determie the axis that the coe opes up alog. For istace, a coe that opes up alog the x-axis will have the equatio, y z x + = b c a For most of the followig surfaces we will ot give the other possible formulas. We will however ackowledge how each formula eeds to be chaged to get a chage of orietatio for the surface. Cylider Here is the geeral equatio of a cylider. x a y b + = This is a cylider whose cross sectio is a ellipse. If a = b we have a cylider whose cross sectio is a circle. We ll be dealig with those kids of cyliders more tha the geeral form so the equatio of a cylider with a circular cross sectio is, x + y = r 007 Paul Dawkis 49

351 Math 40 Here is a sketch of typical cylider with a ellipse cross sectio. The cylider will be cetered o the axis correspodig to the variable that does ot appear i the equatio. Be careful to ot cofuse this with a circle. I two dimesios it is a circle, but i three dimesios it is a cylider. Hyperboloid of Oe Sheet Here is the equatio of a hyperboloid of oe sheet. x y z a b c Here is a sketch of a typical hyperboloid of oe sheet. + - = 007 Paul Dawkis 50

352 Math 40 The variable with the egative i frot of it will give the axis alog which the graph is cetered. Hyperboloid of Two Sheets Here is the equatio of a hyperboloid of two sheets. x y z = a b c Here is a sketch of a typical hyperboloid of two sheets. The variable with the positive i frot of it will give the axis alog which the graph is cetered. 007 Paul Dawkis 5

353 Math 40 Notice that the oly differece betwee the hyperboloid of oe sheet ad the hyperboloid of two sheets is the sigs i frot of the variables. They are exactly the opposite sigs. Elliptic Paraboloid Here is the equatio of a elliptic paraboloid. x y z + = a b c As with cyliders this has a cross sectio of a ellipse ad if a = b it will have a cross sectio of a circle. Whe we deal with these we ll geerally be dealig with the kid that have a circle for a cross sectio. Here is a sketch of a typical elliptic paraboloid. I this case the variable that is t squared determies the axis upo which the paraboloid opes up. Also, the sig of c will determie the directio that the paraboloid opes. If c is positive the it opes up ad if c is egative the it opes dow. Hyperbolic Paraboloid Here is the equatio of a hyperbolic paraboloid. x y z - = a b c Here is a sketch of a typical hyperbolic paraboloid. 007 Paul Dawkis 5

354 Math 40 These graphs are vaguely saddle shaped ad as with the elliptic paraoloid the sig of c will determie the directio i which the surface opes up. The graph above is show for c positive. With the both of the types of paraboloids discussed above the surface ca be easily moved up or dow by addig/subtractig a costat from the left side. For istace z =-x - y + 6 is a elliptic paraboloid that opes dowward (be careful, the - is o the x ad y istead of the z) ad starts at z = 6 istead of z = 0. Here is a couple of quick sketches of this surface. 007 Paul Dawkis 5

355 Math 40 Note that we ve give two forms of the sketch here. The sketch o the right has the stadard set of axes but it is difficult to see the umbers o the axis. The sketch o the left has bee boxed ad this makes it easier to see the umbers to give a sese of perspective to the sketch. I most sketches that actually ivolve umbers o the axis system we will give both sketches to help get a feel for what the sketch looks like. 007 Paul Dawkis 54