Calculus II. Here are a couple of warnings to my students who may be here to get a copy of what happened on a day that you missed

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1 This documet was writte ad copyrighted by Paul Dawkis. Use of this documet ad its olie versio is govered by the Terms ad Coditios of Use located at. The olie versio of this documet is available at At the above web site you will fid ot oly the olie versio of this documet but also pdf versios of each sectio, chapter ad complete set of otes. Preface Here are my olie otes for my Calculus II course that I teach here at Lamar Uiversity. Despite the fact that these are my class otes they should be accessible to ayoe watig to lear Calculus II or eedig a refresher i some of the topics from the class. These otes do assume that the reader has a good workig kowledge of Calculus I topics icludig limits, derivatives ad itegratio by substitutio. Here are a couple of warigs to my studets who may be here to get a copy of what happeed o a day that you missed. Because I wated to make this a fairly complete set of otes for ayoe watig to lear calculus I have icluded some material that I do ot usually have time to cover i class ad because this chages from semester to semester it is ot oted here. You will eed to fid oe of your fellow class mates to see if there is somethig i these otes that was t covered i class.. I geeral I try to work problems i class that are differet from my otes. However, with Calculus II may of the problems are difficult to make up o the spur of the momet ad so i this class my class work will follow these otes fairly close as far as worked problems go. With that beig said I ofte do t have time i class to work all of these problems ad so you will fid that some sectios cotai problems that were t worked i class due to time restrictios.. Sometimes questios i class will lead dow paths that are ot covered here. I try to aticipate as may of the questios as possible i writig these up, but the reality is that I ca t aticipate all the questios. Sometimes a very good questio gets asked i class that leads to isights that I ve ot icluded here. You should always talk to someoe who was i class o the day you missed ad compare these otes to their otes ad see what the differeces are. 4. This is somewhat related to the previous three items, but is importat eough to merit its ow item. THESE NOTES ARE NOT A SUBSTITUTE FOR ATTENDING CLASS!! Usig these otes as a substitute for class is liable to get you i trouble. As already oted ot everythig i these otes is covered i class ad ofte material or isights ot i these otes is covered i class. 005 Paul Dawkis

2 Itegratio Techiques Itroductio I this chapter we are goig to be lookig at itegratio techiques. There are a fair umber of them, some 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 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. 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 a how they ca be used to simplify certai itegrals. Partial Fractios We will use partial fractios to allow us to do itegrals ivolvig 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 itegrals that ivolve quadratics. 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. 005 Paul Dawkis

3 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. 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 x e dx To do this itegral we ll use the followig substitutio. u = x du = x dx x dx = du x u u x xe dx = du = + c = + c e e e 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 to 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. f g = f g+ f g Now, itegrate both sides of this. f g dx= f g+ f g dx The left side is easy eough to itegrate ad we ll split up the right side of the itegral. fg = f gdx+ f g dx 005 Paul Dawkis

4 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 that the itegratio by parts formula. f gdx = 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 we wrote dow above. Example Evaluate the followig itegral. 6x 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 aythig that we choose for u will be differetiated ad so that seems like choosig u=x will be a good choice sice upo differetiatig the x will drop out. Oce u is chose 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. 6x u = x dv= e dx The itegral is the, du = dx v = e dx = e 6 6x 6x 005 Paul Dawkis 4

5 6x x 6x 6x x dx = e 6 e dx 6 e x 6x 6x = e e + c 6 6 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 a look at itegratio by parts for defiite itegrals. I this case the formula is, b b b b = a a a udv uv vdu Note that the uv i the first term is just the stadard itegral evaluatio otatio that a you should be familiar with at this poit. All we do is evaluate at b the subtract off the evaluatio at a. 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 dx 6 6 e x 6x 6x = e e = e + e 6 6 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, through out 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. t ( t+ 5) cos dt 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. 005 Paul Dawkis 5

6 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. t u = t + 5 dv = cos dt 4 t du = dt v = 4si 4 The itegral is the, t t t ( t+ 5) cos dt = 4( t+ 5) si si dt t t = 4 ( t+ 5si ) + 48cos + c 4 4 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. u = w dv= si ( 0w) dw du = wdw v = cos( 0w) 0 The itegral is the, w w si ( 0w) dw= 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. u = w dv= cos( 0w) dw du = dw v = si ( 0 w) 0 So, the itegral becomes, w w w si ( 0w) dw= cos( 0w) + si ( 0w) si ( 0 ) w dw w w = cos0 ( w) + si0 ( w) + cos0 ( w) + c w w = cos( 0w) + si ( 0w) + cos( 0w) + c Paul Dawkis 6

7 Be careful with the coefficiet o the itegral for the secod applicatio of itegratio by parts. Sice the itegral is multiplied by we eed to make sure that the results of 5 actually doig the itegral is also multiplied by. Forgettig to do this is oe of the 5 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 complete a problem. This is somethig that will happe so do t get excited about it whe it happes. 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. Example 5 Evaluate the followig itegral x x+ dx (a) Usig Itegratio by Parts. (b) Usig a stadard Calculus I substitutio. Solutio (a) 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. u = x dv= x+ dx du = dx v = ( x + ) The itegral is the, x x+ dx= x x+ x+ dx 5 4 = x( x+ ) ( x+ ) + c 5 (b) 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, 005 Paul Dawkis 7

8 x x + dx = u u du = u u du 5 = u u + c 5 5 = ( x + ) ( x+ ) + c 5 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 my o more tha a costat if we take a look at the differece of the two x x+ x+ x+ x = ( x+ ) x ( x+ ) ( x+ ) ( 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. 005 Paul Dawkis 8

9 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 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 x dx 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. u = dv= lxdx 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. 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 du = dx v = x x The itegral is the, 005 Paul Dawkis 9

10 Example 7 Evaluate the followig itegral. l x dx = x l x x dx 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 = x dx v = ( x + ) 9 The itegral is the, 5 x x + dx= x ( x + ) x ( x + ) dx = 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. θ e cosθ dθ 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. 005 Paul Dawkis 0

11 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. θ u = cosθ dv= e dθ θ du = siθ dθ v = e The itegral is the, θ θ θ e cosθ dθ = e cosθ + e siθ dθ So, it looks like we ll do itegratio by parts agai. Here are our choices this time. θ u = siθ dv= e dθ θ du = cosθ dθ v = e The itegral is ow, θ θ θ θ e cosθ dθ = e cosθ + e siθ e cosθ dθ 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 its got a mius sig i frot of it. This meas that we ca add the itegral to both sides to get, e θ cosθ dθ = e θ cosθ + e θ siθ All we eed to do ow is divide by ad we re doe. The itegral is, θ θ θ e cosθ dθ = ( cosθ + siθ) + c e e Notice that after dividig by the two we add i the costat of itegratio at that poit. This idea of usig itegratio by parts 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 4 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. 005 Paul Dawkis

12 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 x e dx= ( x ) e ( 4x ) 4e + ( x ) 8e ( 4x) 6e + ( 4) e x x x x x 4 = x e 6x e + 96x e 84xe + 768e + c 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. 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 005 Paul Dawkis

13 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 x dx 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, 5 4 = = si x dx si xsi x dx si x si x dx Now recall the trig idetity, cos x + si x= si x= cos x With this idetity the itegral ca be writte as, 5 = ( ) si x dx cos x si x dx ad we ca ow use the substitutio u = cos x. Doig this gives us, u 5 si xdx= u du 4 = + u du 5 = u u + u + c 5 5 = cos x + cos x cos x+ c 5 So, with a little rewritig o the itegrad we were able to reduce this to a fairly simple 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. 005 Paul Dawkis

14 si x cos 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 xcos xdx= si xcos xcos xdx 6 ( ) 6 u u du xdx = si x si x cos xdx u = si x = 6 8 = u u du si x si x c = + At this poit let s pause for a secod to summarize what we ve leared about itegratig powers of sie ad cosie. m si x cos xdx () I () if the expoet o the sies () is odd we 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 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. 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 is 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. cos x = ( + cos( x) ) si x = ( cos( x) ) si x cos x= si ( x) 005 Paul Dawkis 4

15 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 x cos 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. si x cos xdx= ( cos( x) ) ( + cos( x) ) dx 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 formula give above. si x cos xdx= ( + cos( 4x) ) dx 4 = cos ( 4 x) dx 4 = x si ( 4 x) + c 4 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. xdx 005 Paul Dawkis 5

16 si cos si cos x xdx= x x dx = si ( x) si = x dx 4 Now, we use the double agle formula for sie to reduce to a itegral that we ca do. si x cos 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 sie ad cosie 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. 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. dx siαcos β = si si α β + α + β siα si β = cos( α β) cos( α β) + cosα cos β = cos( α β) + cos( α + β) 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 005 Paul Dawkis 6

17 This itegral requires the last formula listed above. cos( 5x) cos( 4x) dx = cos( ) cos( 9 ) x + x dx = si ( x) + si ( 9x) + c 9 Okay, at this poit we ve covered pretty much all the possible cases ivolvig products of sie ad cosie. 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 x ta 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 x cos x + = cos x cos x cos x 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 = ta x du = sec xdx u = sec x du = sec x ta x dx 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 we ca strip oe out 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 to 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. 005 Paul Dawkis 7

18 Let s take a look at a couple of examples. Example 5 Evaluate the followig itegral. sec x ta 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). xdx Here s the work for this itegral sec x ta x dx = sec x ta x ta xsec x dx 8 8 u u du 0 8 u u u du 9 = + + = sec x sec x ta xsec xdx u = sec x = = + sec x sec x sec x c 9 Example 6 Evaluate the followig itegral. sec x ta 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. xdx Here is the work for this itegral sec xta xdx= sec xta xsec xdx 6 ( u ) 8 6 u u du 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 005 Paul Dawkis 8

19 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. 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 x dx Solutio This itegral is othig more tha a Calculus I substitutio. si x ta x dx = dx u = cos x cos x = du u = l cos x + c rl x= l x = l cos x + c l sec x + c Example 8 Evaluate the followig itegral. ta x dx Solutio The trick to this oe is do the followig maipulatio of the itegrad. ta xdx= ta xta xdx = ta x sec x dx = ta xsec xdx ta 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. r The itegral is the, x dx = 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 x dx ta x sec x dx ta xsec x dx ta x dx So, a quick substitutio ( u = ta x) will give us the first itegral ad the secod itegral will always be the previous odd power. 005 Paul Dawkis 9

20 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. Example 9 Evaluate the followig itegral. sec x dx 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 x dx = = ( + ) sec x sec x ta x dx sec x+ ta x sec x+ ta xsec x dx sec x+ ta x I this form we ca do the itegral usig the substitutio u = sec x+ ta x. Doig this gives, sec x dx = l sec x + ta x + c Example 0 Evaluate the followig itegral. sec x dx 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 = sec x dv= sec xdx du = sec x ta x dx v = ta x Note that usig itegratio by parts o this problem is ot a obvious choice, but it does work very icely here. The itegral is the, sec x dx = sec x ta x sec x ta x dx 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. sec x dx = sec x ta x sec x sec x dx sec ta sec sec = x x xdx+ 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 005 Paul Dawkis 0

21 mius sig i frot. So, add it to both sides to get, sec x dx = sec x ta x + l sec x + ta x Fially divide by a two ad we re doe. sec x dx = sec x ta x + l sec x + ta x + c The two itegrals i the last two examples will arise o occasio i some of the work that we ll be doig i later sectios ad chapters so it would t be a bad idea to make sure you ve got them writte dow somewhere. 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. I 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 4 dx 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 to cosies. We ll the same thig will work i this case. 7 6 si x si x dx = si x dx 4 4 cos x cos x = = ( si x) cos 4 ( cos x) cos x 4 x si x dx si x dx At this poit all we eed to do is use the substitutio u = cos xad we re doe. 005 Paul Dawkis

22 si cos 7 ( u ) x dx = du x u = u u u du = + + u u c + u u = + + cosx cos x c cos x cosx 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, 4 cos x 7 dx si x we would t have bee able to strip out a sie. 4 4 cos x cos x dx = dx 7 6 si x si x si 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. 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 5x 4 dx= 5x 4 + c 75 x dx = 5x 4 + c 5x 4 5 Both of these used the substitutio u = 5x 4. However, let s take a look at the followig itegral. Example Evaluate the followig itegral. 005 Paul Dawkis

23 5x 4 dx x Solutio 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. It would be ice if we could get rid of the square root somehow. The followig substitutio will do that for us. x = sec θ 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. Notice that this is ot the stadard substitutio we are used to workig with. To this poit we ve used substitutio that were i the form u = f ( x). I this case we are goig to explicitly give a substitutio for x. The substitutio will work i pretty much the same maer however. Before we actually do the substitutio however let s verify the claim that this will allow us to get rid of the square root. 4 x θ ( θ ) θ = 5 sec 4 = 4 sec = sec To get rid of the square root all we eed to do is recall the relatioship, ta θ + = sec θ sec θ = ta θ Usig this fact the square root becomes, 5x 4 = ta θ = taθ Note the presece of the absolute value bars there. These are importat. Recall that x There should always be absolute value bars at this stage. If we kew that taθ was always positive or always egative we could elimiate the absolute value bars usig, x if x 0 x = x if x < 0 Without limits we wo t be able to determie this, however, we will eed to elimiate them i order to do the itegral. Therefore, sice we doig a idefiite itegral we will assume that taθ will be positive ad so we ca drop the absolute value bars. This gives, 5x 4 = taθ = x 005 Paul Dawkis

24 So, we were able to elimiate the square root usig this substitutio. Let s go ahead ad put the substitutio ito the itegral ad see what we get. I doig the substitutio do t forget that well also eed to substitute for the dx. This is easy eough to get from the substitutio. x = secθ dx = secθ taθ dθ 5 5 Usig this substitutio the itegral becomes, 5x 4 taθ dx = sec θ ta θ dθ x 5 secθ 5 = ta θ dθ So, 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 dx = sec θ dθ x = ta + c ( θ θ) 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 secθ = = adjacet This gives the followig right triagle. From this we ca see that, 5x 4 taθ = We ca deal with the θ i oe of ay variety of ways. From our substitutio we ca see that, 5x θ = sec While this is a perfectly acceptable method of dealig with the θ we ca use ay of the 005 Paul Dawkis 4

25 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. θ = cos 5x So, with all of this the itegral becomes, 5x 4 5x 4 dx = cos + c x 5x = 5x 4 cos + c 5x 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 x 4 x Solutio The limits here wo t chage the substitutio so that will remai the same. x = sec θ 5 Usig this substitutio the square root still reduces dow to, 5 x = dx 5 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 θ that we re i. Oce we ve got that we ca determie how to drop the absolute value bars. Here s the limits of θ. x = = sec θ θ = π x = = sec θ θ = θ 005 Paul Dawkis 5

26 π I the rage of 0 θ taget is positive ad so i this case we ca just drop the absolute value bars. So, let s do the substitutio. Note that the work is idetical to the previous example ad so most of it is left out x x 5 4 π sec θ 0 dx = ( θ θ) = ta π = 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. Example Evaluate the followig itegral x 4 Solutio Agai, the substitutio ad square root are the same as the first two examples. sec 5 x= θ x 4 = ta θ 5 Let s ext see the limits θ for this problem. x = = sec θ θ = π 4 4 π x = = secθ θ = Note that i determiig the value of θ we used the smallest positive value. Now i the rage of π θ π 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, x x = dx 5 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, θ dθ π Paul Dawkis 6

27 x 4 x π π sec θ dx = ( θ θ) = ta π = π dθ π 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 θ 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. a bx a x= secθ b Let s work a ew ad differet type of example. Example 4 Evaluate the followig itegral. dx 4 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. x = secθ Usig this substitutio the square root becomes, 9 x = 9 9sec θ = sec θ = ta θ So, this will be trouble. Usig this substitutio we will get 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 = siθ dx = cosθ dθ With this substitutio the square root is, 9 x = si θ = cos θ = cosθ = cosθ 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, 005 Paul Dawkis 7

28 x dx = 4 9 x 8si θ = dθ 4 8 si θ csc 4 = θ dθ 8 4 ( cosθ) cosθ dθ 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. dx = csc csc 4 8 θ θ dθ x 9 x = ( cot ) csc cot 8 θ + θ dθ u = θ = u du 8 + cot = θ + cot θ + c 8 Now we eed to go back to x s usig a right triagle. Here is the right triagle for this problem ad trig fuctios for this problem. x 9 x siθ = cotθ = x The itegral is the, x 9 x 9 x dx = + + c 9 x 8 x x 4 ( x ) 9 9 x = + c x x 4 8 Here s the geeral form for this type of square root. 005 Paul Dawkis 8

29 a = siθ b a b x x 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 0 6 ( x + ) Solutio First, otice that there really is a square root i this problem eve though it is t explicitly writte out. dx ( 6x + ) = 6x + = 6x + This square root is ot i the form we saw i the previous examples. Here we will use the substitutio. x = taθ dx = sec θ dθ 6 6 With this substitutio the deomiator becomes, 6x + = ta θ + = sec θ = secθ Now, because we have limits we ll eed to covert them to θ so we ca determie how to drop the absolute value bars. x = 0 0= taθ θ = 0 6 π x = = ta θ θ = I this rage secat is positive ad so we ca drop the absolute value bars. Here is the itegral, π x 7776 ta θ dx = sec θ 0 sec θ 6 0 ( 6x + ) π 5 4 ta θ = d θ secθ 0 dθ There are several ways to proceed from this poit. Normally with a odd expoet o 005 Paul Dawkis 9

30 the taget we would strip o 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. 6 π 5 5 x 4 si dx = cos 0 ( 6x + ) π 4 0 θ d θ θ ( cos θ ) = cos θ siθdθ We ca ow use the substitutio u = cosθ ad we might as well covert the limits as well. θ = 0 u = cos0= π π θ = u = cos = 4 4 The itegral is the, 6 5 x 4 dx = u u du ( 6x + ) = + + u u u = The geeral form for this fial type of square root is a a + b x x= taθ b 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. Example 6 Evaluate the followig itegral. x x 4 7 x 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 dx 005 Paul Dawkis 0

31 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 x x = x x+ = ( x ) = ( x ) 9 So, the root becomes, x x = x 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 = secθ x= + secθ dx= secθ taθ dθ The root reduces to, 4 7 = 9 = 9sec 9 = ta = ta = ta x x x θ θ θ θ Note we could drop the absolute value bars sice we are doig a idefiite itegral. Here is the itegral. x 4 7 x x + secθ dx = sec θ ta θ dθ taθ = secθ + sec θ dθ = l secθ + taθ + taθ + c Ad here is the right triagle for this problem. 4 7 sec x ta x θ θ x = = The itegral is the, 005 Paul Dawkis

32 x 4x 7 ( x ) x x 4x 7 x 4x 7 dx = l c 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. e x = taθ e x dx = sec θ d θ Remember that to compute the differetial all we do is differetiate both sides ad the tack o dx or dθ oto the appropriate side. With this substitutio the square root becomes, x x + e = + e = + ta = sec = sec = sec θ θ θ θ Agai, we ca drop the absolute value bars because we are doig a idefiite itegral. Here s the itegral. e + e dx = e e + e dx x ( e ) x x e ( e ) dx ta ( sec )( sec ) ( ) 4 u u du 4x x x x x = + = θ θ θ dθ = sec θ sec θsecθtaθ dθ u = secθ = sec θ sec θ 5 5 = + Here is the right triagle for this itegral. x x e + e taθ = secθ = = + e c x 005 Paul Dawkis

33 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. a a b x x= siθ b a bx a x= secθ b a a + b x x= taθ b Partial Fractios I this sectio we are goig to take a look a 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 = du usig u x x 6 ad du = + = ( x ) dx x x+ 6 u = l x x+ 6 + 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 + dx x x Paul Dawkis

34 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 x + 4 = x x 6 x x+ the the itegral ca be doe. x + 4 dx = dx x x 6 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. Factor i deomiator ax + b ( ax + b) k ax bx c Term i partial fractio decompositio A ax + b A A Ak k ax + b ( ax + b) ( ax + b) Ax + B ax + bx + c Ax + B Ax + B Ax k + Bk ax + bx + c ax + bx + c ax + bx + c + + ( ax + bx + c) k k 005 Paul Dawkis 4

35 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 with actually doig the itegral above. Example Evaluate the followig itegral. x + dx 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. 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. So, the ext step is to set umerators equal. x+ = A x+ + B x 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) B= x= 0= A 5 + B 0 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. 005 Paul Dawkis 5

36 x + 4 dx = dx = dx dx x x+ = 4l x l x+ + c x x x x 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 itegrals 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. x dx = ta + c 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. Example Evaluate the followig itegral. x + 4 dx x + 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 this step i your head. x + 4= A x+ x + Bx x + Cx x+ 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. 005 Paul Dawkis 6

37 x= 0 4= A A= x= 8= B( )( 8) B= x= = C C = = 9 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 = + + dx x + 4x 4x 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 dx ( x 4) ( x + ) Solutio This time the deomiator is already factored so let s just jump right to the partial fractio decompositio. x 9x+ 5 A B Cx+ D = + + x 4 x + x 4 x 4 x + Settig umerators gives, 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. 005 Paul Dawkis 7

38 : + = 0 : = : = 9 : = 5 x A C x A B C D x A C D 0 x A B D 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 dx 4 x + x x x dx = + ( x 4) ( x + ) x ( x 4) 5 x = + dx x 4 ( x 4) x + x + 5 x = l 4 + l + + ta + x 4 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. x + 0x + x+ 6 dx 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 + 0x + x+ 6 = A x Bx+ C x x Dx+ E x ( 8 4 ) 4 = A+ B x + C B x + A+ B C+ D x + Settig coefficiet equal gives the followig system. 4B + 4C D+ E x+ 6A 4C E 005 Paul Dawkis 8

39 : + = 0 : = : = 0 =, =, =, =, = 0 : = : 6 4 = 6 4 x A B x C B 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. x + 0x + x+ 6 x x dx = + + dx ( x )( x + 4) x x + 4 ( x + 4) x x = + dx x x + 4 x + 4 ( x + 4) x = l l + 4 ta + x + 4 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. Example 5 Evaluate the followig itegral. 4 x 5x + 6x 8 dx x x Solutio So, i this case the degree of the umerator is 4 ad the degree of the umerator 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 4 x x x 5x + 6x 8 4 ( x x ) + + x 6x 8 ( x 6x ) Paul Dawkis 9

40 So, from the log divisio we see that, 4 x 5x + 6x 8 8 = x x x x x ad the itegral becomes, 4 x 5x + 6x 8 8 dx = x dx x x x x 8 = x dx dx 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 the third. x= 0 x= 8= B( ) 8= C( 9) B= 6 C = x= 8= A + B + C = A+ 4 A= The itegral is the, 4 x 5x + 6x 8 6 dx = x dx + dx x x x x x 6 = x x+ l x l x + c x 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, 005 Paul Dawkis 40

41 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 A B C A B C D + + x x x x x x x x 4 4 Let s take a look at oe more example. Example 6 Evaluate the followig itegral. x dx 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 dx = + dx = dx + dx x x x 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( ) B= x= = A A= The itegral is the, x dx = dx + dx x x x+ = x + l x l x+ + c 005 Paul Dawkis 4

42 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 do either 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 + dx 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. x = u + dx= u du Usig this substitutio the itegral is ow, ( u + ) + 4 udu= u 5 udu u = 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. x x+ 0 dx 005 Paul Dawkis 4

43 Solutio We ll do the same thig we did i the previous example. Here s the substitutio, u = x+ 0 x= u 0 dx= udu With this substitutio the itegral is, 4u dx = ( u) du = du x x+ 0 u 0 u 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 = A u+ + 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 = + du x x+ 0 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. Itegrals Ivolvig Quadratics To this poit we ve see quite a few itegrals that ivolve quadratics. A couple of examples are, x x dx = l x ± a + c dx ta = x ± a x + a a a We also saw that itegrals ivolvig doe with a trig substitutio. bx a, a b x ad a + b x could be 005 Paul Dawkis 4

44 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 = 4 4 du u = x + x du = ( x + ) dx 4x + x 4 u l 4 = x + x + c 4 Some itegrals with quadratics ca be doe with partial fractios. For istace, 0x 6 4 dx = 4l 5 l dx = x + x + + c x + 6x+ 5 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 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 = = + + 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 + = taθ x= taθ dx= sec θ dθ x + + = ta θ + = sec θ = secθ = secθ Recall that sice we are doig a idefiite itegral we ca drop the absolute value bars. Usig this substitutio the itegral becomes, 005 Paul Dawkis 44

45 x x dx d = sec θ θ = ( sec θ ta θ + l sec θ + ta θ ) + c We ca fiish the itegral out with the followig right triagle. x θ x x ta = sec = = x + 4x+ 5 θ ( ) = l x x dx x x x x x x 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. b b b b x bx c x bx c x c + + = = 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, b c ax + bx + c = a x + x + a a complete the square o this! 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. 005 Paul Dawkis 45

46 x x+ dx Solutio Okay, this does t factor so partial fractios just wo t work o this. Likewise, sice the umerator is a 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+ = x x+ 9 9 = x x x With this the itegral is, dx = x x+ x 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, dx = du 7 x x+ u + We ca ow see that this is a iverse taget! So, usig the formula from above we get, 4 4u dx = ta + c x x x = ta + c 7 7 Example Evaluate the followig itegral. x dx 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, Paul Dawkis 46

47 x x x x x = = Upo completig the square the itegral becomes, x x dx = dx x + 0x+ 8 x 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 5 dx = du x + 0x+ 8 u + u 6 = du u + u + 6 u = l u + ta c + 6 x + 5 = l ( x + 5) + ta + c So, i geeral whe dealig with a itegral i the form, + B Ax dx ax + bx + c () where the deomiator does t factor 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. 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, Ax + B dx ax + bx + c () Example 4 Evaluate the followig itegral. x dx ( x 6x+ ) Solutio For the most part this itegral will work the same as the previous two with oe exceptio 005 Paul Dawkis 47

48 that will occur dow the road. So, let s start by completig the square o the quadratic i the deomiator. x x x x x 6 + = = + The itegral is the, x x dx = ( x x ) ( 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 x dx = u+ ( x 6x+ ) ( u + ) du u = du + ( u + ) ( 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 = taθ du = sec θ dθ With these two substitutios the itegrals become, du 005 Paul Dawkis 48

49 dx dv dθ x = sec + θ x 6x+ v ta θ + secθ = + dθ 4 v 8( ta θ + ) sec θ = ( u + ) ( sec θ ) (( x ) + ) (( x ) + ) = + dθ sec θ 4 8 dθ cos 4 = + θ dθ 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. cos m θ m d θ si cos m cos m d m θ θ = + m θ θ Let s use this formula to do the itegral. 4 cos θ dθ = siθcos θ + cos 4 4 θ dθ 0 0 = siθcos θ + siθcosθ + cos d cos! 4 4 θ θ θ = = siθcos θ + siθcosθ + θ Next, let s use the followig right triagle to get this back to x s. u x x taθ = = siθ = cosθ = x + x Paul Dawkis 49

50 The cosie itegral is the, 4 x x x cos θ dθ = + + ta ( x ) + ( x ) x x x = + + ta (( x ) + ) ( x ) All told the the origial itegral is, x dx = + x 6x+ 4 x + x x x ta (( x ) 8 ) ( x ) x 9 x 9 x = + + ta + (( x ) + ) ( x ) It s a log ad messy aswer, but there it is. Example 5 Evaluate the followig itegral. x 4 x x Solutio As with the other problems we ll first complete the square o the deomiator. dx 4 x x = x + x 4 = x + x+ 4 = x+ 5 = 5 x+ The itegral is, x x dx = ( 4 x x ) 5 ( x + ) Now, let s do the substitutio. u = x + x = u dx = du ad the itegral is ow, dx c 005 Paul Dawkis 50

51 x u 4 dx = ( 4 x x ) ( 5 u ) du u = du 4 ( 5 u ) ( 5 u ) 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 = 5siθ du = 5cosθ dθ Usig these substitutios the itegral becomes, x 4 dx = 5cos dv d θ θ 4 x x v 5 5si θ 4 5 cosθ = v 5 si ( θ ) 4 5 cosθ = dθ 4 v 5 cos θ 4 5 sec = d v 5 θ θ 5 = ( sec θ ta θ + l sec θ + ta θ ) + c v 5 We ll eed the followig right triagle to fiish this itegral out. u x+ 5 x+ siθ = 5 = 5 secθ = taθ = 5 x+ 5 x+ dθ du So, goig back to x s the itegral becomes, 005 Paul Dawkis 5

52 x 5 5( x + ) 5 x+ dx = + l + + c x x u 5 ( x + ) 5 ( x+ ) 5 ( x+ ) 4x 5 x+ + 5 = + l + c ( x ) ( x ) Ofte the followig formula is eeded whe usig the trig substitutio that we used i the previous example. m sec θ m m m d θ ta sec sec d m θ θ = + m θ θ 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. 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 (5 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. 005 Paul Dawkis 5

53 4x e + e x dx At first glace this looks othig like a trig substitutio problem. However, with the x substitutio u = e we could tur the itegral ito, u + u du which defiitely is a trig substitutio problem ( u = taθ ). We actually did this process x i a sigle step by usig e = taθ, 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. 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, 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 dx = dx dx dx = = x x x x So, we ca ow use the formula with 7 a = x 9 + x 7 dx = + l x x c x x 9 Example Evaluate the followig itegral. cos x dx si x si x 9 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, cos x dx = du si x si x 9 u u Paul Dawkis 5

54 ad this is i the tables. a+ bu a du = l u a + bu a a + bu + a + c if a > 0 = ta a+ bu + c a a 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=-9 ad b= so we ll use the secod formula. cos x u 9 dx ta si x si x 9 ( 9) ( 9) six 9 = ta + c 9 = + This fial example uses a type of formula kow as a reductio formula. Example Evaluate the followig itegral. 4 x cot dx 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, 4 x 4 cot dx = cot u du 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. 4 cot x dx = cot u cot u du At this stage we ca either reuse the reductio formula with = or use the formula cot udu= cot u u+ c c 005 Paul Dawkis 54

55 We ll reuse the reductio formula with = so we ca address somethig that happes o occasio. 4 x 0 0 cot dx = cot cot cot cot! u u u du u = cot = u+ cot u+ du cot = u+ cot u+ u+ c cot x x = + cot + x+ c 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. 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 005 Paul Dawkis 55

56 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 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. 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 x dx 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 dx x x dx x 005 Paul Dawkis 56

57 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 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 b x? 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. cos x + si 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, + u du which is a trig substitutio problem. xdx 005 Paul Dawkis 57

58 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 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 4 dx 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. 005 Paul Dawkis 58

59 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. ta x si x 4 dx = 4 cos x dx sec x cos x = si x cos 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. 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. ta x ta xsec x dx = sec 4 dx u = x 5 sec x sec x = du 5 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. 005 Paul Dawkis 59

60 Example Evaluate the followig itegral. + six 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, six dx = dx + si x + si x si x six = dx si x 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. six dx = dx + six cos x six = dx cos x cos x cos x = sec x ta xsec xdx = ta x 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. u = x x= u dx= udu 005 Paul Dawkis 60

61 With this substitutio the itegral becomes, cos x dx = u cosu du 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 e dx cos( x ) dx dx cos dx x e 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 ad have it do these itegrals here is what it will retur the followig. x π e dx = erf ( x) π cos( x ) dx= FreselC x π si ( x) dx = Si( x ) x x x cos 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 = π e x ( x) 0 Si FreselC ( x) = 0 x t si t dt t dt π x = t dt cos 0 x 005 Paul Dawkis 6

62 x cost Ci( x) = γ + l ( x) + dt 0 t Where γ is the Euler-Mascheroi costat. 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. 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. dx 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 x the iterval [, ) is. We still are t able to do this, however, let s step back a little ad istead ask what the, t were t > ad t is fiite. This is a problem that we area uder f(x) is o the iterval [ ] 005 Paul Dawkis 6

63 ca do. At t = dx= = x x t t Now, we ca get the area uder f(x) o [, ) simply by takig the limit of A t as t goes to ifiity. A= lim At = lim = t t t This is the how we well do the itegral it self. t dx = lim dx t x x t = lim t x = lim 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 a = lim t provided the limit exists ad is fiite. f x dx f x dx t a 005 Paul Dawkis 6

64 b. If f x dx exists for every t < b the, t b b = lim provided the limits exists ad is fiite. c. If f ( x) dx ad f f x dx f x dx t x dx are both coverget the, c = 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. t c Let s take a look at a couple more examples. Example Determie if the follow itegral is coverget or diverget ad if it s coverget fid it s value. dx x Solutio So, the first thig we do is covert the itegral to a limit. t dx = lim dx t x x Now, do the itegral ad the limit. t dx = lim l ( x) t x = lim l t l t = ( () ) So, the limit is ifiite ad so the itegral is diverget. If we go back to thikig i terms of area otice that the area uder g( x) = o the x iterval [, ) is ifiite. This is i cotrast to the area uder f ( x) = which was x 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 005 Paul Dawkis 64

65 p a x is coverget if p > ad diverget if p. dx 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 goes to zero will x 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 dx x 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. 0 0 dx = lim dx x t x = lim x t ( t ) = lim + t = + = So, the limit is ifiite ad so this itegral is diverget. 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. t 0 0 t 005 Paul Dawkis 65

66 0 0 x x e = lim t e t x dx x dx x = lim t e = lim + e t = 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. t x x xe dx = lim xe dx 0 t 0 x = lim t e 0 t = lim e + 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 xe dx= x dx 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 x dx Solutio First covert to a limit. si x dx = lim si x dx t = lim t 0 t t t ( cos x) ( cost) = lim cos t 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 t t 005 Paul Dawkis 66

67 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 [ ab, ) ad ot cotiuous at x=b the, a b t = lim f x dx f x dx t b 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.. If f(x) is cotiuous o the iterval ( ab, ] ad ot cotiuous at x=a the, a b b = lim f x dx f x dx + t a 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. a t c b < < ad ad f. If f(x) is ot cotiuous at x=c where a c b f x dx a are both coverget the, b c b f x dx= f x dx+ f x dx a a c 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=a ad x=b ad if ad f f x dx a both coverget the, b c b f x dx= f x dx+ f x dx a a c c c 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 used oe-sided limits to make sure we stay iside the iterval. Let s do a couple of examples of these kids of itegrals. 005 Paul Dawkis 67

68 Example 6 Determie if the followig itegral is coverget or diverget. If it is coverget fid its value. dx 0 x Solutio The problem poit is the upper limit so we are i the first case above. t dx = lim dx x t x 0 0 t ( x ) ( t ) = lim t 0 = lim t = 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. dx 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 + dx x x x 0 Now we eed to look at each of these itegrals ad see if they are coverget. 0 t dx = lim dx x t 0 x t = lim t 0 x = lim + t 0 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. Example 8 Determie if the followig itegral is coverget or diverget. If it is 005 Paul Dawkis 68

69 coverget fid its value. dx 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 + dx x x 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 above. So, all we eed to do is check the first itegral. dx = lim dx + 0 x t 0 t x = lim+ t 0 x t = lim + + t 0 t = So, the first itegral is diverget ad so the whole itegral is diverget. 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 005 Paul Dawkis 69

70 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 uder the smaller fuctio must also be fiite (i.e. a x dx coverges) the the area a g x dx coverges). Likewise, if the area uder the smaller fuctio is ifiite (i.e. the area uder the larger fuctio must also be ifiite (i.e. f a g x dx diverges) the a 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 coverge. 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. Example Determie if the followig itegral is coverget or diverget. cos x dx 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 005 Paul Dawkis 70

71 dx 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, cos x x x Now, as we ve already oted dx x coverges ad so by the Compariso Test we kow that cos x dx x must also coverge. Example Determie if the followig itegral is coverget or diverget. dx x 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 dx p x a 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 005 Paul Dawkis 7

72 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 dx 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. x < = e x x x + e e Also, t x x e dx = lim e dx So, t ( t e e ) = lim + = e t e x dx is coverget. Therefore, by the Compariso test is also coverget. dx x x + e Example Determie if the followig itegral is coverget or diverget. dx x 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. 005 Paul Dawkis 7

73 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 dx x diverges ad so by the Compariso Test we kow that dx x x e must also diverge. Example 4 Determie if the followig itegral is coverget or diverget. 4 + si ( x) dx x 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) 0 si. I particular, this term is positive ad so if we drop it from the umerator the umerator will get smaller. This gives, 4 + si ( x) > x x ad diverges so by the Compariso Test also diverges. dx x 4 si x dx + 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. x 005 Paul Dawkis 7

74 Let s take a look at a example that works a little differetly so we do t get too locked ito these ideas. Example 5 Determie if the followig itegral is coverget or diverget. x e dx 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. x x e e x < = e x ad we ca show that e x dx 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 x e dx x also coverges. 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. x We ll take advatage of the fact that e is a decreasig fuctio. This meas that x x x > x e < e I other words, plug i a larger umber ad the fuctio gets smaller. From the limits of itegratio we kow that x> ad this meas that if we square x we will get larger. Or, x > x provided x> Note that we ca oly say this sice x>. This wo t be true if x<! We ca ow use the 005 Paul Dawkis 74

75 fact that x e is a decreasig fuctio to get, x e < e x x So, e is a larger fuctio tha e ad we kow that e x dx coverges so by the Compariso Test we also kow that e x dx is coverget. x 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. Example 7 Determie if the followig itegral is coverget or diverget. x e dx 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. 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 = ito the deomiator ad 4 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 dx x x x = We used Maple to get the value of the first itegral. Now, if the secod itegral e x x dx 005 Paul Dawkis 75

76 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. Approximatig Defiite Itegrals I this chapter we ve spet quite a bit to time o computig the values of itegrals. However, ot all itegrals ca be computed. A perfect example is the followig defiite itegral. x e dx 0 We ow eed to talk a little bit about estimatig values of defiite itegrals. We will look at three differet methods, although oe should already be familiar to you from your Calculus I days. We will develop all three methods for estimatig b f ( x ) dx a by thikig of the itegral as a area problem ad usig kow shapes to estimate the area uder the curve. Let s get first develop the methods ad the we ll try to estimate the itegral show above. Midpoit Rule This is the rule that you should be somewhat familiar to you. We will divide the iterval ab, ito subitervals of equal width, [ ] b a Δ x = We will deote each of the itervals as follows, [ x0, x], [ x, x],,[ x, x ] where x0 = a ad x = b 005 Paul Dawkis 76

77 The for each iterval let * x i be the midpoit of the iterval. We the sketch i * rectagles for each subiterval with a height of ( i ) up usig =6. f x. Here is a graph showig the set We ca easily fid the area for each of these rectagles ad so for a geeral we get that, b * * * f ( x ) dx Δ x f ( x ) +Δ x f ( x ) + +Δ x f ( x ) a Or, upo factorig out a Δx we get the geeral Mid Poit Rule. b * * * Δ ( ) f x dx x f x f x f x a Trapezoid Rule For this rule we will do the same set up as for the Midpoit Rule. We will break up the ab, ito subitervals of width, iterval [ ] b a Δ x = The o each subiterval we will approximate the fuctio with a straight lie that is equal to the fuctio values at either edpoit of the iterval. Here is a sketch of this case for = Paul Dawkis 77

78 Each of these objects is a trapezoid (hece the rules ame ) ad as we ca see some of them do a very good job of approximatig the actual area uder the curve ad others do t do such a good job. The area of the trapezoid i the iterval [ x, i xi] Δx Ai = ( f ( xi ) + f ( xi) ) is give by, So, if we use subitervals the itegral is approximately, b Δx Δx Δx f ( x) dx ( f ( x0) + f ( x) ) + f ( x) + f ( x) + + f x a + f x ( ) Upo doig a little simplificatio we arrive at the geeral Trapezoid Rule. b Δx f x dx f x0 f x f x f x f x a Note that all the fuctio evaluatios, with the exceptio of the first ad last, are multiplied by. Simpso s Rule This is the fial method we re goig to take a look at ad i this case we will agai divide up the iterval [ ab, ] ito subitervals. However ulike the previous two methods we eed to require that be eve. The reaso for this will be evidet i a bit The width of each subiterval is, b a Δ x = 005 Paul Dawkis 78

79 I the Trapezoid Rule we approximated the curve with a straight lie. For Simpso s Rule we are goig to approximate the fuctio with a quadratic ad we re goig to require that the quadratic agree with three of the poits from our subitervals. Below is a sketch of this usig =6. Each of the approximatios is colored differetly so we ca see how they actually work. Notice that each approximatio actually covers two of the subitervals. This is the reaso for requirig to be eve. It ca be show that the area uder the approximatio x x x, x + is, o the itervals [, i i] ad [ i i ] Δx Ai = ( f ( xi ) + 4 f ( xi) + f ( xi+ ) ) If we use subitervals the itegral is the approximately, b Δx Δx f ( x) dx ( f ( x0) + 4f ( x) + f ( x) ) + ( f ( x) + 4f ( x) + f ( x4) ) a Δx + + f x + 4 f x + f x Upo simplifyig we arrive at the geeral Simpso s Rule. ( ( ) ( ) ) b Δx f x dx f x f x f x f x f x f x a I this case otice that all the fuctio evaluatios at poits with odd subscripts are multiplied by 4 ad all the fuctio evaluatios at poits with eve subscripts (except for the first ad last) are multiplied by. If you ca remember this, this a fairly easy rule to remember. Okay, it s time to work a example ad see how these rules work. 005 Paul Dawkis 79

80 Example Usig =4 ad all three rules to approximate the value of the followig itegral. e x dx 0 Solutio First, for referece purposes, Maple gives the followig value for this itegral. x e dx = I each case the width of the subitervals will be, 0 Δ x = = 4 ad so the subitervals will be, [ 0, 0.5 ], [ 0.5, ], [,.5 ], [.5, ] Let s go through each of the methods. 0 Midpoit Rule ( 0.5) e x dx = e e e e Remember that we evaluate at the midpoits of each of the subitervals here! The Midpoit Rule has a error of Trapezoid Rule ( 0) e x dx = e e e e e The Trapezoid Rule has a error of Simpso s Rule ( 0) e x dx = e e e e e The Simpso s Rule has a error of Noe of the estimatios i the previous example are all that good. The best approximatio i this case is from the Simpso s Rule ad yet it s still had a error of almost. To get a better estimatio we would eed to use a larger. So, for completeess sake here are the estimates for some larger value of. Midpoit Trapezoid Simpso s Approx. Error Approx. Error Approx. Error Paul Dawkis 80

81 I this case we where able to determie the error for each estimate because we could get our hads o the exact value. Ofte this wo t be the case ad so we d ext like to look at error bouds for each estimate. These bouds will give the largest possible error i the estimate, but it should also be poited out that the actual error may be sigificatly smaller tha the boud. The boud is oly there so we ca say that we kow the actual error will be less tha the boud. So, suppose that f ( x) K ad ( 4 f ) ( x) M for a x b the if E M, E T, ad E S are the actual errors for the Midpoit, Trapezoid ad Simpso s Rule we have the followig bouds, 5 K( b a) K( b a) M ( b a) EM E T E S Example Determie the error bouds for the estimatios i the last example. Solutio We already kow that =4, a=0, ad b= so we just eed to compute K (the largest value of the secod derivative) ad M (the largest value of the fourth derivative). This meas that we ll eed the secod ad fourth derivative of f(x). x f x = e + x Here is a graph of the secod derivative. ( 4 ) x 4 = 4e ( ) f x x x Here is a graph of the fourth derivative. 005 Paul Dawkis 8

82 So, from these graphs it s clear that the largest value of both of these are at x=. So, f = K = 98 f ( 4 ) = M = 56 We rouded to make the computatios simpler. Here are the bouds for each rule. E M E T E S ( ) ( ) ( ) = = = I each case we ca see that the errors are sigificatly smaller tha the actual bouds. Applicatios of Itegrals Itroductio I this sectio we re goig to take a look at some of applicatios of itegratio. It should be oted as well that these applicatios are preseted here, as opposed to Calculus I, simply because may of the itegrals that arise from these applicatios ted to require techiques that we discussed i the previous chapter. Here is a list of applicatios that we ll be takig a look at i this chapter. Arc Legth We ll determie the legth of a curve i this sectio. Surface Area I this sectio we ll determie the surface area of a solid of revolutio. 005 Paul Dawkis 8

83 Ceter of Mass Here we will determie the ceter of mass or cetroid of a thi plate. Hydrostatic Pressure ad Force We ll determie the hydrostatic pressure ad force o a vertical plate submerged i water. Probability Here we will look at probability desity fuctios ad computig the mea of a probability desity fuctio. Arc Legth I this sectio we are goig to look at computig the arc legth of a fuctio. Because it s easy eough to derive the formulas that we ll use i this sectio we will derive oe of them ad leave the other to you to derive. We wat to determie the legth of the cotiuous fuctio y = f ( x) o the iterval [ ab, ]. Iitially we ll eed to estimate the legth of the curve. We ll do this by dividig the iterval up ito equal subitervals each of width Δx ad we ll deote the poit o the curve at each poit by P i. We ca the approximate the curve by a series of straight lies coectig the poits. Here is a sketch of this situatio for =9. Now deote the legth of each of these lie segmets by Pi Pi ad the legth of the curve will the be approximately, L Pi Pi i= ad we ca get the exact legth by takig larger ad larger. I other words, the exact legth will be, L= lim Pi Pi i = 005 Paul Dawkis 8

84 Now, let s get a better grasp o the legth of each of these lie segmets. First, o each segmet let s defie Δ yi = yi yi = f ( xi) f ( xi ). We ca the compute directly the legth of the lie segmets as follows. i i = i i + i i = Δ +Δ i P P x x y y x y By the Mea Value Theorem we kow that o the iterval [ x x ] that, * = ( ) f x f x f x x x i i i i i i * ( i ) Δ y = f x Δx Therefore, the legth ca ow be writte as, P P = x x + y y i i i i i i * ( i ) ( i ) * = Δ x + f x Δx = + f x Δx The exact legth of the curve is the, L= lim P P i = i = i * ( i ) = lim + f x Δx i, i i there is a poit However, usig the defiitio of the defiite itegral, this is othig more tha, b L= + f x dx a A slightly more coveiet otatio (i my opiio ayway) is the followig. b dy L= + dx I a similar fashio we ca also derive a formula for x = h( y) o [, ] d dx L= + h ( y) dy = + dy c dy c Agai, the secod form is probably a little more coveiet. a d dx * x i so cd. This formula is, Note the differece i the derivative uder the square root! Do t get too cofused. With oe we differetiate with respect to x ad with the other we differetiate with respect to y. 005 Paul Dawkis 84

85 Oe way to keep the two straight is to otice that the differetial i the deomiator of the derivative will match up with the differetial i the itegral. This is oe of the reasos why the secod form is a little more coveiet. Before we work ay examples we eed to make a small chage i otatio. Istead of havig two formulas for the arc legth of a fuctio we are goig to reduce it, i part, to a sigle formula. From this poit o we are goig to use the followig formula for the legth of the curve. where, L = ds dy ds = + dx if y = f ( x), a x b dx dx ds = + dy if x = h( y), c y d dy Note that o limits where put o the itegral as the limits will deped upo the ds that we re usig. Usig the first ds will require x limits of itegratio ad usig the secod ds will require y limits of itegratio. Thikig of the arc legth formula as a sigle itegral with differet ways to defie ds will be coveiet whe we ru across arc legths i future sectios. Also, this ds otatio will be a ice otatio for the ext sectio as well. Now that we ve derived the arc legth formula let s work some examples. Example Determie the legth of y l ( sec x) Solutio π = betwee 0 x. 4 I this case we ll eed to use the first ds sice the fuctio is i the form y f ( x) let s get the derivative out of the way. dy sec x ta x ta dy = = x = ta dx sec x dx x =. So, Let s also get the root out of the way sice there is ofte simplificatio that ca be doe ad there s o reaso to do that iside the itegral. dy ta sec sec sec + = + x = x = x = x dx Note that we could drop the absolute value bars here sice secat is positive i the rage give. 005 Paul Dawkis 85

86 The arc legth is the, L= π 4 0 sec xdx = l sec x + ta x = l + π 4 0 Example Determie the legth of x= ( y ) betwee y 4. Solutio There is a very commo mistake that studets make i problems of this type. May studets see that the fuctio is i the form x = h( y) ad they immediately decide that it will be too difficult to work with it i that form so they solve for y to get the fuctio ito the form y = f ( x). While that ca be doe here it will lead to a messier itegral for us to deal with. Sometimes it s just easier to work with fuctios i the form x = h( y). I fact, if you ca work with fuctios i the form y = f ( x) the you ca work with fuctios i the form x = h( y). There really is t a differece betwee the two so do t get excited about fuctios i the form x = h( y). Let s compute the derivative ad the root. dx dx = ( y ) + = + y = y dy dy As you ca see keepig the fuctio i the form x = h( y) is goig to lead to a very easy itegral. To see what would happe if we tried to work with the fuctio i the form y = f x see the ext example. Let s get the legth. L= = = 4 4 y ydy 4 As oted i the last example we really do have a choice as to which ds we use. Provided we ca get the fuctio i the form required for a particular ds we ca use it. However, 005 Paul Dawkis 86

87 as also oted above, there will ofte be a sigificat differece i difficulty i the resultig itegrals. Let s take a quick look at what would happe i the previous example y = f x. if we did put the fuctio ito the form Example Redo the previous example usig the fuctio i the form y = f ( x) istead. Solutio I this case the fuctio ad its derivative would be, x dy y x = + = dx The root i the arc legth formula would the be. x x + x x x dy + = + = = dx All the simplificatio work above was just to put the root ito a form that will allow us to do the itegral. Now, before we write dow the itegral we ll also eed to determie the limits. This particular ds requires x limits of itegratio ad we ve got y limits. They are easy eough to get however. Sice we kow x as a fuctio of y all we eed to do is plug i the origial y limits of itegratio ad get the x limits of itegratio. Doig this gives, 0 x Not easy limits to deal with, but there they are. Let s ow write dow the itegral that will give the legth. L= 0 x x + dx That s a really upleasat lookig itegral. It ca be evaluated however usig the followig substitutio. x x u = + du = dx x = 0 u = x= u = 4 Usig this substitutio the itegral becomes, Paul Dawkis 87

88 L= = u 4 = So, we got the same aswer as i the previous example. Although that should t really be all that surprisig sice we were dealig with the same curve. 4 From a techical stadpoit the itegral i the previous example was ot that difficult. It was just a Calculus I substitutio. However, from a practical stadpoit the itegral was sigificatly more difficult tha the itegral we used i Example. So, the moral of the story here is that we ca use either formula (provided we ca get the fuctio i the correct form of course) however oe will ofte be sigificatly easier to actually evaluate. Okay, let s work oe more example. Example 4 Determie the legth of x = y for 0 x. Assume that y is positive. Solutio We ll use the secod ds for this oe as the fuctio is already i the correct form for that oe. Also, the other ds would agai lead to a particularly difficult itegral. The derivative ad root will the be, udu 4 dx dx = y + = + y dy dy Before writig dow the legth otice that we were give x limits ad we will eed y limits for this ds. With the assumptio that y is positive there are easy eough to get. They are, 0 y The itegral for the arc legth is the, L= + 0 y dy This itegral will require the followig trig substitutio. y = taθ dy = sec θ dθ y = 0 0= taθ θ = 0 y = = taθ π θ = 4 y θ θ θ θ + = + ta = sec = sec = sec 005 Paul Dawkis 88

89 The legth is the, L= π 4 0 sec θ dθ = sec ta + l sec + ta ( θ θ θ θ ) ( l( ) ) = + + π 4 0 The first couple of examples eded up beig fairly simple Calculus I substitutios. However, as this last example had show we will ofte ed up with trig substitutios as well for these itegrals. Surface Area I this sectio we are goig to look oce agai at solids of revolutio. We first looked at them back i Calculus I whe we foud the volume of the solid of revolutio. I this sectio we wat to fid the surface area of this regio. So, for the purposes of the derivatio of the formula, let s look at rotatig the cotiuous fuctio y = f ( x) i the iterval [ ab, ] about the x-axis. Below is a sketch of a fuctio ad the solid of revolutio we get by rotatig the fuctio about the x-axis. We ca derive a formula for the surface area much as we derived the formula for arc legth. We ll start by dividig the itegral ito equal subitervals of width Δ x. O each subiterval we will approximate the fuctio with a straight lie that agrees with the fuctio at the edpoits of the each iterval. Here is a sketch of that for our represetative fuctio usig = Paul Dawkis 89

90 Now, rotate the approximatios about the x-axis ad we get the followig solid. The approximatio o each iterval gives a distict portio of the solid ad to make this clear each portio is colored differetly. Each of these portios are called frustums ad we kow how to fid the surface area of frustums. The surface area of a frustum is give by, A= π rl where, r = ( r+ r) r = radius of right ed r = radius of left ed ad l is the legth of the slat of the frustum. For the frustum o the iterval [ x, i xi] we have, r = f ( xi ) r = f ( xi ) l = P P ( legth of the lie segmet coectig P ad P ) i i i i 005 Paul Dawkis 90

91 ad we kow from the previous sectio that, ( * ) * [ ] P P = + f x Δx where x is some poit i x, x i i i i i i Before writig dow the formula for the surface area we are goig to assume that small ad sice f(x) is cotiuous we ca the assume that, f x f x * ad f x f x * ( i) ( i ) ( i ) ( i ) So, the surface area of the frustum o the iterval [ x, i xi] f ( x ) + f ( x ) i i Ai = π Pi Pi * * i ( i ) π f x + f x Δx The surface area of the whole solid is the approximately, i= * * i i S π f x + f x Δx is approximately, ad we ca get the exact surface area by takig the limit as goes to ifiity. * * S = lim π f ( xi ) + f ( xi ) Δx = i b = π f x + f x dx a Δ x is If we wated to we could also derive a similar formula for rotatig x = h( y) o [ cd, ] about the y-axis. This would give the followig formula. d S = π h( y) + h ( y) dy c These are ot the stadard formulas however. Notice that the roots i both of these formulas are othig more tha the two ds s we used i the previous sectio. Also, we will replace f(x) with y ad h(y) with x. Doig this gives the followig two formulas for the surface area. where, S = π yds rotatio about x axis S = π xds rotatio about y axis 005 Paul Dawkis 9

92 dy ds = + dx if y = f ( x), a x b dx dx ds = + dy if x = h( y), c y d dy There are a couple of thigs to ote about these formulas. First, otice that the variable i the itegral itself is always the opposite variable from the oe we re rotatig about. Secod, we are allowed to use either ds i either formula. This meas that there are, i some way, four formulas here. We will choose the ds based upo which is the most coveiet for a give fuctio ad problem. Now let s work a couple of examples. Example Determie the surface area of the solid obtaied by rotatig x about the x-axis. y = 9 x, Solutio The formula that we ll be usig here is, S = π yds sice we are rotatig about the x-axis ad we ll use the first ds i this case because our fuctio is i the correct form for that ds ad we wo t gai aythig by solvig it for x. Let s first get the derivative ad the root take care of. dy x = ( 9 x ) ( x) = dx 9 x Here s the itegral for the surface area, dy x 9 + = + = = dx 9 x 9 x 9 x S = π y dx 9 x There is a problem however. The dx meas that we should t have ay y s i the itegral. So, before evaluatig the itegral we ll eed to substitute i for y as well. The surface area is the, 005 Paul Dawkis 9

93 S = π 9 x dx 9 x = = 4π 6π dx Previously we made the commet that we could use either ds i the surface area formulas. Let s work a example i which usig either ds wo t create itegrals that are too difficult to evaluate ad so we ca check both ds s. Example Determie the surface area of the solid obtaied by rotatig y about the y-axis. Use both ds s to compute the surface area. y = x, Solutio Note that we ve bee give the fuctio set up for the first ds ad limits that work for the secod ds. Solutio This solutio will use the first ds listed above. We ll start with the derivative ad root. dy = x dx 4 4 dy 9x + 9x + + = + = = 4 4 dx 9x 9x x We ll also eed to get ew limits. That is t too bad however. All we eed to do is plug i the give y s ad get that the rage of x s is x 8. The itegral for the surface area is the, 8 4 9x + S = π x dx x 8 4 π = x 9x + dx Note that this time we did t eed to substitute i for the x as we did i the previous example. I this case we picked up a dx from the ds ad so we do t eed to do a substitutio for the x. I fact if we had substituted for x we would have put y s ito itegral which would have caused problems. Usig the substitutio the itegral becomes, 4 u = 9x + du = x dx 005 Paul Dawkis 9

94 π S = u du π = u 7 0 π = 45 0 = Solutio This time we ll use the secod ds. So, we ll first eed to solve the equatio for x. We ll also go ahead ad get the derivative ad root while we re at it. dx x = y = y dy The surface area is the, dx + = + 9y dy S = π x + y 4 dy 9 We used the origial y limits this time because we picked up a dy from the ds. Also ote that the presece of the dy meas that this time, ulike the first solutio, we ll eed to substitute i for the x. Doig that gives, 4 4 S = π y + 9y dy u = + 9y π 45 = 8 udu 0 π = 45 0 = Note that after the substitutio the itegral was idetical to the first solutio ad so the work was skipped. As this example has show we ca used either ds to get the surface area. It is importat to poit out as well that with oe ds we had to do a substitutio for the x ad with the other we did t. This will always work out that way. Note as well that i the case of the last example it was just as easy to use either ds. That ofte wo t be the case. I may examples oly oe of the ds will be coveiet to work with so we ll always eed to determie which ds is liable to be the easiest to work with before startig the problem Paul Dawkis 94

95 Ceter of Mass I this sectio we are goig to fid the ceter of mass or cetroid of a thi plate with uiform desity ρ. The ceter of mass or cetroid of a regio is the poit i which the regio will be perfectly balaced horizotally if suspeded from that poit. So, let s suppose that the plate is the regio bouded by the two curves f(x) ad g(x) o the iterval [a,b]. So, we wat to fid the ceter of mass of the regio below. We ll first eed the mass of this plate. The mass is, M = ρ Area of plate b = ρ f x g x dx a Next we ll eed the momets of the regio. There are two momets, deoted by M x ad M y. The momets measure the tedecy of the regio to rotate about the x ad y-axis respectively. The momets are give by, ( ) b M x = ρ f x g x dx y a b a ( ) M = ρ x f x g x dx The coordiates of the ceter of mass, ( x, y ), are the, 005 Paul Dawkis 95

96 b ( ) M x f x g x dx y a b x = = = x b ( f ( x) g( x) ) dx M A a f x g x dx a ( ) b f x g x dx b M a x y = = = b ( f ( x) g( x) ) dx M f x g x dx A a a where, b a A= f x g x dx Note that the desity, ρ, of the plate cacels out ad so is t really eeded. Let s work a couple of examples. =, y = 0 π o the iterval 0,. Solutio Here is a sketch of the regio with the ceter of mass deoted with a small box. Example Determie the ceter of mass for the regio bouded by y si( x) Let s first get the area of the regio. A= π 0 si x dx ( x) = cos = Now, the momets (without desity sice it will just drop out) are, π Paul Dawkis 96

97 π 0 M x = π si 0 π 0 = cos 4x dx = x si 4 4 π = x dx ( x) M = xsi x dx itegratig by parts... y 0 0 π 0 = xcos x + cos x dx π = xcos x + si x 0 π = The coordiates of the ceter of mass are the, π π x = = 4 π y π = = 4 π π So, the ceter of mass for this regio is, 4 4. π π π 4 0 Example Determie the ceter of mass for the regio bouded by y =. = x ad y x Solutio The two curves itersect at x=0 ad x= ad here is a sketch of the regio with the ceter of mass marked with a box. 005 Paul Dawkis 97

98 We ll first get the area of the regio. A= x x dx 0 4 = x x 4 5 = Now the momets, agai without desity, are 6 M x = ( x x ) dx 0 7 = x x = 8 M = x x x dx y 0 0 = x 4 x dx 5 5 = x x 5 5 = 5 The coordiates of the ceter of mass is the, Paul Dawkis 98

99 5 x = = y = = 5 7 The coordiates of the ceter of mass are the,, 5 7. Hydrostatic Pressure ad Force I this sectio we are goig to submerge a vertical plate i water ad we wat to kow the force that is exerted o the plate due to the pressure of the water. This force is ofte called the hydrostatic force. There are two basic formulas that we ll be usig here. First, if we are d meters below the surface the the hydrostatic pressure is give by, P= ρgd where, ρ is the desity of the fluid ad g is the gravitatioal acceleratio. We are goig to assume that the fluid i questio is water ad sice we are goig to be usig the metric system these quatities become, ρ = 000 kg/m g = 9.8 m/s The secod formula that we eed is the followig. Assume that a costat pressure P is actig o a surface with area A. The the hydrostatic force that acts o the area is, F = PA Note that we wo t be able to fid the hydrostatic force o a vertical plate usig this formula sice the pressure will vary with depth. We will however eed this for our work. The best way to see how these problems work is to do a example or two. Example Determie the hydrostatic force o the followig triagular plate that is submerged i water as show. 005 Paul Dawkis 99

100 Solutio The first thig to do here is set up a axis system. We ll use the followig axis system. So, we are goig to oriet the x-axis so that positive x is dowward, x=0 correspods to the waters surface ad x=4 correspods to the depth of the tip of the triagle. Next we are break up the triagle ito strips each of equal width Δ x ad i each iterval [, * xi xi] choose ay poit x i. I order to make the computatios easier we are goig to make two assumptios about these strips. First, we will igore the fact that the eds are actually goig to be slated ad assume the strips are rectagular. If Δ x is sufficietly small this will ot affect our computatios much. Secod, we will assume that Δ x is small eough that the hydrostatic pressure o each strip is essetially costat. Below is a represetative strip. 005 Paul Dawkis 00

101 The height of this strip is Δ x ad the width is a. We ca use similar triagles to determie a. a * = a= x * 4 4 x 4 i i Now, sice we are assumig the pressure o this strip is costat, the pressure is give by, P 000( 9.8) * 980 * i = ρgd = xi = xi ad the hydrostatic force is * * * * Fi = Pi A= Pi aδ x = xi xi Δ x= xi xi Δx 4 4 The approximate hydrostatic force o the plate is the the sum of the forces o all the strips or, * * F 960xi xi Δx i= 4 Takig the limit will get the exact hydrostatic force, * * F = lim 960xi xi Δ x i = 4 Usig the defiitio of the defiite itegral this is othig more tha, 005 Paul Dawkis 0

102 4 F 960 = x x dx 0 4 The hydrostatic force is the, 4 F 960 x x dx = 4 0 = 960 x x 4 = N 4 0 Let s take a look at aother example. Example Fid the hydrostatic force o a circular plate of radius that is submerged 6 meters i the water. Solutio First, we re goig to assume that the top of the circular plate is 6 meters uder the water. Next, we will set up the axis system so that the origi of the axis system is at the ceter of the plate. Fially, we will agai split up the plate ito strips each of width Δ y ad * we ll choose a poit y i from each strip. We ll also assume that the strips are rectagular agai to help with the computatios. Here is a sketch of the setup. 005 Paul Dawkis 0

103 The depth below the waters surface of each strip is, * d = 8 y ad that i tur gives us the pressure o the strip, ( * Pi = ρgdi = yi ) i i The area of each strip is, * i A = 4 y Δ y i The hydrostatic force o each strip is, The total force o the plate is, * * F = PA = y 4 y Δ y i i i i i i = ( y) * * ( i ) ( i ) F = lim y 4 y Δy = y dy To do this itegral we ll eed to split it up ito two itegrals. F = y dy 960 y 4 y dy The first itegral requires the trig substitutio y = siθ ad the secod itegral eeds the substitutio v= 4 y. After usig these substitutio we get, π 0 θ θ π 0 π F = cos d vdv π ( θ) = 90 + cos dθ + 0 = 90 θ + si = 90π ( θ) Note that we kow the secod itegral will be zero because the upper ad lower limit is the same. π π 005 Paul Dawkis 0

104 Probability I this last applicatio of itegrals that we ll be lookig at we re goig to look at probability. Before actually gettig ito the applicatios we eed to get a couple of defiitios out of the way. Suppose that we wated to look at the age of a perso, the height of a perso, the amout of time spet waitig i lie, or maybe the lifetime of a battery. Each of these quatities have values that will rage over a iterval of itegers. Because of this these are called cotiuous radom variables. Cotiuous radom variables are ofte represeted by X. Every cotiuous radom variable, X, has a probability desity fuctio, f(x). Probability desity fuctios satisfy the followig coditios.. f ( x) 0 for all x.. f ( x) dx = Probability desity fuctios ca be used to determie the probability that a cotiuous radom variable lies betwee two values, say a ad b. This probability is deoted by P a X b ad is give by, Let s take a look at a example of this. b P a X b = f x dx x Example Let f ( x) = ( 0 x) for 0 x 0 ad f ( x ) = 0 for all other values of 5000 x. Aswer each of the followig questios about this fuctio. (a) Show that f(x) is a probability desity fuctio. P X 4 Solutio (b) Fid (c) Fid P( x 6) (a) To show this is a probability desity fuctio we ll eed to show that f ( x) dx= a 0 x f ( x) dx= ( 0 x) dx x x = = Paul Dawkis 04

105 Note the chage i limits o the itegral. The fuctio is oly o-zero i these rages ad so the itegral ca be reduced dow to oly the iterval where the fuctio is ot zero. (b) I this case we eed to evaluate the followig itegral. 4 x P( X 4) = ( 0 x) dx x x = = So the probability of X beig betwee ad 4 is 8.658%. (c) Note that i this case P( x 6) is equivalet to P( 6 X 0) sice 0 is the largest value that X ca be. So the probability that X is greater tha or equal to 6 is, 0 x P( X 6) = ( 0 x) dx 5000 This probability is the 66.04% x x = = Probability desity fuctios ca also be used to determie the mea of a cotiuous radom variable. The mea is give by, Let s work oe more example. μ = xf x dx Example It has bee determied that the probability desity fuctio for the wait i lie at a couter is give by, 0 if t < 0 f () t = t 0 0.e if t 0 where t is the umber of miutes spet waitig i lie. Aswer each of the followig questios about this probability desity fuctio. (a) Verify that this is i fact a probability desity fuctio. (b) Determie the probability that a perso will wait i lie for at least 6 miutes. (c) Determie the mea wait i lie. Solutio (a) Not much to do here other tha compute the itegral Paul Dawkis 05

106 So it is a probability desity fuctio. t 0 () = 0.e 0 f t dt u = lim 0.e u 0 = lim e u = lim e u = t 0 dt t 0 u 0 u 0 (b) The probability that we re lookig for here is P( x 6). 0 ( 6) = 0.e P X 6 = lim 0.e u 5 6 t 0 t 0 = lim e u = lim e e u 6 6 u 0 0 = e = So the probability that a perso will wait i lie for more tha 6 miutes is 54.88%. (c) Here s the mea wait time. μ = tf t dt = = 0 () 0.te t 0 t u 0 lim 0.t dt itegratig by parts... u e 0 u t 0 lim ( t 0) e u 0 u 0 lim 0 ( u 0) e u = + = + = 0 So, it looks like the average wait time is 0 miutes. dt u t dt u dt dt 005 Paul Dawkis 06

107 Parametric Equatios ad Polar Coordiates Itroductio I this sectio we will be lookig at parametric equatios ad polar coordiates. While the two subjects do t appear to have that much i commo o the surface we will see that several of the topics i polar coordiates ca be doe i terms of parametric equatios ad so i that sese they make a good match i this chapter. We will also be lookig at how to do may of the stadard calculus topics such as tagets ad area i terms of parametric equatios ad polar coordiates. Here is a list of topics that we ll be coverig i this chapter. Parametric Equatios ad Curves A itroductio to parametric equatios ad parametric curves (i.e. graphs of parametric equatios) Tagets with Parametric Equatios Fidig taget lies to parametric curves. Area with Parametric Equatios Fidig the area uder a parametric curve. Arc Legth with Parametric Equatios Determiig the legth of a parametric curve. Surface Area with Parametric Equatios Here we will determie the surface area of a solid obtaied by rotatig a parametric curve about a axis. Polar Coordiates We ll itroduce polar coordiates i this sectio. We ll look at covertig betwee polar coordiates ad Cartesia coordiates. We will also look at some basic graphs i polar coordiates. Tagets with Polar Coordiates Fidig taget lies of polar curves. Area with Polar Coordiates Fidig the area eclosed by a polar curve. Arc Legth with Polar Coordiates Determiig the legth of a polar curve. Surface Area with Polar Coordiates Here we will determie the surface area of a solid obtaied by rotatig a polar curve about a axis. Arc Legth ad Surface Area Revisited I this sectio we will summarize all the arc legth ad surface area formulas from the last two chapters. 005 Paul Dawkis 07

108 Parametric Equatios ad Curves To this poit (i both Calculus I ad Calculus II) we ve looked almost exclusively at fuctios i the form y = f ( x) or x = h( y) ad almost all of the formulas that we ve developed require that fuctios be i oe of these two forms. The problem is that ot all curves or equatios that we d like to look at fall easily ito this form. Take, for example, a circle. It is easy eough to write dow the equatio of a circle cetered at the origi with radius r. x + y = r However, we will ever be able to write the equatio of a circle dow as a sigle equatio i either of the forms above. Sure we ca solve for x or y as the followig two formulas show y =± r x x=± r y but there are i fact two fuctios i each of these. Each formula gives a portio of the circle. ( top) ( right side) ( bottom) ( left side) y = r x x= r y y = r x x= r y Ufortuately we usually are workig o the whole circle, or simply ca t say that we re goig to be workig oly o oe portio of it. Eve if we ca arrow thigs dow to oly oe of these portios the fuctio is still fairly upleasat to work with. There are also a great may curves out there that we ca t eve write dow as a sigle equatio i terms of oly x ad y. To deal with some of these problems we itroduce parametric equatios. Istead of defiig y i terms of x ( y = f ( x) ) or x i terms of y ( x = h( y) ) we defie both x ad y i terms of a third variable called a parameter. x = f ( t) y = g( t) This third variable is usually deoted by t (as we did here) but does t have to be of course. Sometimes we will restrict the values of t that we ll use ad at other times we wo t. This will ofte be depedet o the problem ad just what we are attemptig to do. Each value of t defies a poit ( x, y) ( f ( t), g( t) ) = that we ca plot. The collectio of poits that we get by lettig t be all possible values is the graph of the parametric equatios ad is called the parametric curve. Sketchig a parametric curve is ot always a easy thig to do. Let s take a look at a example to see oe way of sketchig a parametric curve. This example will also illustrate why this method is usually ot the best. 005 Paul Dawkis 08

109 Example Sketch the parametric curve for the followig set of parametric equatios. x = t + t y = t Solutio At this poit our oly optio for sketchig a parametric curve is to pick values of t, plug them ito the parametric equatios ad the plot the poits. So, let s plug i some t s. t x y The first questio that should be asked as this poit is : How did we kow to use the values of t that we did, especially the third choice? Ufortuately there is o real aswer to this questio. We simply pick t s util we are fairly cofidet that we ve got a good idea of what the curve looks like. It is this problem with pickig good values of t that make this method of sketchig parametric curves oe of the poorer choices. Sometimes we have o choice, but if we do have a choice we should avoid it. We ll discuss a alterate graphig method i later examples. We have oe more idea to discuss before we actually sketch the curve. Parametric curves have a directio of motio. The directio of motio is give by icreasig t. So, whe plottig parametric curves we also iclude arrows that show the directio of motio. We will ofte give the value of t that gave specific poits o the graph to make it clear the value of t that have that particular poit. Here is the sketch of this parametric curve. So, we had a parabola that opes to the right. 005 Paul Dawkis 09

110 Before addressig a much easier way to sketch this graph let s first address the issue of limits o the parameter. I the previous example we did t have ay limits o the parameter. This will ot always be the case. So, let s look a slight variatio of the previous example. Example Sketch the parametric curve for the followig set of parametric equatios. x= t + t y = t t Solutio Note that the oly differece here is the presece of the limits o t. All these limits do is tell us that we ca t take ay value of t outside of this rage. Therefore, the parametric curve will oly be a portio of the curve above. Here is the parametric curve for this example. It is ow time to take a look at a easier method of sketchig this parametric curve. This method uses the fact that i may cases we ca actually elimiate the parameter from the parametric equatios ad get a fuctio ivolvig oly x ad y. There will be two small problems with this method, but it will be easy to address those problems. Just how we elimiate the parameter will deped upo the parametric equatios that we ve got. Let s see how to elimiate the parameter for the set of parametric equatios that we ve bee workig with to this poit. Example Elimiate the parameter from the followig set of parametric equatios. x = t + t y = t Solutio Oe of the easiest ways to elimiate the parameter is to simply solve oe of the equatios for the parameter (t, i this case) ad substitute that ito the other equatio. Note that while this may be the easiest, it s usually ot the best as we ll see soo eough. I this case we ca solve y for t. 005 Paul Dawkis 0

111 t = ( y+ ) Pluggig this ito the equatio for x gives, x= ( y+ ) + ( y+ ) = y + y+ 4 4 Sure eough this is a parabola that opes to the right. Gettig a sketch of the parametric curve oce we ve elimiated the parameter is fairly simple. All we eed to do is graph the equatio that we foud by elimiatig the parameter. As oted already however, there are two small problems with this method. The first is directio of motio. The equatio ivolvig oly x ad y will NOT give the directio of motio of the parametric curve. This is a easy problem to fix however. All we eed to do is plug i some values of t ito the parametric equatios ad we ca determie directio of motio from that. How may values of t we plug i will deped upo the parametric equatios. I some cases oly two will be required ad i others we might eed more poits. The secod problem is best illustrated i a example as we ll be seeig this i the remaiig examples. Let s sketch aother parametric curve. Example 4 Sketch the parametric curve for the followig set of parametric equatios. Clearly idicate directio of motio. x= 5cost y = sit 0 t π Solutio I this case we could elimiate the parameter as we did i the previous sectio by solvig oe of these for t ad pluggig this ito the other. For example, x x t = cos y = si cos 5 5 Ca you see the problem with doig this? We have a greater chace of correctly graphig the origial parametric equatios that we do graphig this! There are may ways to elimiate the parameter form the parametric equatios ad solvig for t is usually ot the best way to do it. I this case let s otice that we could do the followig. x y 5cos t 4si t + = + = cos t+ si t = Elimiatig the middle steps gives us, 005 Paul Dawkis

112 x y + = 5 4 ad so it looks like we ve got a ellipse. Next, we eed to determie the directio of motio. Note that i this case we ll eed more tha two poits to do this. Give ay two poits o a ellipse we could get betwee them by goig either clockwise or couter-clockwise about the circle. So, we ll eed at least three poits to accurately determie the directio of motio. While doig this we should also keep i mid that we ve bee give a rage of t s to work with ad as we saw i Example this may mea that we will oly get a portio of the actual ellipse. So, let s choose t s that will cover the whole rage. This will give us the directio of motio ad eough iformatio to determie what portio of the ellipse is i fact traced out. Note that this is the secod problem alluded to above i elimiatig the parameter. Oce we have elimiated the parameter we ve ot oly elimiated the directio of motio, but we ve also elimiated ay iformatio about what portio of the actual graph is traced out by the parametric equatios. We will always eed to keep i mid that this a potetial problem whe elimiatig the parameter. So, here is a table of values for this set of parametric equatios. t x y π 0 π -5 0 π 0 - π 5 0 It looks like we are movig i a couter-clockwise directio about the ellipse ad it also looks like we ll make exactly oe complete trace of the ellipse i the rage give. Here is a sketch of the parametric curve. 005 Paul Dawkis

113 Let s take a look at aother example. Example 5 Sketch the parametric curve for the followig set of parametric equatios. Clearly idicate directio of motio. x= 5cos t y = si t 0 t π Solutio Note that the oly differece i these parametric equatios is that we replaced the t with t. We ca elimiate the parameter here i exactly the same way that we did i the previous example. x y 5cos ( t) 4si ( t) + = + = cos ( t) + si ( t) = So, we get the same ellipse that we did i the previous example. However, we do t get the same parametric curve i some sese. We saw i the previous example that we make oe complete trace of the ellipse i the rage 0 t π. I this set of parametric curves we do t have just a t however. I this set we ve got a t. This meas that we ll complete oe complete trace whe π t = π t = So, while we have the same ellipse that we got i the previous example it will be traced out three times i the rage of t s give istead of oly oce as we got i that example. By pickig values of t we ca see that the directio of motio is t chaged i this case. However, because we re goig aroud faster tha before we should probably use a differet set this time to make sure we get a accurate idea of the directio of motio. t x y π Paul Dawkis

114 π -5 0 Here s the sketch. So, we saw i the last two examples two sets of parametric equatios that i some way gave the same graph. Yet, because they traced out the graph a differet umber of times we really do eed to thik of them as differet parametric curves. This may seem like a differece that we do t eed to worry about, but as we will see i later sectios this ca be a very importat differece. I some of the later sectios we are goig to eed a curve that is traced out exactly oce. Let s take a look at a couple more examples. Example 6 Sketch the parametric curve for the followig set of parametric equatios. Clearly idetify the directio of motio. If the curve is traced out more tha oce give a rage of the parameter for which the curve will trace out exactly oce. x = si t y = cost Solutio We ca elimiate the parameter much as we did i the previous two examples. However, we ll eed to ote that the x already cotais a si t ad so we wo t eed to square the x. We will however, eed to square the y. y y x+ = si t+ cos t = x= 4 4 I this case we get a parabola that opes to the left. We will eed to be very, very careful however i sketchig this parametric curve. We will NOT get the whole parabola. To see this let s ote the followig, sit 0 si t 0 x cost cost y 005 Paul Dawkis 4

115 We will oly get the portio of the parabola that is traced out i this rage of x ad y. Before sketchig let s also get the directio of motio. Here are some poits for a rage of t s. t x y 0 0 π 0 π 0 - π 0 π 0 So, i the rage 0 t π we start at (0,) ad ed up back at that same poit. Recallig that we must travel alog the parabola this meas that we must retrace our path to get back to the startig poit. So, it looks like we ve got a parametric curve that is traced out over ad over i both directios ad we will trace out oce i the rage 0 t π. Here is a sketch of the curve with a few value of t oted o it. To this poit we ve see examples that would trace out the complete graph that we got by elimiatig the parameter if we took a large eough rage of t s. However, i the previous example we ve ow see that this will ot always be the case. It is more tha possible to have a set of parametric equatios which will cotiuously trace out just a portio of the curve. 005 Paul Dawkis 5

116 We ca usually determie this by lookig for limits o x ad y that are imposed up us by the parametric equatio. We will ofte use parametric equatios to describe the path of a object or particle. Let s take a look at a example of that. Example 7 The path of a particle is give by the followig set of parametric equatios. x = cos( t) y = + cos ( t) Completely describe the path of this particle. Do this by sketchig the path, determiig limits o x ad y ad givig a rage of t s for which the path will be traced out exactly oce (provide it traces out more tha oce of course). Solutio Elimiatig the parameter this time will be a little differet. We oly have cosies this time ad we ll use that to our advatage. We ca solve the x equatio for cosie ad plug that ito the equatio for y. This gives, cos( ) x x t = y = + = + x 9 This time we ve got a parabola that opes upward. We also have the followig limits o x ad y. cos( t) cos( t) x 0 cos t + cos t y So, agai we oly trace out a portio of the curve. Here s a set of evaluatios so we ca determie a rage of t s for oe trace of the curve. t x y 0 π 4 0 π - π 4 0 π So, it looks like the particle will agai, cotiuously trace out this portio of the curve π ad will make oe trace i the rage 0 t. Here is a sketch of the particles path with a few value of t o it. 005 Paul Dawkis 6

117 We should give a small warig at this poit. Because of the ideas ivolved i them we cocetrated o parametric curves that retraced portios of the curve more tha oce. Do ot however, get too locked ito the idea that this will always happe. May, if ot most parametric curves will oly trace out oce. The first oe we looked at is a good example of this. That parametric curve will ever repeat ay portio of itself. There is oe fial topic to be discussed i this sectio before movig o. So far we ve started with parametric equatios ad elimiated the parameter to determie the parametric curve. However, there are times i which we wat to go the other way. Give a fuctio or equatio we might wat to write dow a set of parametric equatios for it. I these cases we say that we parameterize the fuctio. If we take Examples 4 ad 5 as examples we ca do this for ellipses (ad hece circles). Give the ellipse x y + = a b a set of parametric equatios for it would be, x = acost y = bsi t There is usually more tha oe set of parametric equatios for the ellipse. Ay of the followig will also parameterize the same ellipse. ( ω ) si ( ω ) ( ω ) cos( ω ) ( ω ) si ( ω ) x = acos t y = b t x = asi t y = b t x = acos t y = b t There are may more of course, but you get the idea. Also ote that while all of these have parametric curves i the shape of the ellipse they will all have differet speeds as they travel aroud the ellipse, they will have differet startig places (if we thik of t=0 as the startig place ayway) ad they have potetially differet directios of motio. 005 Paul Dawkis 7

118 As oted we ca use these to get the parametric equatios for a circle cetered at the origi of radius r as well. Oe possible way to parameterize a circle is, x = rcost y = rsi t Fially, eve though there may ot seem to be ay reaso to, we ca also parameterize fuctios i the form y = f ( x) or x = h( y). I these cases we parameterize them i the followig way, x = t x= h( t) y = f t y = t () Tagets with Parametric Equatios I this sectio we wat to fid the taget lies to the parametric equatios give by, x = f ( t) y = g( t) To do this let s first recall how to fid the taget lie to y F( x) = at x=a. Here the taget lie is give by, dy y = F( a) + m( x a), where m= = F ( a) dx = Now, otice that if we could figure out how to get the derivative dy from the parametric dx equatios we could simply reuse this formula sice we will be able to use the parametric equatios to fid the x ad y coordiates of the poit. So, just for a secod let s suppose that we where able to elimiate the parameter from the parametric form ad write the parametric equatios i the form y = F( x). Now, plug the parametric equatios i for x ad y. Yes, it seem silly to elimiate the parameter, the immediately put it back i, but it s what we eed to do i order to get our hads o the derivative. Doig this gives, g t = F f t ( ) Now, differetiate with respect to t ad otice that we ll eed to use the Chai Rule o the right had side. g t = F f t f t ( ) Let s do aother chage i otatio. We eed to be careful with our derivatives here. Derivatives of the lower case fuctio are with respect to t while derivatives of upper x a 005 Paul Dawkis 8

119 case fuctios are with respect to x. So, to make sure that we keep this straight let s rewrite thigs as follows. dy dx = F ( x) dt dt At this poit we should remid ourselves just what we here after. We eeded a formula for dy F x that we is i terms of the parametric formulas. Notice however that we dx or ca get that from the above equatio. dy dy dx = dt, provided 0 dx dx dt dt Notice as well that this will be a fuctio of t ad ot x. As a aside, otice that we could also get the followig formula with a similar derivatio if we eeded to, dx dx dy = dt, provided 0 dy dy dt dt Why would we wat to do this? Well, recall that i the arc legth sectio of the Applicatios of Itegral sectio we actually eeded this derivative o occasio. So, let s fid a taget lie. Example Fid the taget lie(s) to the parametric curve give by 5 x = t 4t y = t at (0,4). Solutio Note that there is apparetly the potetial for more tha oe taget lie here! We will look ito this more after we re doe with the example. The first thig that we should do is fid the derivative so we ca get the slope of the taget lie. dy dy t = dt = = 4 dx dx 5t t 5t t dt At this poit we ve got a small problem. The derivative is i terms of t ad all we ve got 005 Paul Dawkis 9

120 is a x-y coordiate pair. The ext step the is to determie that value(s) of t which will give this poit. We fid these by pluggig the x ad y values ito the parametric equatios ad solvig for t. 5 0= t 4t = t t 4 t = 0, ± 4= t t =± Ay value of t which appears i both lists will give the poit. So, sice there are two values of t that give the poit we will i fact get two taget lies. That s defiitely ot somethig that happeed back i Calculus I ad we re goig to eed to look ito this a little more. However, before we do that let s actually get the taget lies. t = - Sice we already kow the x ad y-coordiates of the poit all that we eed to do is fid the slope of the taget lie. dy m = = dx t = 8 The taget lie (at t = -) is the, y = 4 x 8 t = Agai, all we eed is the slope. The taget lie (at t = ) is the, dy m = = dx = t y = 4 + x 8 8 Now, let s take a look at just how we could possibly get two tagets lies at a poit. This was defiitely ot possible back i Calculus I where we first ra across taget lies. A quick graph of the parametric curve will explai what is goig o here. 005 Paul Dawkis 0

121 So, the parametric curve crosses itself! That explais how there ca be more tha oe taget lie. There is oe taget lie for each istace that the curve goes through the poit. The ext topic that we eed to discuss i this sectio is that of horizotal ad vertical tagets. We ca easily idetify where these will occur (or at least the t s that will give them) by lookig at the derivative formula. dy dy = dt dx dx dt Horizotal tagets will occur where the derivative is zero ad that meas that we ll get horizotal taget at values of t for which we have, dy dx = 0, provided 0 dt dt Vertical tagets will occur where the derivative is ot defied ad so we ll get vertical tagets at values of t for which we have, dx dy = 0, provided 0 dt dt Example Determie the x-y coordiates of the poits where the followig parametric equatios will have horizotal or vertical tagets. x= t t y = t 9 Solutio We ll first eed the derivatives of the parametric equatios. 005 Paul Dawkis

122 dx dy = t = ( t ) = 6t dt dt Horizotal Tagets We ll have horizotal tagets where, 6t = 0 t = 0 Now, this is the value of t which gives the horizotal tagets ad we were asked to fid the x-y coordiates of the poit. To get these we just eed to plug t ito the parametric equatios. Therefore, the oly horizotal taget will occur at the poit (0,-9). Vertical Tagets I this case we eed to solve, t = 0 t =± The two vertical tagets will occur at the poits (,-6) ad (-,-6). For the sake of completeess ad at least partial verificatio here is the sketch of the parametric curve. The fial topic that we eed to discuss i this sectio really is t related to taget lies, but does fit i icely with the derivatio of the derivative that we eeded to get the slope of the taget lie. Before movig ito the ew topic let s first remid ourselves of the formula for the first derivative ad i the process rewrite it slightly. 005 Paul Dawkis

123 d dy d = ( y) = dt dx dx dx dt Writte i this way we ca see that the formula actually tells us how to differetiate a fuctio y (as a fuctio of t) with respect to x (whe x is also a fuctio of t) whe we are usig parametric equatios. Now let s move oto the fial topic of this sectio. We would also like to kow how to get the secod derivative of y with respect to x. d y dx Gettig a formula for this is fairly simple if we remember the rewritte formula for the first derivative above. d dy d y d dy dt dx = = dx dx dx dx dt ( y ) Note that, d y dx d y dt d x dt Let s work a quick example. Example Fid the secod derivative for the followig set of parametric equatios. 5 x = t 4t y = t Solutio This is the set of parametric equatios that we used i the first example ad so we already have the followig computatios completed. dy = t dt dx 4 = 5t t dt dy = dx 5t t We will first eed the followig, 005 Paul Dawkis

124 The secod derivative is the, 5 ( t ) t d 4 0 = = dt t t t t t t d dy dx dt 4 0t d y dt dx = dx = = ( 5t t) 4 5t t 4 0t 4 ( 5t t )( 5t t) 4 0t = t t t ( 5 ) So, why would we wat the secod derivative? Well, recall from your Calculus I class that with the secod derivative we ca determie where a curve is cocave up ad cocave dow. We could do the same thig with parametric equatios if we wated to. Example 4 Determie the values of t for which the parametric curve give by the followig set of parametric equatios is cocave up ad cocave dow. 7 5 x = t y = t + t Solutio To compute the secod derivative we ll first eed the followig. dy 6 4 = 7t + 5t dt dx = t dt 6 4 dy 7t + 5t 5 = = ( 7 t + 5 t ) dx t Note that we ca also use the first derivative above to get some iformatio about the icreasig/decreasig ature of the curve as well. I this case it looks like the parametric curve will be icreasig if t<0 ad decreasig if t>0. Now let s move o to the secod derivative. 4 ( 5 t + 5 t d y ) = = 5t + 5t dx t Paul Dawkis 4

125 It s clear, hopefully, that the secod derivative will oly be zero at t=0. Usig this we ca see that the secod derivative will be egative if t<0 ad positive if t>0. So the parametric curve will be cocave dow for t<0 ad cocave up for t>0. Here is a sketch of the curve for completeess sake. Area with Parametric Equatios I this sectio we will fid a formula for determiig the area uder a parametric curve give by the parametric equatios, x = f ( t) y = g( t) We will also eed to further add i the assumptio that the curve is traced out exactly oce as t icreases from α to β. We will do this i much the same way that we foud the first derivative i the previous y = F x o a x b. sectio. We will first recall how to fid the area uder b A F( x) dx = We will ow thik of the parametric equatio x f ( t) We will also assume that a = f ( α ) ad b f ( β ) a = as a substitutio i the itegral. = for the purposes of this formula. There is actually o reaso to assume that this will always be the case ad so we ll give a b f α a = f β ). correspodig formula later if it s the opposite case ( = ad So, if this is goig to be a substitutio we ll eed, dx = f t dt 005 Paul Dawkis 5

126 Pluggig this ito the area formula above ad makig sure to chage the limits to their correspodig t values gives us, β A= F( f () t ) f () t dt Sice we do t kow what F(x) is we ll use the fact that y = F( x) = F f ( t) = g t ad we arrive at the formula that we wat. β A= g() t f () t dt α α Now, if we should happe to have b= f ( α ) ad a f ( β ) α A g() t f () t dt Let s work a example. = β = the formula would be, Example Determie the area uder the parametric curve give by the followig parametric equatios. x= 6 θ siθ y = 6 cosθ 0 θ π Solutio First, otice that we ve switched the parameter to θ for this problem. This is to make sure that we do t get too locked ito always havig t as the parameter. Now, we could graph this to verify that the curve is traced out exactly oce for the give rage if we wated to. We are goig to be lookig at this curve i more detail after this example so we wo t sketch its graph here. There really is t too much to this example other tha pluggig the parametric equatios ito the formula. We ll first eed the derivative of the parametric equatio for x however. dx 6 ( cosθ ) dθ = The area is the, 005 Paul Dawkis 6

127 ( θ) π A= 6 cos 0 π = 6 cos + cos 0 π 0 dθ θ θ θ = 6 cosθ + cos( θ) dθ = 6 θ siθ + si ( θ) 4 = 08π The parametric curve (without the limits) we used i the previous example is called a cycloid. I its geeral form the cycloid is, d π ( θ siθ) ( cosθ) x= r y = r The cycloid represets the followig situatio. Cosider a wheel of radius r. Let the poit where the wheel touches the groud iitially be called P. The start rollig the wheel to the right. As the wheel rolls to the right trace out the path of the poit P. The path that the poit P traces out is called a cycloid ad is give by the equatios above. I these equatios we ca thik of θ as the agle through which the poit P has rotated. Here is a sketch of the cycloid used i the previous example. 0 We ca see that oe arch of the cycloid is traced out i the rage 0 θ π. This makes sese whe you cosider that the poit P will be back o the groud after it has rotated through ad agle of π. Arc Legth with Parametric Equatios I the previous two sectios we ve looked at a couple of Calculus I topics i terms of parametric equatios. We ow eed to look at a couple of Calculus II topics i terms of parametric equatios. I this sectio we will look at the arc legth of the parametric curve give by, x= f t y = g t α t β () 005 Paul Dawkis 7

128 We will also be assumig that the curve is traced out exactly oce as t icreases from α to β. Also, for the purposes of the derivatio that we re goig to use we will assume that the curve is traced out from left to right as t icreases. This is equivalet to sayig, dx 0 for α t β dt This is ot actually required for the fial formula, but as oted above we ll eed it for our derivatio. If the curve is t traced out from left to right we would eed to go through a slightly more complicated derivatio. So, let s start out the derivatio by recallig the arc legth formula as we first derived it i the arc legth sectio of the Applicatios of Itegrals chapter. L= ds where, dy ds = + dx if y = f ( x), a x b dx dx ds = + dy if x = h( y), c y d dy We will use the first ds above because we have a ice formula for the derivative i terms of the parametric equatios (see the Tagets with Parametric Equatios sectio). To use this we ll also eed to kow that, dx dx = f () t dt = dt dt The arc legth formula the becomes, β dy dy dx dt dx L dt = dt + = + dt dx dt dx dt dt α dt This is a particularly upleasat formula. However, if we factor out the deomiator from the square root we arrive at, β dx dy dx L= + dt dx dt dt dt α dt β α 005 Paul Dawkis 8

129 Now, makig use of our assumptio that the curve is beig traced out from left to right we ca drop the absolute value bars o the derivative which will allow us to cacel the two derivatives that are outside the square root this gives, β dx dy L= + dt dt α Notice that we could have used the secod formula for ds above is we had assumed istead that dy 0 for α t β dt If we had goe this route i the derivatio we would have gotte the same formula. Let s take a look at a example. Example Determie the legth of the parametric curve give by the followig parametric equatios. x= si t y = cos t 0 t π () Solutio We kow that this is a circle of radius cetered at the origi from our prior discussio about graphig parametric curves. We also kow from this discussio that it will be traced out exactly oce i this rage. So, we ca use the formula we derived above. We ll first eed the followig, dx dy = cos() t = si() t dt dt dt The legth is the, () 9cos () π L= 9si t + t dt 0 π 0 π 0 () () = si t + cos t dt = = 6π dt Sice this is a circle we could have just used the fact that the legth of the circle is just the circumferece of the circle. This is a ice way, i this case, to verify our result. Let s take a look at oe possible cosequece of a curve is traced out more tha oce ad we try to fid the legth of the curve without takig this ito accout. Example Use the arc legth formula for the followig parametric equatios. x= si t y = cos t 0 t π 005 Paul Dawkis 9

130 Solutio Notice that this is the idetical circle that we had i the previous example ad so the legth is still 6π. However, for the rage give we kow trace out the curve three times istead oce as required for the formula. Despite that restrictio let s use the formula ayway ad see what happes. I this case the derivatives are, dx dy = 9cos t = 9si t dt dt ad the legth formula gives, π L= 8si t + 8cos t dt = 0 π 0 = 8π 9 dt () () The aswer we got form the arc legth formula i this example was times the actual legth. Recallig that we also determied that this circle would trace out three times i the rage give, the aswer should make some sese. If we had wated to determie the legth of the circle for this set of parametric equatios we would eed to determie a rage of t for which this circle is traced out exactly oce. π This is, 0 t. Usig this rage of t we get the followig for the legth. which is the correct aswer. π L= 8si () t + 8cos () t dt 0 π = 9 dt 0 = 6π Be careful to ot make the assumptio that this is always what will happe if the curve is traced out more tha oce. Just because the curve traces out times does ot mea that the arc legth formula will give us times the actual legth of the curve! Before movig o to the ext sectio let s otice that we ca put the arc legth formula derived i this sectio ito the same form that we had whe we first looked at arc legth. The oly differece is that we will add i a defiitio for ds whe we have parametric equatios. The arc legth formula ca be summarized as, where, L = ds 005 Paul Dawkis 0

131 dy ds = + dx if y = f ( x), a x b dx dx ds = + dy if x = h( y), c y d dy dx dy ds = + dt if x = f () t, y = g () t, α t β dt dt Surface Area with Parametric Equatios I this fial sectio dealig with calculus usig parametric equatios we will take a look at determiig the surface area of a regio obtaied by rotatig a parametric curve about the x or y-axis. We will rotate the parametric curve give by, x= f () t y = g( t) α t β about the x or y-axis. We are goig to assume that the curve is traced out exactly oce as t icreases from α to β. At this poit there actually is t all that much to do. We kow that the surface area ca be foud by usig oe of the followig two formulas depedig o the axis of rotatio (recall the Surface Area sectio of the Applicatios of Itegrals chapter). S = π yds rotatio about x axis S = π xds rotatio about y axis All that we eed is a formula for ds to use ad from the previous sectio we have, dx dy ds = + dt if x = f () t, y = g () t, α t β dt dt which is exactly what we eed. We will eed to be careful with the x or y that is i the origial surface area formula. Back whe we first looked at surface area we saw that sometimes we had to substitute for the variable i the itegral ad at other times we did t. This was depedet upo the ds that we used. I this case however, we will always have to substitute for the variable. The ds that we use for parametric equatios itroduces a dt ito the itegral ad that meas that everythig eeds to be i terms of t. Therefore, we will eed to substitute the appropriate parametric equatio for x or y depedig o the axis of rotatio. Let s take a quick look at a example. 005 Paul Dawkis

132 Example Determie the surface area of the solid obtaied by rotatig the followig parametric curve about the x-axis. π x= cos θ y = si θ 0 θ Solutio We ll first eed the derivatives of the parametric equatios. dx dy = cos θ siθ = si θcosθ dt dt Before pluggig ito the surface area formula let s get the ds out of the way. 4 4 ds = 9cos θ si θ + 9si θcos θ dt = cosθsiθ cos θ + si θ = cosθsiθ Notice that we could drop the absolute value bars sice both sie ad cosie are positive i this rage of θ give. Now let s get the surface area ad do t forget to also plug i for the y. S = π yds = π si 0 cos si π θ θ θ dθ π 6 4 si 0 cos d u si 4 6π u du 0 = π θ θ θ = θ = = 6π 5 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. 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. 005 Paul Dawkis

133 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, π we also allow r to be egative. Below is a sketch of the two poits, 6 ad, π Paul Dawkis

134 From this sketch we ca see that if r is positive the poit will be i the same quadrat as θ. O the other had if r is egative the poit will ed up i the quadrat exactly π opposite θ. Notice as well that the coordiates, describe the same poit as the 6 7π coordiates, 6 do. The coordiates 7π, 6 tells us to rotate a agle of 7 π from 6 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 , π 5, π 5, π 5, π = = = Here is a sketch of the agles used i these four sets of coordiates. 005 Paul Dawkis 4

135 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, θ ) ca be represeted by ay of the followig coordiate pairs. r, θ + π r, θ + + π, 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 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,θ ). 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. 005 Paul Dawkis 5

136 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. x= rcosθ y = rsiθ Covertig from Cartesia is almost as easy. Let s first otice the followig. x + y = ( rcosθ ) + ( rsiθ ) = r cos θ + r si θ = r cos θ + si θ = r 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 θ is almost as simple. We ll start with, y rsiθ = = taθ x rcosθ Takig the iverse taget of both sides gives, y θ = ta x 005 Paul Dawkis 6

137 We will eed to be careful with this because iverse tagets oly retur values i the π π rage < θ <. Recall that there is a secod possible agle ad that the secod agle is give by θ + π. Summarizig the gives the followig formulas for covertig from Cartesia coordiates to polar coordiates. Let s work a quick example. r = x + y r = x + y y θ = ta x Example Covert each of the followig poits ito the give coordiate system. π (a) 4, ito Cartesia coordiates. (b) (-,-) ito polar coordiates. Solutio (a) This coversio is easy eough. All we eed to do is plug the poits ito the formulas. π x = 4cos = 4 = π y = 4si = 4 = So, i Cartesia coordiates this poit is (, ). (b) Let s first get r. r = ( ) + ( ) = Now, let s get θ. π θ = ta = ta () = 4 This is ot the correct agle however. This value of θ 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 π oto this. Therefore, the actual agle is, π 5π θ = + π = Paul Dawkis 7

138 5π So, i polar coordiates the poit is,. Note as well that we could have used the 4 first θ that we got by usig a egative r. I this case the poit could also be writte i π polar coordiates as, 4. 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. (b) Covert r = 8cosθ ito Cartesia coordiates. Solutio (a) 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 θ (i.e. the polar coordiates). rcosθ 5 rcosθ = + rcosθ rsiθ rcos 5r cos r cos si θ θ = + θ θ (b) 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 right side as well. r = 8rcosθ We ca ow make some substitutios that will covert this ito Cartesia coordiates. + = x y 8x 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 ) y = Paul Dawkis 8

139 So, this was a circle of radius 4 ad ceter (-4,0). This leads us ito the fial topic of this sectio. 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.. θ = β. We ca see that this is a lie by covertig to Cartesia coordiates as follows θ = β y ta = β x y = ta β x y = ta β x This is a lie that goes through the origi ad makes a agle of β with the positive x-axis.. rcosθ = a This is easy eough to covert to Cartesia coordiates to x=a. So, this is a vertical lie.. rsiθ = b Likewise, this coverts to y=b ad so is a horizotal lie. π Example Graph θ =, r cosθ = 4 ad r siθ = 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. 005 Paul Dawkis 9

140 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 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 = acosθ. 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 = bsiθ. This is similar to the previous oe. It is a circle of radius b ad ceter 0,b. 4. r = acosθ + bsiθ. 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 = 4cosθ, ad r = 7siθ o the same axis system. 005 Paul Dawkis 40

141 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 7 0,. Here is the graph of the three equatios. Note that it takes a rage of 0 θ π for a complete graph of r=a ad it oly takes a rage of 0 θ π to graph the other circles give here. Cardioids ad Limacos These ca be broke up ito the followig three cases.. Cardioids : r = a± acosθ ad r = a± asiθ. These have a graph that is vaguely heart shaped ad always cotai the origi.. Limacos with a ier loop : r = a± bcosθ ad r = a± bsiθ with a< b. These will have a ier loop ad will always cotai the origi.. Limacos without a ier loop : r = a± bcosθ ad r = a± bsiθ with a > b. These do ot have a ier loop ad do ot cotai the origi. Example 5 Graph r = 5 5siθ, r = 7 6cosθ, ad r = + 4cosθ. Solutio These will all graph out oce i the rage 0 θ π. Here is a table of values for each followed by graphs of each. θ r = 5 5siθ r = 7 6cosθ r = + 4cosθ Paul Dawkis 4

142 π 0 7 π 5 - π 0 7 π Paul Dawkis 4

143 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 θ 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. 0= + 4cosθ π 4π cos θ = θ =, 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 ( θ ). With the equatio i this form we ca actually use the equatio for the derivative dy we derived whe we looked at taget dx lies with parametric 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= rcosθ y = rsiθ Now, we ll use the fact that we re assumig that the equatio is i the form r f ( θ ) =. Substitutig this ito these equatios gives the followig set of parametric equatios (with θ as the parameter) for the curve. 005 Paul Dawkis 4

144 cos x= f θ θ y = f θ siθ Now, we will eed the followig derivatives. dx dy = f ( θ ) cosθ f ( θ) siθ = f ( θ) siθ + f ( θ) cosθ dθ dθ dr dr = cosθ rsiθ = siθ + rcosθ dθ dθ The derivative dy dx is the, dr siθ + r cosθ dy = dθ dx dr cosθ r siθ dθ 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. π Example Determie the equatio of the taget lie to r = + 8siθ at θ =. 6 Solutio We ll first eed the followig derivative. dr 8cosθ dθ = The formula for the derivative dy dx becomes, dy dx 8cosθsiθ + + 8siθ cosθ 6cosθ siθ + cosθ = = 8cos θ + 8siθ siθ 8cos θ siθ 8si θ The slope of the taget lie is, 4 + dy m = = = dx π θ = π Now, at θ = we have r=7. We ll eed to get the correspodig x-y coordiates so we 6 ca get the taget lie. 005 Paul Dawkis 44

145 π 7 π 7 x= 7cos = y = 7si = 6 6 The taget lie is the, 7 7 y = + x 5 For the sake of completeess here is a graph of the curve ad the taget lie. Area with Polar Coordiates I this sectio we are goig to look at areas eclosed by polar curves. Note as well that we said eclosed by istead of uder as we typically have i these problems. These problems work a little differetly i polar coordiates. Here is a sketch of what the area that we ll be fidig i this sectio looks like. 005 Paul Dawkis 45

146 We ll be lookig for the shaded area i the sketch above. The formula for fidig this area is, β A= r dθ Notice that we use r i the itegral istead of f ( θ ) so make sure ad substitute accordigly whe doig the itegral. Let s take a look at a example. Example Determie the area of the ier loop of r = + 4cosθ. α Solutio We graphed this fuctio back whe we first started lookig at polar coordiates. Here is the sketch agai for the sake of completeess. 005 Paul Dawkis 46

147 So, we will eed the limits that will eclose this area as we trace out the curve. This will be the agles for which the curve passes through the origi. We ca get these by settig the equatio equal to zero ad solvig. 0= + 4cosθ π 4π cos θ = θ =, So, the area is the, 4π A= + 4cos π 4π π 4π π 4π π ( θ) = ( 4 + 6cos θ + 6cos θ) dθ ( ) = + 8cosθ cos θ dθ = 6 + 8cosθ + 4cos θ dθ ( 6θ 8siθ si ( θ) ) = + + = 4π 6 =.74 dθ 4π π So, that s how we determie areas that are eclosed by a sigle curve, but what about situatios like the followig sketch were we wat to fid the area betwee two curves. 005 Paul Dawkis 47

148 I this case we ca use the above formula to fid the area eclosed by both ad the the actual area is the differece betwee the two. The formula for this is, β A= ( r r ) dθ Let s take a look at a example of this. o i α Example Determie the area that lies iside r = + siθ ad outside r =. Solutio Here is a sketch of the regio that we are after. To determie this area we ll eed to kow that value of θ for which the two curves itersect. We ca determie these poits by settig the two equatios ad solvig. 005 Paul Dawkis 48

149 + siθ = 7π π si θ = θ =, 6 6 Here is a sketch of the figure with these agles added. Note as well here that we also ackowledged that aother represetatio for the agle π is. This is importat for this problem. I order to use the formula above the 6π 6 area must be eclosed as we icrease from the smaller to larger agle. So, if we use 7 π 6 to we will ot eclose the shaded area, istead we will eclose the bottom most of 6π the three regios. However if we use the agles we re after. So, the area is the, π to 7 π we will eclose the area that Paul Dawkis 49

150 7π 6 (( θ) ) A= + si π 6 7π 6 = ( 5 + si θ + 4si θ) dθ π 6 7π 6 = ( 7 + si θ cos ( θ) ) dθ π 6 = 7 cos si ( θ θ ( θ) ) 4π = + = 4.87 Let s work a slight modificatio of the previous example. Example Determie the area of the regio outside r = + siθ ad iside r =. Solutio This time we re lookig for the followig regio. dθ 7 6 π π 6 So, this is the regio that we get by usig the limits 7 π π to. The area is the, Paul Dawkis 50

151 π 6 7π 6 π 7π 6 π 6 7π 6 ( ( θ) ) A= + si 6 = ( 5 siθ 4si θ) dθ = ( 7 siθ + cos( θ) ) dθ = + + ( 7θ cosθ si( θ) ) 7π = =.96 Notice that for this area the outer ad ier fuctio where opposite! Let s do oe fial modificatio of this example. Example 4 Determie the area that is iside both r = + siθ ad r =. Solutio Here is the sketch for this example. dθ π π We are ot goig to be able to do this problem i the same fashio that we did the previous two. There is o set of limits that will allow us to eclose this area as we icrease from oe to the other. I this case however, that is ot a major problem. There are two ways to do get the area i this problem. 005 Paul Dawkis 5

152 Solutio Notice that the circle is divided up ito two portios ad we re after the upper portio. Also otice that we foud the area of the lower portio i Example. Therefore, the area is, Area = Area of Circle Area from Example = π.96 = 0.70 Solutio I this case we do pretty much the same thig except this time we ll thik of the area as the other portio of the limaco tha the portio that we were dealig with i Example. So, i this case the area is, Area = Area of Limaco Area from Example π ( si θ) dθ π = + = ( 9 + si θ + 4si θ) dθ π = ( + si θ + cos ( θ) ) dθ π = ( θ cos ( θ) + si ( θ) ) = π 4.87 = 0.70 A slightly loger approach, but sometimes we are forced to take this loger approach. As this last example has show we will ot be able to get all areas i polar coordiates straight from a itegral. Arc Legth with Polar Coordiates We ow eed to move ito the Calculus II applicatios of itegrals ad how we do them i terms of polar coordiates. I this sectio we ll look at the arc legth of the curve give by, r = f ( θ ) α θ β 005 Paul Dawkis 5

153 where we also assume that the curve is traced out exactly oce. Just as we did with the taget lies i polar coordiates we ll first write the curve i terms of a set of parametric equatios. x= rcosθ y = rsiθ = f ( θ ) cosθ = f ( θ) siθ ad we ca ow use the parametric formula for fidig the arc legth. We ll eed the followig derivatives for these computatios. dx dy = f ( θ ) cosθ f ( θ) siθ = f ( θ) siθ + f ( θ) cosθ dθ dθ dr dr = cosθ rsiθ = siθ + rcosθ dθ dθ We ll eed the followig for our ds. dx dy dr dr + = cosθ rsiθ + siθ + rcosθ dθ dθ dθ dθ dr dr = cos θ r cosθsiθ + r si θ dθ dθ dr dr + si θ + r cosθsiθ + r cos θ dθ dθ dr = ( cos θ + si θ) + r ( cos θ + si θ) dθ dr = r + dθ The arc legth formula for polar coordiates is the, L= ds where, Let s work a quick example of this. dr ds = r + dθ dθ Example Determie the legth of r = θ 0 θ. Solutio Okay, let s just jump straight ito the formula sice this is a fairly simple fuctio. L= θ + dθ Paul Dawkis 5

154 We ll eed to use a trig substitutio here. θ = ta x dθ = sec xdx θ = 0 0= tax x = 0 θ = = tax π x= 4 ta x sec x sec x sec x θ + = + = = = The arc legth is the, L = θ + 0 = π 4 0 sec dθ xdx = sec ta + l sec + ta ( x x x x) ( l( ) ) = + + π 4 0 Surface Area with Polar Coordiates We will be lookig at surface area i polar coordiates i this sectio. Note however that all we re goig to do is give the equatios sice most of these itegrals ted to be fairly difficult. We wat to fid the surface area of the regio foud by rotatig, r = f ( θ ) α θ β about the x or y-axis. As we did i the taget ad arc legth sectios we ll write the curve i terms of a set of parametric equatios. x= rcosθ y = rsiθ = f θ cosθ = f θ siθ If we ow use the parametric formula for fidig the surface area we ll get, S = π yds rotatio about x axis where, S = π xds rotatio about y axis dr ds r d r f = + θ = θ, α θ β dθ 005 Paul Dawkis 54

155 Note that because we will pick up a dθ from the ds we ll eed to substitute oe of the parametric equatio i for x or y depedig o the axis of rotatio. This will ofte mea that the itegrals will be somewhat upleasat. Arc Legth ad Surface Area Revisited We wo t be workig ay examples i this sectio. This sectio is here solely for the purpose of summarizig up all the arc legth ad surface area problems. Over the course of the last two chapters the topic of arc legth ad surface area has arise may times ad each time we got a ew formula out of the mix. Studets ofte get a little overwhelmed with all the formulas. However, there really are t as may formulas as it might seem at first glace. There is exactly oe arc legth formula ad exactly two surface area formulas. These are, L= ds S = π yds rotatio about x axis S = π xds rotatio about y axis The problems arise because we have quite a few ds s that we ca use. Agai studets ofte have trouble decidig which oe to use. The examples/problems usually suggest the correct oe to use however. Here is a complete listig of all the ds s that we ve see ad whe they are used. dy ds = + dx if y = f ( x), a x b dx dx ds = + dy if x = h( y), c y d dy dx dy ds = + dt if x = f () t, y = g () t, α t β dt dt dr ds r d r f = + θ if = θ, α θ β dθ Depedig o the form of the fuctio we ca quickly tell which ds to use. There is oly oe other thig to worry about i terms of the surface area formula. The ds will itroduce a ew differetial to the itegral. Before itegratig make sure all the 005 Paul Dawkis 55

156 variables are i terms of this ew differetial. For example if we have parametric equatios well use the third ds ad the we ll eed to make sure ad substitute for the x or y depedig o which axis we rotate about to get everythig i terms of t. Likewise, if we have a fuctio i the form x h( y) = the we ll use the secod ds ad if the rotatio is about the y-axis we ll eed to substitute for the x i the itegral. O the other had if we rotate about the x-axis we wo t eed to do a substitutio for the y. Keep these rules i mid ad you ll always be able to determie which formula to use ad how to correctly do the itegral. 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 simply do t see the use i studyig ad 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. 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. 005 Paul Dawkis 56

157 Itegral Test Usig the Itegral Test to determie if a series coverges or diverges. Compariso Test/Limit Compariso Test Usig the Compariso Test ad Limit Compariso Tests to determie if a series coverges or diverges. Alteratig Series Test Usig the Alteratig Series Test to determie if a series coverges or diverges. Absolute Covergece A brief discussio o absolute covergece ad how it differs from covergece. Ratio Test Usig the Ratio Test to determie if a series coverges or diverges. Root Test Usig the Root Test to determie if a series coverges or diverges. Strategy for Series A set of geeral guidelies to use whe decidig which test to use. Estimatig the Value of a Series Here we will look at estimatig the value of a ifiite series. Power Series A itroductio to power series ad some of the basic cocepts. Power Series ad Fuctios I this sectio we will start lookig at how to fid a power series represetatio of a fuctio. Taylor Series Here we will discuss how to fid the Taylor/Maclauri Series for a fuctio. Applicatios of Series I this sectio we will take a quick look at a couple of applicatios of series. Biomial Series A brief look at biomial series. 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, 005 Paul Dawkis 57

158 a first term a secod term a a + th term ( ) st + term 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! There is a variety of ways of deotig a sequece. Each of the followig are equivalet ways of deotig a sequece. 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. Example Write dow the first few terms of each of the followig sequeces. + (a) (b) = + ( ) (c) { b} = = 0 th, where b = digit of π Solutio (a) 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. 005 Paul Dawkis 58

159 =,,,,, = = = = = 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. (b) This oe is similar to the first oe. The mai differece is that this sequece does t start at =. + ( ),,,,, = = 0 Note that the terms i this sequece alterate i sigs. Sequeces of this kid are sometimes called alteratig sequeces. (c) This sequece is differet from the first two i the sese that it does t have a specific formula for each term. However, it does tell us what each term should be. Each term should be the th digit of π. So we kow that π = The sequece is the, {,, 4,,5,9,,6,5,,5, } 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, + + ( ) f = g = 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 sequece. The first few poits o the graph are = the, (, ),,,,, 4,, 5,, The graph is the, 005 Paul Dawkis 59

160 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, + lim a = lim = 0 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. Defiitios. We say that lim a = L 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 = 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 005 Paul Dawkis 60

161 lim a = 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. We ow have a bit of termiology to get out of the way. If lim a exists ad is fiite we say that the sequece is 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. So just how do we fid the limits of sequeces? We will use the followig theorems ad facts. Theorem Give the sequece { a } if we have a fuctio f(x) such that f a lim a = L f x = L the lim x = ad 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. Theorem If lim a = 0 the lim a = 0. This theorem is coveiet for sequeces that alterate i sigs ad ote that it will oly work if the sequece has a limit of zero. Theorem The sequece { r } = 0 coverges if < r ad diverges for all other value of r. Also, lim r 0 if < r < = if r = Note that the sequece i this theorem will coverge for r = ad diverge for r = Facts 005 Paul Dawkis 6

162 lim a ± b = lim a ± lim b lim ca = c lim a ( ab) = ( a)( b) lim lim lim a lim a lim =, provided lim b 0 b lim b Let s take a look at a couple of examples. Example Determie if the followig sequeces coverge or diverge. If the sequece coverges determie its limit. (a) 0+ 5 = e (b) (c) ( ) (d) ( ) = = { } = 0 Solutio (a) 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 Ivolvig Ifiity 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. lim = lim = lim = So the sequece coverges ad its limit is 5. (b) 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 x f ( x) = e x 005 Paul Dawkis 6

163 ad ote that, f = e Theorem says that all we eed to do is take the limit of the fuctio. x x e e e lim = lim = lim = x x x 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. (c) 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, 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 this 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. ( ) lim = lim = 0 Therefore, sice the limit of the sequece terms with absolute value bars o them goes to zero we kow by Theorem that, ( ) lim = 0 which also meas that the sequece coverges to a value of zero. (d) For this sequece all that we eed to do is ackowledge that lim does't exist to get that the sequece is diverget. If you re ot coviced that this limit does t exist write dow the first few terms of the sequece. =,,,,,,,,,, { } { } = 0 I order for a limit to exist the terms must be settlig dow towards a specific value ad these clearly will ever do that. Note that we could also use Theorem o this sequece if we wated to. Before movig o we 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 005 Paul Dawkis 6

164 ot zero the the theorem will ot hold. The last part of the previous example is a good example of this. Notice that lim = lim = ad yet, lim does t eve exist let aloe equal. So, be careful usig this theorem. We ow eed to take a look at some more termiology ad defiitios for sequeces. 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. a is a decreasig sequece we call it. If { } a is a icreasig sequece or { } 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! 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. We also have the followig theorem about bouded ad mootoic sequeces. Theorem 4 If { a } is bouded ad mootoic the { } a is coverget. 005 Paul Dawkis 64

165 Let s take a look at a couple of examples. Example Determie if the followig sequeces are mootoic ad/or bouded. (a) { } (b) = 0 + = { } (c) = 5 Solutio (a) 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). (b) The sequece terms i this sequece alterate betwee ad - ad so the sequece is ot mootoic. It is bouded however sice it is bouded above by ad bouded below by -. This sequece is also diverget. (c) 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. Therefore, this 5 sequece is bouded. Theorem 4 says that this sequece is the coverget sice it is both bouded ad mootoic. A quick limit ca verify that this sequece is coverget ad its value is zero. 005 Paul Dawkis 65

166 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. Example 4 Determie if the followig sequeces are mootoic ad/or bouded. (a) + = (b) = 0 Solutio (a) 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 problems of this type. 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. To see this 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, + a = = f < f ( + ) = = a+ 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 005 Paul Dawkis 66

167 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. Also, the sequece coverges to, which agai is a verificatio of Theorem 4. Before movig o to the ext part there is a atural questio that may studets will have at this poit. That is this. 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 say thig? The aswer to this questio is the ext part! (b) 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 that oe (sice the umerator is guarateed to be less tha the deomiator) ad so the sequece is bouded. Now, let s move o to the icreasig/decreasig questio. First, 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 = a4 = a5 = a6 = a7 = a8 = a9 = a0 = = The first 0 (at least) 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 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. This sequece is either decreasig or icreasig. The oly sure way to see this is to do 005 Paul Dawkis 67

168 the Calculus I approach to icreasig/decreasig fuctios. I this case we ll eed the followig fuctio ad its derivative. 4 x x x 0000 f ( x) = f 4 ( x) = 4 x x This fuctio will have the followig three critical poits, 4 4 x= 0, x= , x= 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 4 quickly determie that the derivative is positive for 0 < x < 0000 ad so the fuctio is icreasig i this rage. Likewise, we ca see that the derivative is egative for x > ad so the fuctio will be decreasig i this rage. So, our sequece will be icreasig for 0 ad decreasig for. Therefore the fuctio is ot mootoic. Notice however Theorem 4 does NOT say that this sequece will the diverge. I fact, this sequece coverges to a value of zero. Be careful ot to misuse Theorem 4. 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 showed 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 ow that 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 4 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. 005 Paul Dawkis 68

169 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 series. So just what is a series? Well, let s start with a sequece { a} = (ote the = is for coveiece, it ca be aythig) ad the add up all the terms of the sequece. a+ a + a+ + a + This is ot a easy thig to write dow o a regular basis ad so we itroduce the followig otatio. a+ a + a + + a + = a We have just defied what a series is. A series is othig more tha the summatio of a list of umbers or sequece. Sice we started out with a ifiite sequece we will be summig up a ifiite list of umbers. Because of this the series above is sometimes called a ifiite series. The is ofte called a idex of summatio or just idex for short. Note as well that we will use a to represet a ifiite series i which the startig poit for the idex is ot importat. We will do this i quite a few facts ad theorems that we ll be seeig throughout this chapter. I these facts 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. The otatio used for series is called a umber of thigs. The most commo ames are : series otatio, summatio otatio, ad sigma otatio. The series otatio (or summatio otatio or sigma otatio, which ever you prefer) tells us to add all the items from the sequece startig at the value of the idex that is below the sigma. Also ote that the letter that we use for the idex is ot importat. The followig two series are idetical. = = 0 + i= 0 i + As log as we re startig at the same spot (=0 or i=0 respectively) ad we re addig the same terms so the series will be idetical. The poit of this sectio is to cover the basic ideas, cocepts ad maipulatios ivolved i series. We eed to be somewhat familiar with these before we get ito the later sectios. So, let s start off the discussio with some basic facts about arithmetic with series. = 005 Paul Dawkis 69

170 ca = c a c is ay umber a ± b = a ± b = k = k = k a b = c where c = ab i i = 0 = 0 = 0 i= 0 Before discussig these i detail let s agai talk a little bit about the otatio. I the first formula the iitial value of the idex is ot importat to the formula ad so was dropped completely. Note as well that we dropped the ifiity from the top of the sigma i this case as well. We will always drop the ifiity whe we drop the iitial value of the idex. Whe there is o iitial value give for the series the assumptio is that we ca use ay value there ad we eglected puttig oe i the formula to avoid givig the impressio that the startig poit was importat. I the secod series we had a startig poit, but it was t a specific startig poit. I this case we eeded to ackowledge that i order to add or subtract two series we eed the two series to start at the same poit. Just what this value happes to be is t importat, but it must be the same for both series ad to deote this we used the geeral startig poit =k. I the third series we gave a specific startig poit (=0). From this poit o it ca be assumed that if a fact/theorem gives a specific startig poit for a series the that startig poit is required i order of the fact/theorem to be valid. That is the case with this fact about multiplyig series. There are similar facts for series that do t start at zero, however we wo t be doig much i the way of multiplyig series i this chapter ad so we do t really eed them. Okay, ow that the otatioal stuff is out of the way, let s thik about these facts a little. The first fact should make some sese. We ca always factor a costat out of a series. We kow that we ca do this with a fiite sum, x + 4x+ 6= x + x+ ad so it should make some sese that we ca also do it for ifiite sums (i.e. ifiite series). Likewise, the secod fact is t too bad to uderstad. Addig/Subtractig two (or more) series ivolves othig more tha addig/subtractig similar terms The formula that always causes difficulty is the multiplicatio of series formula. First, let s ote the followig, a b ( ab) = 0 = 0 = 0 To covice yourself that this is t true cosider the followig product of two fiite sums. 005 Paul Dawkis 70

171 + x 5x+ x = 6 7x x + x Yeah, it was just the multiplicatio of two polyomials, but it makes the poit. I doig the multiplicatio we did t just multiply the costat terms, them 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. Multiplyig ifiite series eeds to be doe i the same maer. Remember that a series is really a giat summatio ad so here is what we re really askig for multiplicatio, a b = ( a0 + a + a + a + )( b0 + b + b + b + ) = 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. The formula give above is the geeral formula for this, but agai, is usually impossible to use i practice. The reality is that multiplicatio of series is a somewhat difficult process ad i geeral is avoided at all costs 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 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 = Now, whe =, we will get i=0. Notice as well that if = the i = =, so oly the lower limit will chage here. Next, otice that we ca solve this for to get, = i+ 005 Paul Dawkis 7

172 We ca ow completely rewrite the series i terms of the idex i istead of the idex. + 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, = = = = 0 So, sure eough they are the same series. 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 term must icrease by as well. ( + ) = = 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. 005 Paul Dawkis 7

173 ( ) ( ) = = ( ) + + = = = 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 (ote that the = startig poit is oly for coveiece, we ca start aywhere). = a = a + a + a + a + a 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, a = a + a = = I the 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. a = a + a + a = = 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, 4 a = a + a + a + a + a = a + a 4 = = 5 = = 5 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 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 series is a summatio of the list of umbers. So, provided it eve makes sese to do the additio (see the ext sectio), a series will be a sigle value. 005 Paul Dawkis 7

174 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, but we ll always have to remember that they are differet. Series Covergece/Divergece I the previous sectio we spet some time gettig familiar with series, but to be hoest, most of what we did i that sectio wo t be used o a regular basis i this chapter. We covered that material because we eed to be aware of how series work ad ca be maipulated, but we also covered it so that we could start gettig our feet wet i the subject of series. It is ow time to start talkig about a idea ivolved i series that we will deal with to oe extet of aother i almost all of the remaiig sectios of this chapter. There was a very importat questio about series that was oly metioed i passig towards the very ed of the previous sectio. The questio is simply this : Does it eve make sese to add up a ifiite sequece of umbers? Techically we ca always write dow a ifiite series or summatio, but that does t mea that it makes sese to do it. The real questio that we ll be askig here is does the (ifiite) series/summatio yield a fiite value or ifiite value? Of course that also assumes that the series yields a value at all! As we will see it will be possible for a series to ot eve have a value. To aswer this questio we ll eed some more termiology out of the way. Let s start with the followig series. = Note that we re startig at = oly for coveiece. We could start the series aywhere, but the followig otatio ad termiology demads that we start somewhere ad so for the sake of the work we choose to start at =. Now, istead of addig all the terms out to ifiity, let s look at the followig fiite summatios/series. a 005 Paul Dawkis 74

175 s = a s = a + a s = a + a + a s = a + a + a + a 4 4 s = a + a + a + a + + a = a 4 i i= These are called partial sums. Notice that the partial sums will form a ifiite sequece, { s } =, ad that while it might ot make sese to perform a summatio of a ifiite list of umbers these are all summatios of a fiite list of umbers ad so are guarateed to be fiite umbers. Well they will be fiite umbers provided we do t ed up with a divisio by zero error somewhere i the list. I all of the work that we ll be doig i this chapter we will assume that all the sequece terms exist ad are fiite umbers. From the sectio o sequeces we kow how to determie if the sequece of partial sums coverges or diverges. Also otice that as the sequece terms, s, should start lookig more ad more like the ifiite series. I fact it ca be show that if the sequece of partial sums is coverget ad if we defie, s = lim s the, = a I these cases we call the series coverget ad we call s the sum or value of the series. If the sequece of partial sums is diverget (i.e. either the limit does t exist or is ifiite) the we call the series diverget. I other words, the series is coverget if the sequece of partial sums is coverget ad hece has a fiite value. Likewise the series will be diverget if the sequece of partial sums is diverget. I the case of a diverget series, either the series will have a ifiite value or wo t have a value at all depedig o whether or ot the limit of the sequece of partial sums exists or is ifiite. Let s take a look at some series ad see if we ca determie if they are coverget or diverget. Example Determie if the followig series is coverget or diverget. If it coverges determie its sum. Solutio = = s 005 Paul Dawkis 75

176 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 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, ( + ) = is coverget or diverget. That s ot terribly difficult i this case. ( + ) lim = Therefore, the sequece of partial sums diverges to ad so the series also diverges. Note that we ca say that the series has the value of i these cases, although the series is still called diverget. 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 fid the geeral formula. We will cotiue with a few more examples however, sice this is techically how we determie covergece 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. 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. = 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. i= 005 Paul Dawkis 76

177 The geeral formula for the partial sums is, s = = i 4 + ad i this case we have, i= lim s = lim = 4 ( ) + 4 The sequece of partial sums coverges ad so the series coverges as well ad its value is, = 4 = Example Determie if the followig series coverges or diverges. If it coverges determie its sum. = 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. s0 = s = = 0 s = + = s = + = 0 etc. So, it looks like the sequece of partial sums is, { s } = {, 0,, 0,, 0,, 0,, } = 0 ad this sequece diverges sice lim s does t exist. Therefore, the series also diverges. Note as well that ulike the first example, which was also diverget, this series does t eve have a value. Example 4 Determie if the followig series coverges or diverges. If it coverges determie its sum. = Solutio Here is the geeral formula for the partial sums for this series. 005 Paul Dawkis 77

178 s = = i 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, lim s = lim = The sequece of partial sums is coverget ad so the series will also be coverget. The value of the series is, = = 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 term (ot the partial sums!!). lim = ( ) this series diverged lim = 0 this series coverged lim does't exist this series diverged lim = 0 this series coverged 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. Theorem If a coverges the lim a = 0. Be careful to ot misuse this theorem however! This theorem gives us a requiremet for covergece but ot a guaratee of covergece. I other words, the coverse is NOT true. If lim a = 0 the series may actually diverge! Cosider the followig two series. 005 Paul Dawkis 78

179 = = 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 lim a 0 the a will diverge. 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, diverges = coverges = Oe of the more commo mistakes that studets make whe the first get ito series is to assume that if lim a = 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 = Solutio With almost every series 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. 005 Paul Dawkis 79

180 4 lim = 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 about arithmetic of series ad covergece/divergece. If a ad b are both coverget series the so are ca more, these series will have the followig sums or values. ca = c a ( ± ) = ± a b a b = k = k = k ad ( a ± b) = k. Further 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 that fact. However, it is possible to have a ad b are both be diverget series ad yet have ( a ± b) = k be a coverget series. Before leavig this sectio there is oe more topic that we eed to briefly discuss. 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 sometimes called coditioally coverget. At this poit we do t really have the tools at had to properly ivestigate this topic i detail ad so we ll ot say aythig more about this subject for a while. Whe we fially 005 Paul Dawkis 80

181 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 metioed here oly because we were already discussig covergece i this sectio. 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 wrote i the form, or, with a idex shift the geometric series will ofte be writte as, = 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, a( r ) a ar s = = r r r = 0 The series will coverge provided the partial sums form a coverget sequece, so let s take the limit of the partial sums. a ar lim s = lim r r a ar = lim lim r r a a = lim r r r ar ar 005 Paul Dawkis 8

182 Now, from Theorem from the Sequece 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, a lim s = r Therefore, a geometric series will coverge if < r <, which is usually writte r <, its value is, a ar = ar = r = = 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. (a) (b) = ( 4) = 0 5 Solutio (a) 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 ( ) = 9 4 = 9 = = = 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. = = = 005 Paul Dawkis 8

183 = 6 9 = = = = 4 So, this is a geometric series with a=44 ad r = <. Therefore, sice r < we kow 9 the series will coverge ad its value will be, = = ( 44) = 4 = (b) 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) = = 5 = = 0 = 0 = 0 = 0 So, we ve got it ito the correct form ad we ca see that a=5 ad that r ad so this series diverges. 64 r = 5. Also ote 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. (a) (b) = 0 = Solutio (a) I this case we could just ackowledge that this is a geometric series that starts at =0 ad so 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, = = = 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, 005 Paul Dawkis 8

184 = = + = 5 5 (b) 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 = = = = = We ca ow use the value of the series from the previous example to get the value of this series = = 08 = 5 5 = = 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 0 = 0 = a 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 a is also fiite ad so is coverget. = 0 Likewise, suppose that a is coverget. I this case if we subtract three fiite values = 0 a. This is ow a = from this value we will remai fiite ad arrive at the value of fiite value ad so this series will also be coverget. I other words, if we have two series ad they differ oly by the presece, or absece, of a fiite umber of 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. 005 Paul Dawkis 84

185 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. s = i= 0 i+ i+ = = + 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. lim s = lim = + 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. 005 Paul Dawkis 85

186 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 = i i i = = + + i= i+ i+ = = 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 series. The limit of the partial sums is, 5 5 lim s = lim = out of the So, this series is coverget (because the partial sums form a coverget sequece) ad its value is, 5 = = + 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. + = = = 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. 005 Paul Dawkis 86

187 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 = + 4+ Solutio To get the value of this series all we eed to do is rewrite it ad the use the previous results = = + 4+ = + 4+ = + + = = + 4+ = 5 96 = = 5 We did t discuss the covergece of this series because it was the sum of two coverget series 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 chapter. Here is the harmoic series. = 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) = (b) = 4 Solutio 005 Paul Dawkis 87

188 (a) 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 = 5 = = Now, 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) I this case we ll start with the harmoic series ad strip out the first three terms. = = = = 4 = 4 = 6 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. 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. Itegral Test The last topic that we discussed i the previous sectio was the harmoic series. I that discussio we stated that the harmoic series was a diverget series. It is ow time to prove that statemet. This proof will also get us started o the way to our ext test for covergece that we ll be lookig at. So, we will be tryig to prove that the harmoic series, = diverges. 005 Paul Dawkis 88

189 We ll start this off by lookig at a apparetly urelated problem. Let s start off by askig what the area uder f ( x) = is o the iterval [, ). From the sectio o x Improper Itegrals we kow that this is, dx = x ad so we called this itegral diverget (yes, that s the same term we re usig here with series.). So, just how does that help us to prove that the harmoic series diverges? Well, recall that we ca always estimate the area by breakig up the iterval ito segmets ad the sketchig i rectagles ad usig the sum of the area all of the rectagles as a estimate of the actual area. Let s do that for this problem as well ad see what we get. We will break up the iterval ito subitervals of width ad we ll take the fuctio value at the left edpoit as the height of the rectagle. The image below shows the first few rectagles for this area. So, the area uder the curve is approximately, A () + () + () + () + () = Now ote a couple of thigs about this approximatio. First, each of the rectagles overestimates the actual area ad secodly the formula for the area is exactly the harmoic series! Puttig these two facts together gives the followig, A > dx= x = 005 Paul Dawkis 89

190 Notice that this tells us that we must have, > = = = Sice we ca t really be larger tha ifiity the harmoic series must also be ifiite i value. I other words, the harmoic series is i fact diverget. So, we ve maaged to relate a series to a improper itegral that we could compute ad it turs out that the improper itegral ad the series have exactly the same covergece. Let s see if this will also be true for a series that coverges. Whe discussig the Divergece Test we made the claim that = coverges. Let s see if we ca do somethig similar to the above process to prove this. x from the Improper Itegral sectio we kow that, dx = x ad so this itegral coverges. We will try to relate this to the area uder f ( x) = is o the iterval [ ),. Agai, We will oce agai try to estimate the area uder this curve. We will do this i a almost idetical maer as the previous part with the exceptio that we will istead of usig the left ed poits for the height of our rectagles we will use the right ed poits. Here is a sketch of this case, I this case the area estimatio is, 005 Paul Dawkis 90

191 A () + () + () + () = This time, ulike the first case, the area will be a uderestimatio of the actual area ad the estimatio is ot quite the series that we are workig with. Notice however that the oly differece is that we re missig the first term. This meas we ca do the followig, = < + dx = + = = 4 5 x Area Estimatio Or, puttig all this together we see that, = < With the harmoic series this was all that we eeded to say that the series was diverget. With this series however, this is t quite eough. For istace < ad if the series did have a value of the it would be diverget (whe we wat coverget). So, let s do a little more work. First, let s otice that all the series terms are positive (that s importat) ad that the partial sums are, s = i= i Because the terms are all positive we kow that the partial sums must be a icreasig sequece. I other words, + s = < = s + i= i i= i I s + we are addig a sigle positive term oto s ad so must get larger. Therefore, the partial sums form a icreasig (ad hece mootoic) sequece. Also ote that, sice the terms are all positive, we ca say, s = < < s < i= i = ad so the sequece of partial sums is a bouded sequece. I the sectio o Sequeces we gave a theorem that stated that a bouded ad mootoic sequece was guarateed to be coverget. This meas that the sequece of partial sums is a coverget sequece. So, who cares right? Well recall that this meas that the series must the also be coverget! 005 Paul Dawkis 9

192 So, oce agai we were able to relate a series to a improper itegral (that we could compute) ad the series ad the itegral had the same covergece. We wet through a fair amout of work i both of these examples to determie the covergece of the two series. Luckily for us we do t eed to do all this work every time. The ideas i these two examples ca be summarized i the followig test. Itegral Test Suppose that f(x) is a positive, decreasig fuctio o the iterval [, ) f = a the, x dx is coverget so is a.. If f. If f k k = k x dx is diverget so is a. = k k ad that There are a couple of thigs to ote about the itegral test. First, the lower limit o the improper itegral must be the same value that starts the series. Secod, the fuctio does ot actually eed to be decreasig everywhere i the iterval. All that s really required is that evetually the fuctio is decreasig. I other words, it is okay if the fuctio icreases for a while, but evetually the fuctio must start decreasig ad the cotiue to decrease from that poit o. There is oe more very importat poit that must be made about this test. This test does NOT give the value of a series. It will oly give the covergece/divergece of the series. That s it. No value. We ca use the above series as a perfect example of this. All that the test gave us was that, < = So, we got a upper boud o the value of the series, but ot a actual value for the series. I fact, from this poit o we will ot be askig for the value of a series we will oly be askig whether a series coverges or diverges. I a later sectio we look at estimatig values of series, but eve i that sectio still wo t actually be gettig values of series. Just for the sake of completeess the value of this series is kow. π = = < 6 = Let s work a couple of examples. Example Determie if the followig series is coverget or diverget. 005 Paul Dawkis 9

193 Solutio I this case the fuctio we ll use is, l = f ( x) = x l x This fuctio is clearly positive ad if we make x larger the deomiator will get larger ad so the fuctio is also decreasig. Therefore, all we eed to do is determie the covergece of the followig itegral. t dx = lim dx u = l x xl x xl x t t ( ( x) ) ( t) = lim l l t = lim l l l l t = The itegral is diverget ad so the series is also diverget by the Itegral Test. Example Determie if the followig series is coverget or diverget. e = 0 Solutio The fuctio that we ll use i this example is, f ( x) = xe x This fuctio is always positive o the iterval that we re lookig at. Now we eed to check that the fuctio is decreasig. It is ot clear that this fuctio will always be decreasig o the iterval give. We ca use our Calculus I kowledge to help us however. The derivative of this fuctio is, x f ( x) = e ( x ) This fuctio has two critical poits (which will tell us where the derivative chages sig) at x =±. Sice we are startig at =0 we ca igore the egative critical poit. Pickig a couple of test poits we ca see that the fuctio i icreasig o the iterval 0, ad it is decreasig o,. Therefore, evetually the fuctio will be decreasig ad that s all that s required for us to use the Itegral Test. 005 Paul Dawkis 9

194 t x x e lim 0 t e 0 x dx = x dx u = x = lim t e = lim e t = The itegral is coverget ad so the series must also be coverget by the Itegral Test. We ca use the Itegral test to get the followig fact/test for some series. x t t 0 Fact (p series Test) If k>0 the coverges if p > ad diverges if p. p k = Sometimes the series i this fact are called p-series ad so this fact is sometimes called the p-series test. Example Determie if the followig series are coverget or diverget. (a) 7 = 4 (b) = Solutio (a) I this case p=7> ad so by this fact the series is coverget. (b) For this series p = ad so the series is diverget by the fact. It is importat to ote before leavig this sectio that i order to use the Itegral Test the series terms MUST be positive. If they are egative the the test does t work. Also remember that the test oly determies the covergece of a series ad does NOT give the value of the series. Compariso Test / Limit Compariso Test I the previous sectio we saw how to relate a series to a improper itegral to determie the covergece of a series. While the itegral test is a ice test, it does force us to do 005 Paul Dawkis 94

195 improper itegrals which are t always easy ad i some cases may be impossible to evaluate. For istace cosider the followig series. 0 = + I order to use the Itegral Test we would have to itegrate dx 0 x + x ad I m ot eve sure if it s possible to do this itegral. Nicely eough for us there is aother test that we ca use o this series that will be much easier to use. First, let s ote that the series terms are positive. As with the Itegral Test that will be importat i this sectio. Next let s ote that we must have x > 0 sice we are itegratig o the iterval 0 x <. Likewise, regardless of the value of x we will always have x > 0. So, if we drop the x from the deomiator the deomiator will get smaller ad hece the whole fractio will get larger. So, + < Now, is a geometric series ad we kow that sice value will be, 0 = r = = = = 0 < the series will coverge ad it s Now, if we go back to our origial series ad write dow the partial sums we get, s = i + i i= Sice all the terms are positive addig a ew term will oly make the umber larger ad so the sequece of partial sums must be a icreasig sequece. + s = < i = s i + + i + i i= i= The sice, + < 005 Paul Dawkis 95

196 ad because the terms i these two sequeces are positive we ca also say that, s = < i < i = s < + i i= i= = Therefore, the sequece of partial sums is also a bouded sequece. The from the sectio o Sequeces we kow that a mootoic ad bouded sequece is also coverget. So, the sequece of partial sums of our series is a coverget sequece. This meas that the series itself, 0 = + is also coverget. So, what did we do here? We foud a series whose terms were always larger tha the origial series terms ad this ew series was also coverget. The sice the origial series terms were positive (very importat) this meat that the origial series was also coverget. To show that a series (with oly positive terms) was diverget we could go through a similar argumet ad fid a ew diverget series whose terms are always smaller tha the origial series. I this case the origial series would have to take a value larger tha the ew series. However, sice the ew series is diverget its value will be ifiite. This meas that the origial series must also be ifiite ad hece diverget. We ca summarize all this i the followig test. Compariso Test Suppose that we have two series a ad b with a, b 0 for all ad a b for all. The,. If b is coverget the so is a.. If a is diverget the so is b. I other words, we have two series of positive terms ad the terms of oe of the series is always larger tha the terms of the other series. The if the larger series is coverget the smaller series must also be coverget. Likewise, if the smaller series is diverget the the larger series must also be diverget. Do ot misuse this test. Just because the smaller of the two series coverges does ot say aythig about the larger series. The larger series may still diverge. Likewise, just because we kow that the larger of two series diverges we ca t say that the smaller series will also diverge! Be very careful i usig this test 005 Paul Dawkis 96

197 Recall that we had a similar test for improper itegrals back whe we were lookig at itegratio techiques. So, if you could use the compariso test for improper itegrals you ca use the compariso test for series as they are pretty much the same idea. Let s take a look at some examples. Example Determie if the followig series is coverget or diverget. = cos Solutio Sice the cosie term i the deomiator does t get too large we ca assume that the series terms will behave like, = which, as a series, will diverge. So, from this we ca guess that the series will probably diverge ad so we ll eed to fid a smaller series that will also diverge. Recall that from the compariso test with improper itegrals that we determied that we ca make a fractio smaller by either makig the umerator smaller or the deomiator larger. I this case the two terms i the deomiator are both positive. So, if we drop the cosie term we will i fact be makig the deomiator larger sice we will o loger be subtractig off a positive quatity. Therefore, > = cos The, sice = diverges (it s harmoic ad the p-series test) by the Compariso Test our origial series must also diverge. Example Determie if the followig series coverges or diverges. + 4 = + 5 Solutio I this case the + ad the +5 do t really add aythig to the series ad so the series terms should behave pretty much like = 4 which will coverge as a series. Therefore, we will eed to fid a larger series which also coverges. This meas that we ll either have to make the umerator larger or the deomiator smaller. We ca make the deomiator smaller by droppig the +5. Doig this gives, 005 Paul Dawkis 97

198 + + < At this poit, otice that we ca t drop the + from the umerator sice this would make the term smaller ad that s ot what we wat. However, this is actually all the further that we eed to go. Let s take a look at the followig series. + 4 = = = = = + 4 = = As show, we ca write the series as a sum of two series ad both of these series are coverget by the p-series test. Therefore, sice each of these series are coverget we kow that the sum, + 4 = is also a coverget series. Recall that the sum of two coverget series will also be coverget. Now, sice the terms of this series are larger tha the terms of the origial series we kow that the origial series must also be coverget by the Compariso Test. The compariso test is a ice test that allows us to do problems that either we could t have doe with the itegral test or at the best would have bee very difficult to do with the itegral test. That does t mea that it does t have problems of its ow. Cosider the followig series. 0 = This is ot much differet from the first series that we looked at. The origial series coverged because the gets very large very fast ad will be sigificatly larger tha the. Therefore, the does t really affect the covergece of the series i that case. The fact that we are ow subtractig the off ow istead of addig the o really should t chage the covergece. We ca say this because the gets very large very fast ad the fact that we re subtractig off wo t really chage the size of this term for all sufficietly large value of. So, we would expect this series to coverge. However, the compariso test wo t work with this series. To use the compariso test o this series we would eed to fid a larger series that we could easily determie the covergece of. I this case we ca t do what we did with the origial series. If we drop the we will make the deomiator larger (sice the was subtracted off) ad so the fractio will get smaller ad just like whe we looked at the compariso test for improper itegrals kowig that the smaller of two series coverges does ot mea that the larger of the two will also coverge. 005 Paul Dawkis 98

199 So, we will eed somethig else to do help us determie the covergece of this series. The followig variat of the compariso test will allow us to determie the covergece of this series. Limit Compariso Test Suppose that we have two series a ad b with a, b 0 for all. Defie, a c = lim b If c is positive (i.e. c > 0 ) ad is fiite (i.e. c < ) the either both series coverge or both series diverge. Note that it does t really matter which series term is i the umerator for this test, we could just have easily defied c as, b c = lim a ad we would get the same results. To see why this is, cosider the followig two defiitios. a b c= lim c = lim b a Start with the first defiitio ad rewrite it as follows, the take the limit. a c = lim = lim = = b b b lim c a a I other words, if c is positive ad fiite the so is c ad if c is positive ad fiite the so is c. Likewise if c = 0 the c = ad if c = the c = 0. Both defiitios will give the same results from the test so do t worry about which series terms should be i the umerator ad which should be i the deomiator. Choose this to make the limit easy to compute. Also, this really is a compariso test is some ways. If c is positive ad fiite this is sayig that both of the series terms will behave i geerally the same fashio ad so we ca expect the series themselves to also behave i a similar fashio. If c = 0 or c = we ca t say this ad so the test fails to give ay iformatio. The limit i this test will ofte be writte as, c= lim a b sice ofte both terms will be fractios ad this will make the limit easier to deal with. Let s see how this test works. 005 Paul Dawkis 99

200 Example Determie if the followig series coverges or diverges. 0 = Solutio To use the limit compariso test we eed to fid a secod series that we ca determie the covergece of easily ad has what we assume is the same covergece as the give series. O top of that we will eed to choose the ew series i such a way as to give us a easy limit to compute for c. We ve already guessed that this series coverges ad sice it s vaguely geometric let s use 0 = as the secod series. We kow that this series coverges ad there is a chace that sice both series have the i it the limit wo t be too bad. Here s the limit. c = lim = lim Now, we ll eed to use L Hospital s Rule o the secod term i order to actually evaluate this limit. c = lim l = So, c is positive ad fiite so by the Compariso Test both series must coverge sice coverges. 0 = Example 4 Determie if the followig series coverges or diverges = + Solutio Fractios ivolvig oly polyomials or polyomials uder radicals will behave i the same way as the largest power of will behave i the limit. So, the terms i this series should behave as, 005 Paul Dawkis 00

201 = = 7 7 ad as a series this will diverge by the p-series test. I fact, this would make a ice choice for our secod series i the limit compariso test so let s use it lim = lim = lim = = 4 = c So, c is positive ad fiite ad so both limits will diverge sice diverges. = Fially, to see why we eed to c must be positive ad fiite (i.e. c 0 ad c ) cosider the followig two series. = = The first diverges ad the secod coverges. Now compute each of the followig limits. lim i = lim = lim i = lim = 0 I the first case the limit from the limit compariso test yields c = ad i the secod case the limit yields c = 0. Clearly, both series do ot have the same covergece. Note however, that just because we get c = 0 or c = does t mea that the series will have the opposite covergece. To see this cosider the series, = = Both of these series coverge ad here are the two possible limits that the limit compariso test uses. lim i = lim = 0 lim i = lim = 005 Paul Dawkis 0

202 So, eve though both series had the same covergece we got both c = 0 ad c =. The poit of all of this is to remid us that if we get c = 0 or c = from the limit compariso test we will kow that we have chose the secod series icorrectly ad we ll eed to fid a differet choice i order to get ay iformatio about the covergece of the series. Alteratig Series Test The last two tests that we looked at for series covergece have required that all the terms i the series be positive. Of course there are may series out there that have egative terms i them ad so we ow eed to start lookig at tests for these kids of series. The test that we are goig to look ito i this sectio will be a test for alteratig series. A alteratig series is ay series, a, for which the series terms ca be writte i oe of the followig two forms. a = b b 0 + a = b b 0 There are may other ways to deal with the alteratig sig, but they ca all be writte as o of the two forms above. For istace, + ( ) = ( ) ( ) = ( ) ( ) = ( ) ( ) = ( ) + + There are of course may others, but they all follow the same basic patter of reducig to oe of the first two forms give. If you should happe to ru ito a differet form tha the first two, do t worry about covertig it to oe of those forms, just be aware that it ca be ad so the test from this sectio ca be used. Alteratig Series Test Suppose that we have a series b 0 for all. The if, a ad either a = b or + a = b where. lim b = 0 ad,. { b } is evetually a decreasig sequece the series a is coverget. There are a couple of thigs to ote about this test. First, ulike the Itegral Test ad the Compariso/Limit Compariso Test, this test will oly tell us whe a series coverges ad ot if a series will diverge. 005 Paul Dawkis 0

203 Secodly, i the secod coditio all that we eed to require is that the series terms, b will evetually decreasig. It is possible for the first few terms of a series to icrease ad still have the test be valid. All that is required is that evetually we will have b b + for all after some poit. To see why this is cosider the followig series, = ( ) b Let s suppose that for N { } { b } is a decreasig sequece. The series ca the be writte as, b is ot a decreasig sequece ad that for N + N ( ) = ( ) + ( ) b b b = = = N+ The first series is a fiite series (o matter how large N is, it will still be a fiite series) ad so we ca compute its value ad it will be fiite. Therefore, the covergece of the series will deped solely o the covergece of the secod (ifiite) series. If the secod series has a fiite value the the sum of two fiite values is also fiite ad so the origial series will coverge to a fiite value. O the other had if the secod series is diverget either because its value is ifiite or it does t have a value the addig a fiite umber oto this will ot chage that fact ad so the origial series will be diverget. The poit of all this is that we do t eed to require that the series terms be decreasig for all. We oly eed to require that the series terms will evetually be decreasig sice we ca always strip out the first few terms that are t actually decreasig ad look oly at the terms that are actually decreasig. Note that, i practice, we do t actually strip out the terms that are t decreasig. All we do is check that evetually the series terms are decreasig ad the apply the test. Let s work a couple of examples. Example Determie if the followig series is coverget or diverget. + = Solutio First, idetify the b for the test. + ( ) + = ( ) b = = = Now, all that we eed to do is ru through the two coditios i the test. 005 Paul Dawkis 0

204 lim b = lim = 0 b = > = b+ + Both coditios are met ad so by the Alteratig Series Test the series must coverge. The series from the previous example is sometimes called the Alteratig Harmoic Series. I the previous example it was easy to see that the series terms decreased sice icreasig oly icreased the deomiator for the term ad hece made the term smaller. I geeral however we will eed to resort to Calculus I techiques to prove the series terms decrease. We ll see a example of this i a bit. Example Determie if the followig series is coverget or diverget. ( ) = + 5 Solutio First, idetify the b for the test. ( ) = ( ) b = = + 5 = Let s check the coditios. lim b = lim = So, the first coditio is t met ad so there is o reaso to check the secod. Sice this coditio is t met we ll eed to use aother test to check covergece. I these cases where the first coditio is t met it is usually best to use the divergece test. ( ) lim = ( lim ( ) ) lim = lim ( )() = lim does't exist This limit does t exist ad so by the Divergece Test this series diverges. Example Determie if the followig series is coverget or diverget. ( ) = Solutio Notice that i this case the expoet o the - is t or +. That wo t chage how 005 Paul Dawkis 04

205 the test works however so we wo t worry about that. I this case we have, b = + 4 so let s check the coditios. The first is easy eough to check. lim b = lim = The secod coditio requires some work however. It is ot immediately clear that these terms will decrease. Icreasig to + will icrease both the umerator ad the deomiator. Icreasig the umerator says the term should also icrease while icreasig the deomiator says that the term should decrease. Sice its ot clear which of these will wi out we will eed to resort to Calculus I techiques to show that the terms decrease. Let s start with the followig fuctio ad its derivative. x 4 x f ( x) = f ( x) = x + 4 x x+ 4 Now, there are three critical poits for this fuctio, x=-4, x=0, ad x=4. The first is outside the boud of our series so we wo t eed to worry about that oe. Usig the test poits, f () = f ( 5) = ad so we ca see that the fuctio i icreasig o 0 x 4 ad decreasig o x 4. Therefore, sice f = b we kow as well that the b are also icreasig o 0 4 ad decreasig o 4. The b are the evetually decreasig ad so the secod coditio is met. Both coditios are met ad so by the Alteratig Series Test the series must be covergig. As the previous example has show, we sometimes eed to do a fair amout of work to show that the terms are decreasig. Do ot just make the assumptio that the terms will be decreasig ad let it go at that. Let s do oe more example just to make a poit. Example 4 Determie if the followig series is coverget or diverget. cos( π ) = Solutio 005 Paul Dawkis 05

206 The poit of this problem is really just to ackowledge that it is i fact a alteratig series. To see this we eed to ackowledge that, cos( π ) = ( ) ad so the series is really, cos( π ) ( ) = b = = = Checkig the two coditio gives, lim b = lim = 0 b = > = b + The two coditios of the test are met ad so by the Alteratig Series Test the series is coverget. It should be poited out that the rewrite we did i previous example oly works because is a iteger ad because of the presece of the π. Without the π we could t do this ad if was t guarateed to be a iteger we could t do this. + Absolute Covergece Whe we first talked about series covergece we briefly metioed a stroger type of covergece but did t do aythig with it because we did t have ay tools at our disposal that we could use to work problems ivolvig it. We ow have some of those tools so it s ow time to talk about absolute covergece i detail. First, let s go back over the defiitio of absolute covergece. A series a is called absolutely coverget if a is coverget. If a is coverget ad a is diverget we call the series coditioally coverget. We also have the followig fact about absolute covergece. Fact If a is absolutely coverget the it is also coverget. It is this fact that makes absolute covergece a stroger type of covergece. Series that are absolutely coverget are guarateed to be coverget. However, series that are coverget may or may ot be absolutely coverget. 005 Paul Dawkis 06

207 Let s take a quick look at a couple of examples of absolute covergece. Example Determie if each of the followig series are absolute coverget, coditioally coverget or diverget. (a) = ( ) ( ) + (b) = si (c) = Solutio (a) This is the alteratig harmoic series ad we saw i the last sectio that it is a coverget series so we do t eed to check that here. So, let s see if it is a absolutely coverget series. To do this we ll eed to check the covergece of. ( ) = = = This is the harmoic series ad we kow form the itegral test sectio that it is diverget. Therefore, this series is ot absolutely coverget. It is however coditioally coverget sice the series itself does coverge. (b) I this case let s just check absolute covergece first sice if it s absolutely coverget we wo t eed to bother checkig covergece as we will get that for free. + ( ) = = = This series is coverget by the p-series test ad so the series is absolute coverget. Note that this does say as well that it s a coverget series. (c) I this part we eed to be a little careful. First, this is NOT a alteratig series ad so we ca t use ay tools from that sectio. What we ll do here is check for absolute covergece first agai sice that will also give covergece. This meas that we eed to check the covergece of the followig series. si si = = = To do this we ll eed to ote that si si ad so we have, si 005 Paul Dawkis 07

208 Now we kow that = coverges by the p-series test ad so by the Compariso Test we also kow that si coverges. Therefore the origial series is absolutely coverget (ad hece coverget). = Ratio Test I this sectio we are goig to take a look at a test that we ca use to see if a series is absolutely coverget or ot. Recall that if a series is absolutely coverget the we will also kow that it s coverget ad so we will more ofte tha ot use it to simply determie the covergece of a series. We may as well ackowledge that it will give us absolute covergece as well however. Before proceedig with the test let s do a quick remider of factorials. This test will be particularly useful for series that cotai factorials (ad we will see some i the applicatios) so let s make sure we ca deal with them before we ru ito them i a example. If is a iteger such that 0 the factorial is defied as,! = if 0! = by defiitio Let s compute a couple real quick.! =! = () =! = = 6 4! = 4 = 4 5! = 5 4 = 0 I the last computatio above, otice that we could rewrite the factorial i a couple of differet ways. For istace, 005 Paul Dawkis 08

209 5! = 5 4 = 5 4! 4! 5! = 5 4 = 5 4! I geeral we ca always strip out terms from a factorial as follows.! = k k+ ( )!! = k k+! = k k We will eed to do this o occasio so do t forget about it. Also, whe dealig with factorials we eed to be very careful with parethesis. For!! as we ca see if we write each of the followig factorials out. istace, ( )! = ( )( )! = ( )( ) Agai, we will ru across factorials with parethesis so do t drop them. This is ofte oe of the more commo mistakes that studets make whe the first ru across factorials. Okay, we are ow ready for the test. Ratio Test Suppose we have the series a. Defie, a L = lim a The,. if L < the series is absolutely coverget (ad hece coverget).. if L > the series is diverget.. if L = the series may be diverget, coverget, or absolutely coverget. Notice that i the case of L = the ratio test is pretty much worthless ad we would eed to resort to a differet test to determie the covergece of the series. Also, the absolute value bars i the defiitio of L are absolutely required. If they are ot there it will be possible for us to get the icorrect aswer. Let s take a look at some examples. Example Determie if the followig series is coverget or diverget Paul Dawkis 09

210 ( 0) + ( + ) = 4 Solutio With this first example let s be a little careful ad make sure that we have everythig dow correctly. Here are the series terms a. ( 0) a = Recall that to compute a + all that we eed to do is substitute + for all the s i a. Now, to defie L we will use, a ( ) ( ) ( + + ) ( ) ( + ) = = ( ) L= lim a + a sice this will be a little easier whe dealig with fractios as we ve got here. So, + ( ) 0( ) ( + ) + + ( ) L = lim = lim = lim = < 6 So, L < ad so by the Ratio Test the series coverges absolutely ad hece will coverge. As see i the previous example there is usually a lot of cacelig that will happe i these. Make sure that you do this cacelig. If you do t do this kid of cacelig it ca make the limit fairly difficult. Example Determie if the followig series is coverget or diverget.! 0 5 = Solutio Now that we ve worked oe i detail we wo t go ito quite the detail with the rest of these. Here is the limit. ( + ) 5 ( + )!! L = lim = lim + 5! 5! 005 Paul Dawkis 0

211 I order to do this limit we will eed to elimiate the factorials. We simply ca t do the limit with the factorials i it. To elimiate the factorials we will recall from our discussio o factorials above that we ca always strip out terms from a factorial. If we do that with the umerator (i this case because it s the larger of the two) we get, ( + )! L = lim 5! at which poit we ca cacel the! for the umerator a deomiator to get, ( + ) L = lim => 5 So, by the Ratio Test this series diverges. Example Determie if the followig series is coverget or diverget. = ( )! Solutio I this case be careful i dealig with the factorials. ( + ) ( )! L = lim! + ( ) ( + ) ( ) ( + )! = lim! ( + ) ( ) ( + )( ) ( + )! = lim! = lim = 0< ( + )( ) So, by the Ratio Test this series coverges. Example 4 Determie if the followig series is coverget or diverget. 9 + = Solutio Do ot mistake this for a geometric series. The i the deomiator meas that this is t a geometric series. So, let s compute the limit. 005 Paul Dawkis

212 L = lim ( ) ( + ) 9 = lim ( )( + ) 9 = lim + 9 = > ( ) 9 + Therefore, by the Ratio Test this series is diverget. I the previous example the absolute value bars were required to get the correct aswer. 9 If we had t used them we would have gotte L = < which would have implied a coverget series! Now, let s take a look at a couple of examples to see what happes whe we get L =. Recall that the ratio test will ot tell us aythig about the covergece of these series. I both of these examples we will first verify that we get L = ad the use other tests to determie the covergece. Example 5 Determie if the followig series is coverget or diverget. ( ) = 0 + Solutio Let s first get L. + ( ) + + L = lim = lim = So, as implied earlier we get L = which meas the ratio test is o good for determiig the covergece of this series. We will eed to resort to aother test for this series. This series is a alteratig series ad so let s check the two coditios from that test. lim b = lim = 0 + b = > = b The two coditios are met ad so by the Alteratig Series Test this series is coverget. Example 6 Determie if the followig series is coverget or diverget. 005 Paul Dawkis

213 Solutio Here s the limit. = ( + )( + 7) L = lim = lim = Agai, the ratio test tells us othig here. We ca however, quickly use the divergece test o this. I fact that probably should have bee our first choice o this oe ayway. + lim = By the Divergece Test this series is diverget. So, as we saw i the previous two examples if we get L = from the ratio test the series ca be either coverget or diverget. There is oe more thig that we should ote about the ratio test before we move oto the ext sectio. The last series was a polyomial divided by a polyomial ad we saw that we got L = from the ratio test. This will always happe with ratioal expressio ivolvig oly polyomials or polyomials uder radicals. So, i the future it is t eve worth it to try the ratio test o these kids of problems sice we ow kow that we will get L =. Also, i the secod to last example we saw a example of a ratioal expressio times ( ). I this example we also saw that we got L =. Agai, as log as it is mius oe to a power (ay power) this will happe as well ad so agai there really is t ay use i eve tryig the ratio test o these kids of problems. Note however, if it had bee ( ) + or 8 (i.e. aythig other tha - ) we would t have gotte L = ad so the ratio test would work. Root Test This is the last test for series covergece that we re goig to be lookig at. As with the Ratio Test this test will also tell whether a series is absolutely coverget or ot rather tha simple covergece. Root Test Suppose that we have the series a. Defie, L= lim a = lim a The, 4. if L < the series is absolutely coverget (ad hece coverget). 005 Paul Dawkis

214 5. if L > the series is diverget. 6. if L = the series may be diverget, coverget, or absolutely coverget. As with the ratio test, if we get L = the root test will tell us othig ad we ll eed to use aother test to determie the covergece of the series. We will also eed the followig fact i some of these problems. Fact lim = Let s take a look at a couple of examples. Example Determie if the followig series is coverget or diverget. + = Solutio There really is t much to these problems other tha computig the limit ad the usig the root test. Here is the limit for this problem. L + + So, by the Root Test this series is diverget. = lim = lim = => Example Determie if the followig series is coverget or diverget. 5 = Solutio Agai, there is t too much to this series. 5 5 L = lim lim = = = < Therefore, by the Root Test this series coverges absolutely ad hece coverges. Note that we had to keep the absolute value bars o the fractio util we d take the limit to get the sig correct. Example Determie if the followig series is coverget or diverget. = 005 Paul Dawkis 4

215 Solutio Here s the limit for this series. ( ) L = lim = lim = = > After usig the fact from above we ca see that the Root Test tells us that this series is diverget. Note that if we get L = from the ratio test the the root test will also give also L = ad so there is t ay reaso to try the root test o aythig that gives L = o the ratio test. Strategy for Series Now that we ve got all of our tests out of the way it s time to thik about orgaizig all of them ito a geeral set of guidelies to help us determie the covergece of a series. Note that these are a geeral set of guidelies ad because some series ca have more tha oe test applied to them we will get a differet result depedig o the path that we take through this set of guidelies. I fact, because more tha oe test may apply, you should always go completely through the guidelies ad idetify all possible tests that ca be used o a give series. Oce this has bee doe you ca idetity the test that you feel will be the easiest for you to use. With that said here is the set of guidelies for determiig the covergece of a series.. With a quick glace does it look like the series terms do t coverge to zero i the limit, i.e. does lim a 0? If so, use the Divergece Test. Note that you should oly do the divergece test if a quick glace suggests that the series terms may ot coverge to zero i the limit.. Is the series a p-series ( ) or a geometric series ( ar p or ar )? If so = 0 = use the fact that p-series will oly coverge if p > ad a geometric series will oly coverge if r <. Remember as well that ofte some algebraic maipulatio is required to get a geometric series ito the correct form.. Is the series similar to a p-series or a geometric series? If so, try the Compariso Test. 4. Is the series a ratioal expressio ivolvig oly polyomials or polyomials uder radicals (i.e. a fractio ivolvig oly polyomials or polyomials uder 005 Paul Dawkis 5

216 radicals)? If so, try the Compariso Test ad/or the Limit Compariso Test. Remember however, that i order to use the Compariso Test ad the Limit Compariso Test the series terms all eed to be positive. 5. Does the series cotai factorials or costats raised to powers ivolvig? If so, the the Ratio Test may work. Note that if the series term cotais a factorial the the oly test that we ve got that will work is the Ratio Test. 6. Ca the series terms be writte i the form a = ( ) b or + a the the Alteratig Series Test may work. 7. Ca the series terms be writte i the form a ( b ) may work. = b? If so, =? If so, the the Root Test 8. If a = f for some positive, decreasig fuctio ad f evaluate the the Itegral Test may work. x dx is easy to a Agai, remember that these are oly a set of guidelies ad ot a set of hard ad fast rules to use whe tryig to determie the best test to use o a series. If more tha oe test ca be used try to use the test that will be the easiest for you to use ad remember that what is easy for someoe else may ot be easy for you! Also just so we ca put all the tests ito oe place here is a quick listig of all the test that we ve got. Divergece Test If lim a 0 the a will diverge Itegral Test Suppose that f(x) is a positive, decreasig fuctio o the iterval [, ) f = a the,. If f ( x) dx a is coverget so is.. If f k k = k x dx is diverget so is a. Compariso Test Suppose that we have two series all. The, = k a ad b k ad that with a, b 0 for all ad a b for 005 Paul Dawkis 6

217 . If b is coverget the so is. If a is diverget the so is b. a. Limit Compariso Test Suppose that we have two series a ad b with a, b 0 for all. Defie, a c = lim b If c is positive (i.e. c > 0 ) ad is fiite (i.e. c < ) the either both series coverge or both series diverge. Alteratig Series Test Suppose that we have a series b 0 for all. The if, a ad either a + = b or a = ( ) b where. lim b = 0 ad,. { b } is evetually a decreasig sequece the series a is coverget Ratio Test Suppose we have the series a. Defie, a L = lim a The,. if L < the series is absolutely coverget (ad hece coverget).. if L > the series is diverget.. if L = the series may be diverget, coverget, or absolutely coverget. Root Test Suppose that we have the series a. Defie, + L= lim a = lim a The,. if L < the series is absolutely coverget (ad hece coverget).. if L > the series is diverget.. if L = the series may be diverget, coverget, or absolutely coverget. 005 Paul Dawkis 7

218 Estimatig the Value of a Series We have ow spet quite a few sectios determiig the covergece of a series, however, with the exceptio of geometric ad telescopig series, we have ot talked about fidig the value of a series. This is usually a very difficult thig to do ad we still are t goig to talk about how to fid the value of a series. What we will do is talk about how to estimate the value of a series. Ofte that is all that you eed to kow. Before we get ito how to estimate the value of a series let s remid ourselves how series covergece works. It does t make ay sese to talk about the value of a series that does t coverge ad so we will be assumig that the series we re workig with coverges. Also, as well see the mai method of estimatig the value of series will come out of this discussio. So, let s start with the series a (the startig poit is ot importat, but we eed oe to = do the work) ad let s suppose that it coverges to s. Recall that this meas that if we get the partial sums, s = ai the they will form a coverget sequece ad its limit is s. I other words, lim s = s Now, just what does this mea for us? Well, sice this limit coverges it meas that we ca make the partial sums, s, as close to s as we wat simply by takig large eough. I other words, if we take large eough the we ca say that, i= s s This is oe method of estimatig the value of a series. We ca just take a partial sum ad use that as a estimatio of the value of the series. There are ow two questios that we should ask about this. First, how good is the estimatio? If we do t have a idea of how good the estimatio is the it really does t do all that much for us as a estimatio. Secodly, is there ay way to make the estimate better? Sometimes we ca use this as a startig poit ad make the estimatio better. We wo t always be able to do this, but if we ca that will be ice. So, let s start with a geeral discussio about the determiig how good the estimatio is. Let s first start with the full series ad strip out the first terms. 005 Paul Dawkis 8

219 a = a + a () i i i i= i= i= + Note that we coverted over to a idex of i i order to make the otatio cosistet with prior otatio. Recall that we ca use ay letter for the idex ad it wo t chage the value. Now, otice that the first series (the terms that we ve stripped out) is othig more tha the partial sum s. The secod series o the right (the oe startig at i = + ) is called the remaider ad deoted by R. Fially let s ackowledge that we also kow the value of the series sice we are assumig it s coverget. Takig this otatio ito accout we ca rewrite () as, s = s + R We ca solve this for the remaider to get, R = s s So, the remaider tells us the differece, or error, betwee the exact value of the series ad the value of the partial sum that we are usig as the estimatio of the value of the series. Of course we ca t get our hads o the actual value of the remaider because we do t have the actual value of the series. However, we ca use some of the tests that we ve got for covergece to get a pretty good estimate of the remaider provided we make some assumptios about the series. Oce we ve got a estimate o the value of the remaider we ll also have a idea o just how good a job the partial sum does of estimatig the actual value of the series. There are several tests that will allow us to get estimates of the remaider. We ll go through each oe separately. Itegral Test Recall that i this case we will eed to assume that the series terms are all positive ad will evetually be decreasig. We origially derived the itegral test by usig the fact that the series could be thought of as a estimatio of the area uder the curve of f(x) f = a. We ca do somethig similar with the remaider. where First, let s recall that the remaider is, R = a = a + a + a + a + i i= + Now, if we start at x=+, take rectagles of width ad use the left edpoit as the height of the rectagle we ca estimate the area uder f(x) o the iterval [ +, ) as show i the sketch below. 005 Paul Dawkis 9

220 We ca see that the remaider, R, is exactly this area estimatio ad it will over estimate the exact area. So, we have the followig iequality. + R f x dx () Next, we could also estimate the area by startig at x=, takig rectagles of width agai ad the usig the right edpoit as the height of the rectagle. This will give a estimatio of the area uder f(x) o the iterval [, ). This is show i the followig sketch. Agai, we ca see that the remaider, R, is agai this estimatio ad i this case it will uderestimate the area. This leads to the followig iequality, Combiig () ad () gives, R f x dx () 005 Paul Dawkis 0

221 + f x dx R f x dx So, provided we ca do these itegrals we ca get both a upper ad lower boud o the remaider. This will i tur give us a upper boud ad a lower boud o just how good the partial sum, s, is as a estimatio of the actual value of the series. I this case we ca also use these results to get a better estimate for the actual value of the series as well. First, we ll start with the fact that Now, if we use () we get, Likewise if we use () we get, Puttig these two together gives us, s = s + R = s s R s f x dx = + + s s R s f x dx s f x dx s s + + f x dx + (4) This gives a upper ad a lower boud o the actual value of the series. We could the use as a estimate of the actual value of the series the average of the upper ad lower boud. Let s work a example with this. Example Usig =5 to estimate the value of. = Solutio First, for compariso purposes, we ll ote that the actual value of this series is kow to be, π = = = Usig =5 let s first get the partial sum. 5 s5 = = i i= 005 Paul Dawkis

222 Note that this is close to the actual value i some sese, but is t really all that close either. Now, let s compute the itegrals. These are fairly simple itegrals so we ll leave it to you to verify the values. dx = dx = x 5 x Pluggig these ito (4) gives us, s s Both the upper ad lower boud are ow very close to the actual value ad if we take the average of the two we get the followig estimate of the actual value. That is pretty dar close to the actual value. s So, that is how we ca use the Itegral Test to estimate the value of a series. Let s move o to the ext test. Compariso Test I this case, ulike with the itegral test, we may or may ot be able to get a idea of how good a particular partial sum will be as a estimate of the exact value of the series. Much of this will deped o how the compariso test is used. First, let s remid ourselves o how the compariso test actually works. Give a series a let s assume that we ve used the compariso test to show that it s coverget. Therefore, we foud a secod series b that coverged ad a b for all. What we wat to do is determie how good of a job the partial sum, s = ai will do i estimatig the actual value of the series a. Agai, we will use the remaider to do this. Let s actually write dow the remaider for both series. i= R = a T = b i i i= + i= + Now, sice a b we also kow that R T 005 Paul Dawkis

223 Whe usig the compariso test it is ofte the case that the b are fairly ice terms ad that we might actually be able to get a idea o the size of T. For istace, if our secod series is a p-series we ca use the results from above to get a upper boud o T as follows, where = + R T g x dx g b Also, if the secod series is a geometric series the we will be able to compute T exactly. If we are uable to get a idea of the size of T the usig the compariso test to help with estimates wo t do us much good. Let s take a look at a example. Example Usig =5 to estimate the value of. = Solutio To do this we ll first eed to go through the compariso test so we ca get the secod series. So, = ad = 0 is a geometric series ad coverges because r = <. Now that we ve gotte our secod series let s get the estimate. 5 s5 = = = 0 So, how good is it? Well we kow that, R5 T5 = = 6 will be a upper boud for the error betwee the actual value ad the estimate. Sice our secod series is a geometric series we ca compute this directly as follows. 5 = + = 0 = 0 = 6 The series o the left is i the stadard form ad so we ca compute that directly. The first series o the right has a fiite umber of terms ad so ca be computed exactly ad 005 Paul Dawkis

224 the secod series o the right is the oe that we d like to have the value for. Doig the work gives, 5 = = 6 = 0 = 0 = ( ) = So, accordig to this if we use s as a estimate of the actual value we will be off from the exact value by o more tha ad that s ot too bad. I this case it ca be show that = = ad so we ca see that the actual error i our estimatio is, Error = Actual Estimate = = Note that i this case the estimate of the error is actually fairly close (ad i fact exactly the same) as the actual error. This will ot always happe ad so we should t expect that to happe i all cases. The error estimate above is simply the upper boud o the error ad the actual error will ofte be less tha this value. Before movig o to the fial part of this sectio let s agai ote that we will oly be able to determie how good the estimate is usig the compariso test if we ca easily get our hads o the remaider of the secod term. The reality is that we wo t always be able to do this. Alteratig Series Test Both of the methods that we ve looked at so far have required the series to cotai oly positive terms. If we allow series to have egative terms i it the process is usually more difficult. However, with that said there is oe case where it is t too bad. That is the case of a alteratig series. Oce agai we will start of with a coverget series a = ( ) case happes to be a alteratig series, so we kow that 0 b which i this b for all. Also ote that we could have ay power o the - we just used for the sake of coveiece. We wat to kow how good of a estimatio of the actual value will the partial sum s be. As with the prior cases we kow that the remaider, R, will be the error i the estimatio ad so if we ca get a hadle o that we ll kow approximately how good the estimatio is. 005 Paul Dawkis 4

225 I this case the work is a little harder to justify so we ll just give the relevat fact. It ca be show that, R = s s b + We eeded absolute value bars because we wo t kow ahead of time if the estimatio is larger or smaller tha the actual value ad we kow that the b s are positive. Let s take a look at a example. Example Usig =5 to estimate the value of ( ). = Solutio This is a alteratig series ad is does coverge. I this case the exact value is kow ad so for compariso purposes, = ( ) π = = Now, the estimatio is, s 5 ( ) = = = From the fact above we kow that R5 = s s5 b6 = = So, our estimatio will have a error of o more tha I this case the exact value is kow ad so the actual error is, R5 = s s5 = I the pervious example the estimatio had oly half the estimated error. It will ofte be the case the actual error will be less tha the estimated error. Remember that this is oly a upper boud for the actual error. Ratio Test This will be the fial case that we re goig to look at for estimatig series values ad we are goig to have to put a couple of fairly striget restrictio o the series terms i order to do the work. Oe of the mai restrictios we re goig to make is to assume that the series terms are positive. Well also be addig o aother restrictio i a bit. I this case we ve used the ratio test to show that computed a is coverget. To do this we 005 Paul Dawkis 5

226 ad foud that L <. L = lim a a + As with the previous cases we are goig to use the remaider, R, to determie how good of a estimatio of the actual value the partial sum, s, is. To get a estimate of the remaider let s first defie the followig sequece, a+ r = a We ow have two possible cases.. If { } r is a decreasig sequece ad r + < the, a+ R r. If { r } is a icreasig sequece the, R + a+ L Note that there are some restrictios o the sequece { r } ad at least oe of its terms i order to use these formulas. If the restrictios are t met the the formulas ca t be used. Let s take a look at a example of this. Example 4 Usig =5 to estimate the value of. 0 = Solutio First, let s use the ratio test to verify that this is a coverget series. + + L = lim = lim = < + So, it is coverget. Now let s get the estimate. 5 s5 = = To determie a estimate o the remaider, ad hece the error, let s first get the r. sequece { } 005 Paul Dawkis 6

227 r + + = = = + + The last rewrite was just to simplify some of the computatios a little. Now, otice that, f ( x) = + f ( x) = < 0 x x Sice this fuctio is always decreasig ad f r decreasig. Also ote that ( ) the fact above to get, R 6 6 = ad so, this sequece is r = + <. Therefore we ca use the first case from a = = r6 + 6 So, it looks like our estimate is probably quite good. I this case the exact value is kow. = = 0 4 ad so we ca compute the actual error. R5 = s s5 = This is less tha the upper boud, but ulike i the previous example this actual error is quite close to the upper boud. I the last two examples we ve see that the upper boud computatios o the error ca sometimes be quite close to the actual error ad at other times they ca be off by quite a bit. There is usually o way of kowig ahead of time which it will be ad without the exact value i had there will ever be a way of determiig which it will be. Notice that this method did require the series terms to be positive, but that does t mea that we ca t deal with ratio test series if they have egative terms. Ofte series that we used ratio test o are also alteratig series ad so if that is the case we ca always resort to the previous material to get a upper boud o the error i the estimatio, eve if we did t use the alteratig series test to show covergece. Note however that if the series does have egative terms, but does t happe to be a alteratig series the we ca t use ay of the methods discussed i this sectio to get a upper boud o the error. Power Series We ve spet quite a bit of time talkig about series ow ad with oly a couple of exceptios we ve spet most of that time talkig about how to determie if a series will 005 Paul Dawkis 7

228 coverge or ot. It s ow time to start lookig at some specific kids of series ad we ll evetually reach the poit where we ca talk about applicatios of series. I this sectio we are goig to start talkig about power series. A power series about a, or just power series, is ay series that ca be writte i the form, = 0 ( ) c x a where a ad c are umbers. The c s are ofte called the coefficiets of the series. The first thig to otice about a power series is that it is a fuctio of x. That is differet from ay other kid of series that we ve looked at to this poit. I all the prior sectios we ve oly allowed umbers i the series ad ow we are allowig variables to be i the series as well. This will ot chage how thigs work however. Everythig that we kow about series still holds. I the discussio of power series covergece is still a major questio that we ll be dealig with. The differece is that the covergece of the series will ow deped upo the value of x that we put ito the series. A power series may coverge for some values of x ad ot for other values of x. Before we get too far ito power series there is some termiology that we eed to get out of the way. First, as we will see i our examples, we will be able to show that there is a umber R so that the power series will coverge for, x a < Rad will diverge for x a > R. This umber is called the radius of covergece for the series. Note that the series may or may ot coverge if x a = R. What happes at these poits will ot chage the radius of covergece. Secodly, the iterval of all x s, icludig the ed poits if eed be, for which the power series coverges is called the iterval of covergece of the series. These two cocepts are fairly closely tied together. If we kow that the radius of covergece of a power series is R the we have the followig. a R< x< a+ R x< a R ad x> a+ R power series coverges power series diverges The iterval of covergece must the cotai the iterval a R< x< a+ R sice we kow that the power series will coverge for these values. We also kow that the iterval of covergece ca t cotai x s i the rages x < a R ad x > a+ R sice we kow the power series diverges for these value of x. Therefore, to completely idetify the iterval of covergece all that we have to do is determie if the power series will coverge for x = a R or x = a+ R. If the power series coverges for oe or both of these values the we ll eed to iclude those i the iterval of validity. 005 Paul Dawkis 8

229 Before gettig ito some examples let s take a quick look at the covergece of a power series for the case of x = a. I this case the power series becomes, 0 c a a = c 0 = c 0 + c 0 = c + 0= c + 0= c = 0 = 0 = = ad so the power series coverges. Note that we had to strip out the first term sice it was the oly o-zero term i the series. It is importat to ote that o matter what else is happeig i the power series we are guarateed to get covergece for x = a. The series may ot coverge for ay other value of x, but it will always coverge for x = a. Let s work some examples. We ll put quite a bit of detail ito the first example ad the ot put quite as much detail i the remaiig examples. Example Determie the radius of covergece ad iterval of covergece for the followig power series. ( ) ( x + ) = 4 Solutio Okay, we kow that this power series will coverge for x =, but that s it at this poit. To determie the remaider of the x s for which we ll get covergece we ca use ay of the tests that we ve discussed to this poit. After applicatio of the test that we choose to work with we will arrive at coditio(s) o x that we ca use to determie which values of x for which the power series will coverge ad which values of x for which the power series will diverge. From this we ca get the radius of covergece ad most of the iterval of covergece (with the possible exceptio of the edpoits. With all that said, the best tests to use here are almost always the ratio or root test. Most of the power series that we ll be lookig at are set up for oe or the other. I this case we ll use the ratio test. L = lim = lim ( ) ( + )( x+ ) ( )( x ) ( ) ( x+ ) + 4 Before goig ay farther with the limit let s otice that sice x is ot depedet o the limit ad so it ca be factored out of the limit. Notice as well that i doig this well eed to keep the absolute value bars o it sice we eed to make sure everythig stays positive ad x could well be a value that will make thigs egative. The limit is the, 005 Paul Dawkis 9

230 + L= x+ lim 4 = x + 4 So, the ratio test tells us that if L < the series will coverge, if L > the series will diverge, ad if L = we do t kow what will happe. So, we have, x+ < 4 x+ < 4 series coverges x+ > 4 x+ > 4 series diverges We ll deal with the L = case i a bit. Notice that we ow have the radius of covergece for this power series. These are exactly the coditios required for the radius of covergece. The radius of covergece for this power series is R = 4. Now, let s get the iterval of covergece. We ll get most (if ot all) of the iterval by solvig the first iequality from above. 4< x + < 4 7< x < So, most of the iterval of validity is give by 7< x <. All we eed to do is determie if the power series will coverge or diverge at the edpoits of this iterval. Note that these values of x will correspod to the value of x that will give L =. The way to determie covergece at these poits is to simply plug them ito the origial power series ad see if the series coverges or diverges usig ay test ecessary. x = 7 : I this case the series is, ( ) ( ) ( 4) = = = = = = = = = This series is diverget by the Divergece Test sice lim = 0. x = : I this case the series is, ( ) 4 ( 4) = ( ) = = 005 Paul Dawkis 0

231 This series is also diverget by the Divergece Test sice lim does t exist. So, i this case the power series will ot coverge for either edpoit. The iterval of covergece is the, 7< x < I the previous example the power series did t coverge for either ed poit of the iterval. Sometimes that will happe, but do t always expect that to happe. The power series could coverge at either both of the ed poits or oly oe of the ed poits. Example Determie the radius of covergece ad iterval of covergece for the followig power series. ( 4x 8) = Solutio Let s jump right ito the ratio test. L = lim ( x ) ( x ) 4 8 = lim + = 4x 8 lim + = 4x ( x ) So we will get the followig covergece/divergece iformatio from this. 4x 8 < series coverges 4x 8 > series diverges We eed to be careful here i determiig the iterval of covergece. The iterval of covergece requires x a < Rad x a > R. I other words, we eed to factor a 4 out of the absolute value bars i order to get the correct radius of covergece. Doig this gives, 8 x < x < 8 series coverges 8 x > x > 8 series diverges So, the radius of covergece for this power series is R = Paul Dawkis

232 Now, let s fid the iterval of covergece. Agai, we ll first solve the iequality that gives covergece above. < x < < x < 8 8 Now check the ed poits. 5 x = : 8 The series here is, 5 8 = = = = = = = ( ) ( ) This is the alteratig harmoic series ad we kow that it coverges. 7 x = : 8 The series here is, 7 8 = = = = = = = This is the harmoic series ad we kow that it diverges. So, the power series coverges for oe of the ed poits, but ot the other. This will ofte happe so do t get excited about it whe it does. The iterval of covergece for this power series is the, 5 7 x < 8 8 We ow eed to take a look at a couple of special cases with radius ad itervals of covergece. 005 Paul Dawkis

233 Example Determie the radius of covergece ad iterval of covergece for the followig power series. = 0 ( x+ )! Solutio Well start this example with the ratio test as we have for the previous oes. L = lim + ( +! ) ( x+ )! ( x+ ) ( + )! ( x+ ) = lim! = x+ lim + ( ) At this poit we eed to be careful. The limit is ifiite, but there is that term with the x s i frot of the limit. We ll have L = > provided x. So, this power series will oly coverge if x =. If you thik about it we actually already kew that however. From our iitial discussio we kow that every power series will coverge for x = a ad i this case a =. Remember that we get a from ( x a), ad otice the coefficiet of the x must be a oe!. I this case we say the radius of covergece is R = 0 ad the iterval of covergece is x =, ad yes we really did mea iterval of covergece eve though it s oly a poit. Example 4 Determie the radius of covergece ad iterval of covergece for the followig power series. ( x 6) = Solutio I this example the root test seems more appropriate. So, 005 Paul Dawkis

234 L = lim ( x 6) x 6 = lim = x 6lim = 0 So, sice L = 0< regardless of the value of x this power series will coverge for every x. I these cases we say that the radius of covergece is R = ad iterval of covergece is < x <. So, let s summarize the last two examples. If the power series oly coverges for x = a the the radius of covergece is R = 0 ad the iterval of covergece is x = a. Likewise if the power series coverges for every x the radius of covergece is R = ad iterval of covergece is < x <. Let s work oe more example. Example 5 Determie the radius of covergece ad iterval of covergece for the followig power series. x = Solutio First otice that a=0 i this problem. That s ot really importat to the problem, but it s worth poitig out so people do t get excited about it. The importat differece i this problem is the expoet o the x. I this case it is rather tha the stadard. As we will see some power series will have expoets other tha a ad so we still eed to be able to deal with these kids of problems. This oe seems set up for the root test agai so let s use that. x L = lim ( ) x = lim x = 005 Paul Dawkis 4

235 So, we will get covergece if x < x < The radius of covergece is NOT however. The radius of covergece requires a expoet of o the x. Therefore, x < x < Be careful with the absolute value bars! I this case it looks like the radius of covergece is R =. Notice that we did t bother to put dow the iequality for divergece this time. The iequality for divergece is just the iterval for covergece that the test gives with the iequality switched ad geerally is t eeded. We will usually skip that part. Now let s get the iterval of covergece. First from the iequality we get, < x < Now check the ed poits. x = : Here the power series is, ( ) ( ) = = = = (( ) ) = = ( ) ( ) = This series is diverget by the Divergece Test sice lim does t exist. x = : Because we re squarig the x this series will be the same as the previous step. which is diverget. ( ) = = = The iterval of covergece is the, < x < 005 Paul Dawkis 5

236 Power Series ad Fuctios We opeed the last sectio by sayig that we were goig to start thikig about applicatios of series ad the promptly spet the sectio talkig about covergece agai. It s ow time to actually start with the applicatios of series. With this sectio we will start talkig about how to represet fuctios with power series. The atural questio of why we might wat to do this will be aswered i a couple of sectios oce we actually lear how to do this. Let s start off with oe that we already kow how to do, although whe we first ra across this series we did t thik of it as a power series or did we ackowledge that it represeted a fuctio. Recall that the geometric series is a ar = provided r < r = 0 Do t forget as well that if r the series diverges. Now, if we take a= ad r=x this becomes, x = x provided x < () = 0 Turig this aroud we ca see that we ca represet the fuctio f ( x) = x () with the power series x provided x < () = 0 This provisio is importat. We ca clearly plug ay umber other tha x= ito the fuctio, however, we will oly get a coverget power series if x <. This meas the equality i () will oly hold if x <. For ay other value of x the equality wo t hold. Note as well that we ca also use this to ackowledge that the radius of covergece of this power series is R = ad the iterval of covergece is x <. This idea of covergece is importat here. We will be represetig may fuctios as power series ad it will be importat to recogize that the represetatios will ofte oly be valid for a rage of x s ad that there may be values of x that we ca plug ito the fuctio that we ca t plug ito the power series represetatio. 005 Paul Dawkis 6

237 I this sectio we are goig to cocetrate o represetig fuctios with power series where the fuctios ca be related back to (). I this way we will hopefully become familiar with some of the kids of maipulatios that we will sometimes eed to do whe workig with power series. So, let s jump ito a couple of examples. Example Fid a power series represetatio for the followig fuctio ad determie it s iterval of covergece. g( x) = + x Solutio What we eed to do here is to relate this fuctio back to (). This is actually easier tha it might look. Recall that the x i () is simply a variable ad ca represet aythig. So, a quick rewrite of g(x) gives, g( x) = x ad so the x i g(x) holds the same place as the x i (). Therefore, all we eed to do is replace the x i () ad we ve got a power series represetatio for g(x). g x = x provided x < = 0 Notice that we replaced both the x i the power series ad i the iterval of covergece. All we eed to do ow is a little simplificatio. = 0 provided g x = x x < x < So, i this case the iterval of covergece is the same as the origial power series. This usually wo t happe. More ofte tha ot the ew iterval of covergece will be differet form the origial iterval of covergece. Example Fid a power series represetatio for the followig fuctio ad determie it s iterval of covergece. x h( x) = + x Solutio This fuctio is similar to the previous fuctio. The differece is the umerator ad at first glace that looks to be a importat differece. Sice () does t have a x i the umerator it appears that we ca t relate this fuctio back to that. However, ow that we ve worked the first example this oe is actually very simple sice we ca use the result of the aswer from that example. To see how to do this let s first rewrite the fuctio a little. 005 Paul Dawkis 7

238 h( x) = x + x Now, from the first example we ve already got a power series for the secod term so let s use that to write the fuctio as, provided = 0 h x = x x x < Notice that the presece of x s outside of the series will NOT affect its covergece ad so the iterval of covergece remais the same. The last step is to brig the coefficiet ito the series ad we ll be doe. Whe we do this make sure ad combie the x s as well. We typically oly wat a sigle x i a power series. + h x = x provided x < = 0 As we saw i the previous example we ca ofte use previous results to help us out. This is a importat idea to remember as it ca ofte greatly simplify our work. Example Fid a power series represetatio for the followig fuctio ad determie it s iterval of covergece. x f ( x) = 5 x Solutio So, agai, we ve got a x i the umerator. So, as with the last example let s factor that out ad see what we ve got left. f ( x) = x 5 x If we had a power series represetatio for g( x) = 5 x we could get a power series represetatio for f(x). So, let s fid oe. We ll first otice that i order to use (4) we ll eed the umber i the deomiator to be a oe. That s easy eough to get. g( x) = 5 x 5 Now all we eed to do to get a power series represetatio is to replace the x i () with x. Doig this gives, 5 g x x x = provided < 5 = Paul Dawkis 8

239 Now let s do a little simplificatio o the series. x g( x) = 5 = 0 5 x = + 5 The iterval of covergece for this series is, x < x < 5 5 x < 5 Okay, this was the work for the power series represetatio for g(x) let s ow fid a power series represetatio for the origial fuctio. All we eed to do for this is to multiply the power series represetatio for g(x) by x ad we ll have it. f ( x) = x 5 x x = x + = x = + 5 The iterval of covergece does t chage ad so it will be x < 5. So, hopefully we ow have a idea o how to fid the power series represetatio for some fuctios. Admittedly all of the fuctios could be related back to () but it s a start. We ow eed to look at some further maipulatio of power series that we will eed to do o occasio. We eed to discuss differetiatio ad itegratio of power series. = 0 = 0 Let s start with differetiatio of the power series, = 0 0 f x = c x a = c + c x a + c x a + c x a + Now, we kow that if we differetiate a fiite sum of terms all we eed to do is differetiate each of the terms ad the add them back up. With ifiite sums there are some subtleties ivolved that we eed to be careful with, but are somewhat beyod the scope of this course. While, we ca always just differetiate all the terms i a ifiite series is it ot always guarateed to be the power series represetatio of the derivative of the origial fuctio. Nicely eough for us however, it is kow that if the power series represetatio of f(x) has a radius of covergece of R > 0 the the term by term differetiatio of the power 005 Paul Dawkis 9

240 series will also have a radius of covergece of R ad (more importatly) will i fact be f x provided x is i the iterval of covergece of the power series represetatio of the origial fuctio. I other words, f x = c + c x a + c x a + = c x a = Note the iitial value of this series. It has bee chaged from =0 to =. This is a ackowledgemet of the fact that the derivative of the first term is zero ad hece is t i the derivative. Notice however, that sice the =0 term of the above series is also zero, we could start the series at =0 if it was required for a particular problem. I geeral however, this wo t be doe i this class. We ca ow fid formulas for higher order derivatives as well ow. etc. f x = c x a = f x = c x a = Oce agai, otice that the iitial value of chages with each differetiatio i order to ackowledge that a term from the origial series differetiated to zero. Let s ow briefly talk about itegratio. Just as with the differetiatio, whe we ve got a ifiite series we eed to be careful about just itegratio term by term. As log as we are i the iterval of covergece for the origial fuctio we ca do the itegratio. I this case we get, + ( x a) f ( x) dx= C+ c + = 0 Notice that we pick up a costat of itegratio, C, that is outside the series here. Let s summarize the differetiatio ad itegratio ideas before movig o to a example or two. Fact has a radius of covergece of 0 = 0 If f ( x) = c ( x a) R > the, 005 Paul Dawkis 40

241 f x = c x a = f x dx= C+ c ( x a) = 0 + ad both of these also have a radius of covergece of R. Now, let s see how we ca use these facts to geerate some more power series represetatios of fuctios. Example 4 Fid a power series represetatio for the followig fuctio ad determie it s iterval of covergece. g( x) = x Solutio To do this problem let s otice that ( x) d = dx x The sice we ve got a power series represetatio for x all that we ll eed to do is differetiate that power series to get a power series represetatio for g(x). g( x) = ( x) d = dx x d = x dx = 0 = = The sice the origial power series had a radius of covergece of R = the derivative, ad hece g(x), will also have a radius of covergece of R =. Example 5 Fid a power series represetatio for the followig fuctio ad determie it s iterval of covergece. h( x) = l ( 5 x) Solutio I this case we eed to otice that x Paul Dawkis 4

242 dx = l ( 5 x) 5 x ad the recall that we have a power series represetatio for 5 x Remember we foud a represetatio for this i Example. So, l ( 5 x) = dx 5 x x = dx x = C 5 = 0 = 0 ( + ) We ca fid the costat of itegratio, C, by pluggig i a value of x. A good choice is x=0 sice that will make the series easy to evaluate. + 0 l ( 5 0) = C + = 0 ( + 5 ) l 5 = C + So, the fial aswer is, ( x) l 5 = l 5 = 0 + x 5 ( + ) + Note that it is okay to have the costat sittig outside of the series like this. I fact, there is o way to brig it ito the series so do t get excited about it. Taylor Series I the previous sectio we started lookig at writig dow a power series represetatio of a fuctio. The problem with the approach i that sectio is that everythig came dow to eedig to be able to relate the fuctio i some way to x ad while there are may fuctios out there that ca be related to this fuctio there are may more that simply ca t be related to this. So, without takig aythig away from the process we looked at i the previous sectio, we eed to do is come up with a more geeral method for writig a power series represetatio for a fuctio. 005 Paul Dawkis 4

243 So, for the time beig, let s make two assumptios. First, lets assume that the fuctio f(x) does i fact have a power series represetatio about x=a, = f x = c x a = c + c x a + c x a + c x a + c x a + Next, we will eed to assume that the fuctio, f(x), has derivatives of every order ad that we ca i fact fid them all. Now that we ve assumed that a power series represetatio exists we eed to determie what the coefficiets, c, are. This is easier tha it might at first appear to be. Let s first just evaluate everythig at x=a. This gives, = c0 f a So, all the terms except the first are zero ad we ow kow what c 0 is. Ufortuately, there is t ay other value of x that we ca plug ito the fuctio that will allow us to quickly fid ay of the other coefficiets. However, if we take the derivative of the fuctio (ad its power series) the plug i x=a we get, f ( x) = c+ c( x a) + c( x a) + 4c4( x a) + f a = c ad we ow kow c. Lets cotiue with this idea ad fid the secod derivative. f x = c + c x a + 4 c x a + 4 f a = c So, it looks like, f ( a) c = Usig the third derivative gives, f x = c + 4 c x a + 4 f a = c c = f ( a) Usig the fourth derivative gives, 4 f x = 4 c c x a 4 5 ( 4 ) ( a) ( 4 ) f f ( a) = 4 c4 c4 = Paul Dawkis 4

244 Hopefully by this time you ve see the patter here. It looks like, i geeral, we ve got the followig formula for the coefficiets. ( f ) ( a) c =! This eve works for =0 if you recall that 0! = ad defie ( 0 ) f x = f x. So, provided a power series represetatio for the fuctio f(x) about x=a exists it will have the form, = 0 ( ) ( a) f f ( x) = x a! f a f a = f ( a) + f ( a)( x a) + x a + x a +!! This is called the Taylor Series for f(x) about x=a. I the case that a=0, so the Taylor Series about x=0, we have, = 0 ( ) ( 0) f f ( x) = x! This is called a Maclauri Series for f(x). ( 0) f ( 0) f f ( 0) f ( 0) x x x!! = Before workig ay examples of Taylor Series we first eed to address the assumptio that a Taylor Series will i fact exist for a give fuctio. Let s first get some otatio out of the way first. We ll first split the Taylor Series formula as follows, i i f a f a f a x a = x a + x a = 0! i= 0 i! i= + i! = T( x) + R( x) where, ( i f ) ( a) i T ( x) = ( x a ) i= 0 i! is called the th degree Taylor Polyomial of f(x) ad ( i f ) ( a) i R ( x) = ( x a ) i= + i! is called the remaider. () () i i 005 Paul Dawkis 44

245 Note that the Taylor polyomial really is a polyomial ad its degree will be at most. We ll see a ice applicatio of Taylor polyomials i the ext sectio. We ow have the followig Theorem. Theorem Suppose that f ( x) T ( x) R ( x) for x a < R the, o x a < R. = +. The if, I geeral showig that R ( x) ( x) lim R = 0 = 0! ( ) ( a) f f ( x) = x a lim = 0 is a somewhat difficult process ad so we will be assumig that this ca be doe for some R i all of the examples that we ll be lookig at. Now let s look at some examples. x Example Fid the Taylor Series for f ( x ) = e about x=0. Solutio This is actually oe of the easier Taylor Series that we ll be asked to compute. To fid ( the Taylor Series for a fuctio we will eed to determie a geeral formula for f ) ( a ). This is oe of the few fuctios where this is easy to do right from the start. To get a formula for ad so, ( ) ( 0) f all we eed to do is recogize that, f ( ) x = = 0,,,, f x e ( ) 0 0 = e = = 0,,,, x Therefore, the Taylor series for f ( x ) = e about x=0 is, e x x = x =!! = 0 = 0 x Example Fid the Taylor Series for f ( x) = e about x=0. Solutio There are two ways to do this problem. Both are fairly simple, however oe of them requires sigificatly less work. We ll work both solutios sice the loger oe has some ice ideas that we ll see i other examples. 005 Paul Dawkis 45

246 Solutio As with the first example we ll eed to get a formula for ( ) ( 0) f. However, ulike the first oe we ve got a little more work to do. Let s first take some derivatives ad evaluate them at x=0. 0 x 0 f x = e f 0 = () ( ) () x f x = e f 0 = x f x = e f 0 = x f x = e f 0 = ( ) x f x = e f 0 = = 0,,, After a couple of computatios we were able to get geeral formulas for both ( ) ( 0) f ( ) ( x ) ad f. We ofte wo t be able to get a geeral formula for f x so do t get too excited about gettig that formula. Also, as we will see it wo t always be easy to get a geeral formula for f a. ( ) So, i this case we ve got geeral formulas so all we eed to do is plug these ito the Taylor Series formula ad be doe with the problem. x ( ) x e =! = 0 Solutio The previous solutio was t too bad ad we ofte have to do thigs i that maer. However, i this case there is a much shorter solutio method. I the previous sectio we used series that we ve already foud to help us fid a ew series. Let s do the same thig x with this oe. We already kow a Taylor Series for e about x=0 ad i this case the oly differece is we ve got a -x i the expoet istead of just a x. So, all we eed to do is replace the x i the Taylor Series that we foud i the first example with -x. x ( x) ( ) x e = =!! = 0 = 0 This is a much shorter method of arrivig at the same aswer so do t forget about usig previously computed series where possible (ad allowed of course). 4 x Example Fid the Taylor Series for f ( x) x = e about x= Paul Dawkis 46

247 Solutio For this example we will take advatage of the fact that we already have a Taylor Series x for e about x=0. I this example, ulike the previous example, doig this directly would be sigificatly loger ad more difficult. ( x ) 4 x 4 x e = x = x = 4 = 0 = 0 = 0 ( ) ( )! x! x! To this poit we ve oly looked at Taylor Series about x=0 (also kow as Maclauri Series) so let s take a look at a Taylor Series that is t about x=0. Also, we ll pick o the expoetial fuctio oe more time sice it makes some of the work easier. This will be the fial Taylor Series for expoetials i this sectio. x Example 4 Fid the Taylor Series for f ( x) Solutio Fidig a geeral formula for ( ) ( 4) ( ) x e + 4 = e about x = 4. f is fairly simple. ( ) 4 f x = f 4 = e The Taylor Series is the, e x = 0 ( ) 4 e = +! ( x 4) Okay, we ow eed to work some examples that do t ivolve the expoetial fuctio sice these will ted to require a little more work. Example 5 Fid the Taylor Series for f ( x) cos( x) = about x=0. Solutio First we ll eed to take some derivatives of the fuctio ad evaluate them at x= Paul Dawkis 47

248 () () 0 0 f x = cos x f 0 = f x = si x f 0 = 0 f x = cos x f 0 = f x = si x f 0 = f x = cos x f 0 = 5 5 f x = si x f 0 = 0 6 f x = cos x 6 f 0 = I this example, ulike the previous oes, there is ot a easy formula for either the geeral derivative or the evaluatio of the derivative. However, there is a clear patter to the evaluatios. So, let s plug what we ve got ito the Taylor series ad see what we get, cos x= = 0 f ( ) ( 0 )! x 4 5 f 0 f 0 f 0 4 f 0 5 = f ( 0) + f ( 0) x+ x + x + x + x +!! 4! 5! 4 6 = + 0 x x + 0 x +! 4! 6! = 0 = = = 5 = = 4 = 6 So, we oly pick up terms with eve powers o the x s. This does t really help us to get a geeral formula for the Taylor Series. However, let s drop the zeroes ad reumber the terms as follows to see what we ca get. 4 6 cos x= x + x x +! 4! 6! = 0 = = = By reumberig the terms as we did we ca actually come up with a geeral formula for the Taylor Series ad here it is, cos x = = 0 ( ) This idea of reumberig the series terms as we did i the previous example is t used all that ofte, but occasioally is very useful. There is oe more series where we eed to do it so let s take a look at that so we ca get oe more example dow of reumberig series terms. x! Example 6 Fid the Taylor Series for f ( x) si ( x) = about x=0. Solutio As with the last example we ll start off i the same maer. 005 Paul Dawkis 48

249 () () 0 0 f x = si x f 0 = 0 f x = cos x f 0 = f x = si x f 0 = 0 f x = cos x f 0 = 4 4 f x = si x f 0 = f x = cos x f 0 = 6 f x = si x 6 f 0 = 0 So, we get a similar patter for this oe. Let s plug the umbers ito the Taylor Series. ( f ) ( 0 ) si x= x = 0! 5 7 = x x + x x +!! 5! 7! I this case we oly get terms that have a odd expoet o x ad as with the last problem oce we igore the zero terms there is a clear patter ad formula. So reumberig the terms as we did i the previous example we get the followig Taylor Series. si x = = 0 ( ) x ( + ) +! We really eed to work aother example or two i which is t about x=0. Example 7 Fid the Taylor Series for f ( x) l ( x) Solutio Here are the first few derivatives ad the evaluatios. = about x=. 005 Paul Dawkis 49

250 0 0 f x = l x f = l () () f ( x) = f = x ( ) ( ) f x = f = x ( ) ( ) f x = f = x ( 4 ) ( 4 ) f x = f 4 = 4 x ( 5 ) ( 4) ( 5 ) 4 f x = f 5 = 5 x ( ) ( ) ( ) + + ( )!! f ( x) = f = =,,, x Note that while we got a geeral formula here it does t work for =0. This will happe o occasio so do t worry about it whe it does. I order to plug this ito the Taylor Series formula we ll eed to strip out the =0 term first. ( f ) l ( x) = ( x ) = 0! ( f ) = f + ( x ) =! + ( ) ( )! = l + ( x ) =! + ( ) = l + ( x ) = Notice that we simplified the factorials i this case. You should always simplify them if there are more tha oe ad it s possible to simplify them. Also, do ot get excited about the term sittig i frot of the series. Sometimes we eed to do that whe we ca t get a geeral formula that will hold for all values of. Example 8 Fid the Taylor Series for f ( x) Solutio Agai, here are the derivatives ad evaluatios. = about x =. x 005 Paul Dawkis 50

251 ( 0 ) ( 0 ) f x = f ( ) = = x ( ) ( ) ( ) () () f ( x) = f ( ) = = x ( ) 4 ( ) 5 ( ) ( ) f ( x) = f = = 4 x ( ) 4 4 f ( x) = f = = 4 5 x ( ) ( ) + ( ) ( ) +! +! f ( x) = f = = ( + )! + x Notice that all the egatives sigs will cacel out i the evaluatio. Also, this formula will work for all, ulike the previous example. Here is the Taylor Series for this fuctio. f = x + x = 0! ( +! ) = ( x + )! = 0 = 0 ( )( x ) = + + Now, let s work oe of the easier examples i this sectio. The problem for most studets is that it may ot appear to be that easy (or maybe it will appear to be too easy) at first glace. Example 9 Fid the Taylor Series for f ( x) x x = about x =. Solutio Here are the derivatives for this problem. 0 0 f x = x 0x + 6 f = 57 () ( ) () ( 4 ) f x = x 0x f = f x = 6x 0 f = f x = 6 f = 6 f x = 0 f = 0 4 This Taylor series will termiate after =. This will always happe whe we are fidig the Taylor Series of a polyomial. Here is the Taylor Series for this oe. 005 Paul Dawkis 5

252 ( ) f = x x x = 0! ( x ) ( x ) ( x ) f f = f + f ( x ) + x + x + 0!! = 57 + Whe fidig the Taylor Series of a polyomial we do t do ay simplificatio of the right had side. We leave it like it is. I fact, if we were to multiply everythig out we just get back to the origial polyomial! While it s ot apparet that writig the Taylor Series for a polyomial is useful there are times where this eeds to be doe. The problem is that they are beyod the scope of this course ad so are t covered here. For example, there is oe applicatio to series i the field of Differetial Equatios where this eeds to be doe o occasio. So, we ve see quite a few examples of Taylor Series to this poit ad i all of them we where able to fid geeral formulas for the series. This wo t always be the case. To see a example of oe that does t have a geeral formula check out the last example i the ext sectio. Before leavig this sectio there are three importat Taylor Series that we ve derived i this sectio that we should summarize up i oe place. I my class I will assume that you kow these formulas from this poit o. e x = cos x = si x = = 0 = 0 = 0 x! ( ) x ( )! ( ) x ( + ) +! Applicatios of Series Now, that we kow how to represet fuctio as power series we ca ow talk about at least a couple of applicatios of series. There are i fact may applicatios of series, ufortuately most of them are beyod the scope of this course. Oe applicatio of power series (with the occasioal use of Taylor Series) is i the field of Ordiary Differetial Equatios whe fidig Series Solutios to Differetial Equatios. If you are iterested i seeig how that works you ca check out that chapter of my Differetial Equatios otes. 005 Paul Dawkis 5

253 Aother applicatio of series arises i the study of Partial Differetial Equatios. Oe of the more commoly used methods i that subject makes use of Fourier Series. May of the applicatios of series, especially those i the differetial equatios fields, rely o the fact that fuctios ca be represeted as a series. I these applicatios it is very difficult, if ot impossible, to fid the fuctio itself. However, there are methods of determiig the series represetatio for the ukow fuctio. While the differetial equatios applicatios are beyod the scope of this course there are some applicatios from a Calculus settig that we ca look at. Example Determie a Taylor Series about x=0 for the followig itegral. si x dx x Solutio To do this we will first eed to fid a Taylor Series about x=0 for the itegrad. This however is t terribly difficult. We already have a Taylor Series for sie about x=0 so we ll just use that as follows, We ca ow do the problem. si x dx = x ( ) x = 0 = 0 ( ) x + si x = = x x +! +! = 0 = C + ( ) x dx ( + )! + ( ) x = 0 ( + ) ( + )! So, while we ca t itegrate this fuctio i terms of kow fuctios we ca come up with a series represetatio for the itegral. This idea of derivig a series represetatio for a fuctio istead of tryig to fid the fuctio itself is used quite ofte i several fields. I fact, there are some fields where this is oe of the mai ideas used ad without this idea it would be very difficult to accomplish aythig i those fields. Aother applicatio of series is t really a applicatio of ifiite series. It s more a applicatio of partial sums. I fact, we ve already see this applicatio i use oce i this chapter. I the Estimatig the Value of a Series we used a partial sum to estimate the value of a series. We ca do the same thig with power series ad series represetatios of fuctios. The mai differece is that we will ow be usig the partial sum to approximate a fuctio istead of a sigle value. 005 Paul Dawkis 5

254 We will look at Taylor series for our examples, but we could just as easily use ay series represetatio here. Recall that the th degree Taylor Polyomial of f(x) is give by, ( i f ) ( a) i T ( x) = ( x a ) i Let s take a look at example of this. i= 0! Example For the fuctio f ( x) = cos( x) plot the fuctio as well as T ( x ), 4 ad T8 ( x ) o the same graph for the iterval [-4,4]. T x, Solutio Here is the geeral formula for the Taylor polyomials for cosie. T ( x) = i= 0 i ( ) ( i) The three Taylor polyomials that we ve got are the, x T ( x) = 4 x x T4 ( x) = x x x x T8 ( x) = Here is the graph of these three Taylor polyomials as well as the graph of cosie. x! i 005 Paul Dawkis 54

255 As we ca see from this graph as we icrease the degree of the Taylor polyomial it starts to look more ad more like the fuctio itself. I fact by the time we get to T8 ( x ) the oly differece is right at the eds. The higher the degree of the Taylor polyomial the better it approximates the fuctio. Also the larger the iterval the higher degree Taylor polyomial we eed to get a good approximatio for the whole iterval. Before movig o let s otice write dow a couple more Taylor polyomials from the previous example. Notice that because the Taylor series for cosie does t cotai ay terms with odd powers o x we get the followig Taylor polyomials. T 9 ( x) x T ( x) = 4 x x T5 ( x) = x x x x = These are idetical to those used i the example. Sometimes this will happe although that was ot really the poit of this. The poit is to otice that the th degree Taylor polyomial may actually have a degree that is less tha. It will ever be more tha, but it ca be less tha. The fial example i this sectio really is t a applicatio of series ad probably beloged i the previous sectio. However, the previous sectio was gettig too log so the example is i this sectio. This is a example of how to multiply series together ad while this is t a applicatio of series it is somethig that does have to be doe o occasio i the applicatios. So, i that sese it does belog i this sectio. x Example Fid the first three o-zero terms i the Taylor Series for = e cos f x x about x=0. Solutio Before we start let s ackowledge that the easiest way to do this problem is to simply compute the first -4 derivatives, evaluate them at x=0, plug ito the formula ad we d be doe. However, as we oted prior to this example we wat to use this example to illustrate how we multiply series. We will make use of the fact that we ve got Taylor Series for each of these so we ca use them i this problem. x x ( ) x e cos x = = 0! = 0 ( )! 005 Paul Dawkis 55

256 We re ot goig to completely multiply out these series. We re goig to eough of the multiplicatio to get a aswer. The problem statemet says that we wat the first three o-zero terms. That o-zero bit is importat as it is possible that some of the terms will be zero. If oe of the terms are zero this would mea that the first three o-zero terms would be the costat term, x term, ad x term. However, because some might be zero let s assume that if we get all the terms up through x 4 we ll have eough to get the aswer. If we ve assumed wrog it will be very easy to fix so do t worry about that. Now, let s write dow the first few terms of each series ad we ll stop at the x 4 term i each. 4 4 cos x x x x x x e x= + x Note that we do eed to ackowledge that these series do t stop. That s the purpose of the + at the ed of each. Just for a secod however, let s suppose that each of these did stop ad ask ourselves how we would multiply each out. If this were the case we would take every term i the secod ad multiply by every term i the first. I other words, we would first multiply every term i the secod series by, the every term i the secod series by x, the by x etc. By stoppig each series at x 4 we have ow guarateed that we ll get all terms that have a expoet of 4 or less. Do you see why? Each of the terms that we eglected to write dow have a expoet of at least 5 ad so multiplyig by or ay power of x will result i a term with a expoet that is at a miimum 5. Therefore, oe of the eglected terms will cotribute terms with a expoet of 4 or less ad so were t eeded. So, let s start the multiplicatio process. 4 4 x x x x x x e cos x= + x x x x x x x x = x Secod Series Secod Series x Secod Series x x x x x x x Secod Series x 6 4 Secod Series x 4 Now, collect like terms igorig everythig with a expoet of 5 or more sice we wo t have all those terms ad do t wat them either. Doig this gives, 005 Paul Dawkis 56

257 e x cos x x x x x x x = + x = There we go. It looks like we over guessed ad eded up with four o-zero terms, but that s okay. If we had uder guessed ad it tured out that we eeded terms with x 5 i them all we would eed to do at this poit is go back ad add i those terms to the origial series ad do a couple quick multiplicatios. I other words, there is o reaso to completely redo all the work. Biomial Series I this fial sectio of this chapter we are goig to look at aother series represetatio for a fuctio. Before we do this let s first recall the followig theorem. Biomial Theorem If is ay positive iteger the, i i ( a+ b) = a b i= 0 i ( ) = a + a b + a b + + ab + b! where, ( )( ) ( i+ ) = i =,,, i i! = 0 This is useful for expadig ( a+ b) for large whe straight forward multiplicatio would t be easy to do. Let s take a quick look at a example. Example Use the Biomial Theorem to expad ( x ) 4 Solutio There really is t much to do other tha pluggig ito the theorem. 005 Paul Dawkis 57

258 4 4 i i= 0 i = = ( x) + 4( x) ( ) + ( x ) ( ) + 4 ( x )( ) + ( ) = x x + x x+ 4 4 i ( x ) = ( x) ( ) ( x) ( x) ( x) ( x) Now, the Biomial Theorem required that be a positive iteger. There is a extesio to this however that allows for ay umber at all. Biomial Series If k is ay umber ad x < the, k k ( + x) = = 0 where, x ( ) ( )( ) k k k k k kx x x!! = ( )( ) ( + ) k k k k k = =,,,! k = 0 So, similar to the biomial theorem except that it s a ifiite series ad we must have x < i order to get covergece. Let s check out a example of this. Example Write dow the first four terms i the biomial series for 9 x Solutio So, i this case required. k = ad we ll eed to rewrite the term a little to put it ito the form x x 9 x = = The first four terms i the biomial series is the, 005 Paul Dawkis 58

259 x 9 x = + 9 x = = 0 9 ( ) ( )( ) x x x = x x x = 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. 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 005 Paul Dawkis 59

260 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, v =,5 ad each of the directed lie segmets i the sketch are called represetatios of the vector. Be careful to distiguish vector otatio,,5, from the otatio we use for 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! A represetatio of the vector v = a, a segmet, AB A x, y i two dimesioal space is ay directed lie, from the poit = to the poit B ( x a, y a ) = + +. Likewise a 005 Paul Dawkis 60

261 represetatio of the vector v = a, a, a segmet, AB A x, y, z i three dimesioal space is ay directed lie, from the poit = 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 vector v = a, a, a that starts at the poit A = ( 0,0,0) ad eds at the poit = (,, ) is called the positio vector of the poit (,, ) B a a a a a a. So, whe we talk about positio vectors we are specifyig the iitial ad fial poit of the vector. These are useful if we ever eed to (ad we will o occasio eed to) represet a poit as a vector. 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) the vector with the represetatio AB is, v = b a, b a, b a Note that we have to be very careful with directio here. This is the vector that starts at A ad eds at B. The vector that starts at B ad eds at A, i.e. with represetatio BA is, w= a b, a b, a b Whe determiig the vector from the iitial ad fial poits we always subtract the iitial poit from the fial poit. 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. 005 Paul Dawkis 6

262 (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 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. (a) a =, 5,0 (b) u =, 5 5 (c) w = 0,0 (d) i =, 0, 0 Solutio There is t too much to these other tha plug ito the formula. (a) a = = 4 (b) 4 u = + = = 5 5 (c) w = 0+ 0 = 0 (d) i = = We also have the followig fact about the magitude. If a = 0 the a = 0 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 Ay vector with magitude of, i.e. u =, is called a uit vector. 005 Paul Dawkis 6

263 Example Which of the vectors from the first 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 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. Be careful to distiguish 0 (the umber) from 0 (the vector). The umber 0 deote the origi i space, while the vector 0 deotes a vector that has o magitude or directio. Stadard Basis Vectors The fourth vector from the secod example, i =, 0, 0, is called a stadard basis vector. I three dimesioal space there are three stadard basis vectors, i =,0,0 j = 0,,0 k = 0,0, I two dimesioal space there are two stadard basis vectors, i =, 0 j = 0, Note that stadard basis vectors are also uit vectors. 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, v = a, a, a,, 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. 005 Paul Dawkis 6

264 Vector Arithmetic I this sectio we eed to have a brief discussio of vector arithmetic. We ll start with additio of two vectors. So, give the vectors a = a, a, a b = b, b, b the additio of the two vectors is give by, a+ b = a + b, a + b, a + b ad The followig figure gives the geometric iterpretatio of the additio of two vectors. This is sometimes called the parallelogram law or triagle law. Computatioally, subtractio is very similar. Give the vectors a = a, a, a b = b, b, b the differece of the two vectors is give by, a b = a b, a b, a b ad Here is the geometric iterpretatio of the differece of two vectors. 005 Paul Dawkis 64

265 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 a = a, a, a ad ay umber c the scalar multiplicatio is, ca = ca, ca, ca 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. Example For the vector a =, 4 compute a, a ad a. Graph all four vectors o the same axis system. Solutio Here are the three scalar multiplicatios. a = 6, a =, a = 4, 8 Here is the graph. 005 Paul Dawkis 65

266 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 ad b are parallel vectors. If they are parallel the there must be a umber c so that, a = cb So, two vectors are parallel if oe is a scalar multiple of the other. Example Determie if the sets of vectors are parallel or ot. (a) a =, 4,, b = 6,, 005 Paul Dawkis 66

267 (b) a = 4,0, b =, 9 Solutio (a) These two vectors are parallel sice, b = a (b) These two vectors are t parallel. This ca be see by oticig that 4 = 0 = 5 9. I other words we ca t make a be a scalar multiple of b. ad yet The ext applicatio is best see i a example. 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. w = = 0 Now, let s form the followig ew vector, 5 u = w= 5,, =,, w The claim is that this is a uit vector. That s easy eough to check u = + + = = This vector also poits i the same directio as w sice it is oly a scalar multiple of w ad we used a positive multiple. So, i geeral, give a vector w, u = directio as w. w will be a uit vector that poits i the same w 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 a = a, a, a 005 Paul Dawkis 67

268 We ca use the additio of vectors to break this up as follows, a = a, a, a = a,0,0 + 0, a,0 + 0,0, a Usig scalar multiplicatio we ca further rewrite the vector as, 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. a, a, a = ai + a j + a k 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. Example 4 If a =, 9, ad w= i + 8k compute a w. Solutio I order to do the problem we ll covert to oe otatio ad the perform the idicated operatios. 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, w ad u are vectors (each with the same umber of compoets) ad a ad b are two umbers the we have the followig properties. v+ w= w+ v u+ ( v+ w) = ( u+ v) + w v+ 0= v v = v a v + w = av + aw a + b v = av + bv 005 Paul Dawkis 68

269 Dot Product The ext topic for discussio is that of the dot product. Let s jump right ito the defiitio of the dot product. Give the two vectors a = a, a, a ad b = b, b, b the dot product is, ab i = 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. (a) v = 5i 8 j, w= i + j (b) a = 0,, 7, b =,, Solutio Not much to do with these other tha use the formula. (a) vw= i 5 6= (b) ab= i = Here are some properties of the dot product. Properties ui v w uiv uiw cv iw vi cw c viw vw i = wv i vi0= 0 vv i = v If vv i = 0 the v= 0 ( + ) = + = = There is also a ice geometric iterpretatio to the dot product. First suppose that θ is the agle betwee a ad b such that 0 θ π as show i the image below. 005 Paul Dawkis 69

270 the, ab i = a b cosθ () This is, more ofte tha ot, used to determie the agle betwee two vectors as show i the followig example. Example Determie the agle betwee a =, 4, ad b = 0,5, Solutio We will eed the dot product as well as the magitudes of each vector. ab i = a = 6 b = 9 The agle is the, ab i cosθ = = = a b 6 9. θ = = 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, ab= i Paul Dawkis 70

271 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. ab i = a b θ = 0 OR ab i = a b θ = 80 Example Determie if the followig vectors are parallel, orthogoal, or either. (a) a = 6,,, b =,5, (b) u = i j, v = i + j 4 Solutio (a) First get the dot product to see if they are orthogoal. ab= i 0 = 0 The two vectors are orthogoal. (b) Agai, let s get the dot product first. 5 uv= i = 4 4 So, they are t orthogoal. Let s get the magitudes ad see if they are parallel. 5 5 u = 5 v = = 6 4 Now, otice that, 5 5 uv i = = 5 = u v 4 4 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 ad b we wat to determie the projectio of b oto a. The projectio is deoted by proj a b. Here are a couple of sketches illustratig the projectio. 005 Paul Dawkis 7

272 So, to get the projectio of b oto a we drop straight dow from the ed of b util we hit (ad form a right agle) with the lie that is parallel to a. The projectio is the the vector that is parallel to a, starts at the same poit both of the origial vectors started at ad eds where the dashed lie hits the lie parallel to a. There is a ice formula for fidig the projectio of b oto a. Here it is, i proj a b = a ab a Note that we also eed to be very careful with otatio here. The projectio of a oto b is give by i proj b a = b ab b We ca see that this will be a totally differet vector. This vector is parallel to b, while proj a b is parallel to a. So, be careful with otatio ad make sure you are fidig the correct projectio. Here s a example. Example 4 Determie the projectio of b =,, oto a =, 0,. Solutio We eed the dot product ad the magitude of a. ab i = 4 a = 5 The projectio is the, 005 Paul Dawkis 7

273 proj a ab i b = a a 4 =, 0, =,0, 5 5 For compariso purposes let s do it the other way aroud as well. Example 5 Determie the projectio of a =, 0, otob =,,. Solutio We eed the dot product ad the magitude of b. ab i = 4 b = 6 The projectio is the, proj b a = ab i b b 4 =,, 6 4 =,, As we ca see form 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, i three dimesioal space. This vector will form agles with the x-axis (α ), the y-axis (β ), ad the z-axis (γ ). 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. 005 Paul Dawkis 7

274 The formulas for the directio cosies are, ai i a aij a ak i a α = = β = = γ = = a a a a a a cos cos cos where i, j ad k are the stadard basis vectors. Let s verify the first dot product above. We ll leave the rest to you to verify. ai i = a, a, a i,0,0 = a Here are a couple of ice facts about the directio cosies.. The vector u = cos α,cos β,cosγ is a uit vector.. α β γ. a = a cos α,cos β,cosγ cos + cos + cos = Example 6 Determie the directio cosies ad directio agles for a =,, 4. Solutio We will eed the magitude of the vector. a = = The directio cosies ad agles are the, 005 Paul Dawkis 74

275 cosα = α =.9 radias = 64. degrees cos β = β =.5 radias = degrees cosγ = 4 γ =.6 radias = degrees 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. So, let s start with the two vectors a = a, a, a product is give by the formula, ad b = b, b, b a b= ab ab, ab ab, ab ab the the cross 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, i j k 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 ad the third row is the compoets of b. 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. a a a a a a a b = i j + k b b b b b b 005 Paul Dawkis 75

276 where, a b ad bc c d = 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. 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. i j k i j 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. Example If a =,, ad b =, 4, compute each of the followig. (a) a b (b) b a Solutio (a) Here is the computatio for this oe. i j k i j a b = 4 4 = i + j + k 4 j i 4 k = 5i + j + k (b) Ad here is the computatio for this oe. i j k i j b a = 4 4 = i 4 + j + k j i k = 5i j k Paul Dawkis 76

277 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 θ be the agle betwee the two vectors a ad b ad assume that 0 θ π, the we have the followig fact, a b = a b siθ () ad the followig figure. 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 ad rotate our figers towards b our thumb 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 poits P = (, 0, 0), Q = (,,) ad R = (,,) fid a vector that is orthogoal to the plae. 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 where 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. 005 Paul Dawkis 77

278 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, PQ =, 0, 0 = 0,, PR =, 0, 0 =,, The cross product of these two vectors will be orthogoal to the plae. So, let s fid the cross product. i j k i j PQ PR = 0 0 = 4i + j k So, the vector 4i + j k 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 ad b, are parallel the the agle betwee them is either 0 or 80 degrees. From () this implies that, a b = 0 From a fact about the magitude we saw i the first sectio we kow that this implies 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 If a b = 0 the a ad b will be parallel vectors. Let s also formalize up the fact about the cross product beig orthogoal to the origial vectors. Fact Provided a b 0 the a b is orthogoal to both a ad b. Here are some ice properties about the cross product. Properties If u, v ad w are vectors ad c is a umber the, 005 Paul Dawkis 78

279 u v = v u cu v = u cv = c u v u v+ w = u v+ u w u v w = u v w u u u ui( v w) = v v v w w w i i 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. There are a couple of geometric applicatios to the cross product as well. Suppose we have three vectors a, b ad c ad we form the three dimesioal figure show below. The area of the parallelogram (two dimesioal frot of this object) is give by, Area = a b ad the volume of the parallelpiped (the whole three dimesioal object) is give by, Volume = ai b c 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. Example Determie if the three vectors a =, 4, 7, b =,, 4 ad c = 0, 9,8 005 Paul Dawkis 79

280 lie i the same plae or ot. 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 ai( b 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. 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 i 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. 005 Paul Dawkis 80

281 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. Fuctios of Several Variables A quick review of some importat topics about fuctios of several variables. Vector Fuctios We itroduce the cocept of vector fuctios i this sectio. We cocetrate primarily o curves i three dimesioal space. We will however, touch briefly o surfaces as well. Calculus with Vector Fuctios Here we will take a quick look at limits, derivatives, ad itegrals with vector fuctios. Taget, Normal ad Biormal Vectors We will defie the taget, ormal ad biormal vectors i this sectio. Arc Legth with Vector Fuctios I this sectio we will fid the arc legth of a vector fuctio. Velocity ad Acceleratio I this sectio we will revisit a stadard applicatio of derivatives. We will look at the velocity ad acceleratio of a object whose positio fuctio is give by a vector fuctio. Curvature We will determie the curvature of a fuctio i this sectio. Cylidrical Coordiates We will defie the cylidrical coordiate system i this sectio. The cylidrical coordiate system is a alterate coordiate system for the three dimesioal coordiate system. Spherical Coordiates I this sectio we will defie the spherical coordiate system. The spherical coordiate system is yet aother alterate coordiate system for the three dimesioal coordiate system. 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. 005 Paul Dawkis 8

282 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 that 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. May of the formulas that you are used to workig with i have atural extesios i. For istace the distace betwee two poits i is give by, (, ) = ( ) + ( ) d P P x x y y While the distace betwee ay two poits i is give by, (, ) = ( ) + ( ) + ( ) d P P x x y y z z 005 Paul Dawkis 8

283 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 is a vertical lie i a -D coordiate system. x = tells us to graph all the poits that are i the form, y. This I the equatio, 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 Here are the graphs of each of these. x = tells us to graph all the poits that are i the form 005 Paul Dawkis 8

284 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. 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 are the graphs for this example. 005 Paul Dawkis 84

285 Notice that if we look to where the plae itersect the xy-plae we will get the graph of the lie i as oted i the above graph. 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 005 Paul Dawkis 85

286 cetered o the z-axis. Here are the graphs for this example. 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 ad yet that does t really make sese. There is o reaso for the graph of a lie or a circle i. Let s thik about the example of the circle. To graph a circle i we would eed to do somethig like 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. Aother quick poit to make here is that, as we ve see i the above examples, may graphs of equatios i are surfaces. That does t mea that we ca t graph curves i. We ca ad will graph curves i as well as we ll see later i this chapter. Equatios of Lies I this sectio we eed to take a look at the equatio of a lie i. As we saw i the previous sectio the equatio y = mx + b does ot describe a lie i, istead it describes a plae. 005 Paul Dawkis 86

287 This does t mea however that we ca t write dow a equatio for a lie i -D space. To see how 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, P0 = ( x0, y0, z0), ad that v= abc,, is some vector that is parallel to the lie. Note, i all likelihood, v will ot be o the lie itself. We oly eed v to be parallel to P = x, y, z be ay poit o the lie. the lie. Fially, let Now, sice our slope is a vector let s also tur the two poits ito vectors as well. Of course, we do t actually tur them ito vectors, we istead use positio vectors to represet them. So, let r 0 ad r be the positio vectors for P 0 ad P respectively. Also, for o apparet reaso, let s defie a to be the vector with represetatio PP 0. We ow have the followig sketch with all these vectors. 005 Paul Dawkis 87

288 At this poit, otice that we ca write r as follows, 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 ad v are parallel. Therefore there is a umber, t, such that a = tv We ow have, 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 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 to the right of the origial poit ad if t is egative we move to the left of the origial poit. As t varies over all possible values we will completely cover the lie. 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= x0, y0, z0 + t abc,, x, yz, = x+ tay, + tbz, + tc The oly way for two vectors to be equal is for the compoets to be equal. I other words, 005 Paul Dawkis 88

289 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 that will be parallel to the lie. This ca be ay vector as log as it s parallel to the lie. I geeral, v wo t lie o the lie itself. However, i this case it will. All we eed to do is let v be the vector that starts at the secod poit ad eds at the first poit. Sice these two poits are oe the lie the vector betwee them will also lie o the lie ad will hece be parallel to the lie. So, v =, 5, 6 Note that the order of the poits was chose to reduce the umber of mius sigs i the 005 Paul Dawkis 89

290 vector. We just have easily goe the other way. Oce we ve got v 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, 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 we ca, this will give the value of t for which the poit will pass through the xz-plae. + t = 0 t = 4 So, the lie does pass through the xz-plae. To get the complete coordiates of the poit 005 Paul Dawkis 90

291 all we eed to do is plug t = ito ay of the equatios. We ll use the vector form. 4 r =, +,8 =,0, Recall that this vector is the positio vector for the poit o the lie ad so the coordiates of the poit here the lie will pass through the xz-plae are,0, 4 4. Equatios of Plaes I the first sectio of this chapter we saw some 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 suppose that we have a vector that is orthogoal (perpedicular) to the plae, = a, b, c P = x, y, z is. This vector is called the ormal vector. Now, assume that ay poit i the plae. Fially, sice we are goig to be workig with vectors iitially we ll let r 0 ad r be the positio vectors for P 0 ad P respectively. Here is a sketch of all these vectors. 005 Paul Dawkis 9

292 Notice that we added i the vector r r 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 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 0. Recall from the Dot Product sectio that two orthogoal vectors will have a dot product of zero. I other words, i r r = 0 ir = i r This is called the vector equatio of the plae. 0 0 The vector equatio of the plae is ot a very useful equatio i some ways. Let s get a much more useful form of the equatios. Let s start with the first form of the vector equatio. abc,, i xyz,, x, y, z = 0 ( ) abc,, i x x, y y, z z = Now, actually compute the dot product. 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, = a, b, c 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. PQ =,, 4 PR =,, 005 Paul Dawkis 9

293 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. i j k i j = 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 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 ad v are parallel, the v is orthogoal to the plae, but v is also parallel to the lie. So, if the two vectors are parallel the lie ad plae will be orthogoal. Let s check this. i j k i j 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 ad v are orthogoal the the lie ad the plae will be parallel. 005 Paul Dawkis 9

294 Let s check this. v= i = 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. 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. x y z + + = a b c Here is a sketch of a typical ellipsoid. If a=b=c the we will have a sphere. 005 Paul Dawkis 94

295 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 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. 005 Paul Dawkis 95

296 Here is a sketch of typical cylider. x + y = r 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. 005 Paul Dawkis 96

297 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. 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 Here is a sketch of a typical elliptic paraboloid. 005 Paul Dawkis 97

298 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. 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 paraboloids the surface ca be easily moved up or dow by addig/subtractig a costat from the left side. 005 Paul Dawkis 98

299 For istace z = x y + 6 is a elliptic paraboloid that opes dowward ad starts at z=6 istead of z=0. Here is a sketch of this surface. Fuctios of Several Variables I this sectio we wat to go over some of the basic ideas about fuctios of more tha oe variable. First, remember that graphs of fuctios of two variables, z f ( x, y) three dimesioal space. For example here is the graph of = are surfaces i z = x + y Paul Dawkis 99

300 This is a elliptic parabaloid ad is a example of a quadric surface. We saw several of these i the previous sectio. We will be seeig quadric surfaces fairly regularly later o i the semester. Aother commo graph that we ll be seeig quite a bit i this course is the graph of a plae. We have a covetio for graphig plaes that will make them a little easier to graph ad hopefully visualize. Recall that the equatio of a plae is give by ax + by + cz = d or i terms of fuctio otatio this would be give by, f ( x, y) = ax + by + c To graph a plae we will geerally fid the itersectio poits with the three axes ad the graph the triagle that coects those three poits. This triagle will be a portio of the plae ad it will give us a fairly decet idea o what the plae itself should look like. For example let s graph the plae give by, f xy, = x 4y For purposes of graphig this it would probably be easier to write this as, z = x 4y x+ 4y+ z = Now, each of the itersectio poits with the three mai coordiate axes is defied by the fact that two of the coordiates are zero. For istace, the itersectio with the z-axis is defied by x= y = 0. So, the three itersectio poits are, 005 Paul Dawkis 00

301 Here is the graph of the plae. x axis : 4,0,0 y axis : 0,,0 z axis : 0,0, Now, to exted this out, graphs of fuctios of the form w f ( x, y, z) = would be four dimesioal surfaces. Of course we ca t graph them, but it does t hurt to poit this out. We ext wat to talk about the domais of fuctios of more tha oe variable. Recall that domais of fuctios of a sigle variable, y = f ( x), cosisted of all the values of x that we could plug ito the fuctio ad get back a real umber. Now, if we thik about it, this meas that the domai of a fuctio of a sigle variable is a iterval (or itervals) of values from the umber lie, or oe dimesioal space. The domai of fuctios of two variables, y f ( x, y) space ad cosist of all the coordiate pairs, ( x, y ), that we could plug ito the fuctio ad get back a real umber. Example Determie the domai of each of the followig. f xy, = x+ y (a) (b) f ( xy, ) = x+ y (c) f ( xy, ) = l( 9 x 9y ) =, are regios from two dimesioal Solutio (a) I this case we kow that we ca t take the square root of a egative umber so this meas that we must require, x+ y Paul Dawkis 0

302 Here is a sketch of the graph of this regio. (b) This fuctio is differet from the fuctio i the previous part. Here we must require that, x 0 ad y 0 ad they really do eed to be separate iequalities. There is oe for each square root i the fuctio. Here is the sketch of this regio. (c) I this fial part we kow that we ca t take the logarithm of a egative umber or zero. Therefore we eed to require that, x 9 x 9y > 0 + y < 9 ad upo rearragig we see that we eed to stay iterior to a ellipse for this fuctio. Here is a sketch of this regio. 005 Paul Dawkis 0

303 Note that domais of fuctios of three variables, w f ( x, y, z) dimesioal space. Example Determie the domai of the followig fuctio, f ( x, y, z) = x + y + z 6 =, will be regios i three Solutio I this case we have to deal with the square root ad divisio by zero issues. These will require, x + y + z 6 > 0 x + y + z > 6 So, the domai for this fuctio is the set of poits that lies completely outside a sphere of radius 4 cetered at the origi. The ext topic that we should look at is that of level curves or cotour curves. The f xy, are two dimesioal curves with equatio level curves of the fuctio f ( xy, ) = kwhere k is ay umber. You ve probably see level curves (or cotour curves, whatever you wat to call them) before. If you ve ever see the elevatio map for a piece of lad, this is othig more tha the cotour curves for the fuctio that gives the elevatio of the lad i that area. Of course, we probably do t have the fuctio that gives the elevatio, but we ca at least graph the cotour curves. Example Idetify the level curves of f ( xy, ) x y Solutio = +. Sketch a few of them. First, for the sake of practice, let s idetify what this surface give by f (, ) this let s rewrite it as, xy is. To do 005 Paul Dawkis 0

304 z = x + y Now, this equatio is ot listed i the Quadric Surfaces sectio, but if we square both sides we get, z = x + y ad this is listed i that sectio. So, we have a coe, or at least a portio of a coe. Sice we kow that square roots will oly retur positive umbers, it looks like we ve oly got the upper half of a coe. Note that this was ot required for this problem. It was doe for the practice of idetifyig the surface ad this may come i hady dow the road. Now o to the real problem. The level curves (or cotour curves) for this surface are give by the equatio, k = x + y x + y = k where k is ay umber. This is the equatio of a circle of radius k with ceter at the origi. We ca graph these i oe of two ways. We ca either graph them o the surface itself or we ca graph them i a two dimesioal axis system. Here is each graph for some values of k. 005 Paul Dawkis 04

305 Note that we ca thik of cotours i terms of the itersectio of the surface that is give by z = f ( x, y) ad the plae z = k. The cotour will represet the itersectio of the surface ad the plae. For fuctios of the form f (,, ) equatios of level surfaces are give by f ( xyz,, ) = kwhere k is ay umber. xyz we will occasioally look at level surfaces. The The fial topic i this sectio is that of traces. I some ways these are similar to cotours. As oted above we ca thik of cotours as the itersectio of the surface give by z = f ( x, y) ad the plae z = k. Traces of surfaces are curves that represet the itersectio of the surface ad the plae give by x = a or y = b. Let s take a quick look at a example of traces. Example 4 Sketch the traces of f xy, = 0 4x y for the plae x = ad y =. Solutio We ll start with x =. We ca get a equatio for the trace by pluggig x = ito the equatio. Doig this gives,, z = f y = y z = y ad this will be graphed i the plae give by x =. Below are two graphs. The graph o the left is a graph showig the itersectio of the surface ad the plae give by x =. O the right is a graph of the surface ad the trace that we are after i this part. 005 Paul Dawkis 05

306 For y = we will do pretty much the same thig that we did with the first part. Here is the equatio of the trace, ad here are the sketches for this case. z = f x, = 0 4x z = 6 4x Vector Fuctios To this poit, with the exceptio of lies, we oly looked at graphig surfaces i. However, as we saw with lies, ot every graph i eeds to be a surface. We ca graph curves (sometimes called space curves) that are three dimesioal as well. To do this we use vector-valued fuctio or vector fuctios. Note that we ca also use vector fuctios to represet surfaces as well as we ll see at the ed of this sectio. With that beig said however we will sped most of this sectio talkig about curves istead of surfaces. The vector form of the equatio of a lie is a good example a vector fuctio. 005 Paul Dawkis 06