Chapter 6: Transcendental functions: Table of Contents: 6.3 The natural exponential function. 6.2 Inverse functions and their derivatives

Save this PDF as:
 WORD  PNG  TXT  JPG

Size: px
Start display at page:

Download "Chapter 6: Transcendental functions: Table of Contents: 6.3 The natural exponential function. 6.2 Inverse functions and their derivatives"

Transcription

1 Chpter 6: Trnscendentl functions: In this chpter we will lern differentition nd integrtion formuls for few new functions, which include the nturl nd generl eponentil nd the nturl nd generl logrithmic function We will lso look t simple differentil equtions tht involve the nturl eponentil function Tble of Contents: 63 The nturl eponentil function 6 Inverse functions nd their derivtives 6 The nturl logrithm function 64 Generl Eponentil nd Logrithmic Functions 65 Eponentil Growth nd Dec 66 First order liner differentil equtions 67 Approimtions for differentil equtions (direction fields nd Euler s method) (optionl) 68 The inverse trigonometric functions nd their derivtives 69 The hperbolic functions, their inverses nd their derivtives

2 63 The nturl eponentil function: Consider the function :, here nd tht we hve fied positive bse Define We define : s follows: f R, f ( ), R,, note tht the rgument is n eponent N: n, if n times, then define (P) ; Similrl, n, for ever n N(P) Note tht (P) nd (P) follow s nturl properties for n m m n m n powers ( mens multiplied with itself times, nd since mn, then (cross n multipl), which genertes P) Therefore, if Z, then either, then tke the definition bove, or, nd use (P) If p Q( generic rtionl number), then define q the propert m n m n 3 Emple: 9, 8 p q, such tht q p (here we keep in mind However, if R Q, wht is? For emple, let monotonicit (continuit) of, strt with 3 73? In order to mintin the nd we cn improve this bound of between to rtionl eponents s close s we wnt In other words, strt with the grph of which is defined onl for rtionl eponents until now, nd fill the holes with to irrtionl powers to fill the holes (lws respecting tht the function is incresing (for is unique vlue of 3, nd we cn get s close s we wnt to 3 ) This definition of, is unique, since there 3 this w Properties of : Bsed on these definitions, we collect the properties of eponentil (strt to prove these with, N nd continue with the other sets):

3 (PE),, R,, R,, R *, b b,, b R, R Also note wh we hve the necessr restriction, since is ssumed rel here, from where we find tht the domin nd rnges of f ( ) 3 Grphs of : re :, f R Consider the grphs of f ( ) for few vlues of : Figure 63 Grphs of,,,, 3, We see tht the function f ( ) is strictl decresing for, constnt for, nd strictl incresing for Moreover, (<) increses more rpidl for lrger (>), nd lso decreses more rpidl for smller 3

4 4 Define the number e : We consider prticulr eponentil function ( prticulr bse ) with prticulr propert of More precisel, we clculte the slopes of the tngent lines to for for (we do not hve formul for ' et, so we need to drw tngent lines nd mesure slopes) We find tht the slope of the tngent line to for when is m 7 nd the sme slope for 3is m3 We consider the number e, e 3, s the bse of tht eponentil function for which the slope of the tngent line to e for is We will see tht this number e, s the bse of this prticulr eponentil function (with this propert), is ver importnt, nd it will be used etensivel Alternte properties (definitions) cn be given For now, from the bove propert, we hve tht: ( P ) f h f ( h) f () e '() lim lim h h h h We cn plug in different vlues for e to clculte this limit, nd we find tht e 788 (irrtionl number) From ( P ) we cn non-rigorousl derive the more direct reltion: P ) elim h h ( h (we will give rigorous proof of this result lter, let s keep this in mind for now) Two importnt properties of the eponentil follow from ( P ): 5 Find e ' nd First, e d : P ) h h ( h h e e e e ' lim e lim e h h Therefore, we hve found e ' for ever, bsed on e ' for (dd this to formuls of derivtives nd memorize, since we ll use it often) It is remrkble tht the eponentil is itself However, remember tht e d becuse of ( P ), we hve tht (the nti-derivtive of e is the onl function whose derivtive e ) is function whose derivtive is e, so in this cse, ( P 3 ) e d e C Add this to the tble with formuls for integrls nd memorize s it will be used mn times 4

5 d e B ppling the Chin Rule, obtin tht d 6 Emples: Find d e d u( ) u( ) e u '( ) Let f ( ) e Find where f is incresing nd decresing nd where it is concve upwrd nd concve downwrd Also, identif ll etreme vlues nd points of inflection Then sketch grph of f( ) Evlute 4 e d Evlute e 3 d Evlute 3 3 e d Evlute 6e d Find the derivtive of: f ( ) e () t t g t e e g() t cost e f ( ) sin e e e 5

6 e Evlute the following indefinite integrls: e 3 d e e d ep e d (86 AB5, BC) Let A() be the re of the rectngle inscribed under the curve nd (, ), > e with vertices t (-, ) () Find A() (b) Wht is the gretest vlue of A()? Justif our nswer (c) Wht is the verge vlue of A() on the intervl [, ]? Homework: Choose from 4,9,,,4,6,39,44,46,47,5,53 from the end of section 63 + from bove 6 Inverse functions nd their derivtives: Definition nd min theorem: Let f : D R be one to one function on D (psses the horizontl line test: ever horizontl line intersects the grph of f( ) t most once) We define g : R D to be the inverse of the function f if it stisfies: (DI) g f ( ) D f g( ) R Theorem : A (surjective) function f : D R is invertible (it hs n inverse) if nd onl if f is one to one (ie it psses the horizontl line test) 6

7 Proof: (tr the proof) Consider s homework too Defn: f : D R is function if (PF) D! R s t f ( ) (note the uniqueness here, otherwise the reltion would be mbiguous (not D f R function), Figure 6: Emple of non-surjective function Emple: is f : R R, f ( ), function?, Wht bout f : R R, f ( ),,? Wht bout f : R R, f ( ),? Defn: A function f : D R is surjective (onto) function if (PS) R D s t f ( ) D f R f : R,, f ( ) surjective? Figure 6: Emple of surjective function Emple: Is,, Wht bout f : R R, f ( )? Wht bout f : R R, f ( ),, Defn: A function f : D R is one to (injective) function if (PI) f ( ) f ( ),, D, tht is the in (PS) bove is unique D f R Figure 63: Emple of one to one surjective (tht is, bijective) function 7

8 Emple: Is f : R [, ), f ( ) bijective function?, Wht bout f : R R, f ( )? (solution b discussion fter nd ), Note: Think lso bout these definitions in terms of the grph of f Bsed on these definitions, let s prove first, tht is f bijective, implies f invertible Use (PS) with unique (since injective), tke g( ) Due to surjectivit nd injectivit of f, g is function: R! D s t f ( ) g( ) (b choice of ) Since f ( ) Let us ppl g to f ( ) invertible then f ( g( )), R to get g( f ( )) g( ), D Therefore, when f is bijective, f is Now, tht is, we hve to show tht if f is n invertible function, then f is bijective, tht is (PS) with unique Since g is function, then R! D s t g( ) f ( ) (b ppling f nd using tht f nd g re inverses to ech other) QED Method to find the inverse nd emples: To find the inverse of function f, solve the eqution () f ( ) This is obvious from the proof bove: If f is given b f ( ) for, then to find the inverse g( ), we need to reverse the reltion bove, tht is, we need to find from which cme This reltion is cler in Figure 3: we hve to reverse the rrow (the effect of f ) in order to undo its effect over (give numbers here) Therefore, in order to find the inverse of function f, solve the eqution () f ( ) Emple : for Find the inverse of f( ) First, ou need to find the domin nd rnge of f, nd mke sure tht f is one to one (bijective) over its domin 8

9 Note tht the grphs of f ( ) nd set of points, nd, (This is becuse f ( ) re smmetric versus the line, since the represent the f ( ) f ( ) re the sme grph, but we needed to switch the roles of nd to plot the inverse) This is n importnt propert which gives n es w to plot the inverse of n function A simpler criterion (thn one-to-one) for deciding if function is invertible: Theorem : If f : D R is strictl monotonic on D, then f is invertible on D Proof: Here, like usul, we ssume tht f is surjective on D But f strictl monotonic on D implies tht f is one to one on D (es to prove tht f ( ) f ( ) ) Therefore f is invertible on D from Theorem Emple : Show tht 5 f ( ) is invertible Emple 3: ) Let f( ) Find the domin nd the rnge of f( ), show tht f( ) is invertible over its domin, find formul for f ( ), nd verif (DI) nd the propert of the grphs b) Let f ( ) 4,, f :[, ) [ 4, ) Show tht f( ) is invertible over its domin, find formul for f ( ), nd verif (DI) nd the propert of the grphs 3 The inverse function theorem: Theorem 3 (inverse function theorem): Let f : D Rbe differentible one to one (or strictl monotonic) function on D Then: () d f ( ), R d d f ( ) d Note tht the derivtives bove re with respect to different vribles, first with respect to, the second with respect to Also, to find this useful ( formul for of, fter we found the derivtive) d f ( ) d, we need to epress the right hnd side in terms 9

10 Proof: d f ( ) chin rule d f ( ) f f ( ) () d d Note: This inverse function theorem bove will be useful in mn instnces, when we wnt to clculte the derivtive of n inverse of function, nd we cn find the derivtive of the originl function, nd epress somehow in terms of fter tht Emples: Problems 6,,, 37 tetbook Emple: Let f Homework: 5 ( ) Find f '(4) Choose from:, 5,,,,, 5, 9, 3, 35, 38, 4, 43 from end of section 6 Etr Credit: 45 6 The nturl logrithm function: Definition: Consider f ( ) e, :, f R, e s defined before Then f '( ) e, R, nd therefore f( ) is strictl incresing f( ) is one to one so invertible (cn esil check tht the function is lso surjective) Therefore, there eists function f :, f ( ) ln( ), such tht: R, which we cll the nturl logrithm function, () ln e, R ln( ) e,, Emple: ln(3)? (from wht, =3 cme?), solve e 3 nd so on ln( e)?

11 Figure 6 Grph of e (red), ln( ) (mgent) nd their line of smmetr (blue) From the grph, we note ln(), ln(), ln( ) Properties of ln : Remember the properties of e : () e e e e,, R e e,, R e e e,, R (3) these produce ln() ln( ) ln( b) ln( b),, b(, ) ln( ) ln( b) ln,, b(, ) b ln b b ln( ),, b(, ) Prove some of these First, b definition: ln() is number such tht e For the second, cll e nd e bin nd ppl ln using the inverse propert Similr for 3 For 4 let 3 e nd b ln( ) ' nd d : b, then ln ln( ) b ln( ) Let us look now t ln( ) '

12 Using the inverse function theorem for derivtives for f ( ) e : d Therefore, we hve found ln( ) d d e e ln( ) ', for d This cn be generlized to: ln ', for (use chin rule when ) Therefore, we hve found new integrl: d ln C 4 Logrithmic differentition: Logrithmic differentition is ver useful (nd sometimes indispensble tool) It mkes use of the properties of ln to simplif function first, fter which we clculte the derivtive Emples: ) f ( ) ln 3, : introductor emple, works both ws but suggests log differentition much esier b) f( ) /3 (cn compre methods nd get domin nd rnge on the w) c) f ( ) e Therefore, log differentition is useful when there re lrge products or frction, or powers, which reduce to sums, differences, coefficients, fter ppling ln Don t forget to multipl bck b f ( ) t end However, there re other cses in which log differentition is ver useful (nd prett much indispensble): d) f ( ) find domin, rnge nd derivtive e) f ( ) find domin, rnge nd derivtive Other emples: ) Find d ln d b) Find d ln d

13 c) Find 5 7 d d) Evlute 3 d e) tn d f) Given the function f defined b f ( ) ln( 9) ) Determine the smmetr of the grph of f b) Find the domin of f c) Find ll vlues of such tht f() = d) Grph the function g) Find the derivtive of f ( ) ln( ) ln( ) (discuss the domin nd the rnge first) -4 h) Find '( ) if e i) Find '( ) if sin e j) Sketch the grph of ln f ( ), - - k) Evlute: e e d l) Evlute: ep e d 3

14 m) Evlute: d (85 BC5) Let f be the function defined b f ( ) ln for nd let R be the region between the grph of f nd the -is () Determine whether the region R hs finite re Justif our nswer (b) Determine whether the solid generted b revolving the region R bout the -is hs finite volume Justif our nswer (9 AB) Let f be the function given b f( ) ln () Wht is the domin of f? (b) Find the vlue of the derivtive of f t (c) Write n epression for f ( ) where f denotes the inverse function of f Homework: Choose from problems 3,5,,6,,,33,4,45 /end of section 6 nd from bove Etr credit: 55,56 64 Generl Eponentil nd Logrithmic Functions ' nd d We wnt to generlize bit the formuls for derivtives nd integrls which we found to generl eponentil nd logrithmic functions Consider f ( ),, f : (, ) In order to find ' bsed on wht formuls we lred hve, write e ln( ) ln( ) ln( ) ' e ' ln( ) ' e ln( ) Then Therefore () ' ln( ), for (dd this to tble of derivtives nd memorize) Cn we then find d? From bove, () d = ln( ) C ( dd to tble of integrls nd memorize) log ( ) nd its derivtive 4

15 Let s consider now f ( ),, f : (, ), remember the grphs of Note tht ll these functions re strictl monotonic, therefore f ( ) f( ),,,,3,5 5 3 is strictl monotonic if Therefore, f ( ),,, f : R (, ) is invertible nd we cll its inverse f R f :,, ( ) log ( ) with properties (3) log, R,, log ( ),,,, For emple, log 3 such tht 3 New properties for f ( ) log ( ) follow, which correspond nd cn be shown in similr w with the properties (3) in 6, usull b replcing e (4) log log blog Proof: b Let log nd log z Then nd z b which shows () b replcing b b A new propert is: z z Therefore b log b z log b (this needs to be proven s indicted bove), nd z The derivtive log ( ) ' follows from the inverse function theorem, like before: For f ( ), log ( ) ' ( )' ln( ) ln( ) You cn re-write this s (5) log ( ) ', for ln( ), dd to tble of derivtives, memorize nd use in this form No new integrl forms, since (5) still involves just the function on the right hnd side (cn write d ln( )log ( ) C but this is known formul, using (4)) 5

16 Emples: ) Find d 3 d d if 5 d 4 b) Find 5 4 c) 3, 4find d d 3 4 d d) If log e) Compre the grphs of,, nd is eponentil function, is n power function D ln but D 6 d C nd d C ln Find D, f) 5 d g) Find if sin d d h) Find the derivtive of ech of the following: f( ) 3 cos f( ) 7 f ( ) ep log f 3 f( ) 4 f ( ) 3 ln f( ) f( ) log3 ( ) log i) Find d log Homework: From bove nd from problems 5,4,6,7,,6,7,33,34(EC),38,4,49 6

17 65 Eponentil Growth nd Dec: Eponentil growth nd dec: Since the derivtive of function represents its instntneous rte of chnge, mn mthemticl nd phsicl models for different lws in nture involve equtions which contin the derivtive of function (these re clled differentil equtions) The eponentil function is ver importnt nd omnipresent in mthemtics lso becuse it ppers often in mthemticl models We often her the term of quntit growing (or decing) eponentill, but wht does it men? For n eponentil growth of popultion, we ssume tht the rte of growth (births minus deths) of given popultion t n certin fied time is proportionl with the size of the current popultion (no eponentil in our model et!) Although idelistic (neglects limittion problems) this model is bsed on resonble ssumption If t () is the size of the popultion t time t, this model gives: () d k dt, where k is just proportionlit constnt nd it cn be considered popultion dependent Note tht for k we hve popultion growth nd for k we hve popultion dec Solving () for t t with ( t) gives kt () () t e, from where the nme of eponentil growth (nd dec) Therefore, the ssumption () produced solution of eponentil growth/dec Emple: If the popultion of the world ws 64 billion t the beginning of 4 nd k for the world popultion is pproimtel 3 (ssuming t is mesured in ers), how long will it tke for the world popultion to double? The time intervl needed for popultion (growing eponentill) to double is clled the doubling time Note tht if n eponentill growing quntit doubles from to in n initil intervl of length T, it will double in n intervl of length T ktt t T e e () t e kt kt Logistic model: As mentioned before, the eponentil growth model is unrelistic, since it does not include n limittion (due to resources, etc), nd it projects fster nd fster growth indefinitel into the future A more relistic model, clled the 7

18 logistic model, ssumes tht the rte of growth is proportionl to both the popultion size nd to the difference L, where L is the mimum popultion tht cn be supported This leds to the differentil eqution d dt (3) k L Note tht for smll, d kl (second fctor not essentil model-wise) which suggests eponentil-tpe growth But dt s pproches L, the second fctor becomes importnt, growth is curtiled, nd d gets smller nd smller (until it dt gets ver close to zero so there is lmost no chnge in t ()), producing growth curve tht flttens out Eqution (3) cn be solved ectl to obtin this tpe of solution (indicte steps) 3 Eponentil dec: Some quntities dec t rte proportionl to their current size, one such dec model is the rdioctive dec Decing popultions re modeled b () but with k Emple: Problem 6 or 7 t end of section 4 Newton s lw of cooling: Newton s lw of cooling ssumes tht the rte t which n object cools (or wrms) is proportionl with the difference in temperture between the objects nd the temperture of the surrounding medium: dt (4) k( T T ) dt (note tht k is negtive in this model) This is similr model with (), the onl difference is tht the solution increses (or decreses) to T (s T gets close to T the rte of chnge gets close to ) Wht is the ect solution of (4)? kt T() t T T T e 5 A better definition of e : We hinted in 63 tht (5) elim h h h 8

19 but we defined e from (*) h e lim h h Now we hve ll tools (properties of eponentil nd logrithms) to prove (5) As mentioned, (5) is preferble method nd we wnt to memorize it, s it will be useful for mn limits (nd test problems!!!) Proof: Remember tht for (*) we strted with the slope of the tngent line to e t (,) is We see tht we need some form of n inverse, in order to subtrct (solve for e) in (*) Therefore, we consider now the slope of the tngent line t the corresponding point ((,)) for the inverse of e : ln( ) From the inverse function theorem, or using ln( ) '/ Using this, tr to find (5) formul: Let f ( ) ln( ) Since '() Therefore: h h ln( h) ln() ln( h) h lim lim lim ln h h h h h f, then h / / h ln lim ln( e) (since f ( ) ln( ) is continuous function, wh?, since it is the inverse of continuous function) But f ( ) ln( ) is one to one function, which produces the result Let us memorize (5) s it is the preferred w to define e From (5) we cn derive tht: (6) lim n n n e b mking chnge of vrible in limit Using (5) or (6), solve the following limits: lim h h h n lim n 3 n lim n n lim n n n n h h lim h lim n n n lim n n lim n n n3 lim n n n lim n n n lim n n n lim n n n 3 lim n n 3n lim n n n3 Homework: Problems 3, 8,, 6, 7, 8, 9, 36, 37 (importnt) 9

20 66 First order liner differentil equtions: d Method for solving P( ) Q( ) d : The differentil eqution () d k dt which we solved in the previous section is one of the simplest tpes of differentil equtions Let us look now t the more generl differentil eqution: d () P( ) Q( ) This eqution is clled liner (becuse, where is the eqution d opertor), first order with vrible coefficients differentil eqution (usull for liner differentil eqution the rhs is of the liner form P( ) Q( ) ) Eqution () is not seprble nmore, so we cnnot seprte the vribles s we did for () Insted, we ppl ver prticulr method for this eqution (mke sure tht ou hve ectl this form before solving the eqution, especill mke sure tht d d hs coefficient nd tht the term P( ) is on the sme side with d d ): In order to obtin seprble eqution, we tr to get rid of the P( ) term b multipling both sides of () b P( ) d e After this, () becomes: (3) d e Q ( ) e d P( ) d P( ) d such tht P( ) d P( ) d e Q( ) e d C Which gives tht: e Q( ) e d Ce P( ) d P( ) d P( ) d (creful with the plce t which C pops up in the problem) As with ll first order differentil equtions, n initil condition should be given (quntit chnges, but t wht vlue does it strt?) The vlue of C will be determined from this initil condition Emple: Problem 3/pp 358: Solve the differentil eqution: (4) Is this of the form ()? Not et, we need to divide first b ', < After this, we hve:

21 ' (5) P( ) nd Q( ) which is of the form () with (note tht the division is llowed in domin) First, P( ) d d ln C, P ( ) d so e Multipling (5) b this integrtion fctor, we obtin: ' d d 3/ 3/ 3/ This is the seprble form, which, fter integrtion, gives: d C 3/ (ver importnt to put the C here!) Therefore: C (ou should check this b plugging it bck into (4)) 3 Emple 3/356 (tnk problems): Initil: gllons brine 75 lbs slt Rte in = gls brine/min with lbs of slt/gll Rte out=gls brine /min with / slt/gll Find () - the totl mount of slt in tnk fter hour (6 mins) Let (t) the totl quntit of slt in tnk t time t d d Then rte in - rte out gl/min slt/gl gl / min slt/gl dt dt d So 4 dt, where time is mesured in minutes Solving, we find 6 ( t) 44 69e 6 t

22 So (6) 44 69e 86 lbs Red other problems with tnk Cn lso do problem 8/358 (we hve to solve two eqution there, for (t): d 3 gl / min lbs / gl gl / min lbs / gl dt z nd for z(t): dz 3 dt ) Homework: Problems,6,8,3,4,6,8, 68 The inverse trigonometric functions nd their derivtives: Let us turn now to the si trigonometric functions: sin( ), cos( ), tn( ), cot( ), sec( ), csc( ) Gols: sin ( ) rcsin( ), cos ( ) rccos( ) Remember the inverses tn ( ) rctn( ), cot ( ) rc cot( ) d(rcsin( )) d(rccos( )), d d Find nd memorize the formuls for d(rctn( )) d( rc cot( )), d d Use the formuls bove to find, memorize nd use formuls for new integrls: d d d d

23 The si trigonometric functions listed bove re not es to work with, with respect to their inverses Since the re ll periodic, the re not one to one (do not pss the horizontl line test) Therefore we will need to restrict the domin for ech of them, choosing the lrgest possible restricted domin such tht the function is one-to-one (invertible) on this restricted domin Some freedom is possible with this choice of the restricted domin, nd we will then choose the most nturl restricted domin Let us strt with the inverse of sin( ) : sin : R [,] is not one-to-one, restrict the domin to, (look t the grph) Therefore, sin :, [,] is one-to-one (invertible) So, there eists () rc sin :, [, ] such tht () rcsin(sin( )),, sin(rcsin( )),, Plot sin( ) nd sin ( ) rcsin( ) on the sme grph: Figure 68: Grph of sin( ) (green) (blue) nd sin ( ) (red) Emple: Clculte ) sin b) sin c) sin 3 d) sin sin e) sin 3

24 Similrl, cos : [,] is not one-to-one, nd to mke it one to one we restrict its domin to, (there re more choices here but this is the most nturl ) Therefore cos :,, is invertible, so there eists cos rccos :, [, ] such tht (4) (3) rc rccos cos( ),, cos cos( ),, Plot cos( ) nd cos ( ) rccos( ) on the sme grph Figure 68: Grph of cos( ) (green) (blue) nd cos ( ) (red) (ignore the left to zero prt for cos( ) ) Creful with the domin of the restricted cos, nd the rnge of rccos, the will be importnt when finding vlues Emple: Clculte: ) rccos 3 b) rccos 3 c) rccos d) 7 rccos cos 6 e) rccos cos 3 Similrl, is not one-to-one, nd we restrict its domin to, to mke it one- to-one (k ) tn : R \, k Z R So, there eists 4

25 (5) tn rctn : R, such tht (6) rctn tn( ),, tn rctn( ), R Plot tn( ) nd tn ( ) rctn( ) on sme grph: Figure 683: Grph of tn( ) (green) (blue) nd tn ( ) (red) Emple: Clculte tn ( ) Similrl, cot : R \ k, k Z R is not one-to-one, nd we restrict its domin to Eercise: Prove the following identities (eventull using the fundmentl trigonometric identit sin ( ) cos ( ) :, to mke it one-to-one ) sin cos b) cos sin c) sectn d) Let us find now the derivtives of the four inverse trigonometric function bove tn sec Using the inverse function theorem (note tht ll these restricted trigonometric functions re strictl monotonic nd differentible), we hve: For sin( ), d sin d dsin( ) cos( ) sin ( ) d (the sign in the identit before the lst comes from,, for which cos( ) ) 5

26 Therefore, the new formul is: (7) sin ( ) ' (dd to tble of derivtives nd memorize) Similrl: (8) cos ( ) ' derivtives nd memorize) (the minus comes becuse cos( ) ' sin( ) now, dd to tble of Also, for tn( ), d tn cos ( ) d d tn( ) sec ( ) tn ( ) d, which gives tht: (9) tn ( ) ' You cn lso derive (on our own) in similr mnner tht: cot ( ) ' () Like for ll the previous derivtives tht we obtined in section 6, 63 nd 64, formuls (7) () produce new integrls: () d rcsin( ) C, which generlizes to () rcsin of the right side or tr to obtin the integrl in the left with chnge of vribles ) d C (tke the derivtive nd: (3) rctn( ) d C, which generlizes to (4) rctn d C The other two functions do not give new integrls Add the boed formuls to tble of integrls nd memorize These new formuls for derivtives nd integrls m pper comple, but the will be ver useful in solving more complicted integrls (frctions nd rdicls) Eercises: 6

27 Let s remember (or clculte) the following integrls: ) sec ( ) d b) csc ( ) d c) tn( ) d d) cot( ) d Lter on (net section) we ll lern three more importnt formuls for integrls: d, d nd d which gives us bsic set of formuls, which, if we memorize, help us solve ll integrl problems in the tet Eercises: Find: sin 3 ' tn ' d A mn is stnding on top of verticl cliff feet bove lke As he wtches, motorbot moves directl w from the foot of the cliff t rte of 5 feet per second How fst is the ngle of depression of his line of sight chnging when the bot is 5 feet from the foot of the cliff? 3 Differentite: f ( ) sin sec e f( ) sin sin t P, 6 4 Evlute: f ( ) tn ln d d 9 d 4 d 3 d 6 5 Homework: From bove nd problems 4,7,8,4,7,,4,5,3,3,36,46,48,57,6,67,7,7,7, nd 77(EC) 7

28 69 The hperbolic functions, their inverses nd their derivtives: We wnt to prove, memorize nd use the following formuls for integrls, which will complete our tble of integrtion for now: () ln d C () d ln C (3) d ln C These formuls will help us solve integrls with qudrtics in the denomintor, under or not under squre root Tking the derivtive of the right side we obtin the left side, which is enough for proof Wh? Solve few integrl problems here: ) 5 d b) 4 c) 4 d d) 3 t dt rctn( ) d e) d f) 4 6 sec ( ) tn ( ) d However, these functions bove cn be relted to new set of functions clled hperbolic functions Therefore, we define: e e e e (4) sinh( ), cosh( ), sinh( ) e e tnh( ) cosh( ) e e, cosh( ) e e coth( ) sinh( ) e e These functions re clled hperbolic since prmetriztion : cosh( t), sinh( t) describes the right brnch of the unit hperbol: Therefore, we hve : (5) cosh ( ) sinh ( ) Other identities, similr with the trigonometric identities, cn be derived 8

29 For emple: (6) d sinh( ) d cosh( ), however: (7) d cosh( ) d sinh( ) sinh : R Ris one to one (invertible) (plot or tr to prove injective directl) Also, the restriction cosh :, [, ) is one to one (invertible) (plot) We cn esil find tht (give this s n eercise in clss): sinh : R R is sinh ( ) ln Also, cosh :[, ) [, ) is cosh ( ) ln We lso hve tht tnh :[,) R is tnh ( ) ln We cn find the derivtive of these functions either directl or using the inverse function theorem to find the prticulr form of the formuls () (3) A generliztion to get () (3) follows b chnge of vribles in the corresponding integrls Homework: Problems,8,6,8,5,4,4,5 t the end of section Preprtion for test: All problems t the end of chpter, strting with the true nd flse tpe questions 9

AQA Further Pure 2. Hyperbolic Functions. Section 2: The inverse hyperbolic functions

AQA Further Pure 2. Hyperbolic Functions. Section 2: The inverse hyperbolic functions Hperbolic Functions Section : The inverse hperbolic functions Notes nd Emples These notes contin subsections on The inverse hperbolic functions Integrtion using the inverse hperbolic functions Logrithmic

More information

Topic 1 Notes Jeremy Orloff

Topic 1 Notes Jeremy Orloff Topic 1 Notes Jerem Orloff 1 Introduction to differentil equtions 1.1 Gols 1. Know the definition of differentil eqution. 2. Know our first nd second most importnt equtions nd their solutions. 3. Be ble

More information

approaches as n becomes larger and larger. Since e > 1, the graph of the natural exponential function is as below

approaches as n becomes larger and larger. Since e > 1, the graph of the natural exponential function is as below . Eponentil nd rithmic functions.1 Eponentil Functions A function of the form f() =, > 0, 1 is clled n eponentil function. Its domin is the set of ll rel f ( 1) numbers. For n eponentil function f we hve.

More information

Logarithms. Logarithm is another word for an index or power. POWER. 2 is the power to which the base 10 must be raised to give 100.

Logarithms. Logarithm is another word for an index or power. POWER. 2 is the power to which the base 10 must be raised to give 100. Logrithms. Logrithm is nother word for n inde or power. THIS IS A POWER STATEMENT BASE POWER FOR EXAMPLE : We lred know tht; = NUMBER 10² = 100 This is the POWER Sttement OR 2 is the power to which the

More information

Chapter 3 Exponential and Logarithmic Functions Section 3.1

Chapter 3 Exponential and Logarithmic Functions Section 3.1 Chpter 3 Eponentil nd Logrithmic Functions Section 3. EXPONENTIAL FUNCTIONS AND THEIR GRAPHS Eponentil Functions Eponentil functions re non-lgebric functions. The re clled trnscendentl functions. The eponentil

More information

13.3. The Area Bounded by a Curve. Introduction. Prerequisites. Learning Outcomes

13.3. The Area Bounded by a Curve. Introduction. Prerequisites. Learning Outcomes The Are Bounded b Curve 3.3 Introduction One of the importnt pplictions of integrtion is to find the re bounded b curve. Often such n re cn hve phsicl significnce like the work done b motor, or the distnce

More information

Math 113 Exam 2 Practice

Math 113 Exam 2 Practice Mth Em Prctice Februry, 8 Em will cover sections 6.5, 7.-7.5 nd 7.8. This sheet hs three sections. The first section will remind you bout techniques nd formuls tht you should know. The second gives number

More information

1 Part II: Numerical Integration

1 Part II: Numerical Integration Mth 4 Lb 1 Prt II: Numericl Integrtion This section includes severl techniques for getting pproimte numericl vlues for definite integrls without using ntiderivtives. Mthemticll, ect nswers re preferble

More information

3.1 Exponential Functions and Their Graphs

3.1 Exponential Functions and Their Graphs . Eponentil Functions nd Their Grphs Sllbus Objective: 9. The student will sketch the grph of eponentil, logistic, or logrithmic function. 9. The student will evlute eponentil or logrithmic epressions.

More information

Calculus AB. For a function f(x), the derivative would be f '(

Calculus AB. For a function f(x), the derivative would be f '( lculus AB Derivtive Formuls Derivtive Nottion: For function f(), the derivtive would e f '( ) Leiniz's Nottion: For the derivtive of y in terms of, we write d For the second derivtive using Leiniz's Nottion:

More information

50. Use symmetry to evaluate xx D is the region bounded by the square with vertices 5, Prove Property 11. y y CAS

50. Use symmetry to evaluate xx D is the region bounded by the square with vertices 5, Prove Property 11. y y CAS 68 CHAPTE MULTIPLE INTEGALS 46. e da, 49. Evlute tn 3 4 da, where,. [Hint: Eploit the fct tht is the disk with center the origin nd rdius is smmetric with respect to both es.] 5. Use smmetr to evlute 3

More information

ARITHMETIC OPERATIONS. The real numbers have the following properties: a b c ab ac

ARITHMETIC OPERATIONS. The real numbers have the following properties: a b c ab ac REVIEW OF ALGEBRA Here we review the bsic rules nd procedures of lgebr tht you need to know in order to be successful in clculus. ARITHMETIC OPERATIONS The rel numbers hve the following properties: b b

More information

Definition of Continuity: The function f(x) is continuous at x = a if f(a) exists and lim

Definition of Continuity: The function f(x) is continuous at x = a if f(a) exists and lim Mth 9 Course Summry/Study Guide Fll, 2005 [1] Limits Definition of Limit: We sy tht L is the limit of f(x) s x pproches if f(x) gets closer nd closer to L s x gets closer nd closer to. We write lim f(x)

More information

Properties of Integrals, Indefinite Integrals. Goals: Definition of the Definite Integral Integral Calculations using Antiderivatives

Properties of Integrals, Indefinite Integrals. Goals: Definition of the Definite Integral Integral Calculations using Antiderivatives Block #6: Properties of Integrls, Indefinite Integrls Gols: Definition of the Definite Integrl Integrl Clcultions using Antiderivtives Properties of Integrls The Indefinite Integrl 1 Riemnn Sums - 1 Riemnn

More information

Topics Covered AP Calculus AB

Topics Covered AP Calculus AB Topics Covered AP Clculus AB ) Elementry Functions ) Properties of Functions i) A function f is defined s set of ll ordered pirs (, y), such tht for ech element, there corresponds ectly one element y.

More information

MATH 144: Business Calculus Final Review

MATH 144: Business Calculus Final Review MATH 144: Business Clculus Finl Review 1 Skills 1. Clculte severl limits. 2. Find verticl nd horizontl symptotes for given rtionl function. 3. Clculte derivtive by definition. 4. Clculte severl derivtives

More information

ES.182A Topic 32 Notes Jeremy Orloff

ES.182A Topic 32 Notes Jeremy Orloff ES.8A Topic 3 Notes Jerem Orloff 3 Polr coordintes nd double integrls 3. Polr Coordintes (, ) = (r cos(θ), r sin(θ)) r θ Stndrd,, r, θ tringle Polr coordintes re just stndrd trigonometric reltions. In

More information

Main topics for the Second Midterm

Main topics for the Second Midterm Min topics for the Second Midterm The Midterm will cover Sections 5.4-5.9, Sections 6.1-6.3, nd Sections 7.1-7.7 (essentilly ll of the mteril covered in clss from the First Midterm). Be sure to know the

More information

Chapter 6 Techniques of Integration

Chapter 6 Techniques of Integration MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi Chpter 6 Techniques of Integrtion Recll: Some importnt integrls tht we hve lernt so fr. Tle of Integrls n+ n d = + C n + e d = e + C ( n ) d = ln

More information

Anti-derivatives/Indefinite Integrals of Basic Functions

Anti-derivatives/Indefinite Integrals of Basic Functions Anti-derivtives/Indefinite Integrls of Bsic Functions Power Rule: In prticulr, this mens tht x n+ x n n + + C, dx = ln x + C, if n if n = x 0 dx = dx = dx = x + C nd x (lthough you won t use the second

More information

Chapter 5 1. = on [ 1, 2] 1. Let gx ( ) e x. . The derivative of g is g ( x) e 1

Chapter 5 1. = on [ 1, 2] 1. Let gx ( ) e x. . The derivative of g is g ( x) e 1 Chpter 5. Let g ( e. on [, ]. The derivtive of g is g ( e ( Write the slope intercept form of the eqution of the tngent line to the grph of g t. (b Determine the -coordinte of ech criticl vlue of g. Show

More information

Goals: Determine how to calculate the area described by a function. Define the definite integral. Explore the relationship between the definite

Goals: Determine how to calculate the area described by a function. Define the definite integral. Explore the relationship between the definite Unit #8 : The Integrl Gols: Determine how to clculte the re described by function. Define the definite integrl. Eplore the reltionship between the definite integrl nd re. Eplore wys to estimte the definite

More information

Goals: Determine how to calculate the area described by a function. Define the definite integral. Explore the relationship between the definite

Goals: Determine how to calculate the area described by a function. Define the definite integral. Explore the relationship between the definite Unit #8 : The Integrl Gols: Determine how to clculte the re described by function. Define the definite integrl. Eplore the reltionship between the definite integrl nd re. Eplore wys to estimte the definite

More information

Unit 1 Exponentials and Logarithms

Unit 1 Exponentials and Logarithms HARTFIELD PRECALCULUS UNIT 1 NOTES PAGE 1 Unit 1 Eponentils nd Logrithms (2) Eponentil Functions (3) The number e (4) Logrithms (5) Specil Logrithms (7) Chnge of Bse Formul (8) Logrithmic Functions (10)

More information

Optimization Lecture 1 Review of Differential Calculus for Functions of Single Variable.

Optimization Lecture 1 Review of Differential Calculus for Functions of Single Variable. Optimiztion Lecture 1 Review of Differentil Clculus for Functions of Single Vrible http://users.encs.concordi.c/~luisrod, Jnury 14 Outline Optimiztion Problems Rel Numbers nd Rel Vectors Open, Closed nd

More information

13.3. The Area Bounded by a Curve. Introduction. Prerequisites. Learning Outcomes

13.3. The Area Bounded by a Curve. Introduction. Prerequisites. Learning Outcomes The Are Bounded b Curve 3.3 Introduction One of the importnt pplictions of integrtion is to find the re bounded b curve. Often such n re cn hve phsicl significnce like the work done b motor, or the distnce

More information

5.7 Improper Integrals

5.7 Improper Integrals 458 pplictions of definite integrls 5.7 Improper Integrls In Section 5.4, we computed the work required to lift pylod of mss m from the surfce of moon of mss nd rdius R to height H bove the surfce of the

More information

Operations with Polynomials

Operations with Polynomials 38 Chpter P Prerequisites P.4 Opertions with Polynomils Wht you should lern: How to identify the leding coefficients nd degrees of polynomils How to dd nd subtrct polynomils How to multiply polynomils

More information

DERIVATIVES NOTES HARRIS MATH CAMP Introduction

DERIVATIVES NOTES HARRIS MATH CAMP Introduction f DERIVATIVES NOTES HARRIS MATH CAMP 208. Introduction Reding: Section 2. The derivtive of function t point is the slope of the tngent line to the function t tht point. Wht does this men, nd how do we

More information

and that at t = 0 the object is at position 5. Find the position of the object at t = 2.

and that at t = 0 the object is at position 5. Find the position of the object at t = 2. 7.2 The Fundmentl Theorem of Clculus 49 re mny, mny problems tht pper much different on the surfce but tht turn out to be the sme s these problems, in the sense tht when we try to pproimte solutions we

More information

Chapter 8: Methods of Integration

Chapter 8: Methods of Integration Chpter 8: Methods of Integrtion Bsic Integrls 8. Note: We hve the following list of Bsic Integrls p p+ + c, for p sec tn + c p + ln + c sec tn sec + c e e + c tn ln sec + c ln + c sec ln sec + tn + c ln

More information

Loudoun Valley High School Calculus Summertime Fun Packet

Loudoun Valley High School Calculus Summertime Fun Packet Loudoun Vlley High School Clculus Summertime Fun Pcket We HIGHLY recommend tht you go through this pcket nd mke sure tht you know how to do everything in it. Prctice the problems tht you do NOT remember!

More information

Before we can begin Ch. 3 on Radicals, we need to be familiar with perfect squares, cubes, etc. Try and do as many as you can without a calculator!!!

Before we can begin Ch. 3 on Radicals, we need to be familiar with perfect squares, cubes, etc. Try and do as many as you can without a calculator!!! Nme: Algebr II Honors Pre-Chpter Homework Before we cn begin Ch on Rdicls, we need to be fmilir with perfect squres, cubes, etc Try nd do s mny s you cn without clcultor!!! n The nth root of n n Be ble

More information

Polynomial Approximations for the Natural Logarithm and Arctangent Functions. Math 230

Polynomial Approximations for the Natural Logarithm and Arctangent Functions. Math 230 Polynomil Approimtions for the Nturl Logrithm nd Arctngent Functions Mth 23 You recll from first semester clculus how one cn use the derivtive to find n eqution for the tngent line to function t given

More information

Calculus 2: Integration. Differentiation. Integration

Calculus 2: Integration. Differentiation. Integration Clculus 2: Integrtion The reverse process to differentition is known s integrtion. Differentition f() f () Integrtion As it is the opposite of finding the derivtive, the function obtined b integrtion is

More information

Calculus Module C21. Areas by Integration. Copyright This publication The Northern Alberta Institute of Technology All Rights Reserved.

Calculus Module C21. Areas by Integration. Copyright This publication The Northern Alberta Institute of Technology All Rights Reserved. Clculus Module C Ares Integrtion Copright This puliction The Northern Alert Institute of Technolog 7. All Rights Reserved. LAST REVISED Mrch, 9 Introduction to Ares Integrtion Sttement of Prerequisite

More information

(i) b b. (ii) (iii) (vi) b. P a g e Exponential Functions 1. Properties of Exponents: Ex1. Solve the following equation

(i) b b. (ii) (iii) (vi) b. P a g e Exponential Functions 1. Properties of Exponents: Ex1. Solve the following equation P g e 30 4.2 Eponentil Functions 1. Properties of Eponents: (i) (iii) (iv) (v) (vi) 1 If 1, 0 1, nd 1, then E1. Solve the following eqution 4 3. 1 2 89 8(2 ) 7 Definition: The eponentil function with se

More information

Main topics for the First Midterm

Main topics for the First Midterm Min topics for the First Midterm The Midterm will cover Section 1.8, Chpters 2-3, Sections 4.1-4.8, nd Sections 5.1-5.3 (essentilly ll of the mteril covered in clss). Be sure to know the results of the

More information

AB Calculus Review Sheet

AB Calculus Review Sheet AB Clculus Review Sheet Legend: A Preclculus, B Limits, C Differentil Clculus, D Applictions of Differentil Clculus, E Integrl Clculus, F Applictions of Integrl Clculus, G Prticle Motion nd Rtes This is

More information

( ) as a fraction. Determine location of the highest

( ) as a fraction. Determine location of the highest AB Clculus Exm Review Sheet - Solutions A. Preclculus Type prolems A1 A2 A3 A4 A5 A6 A7 This is wht you think of doing Find the zeros of f ( x). Set function equl to 0. Fctor or use qudrtic eqution if

More information

Math 113 Exam 1-Review

Math 113 Exam 1-Review Mth 113 Exm 1-Review September 26, 2016 Exm 1 covers 6.1-7.3 in the textbook. It is dvisble to lso review the mteril from 5.3 nd 5.5 s this will be helpful in solving some of the problems. 6.1 Are Between

More information

( ) where f ( x ) is a. AB Calculus Exam Review Sheet. A. Precalculus Type problems. Find the zeros of f ( x).

( ) where f ( x ) is a. AB Calculus Exam Review Sheet. A. Precalculus Type problems. Find the zeros of f ( x). AB Clculus Exm Review Sheet A. Preclculus Type prolems A1 Find the zeros of f ( x). This is wht you think of doing A2 A3 Find the intersection of f ( x) nd g( x). Show tht f ( x) is even. A4 Show tht f

More information

HOMEWORK SOLUTIONS MATH 1910 Sections 7.9, 8.1 Fall 2016

HOMEWORK SOLUTIONS MATH 1910 Sections 7.9, 8.1 Fall 2016 HOMEWORK SOLUTIONS MATH 9 Sections 7.9, 8. Fll 6 Problem 7.9.33 Show tht for ny constnts M,, nd, the function yt) = )) t ) M + tnh stisfies the logistic eqution: y SOLUTION. Let Then nd Finlly, y = y M

More information

Calculus - Activity 1 Rate of change of a function at a point.

Calculus - Activity 1 Rate of change of a function at a point. Nme: Clss: p 77 Mths Helper Plus Resource Set. Copright 00 Bruce A. Vughn, Techers Choice Softwre Clculus - Activit Rte of chnge of function t point. ) Strt Mths Helper Plus, then lod the file: Clculus

More information

Chapter 6 Notes, Larson/Hostetler 3e

Chapter 6 Notes, Larson/Hostetler 3e Contents 6. Antiderivtives nd the Rules of Integrtion.......................... 6. Are nd the Definite Integrl.................................. 6.. Are............................................ 6. Reimnn

More information

First midterm topics Second midterm topics End of quarter topics. Math 3B Review. Steve. 18 March 2009

First midterm topics Second midterm topics End of quarter topics. Math 3B Review. Steve. 18 March 2009 Mth 3B Review Steve 18 Mrch 2009 About the finl Fridy Mrch 20, 3pm-6pm, Lkretz 110 No notes, no book, no clcultor Ten questions Five review questions (Chpters 6,7,8) Five new questions (Chpters 9,10) No

More information

f(a+h) f(a) x a h 0. This is the rate at which

f(a+h) f(a) x a h 0. This is the rate at which M408S Concept Inventory smple nswers These questions re open-ended, nd re intended to cover the min topics tht we lerned in M408S. These re not crnk-out-n-nswer problems! (There re plenty of those in the

More information

Appendix 3, Rises and runs, slopes and sums: tools from calculus

Appendix 3, Rises and runs, slopes and sums: tools from calculus Appendi 3, Rises nd runs, slopes nd sums: tools from clculus Sometimes we will wnt to eplore how quntity chnges s condition is vried. Clculus ws invented to do just this. We certinly do not need the full

More information

MA Lesson 21 Notes

MA Lesson 21 Notes MA 000 Lesson 1 Notes ( 5) How would person solve n eqution with vrible in n eponent, such s 9? (We cnnot re-write this eqution esil with the sme bse.) A nottion ws developed so tht equtions such s this

More information

FUNCTIONS: Grade 11. or y = ax 2 +bx + c or y = a(x- x1)(x- x2) a y

FUNCTIONS: Grade 11. or y = ax 2 +bx + c or y = a(x- x1)(x- x2) a y FUNCTIONS: Grde 11 The prbol: ( p) q or = +b + c or = (- 1)(- ) The hperbol: p q The eponentil function: b p q Importnt fetures: -intercept : Let = 0 -intercept : Let = 0 Turning points (Where pplicble)

More information

The Trapezoidal Rule

The Trapezoidal Rule _.qd // : PM Pge 9 SECTION. Numericl Integrtion 9 f Section. The re of the region cn e pproimted using four trpezoids. Figure. = f( ) f( ) n The re of the first trpezoid is f f n. Figure. = Numericl Integrtion

More information

SUMMER KNOWHOW STUDY AND LEARNING CENTRE

SUMMER KNOWHOW STUDY AND LEARNING CENTRE SUMMER KNOWHOW STUDY AND LEARNING CENTRE Indices & Logrithms 2 Contents Indices.2 Frctionl Indices.4 Logrithms 6 Exponentil equtions. Simplifying Surds 13 Opertions on Surds..16 Scientific Nottion..18

More information

Section 4: Integration ECO4112F 2011

Section 4: Integration ECO4112F 2011 Reding: Ching Chpter Section : Integrtion ECOF Note: These notes do not fully cover the mteril in Ching, ut re ment to supplement your reding in Ching. Thus fr the optimistion you hve covered hs een sttic

More information

The First Fundamental Theorem of Calculus. If f(x) is continuous on [a, b] and F (x) is any antiderivative. f(x) dx = F (b) F (a).

The First Fundamental Theorem of Calculus. If f(x) is continuous on [a, b] and F (x) is any antiderivative. f(x) dx = F (b) F (a). The Fundmentl Theorems of Clculus Mth 4, Section 0, Spring 009 We now know enough bout definite integrls to give precise formultions of the Fundmentl Theorems of Clculus. We will lso look t some bsic emples

More information

Math Calculus with Analytic Geometry II

Math Calculus with Analytic Geometry II orem of definite Mth 5.0 with Anlytic Geometry II Jnury 4, 0 orem of definite If < b then b f (x) dx = ( under f bove x-xis) ( bove f under x-xis) Exmple 8 0 3 9 x dx = π 3 4 = 9π 4 orem of definite Problem

More information

Logarithmic Functions

Logarithmic Functions Logrithmic Functions Definition: Let > 0,. Then log is the number to which you rise to get. Logrithms re in essence eponents. Their domins re powers of the bse nd their rnges re the eponents needed to

More information

Chapter 0. What is the Lebesgue integral about?

Chapter 0. What is the Lebesgue integral about? Chpter 0. Wht is the Lebesgue integrl bout? The pln is to hve tutoril sheet ech week, most often on Fridy, (to be done during the clss) where you will try to get used to the ides introduced in the previous

More information

Review Exercises for Chapter 4

Review Exercises for Chapter 4 _R.qd // : PM Pge CHAPTER Integrtion Review Eercises for Chpter In Eercises nd, use the grph of to sketch grph of f. To print n enlrged cop of the grph, go to the wesite www.mthgrphs.com... In Eercises

More information

A-Level Mathematics Transition Task (compulsory for all maths students and all further maths student)

A-Level Mathematics Transition Task (compulsory for all maths students and all further maths student) A-Level Mthemtics Trnsition Tsk (compulsory for ll mths students nd ll further mths student) Due: st Lesson of the yer. Length: - hours work (depending on prior knowledge) This trnsition tsk provides revision

More information

Math 153: Lecture Notes For Chapter 5

Math 153: Lecture Notes For Chapter 5 Mth 5: Lecture Notes For Chpter 5 Section 5.: Eponentil Function f()= Emple : grph f ) = ( if = f() 0 - - - - - - Emple : Grph ) f ( ) = b) g ( ) = c) h ( ) = ( ) f() g() h() 0 0 0 - - - - - - - - - -

More information

f(x) dx, If one of these two conditions is not met, we call the integral improper. Our usual definition for the value for the definite integral

f(x) dx, If one of these two conditions is not met, we call the integral improper. Our usual definition for the value for the definite integral Improper Integrls Every time tht we hve evluted definite integrl such s f(x) dx, we hve mde two implicit ssumptions bout the integrl:. The intervl [, b] is finite, nd. f(x) is continuous on [, b]. If one

More information

ACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS 1 MATH00030 SEMESTER /2019

ACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS 1 MATH00030 SEMESTER /2019 ACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS MATH00030 SEMESTER 208/209 DR. ANTHONY BROWN 7.. Introduction to Integrtion. 7. Integrl Clculus As ws the cse with the chpter on differentil

More information

Math& 152 Section Integration by Parts

Math& 152 Section Integration by Parts Mth& 5 Section 7. - Integrtion by Prts Integrtion by prts is rule tht trnsforms the integrl of the product of two functions into other (idelly simpler) integrls. Recll from Clculus I tht given two differentible

More information

5.2 Exponent Properties Involving Quotients

5.2 Exponent Properties Involving Quotients 5. Eponent Properties Involving Quotients Lerning Objectives Use the quotient of powers property. Use the power of quotient property. Simplify epressions involving quotient properties of eponents. Use

More information

Chapters 4 & 5 Integrals & Applications

Chapters 4 & 5 Integrals & Applications Contents Chpters 4 & 5 Integrls & Applictions Motivtion to Chpters 4 & 5 2 Chpter 4 3 Ares nd Distnces 3. VIDEO - Ares Under Functions............................................ 3.2 VIDEO - Applictions

More information

6.2 CONCEPTS FOR ADVANCED MATHEMATICS, C2 (4752) AS

6.2 CONCEPTS FOR ADVANCED MATHEMATICS, C2 (4752) AS 6. CONCEPTS FOR ADVANCED MATHEMATICS, C (475) AS Objectives To introduce students to number of topics which re fundmentl to the dvnced study of mthemtics. Assessment Emintion (7 mrks) 1 hour 30 minutes.

More information

Mathematics Extension 1

Mathematics Extension 1 04 Bored of Studies Tril Emintions Mthemtics Etension Written by Crrotsticks & Trebl. Generl Instructions Totl Mrks 70 Reding time 5 minutes. Working time hours. Write using blck or blue pen. Blck pen

More information

AP Calculus Multiple Choice: BC Edition Solutions

AP Calculus Multiple Choice: BC Edition Solutions AP Clculus Multiple Choice: BC Edition Solutions J. Slon Mrch 8, 04 ) 0 dx ( x) is A) B) C) D) E) Divergent This function inside the integrl hs verticl symptotes t x =, nd the integrl bounds contin this

More information

The semester B examination for Algebra 2 will consist of two parts. Part 1 will be selected response. Part 2 will be short answer. is a real number.

The semester B examination for Algebra 2 will consist of two parts. Part 1 will be selected response. Part 2 will be short answer. is a real number. The semester B emintion for Algebr will consist of two prts. Prt will be selected response. Prt will be short nswer. Students m use clcultor. If clcultor is used to find points on grph, the pproprite clcultor

More information

MAT187H1F Lec0101 Burbulla

MAT187H1F Lec0101 Burbulla Chpter 6 Lecture Notes Review nd Two New Sections Sprint 17 Net Distnce nd Totl Distnce Trvelled Suppose s is the position of prticle t time t for t [, b]. Then v dt = s (t) dt = s(b) s(). s(b) s() is

More information

than 1. It means in particular that the function is decreasing and approaching the x-

than 1. It means in particular that the function is decreasing and approaching the x- 6 Preclculus Review Grph the functions ) (/) ) log y = b y = Solution () The function y = is n eponentil function with bse smller thn It mens in prticulr tht the function is decresing nd pproching the

More information

Advanced Functions Page 1 of 3 Investigating Exponential Functions y= b x

Advanced Functions Page 1 of 3 Investigating Exponential Functions y= b x Advnced Functions Pge of Investigting Eponentil Functions = b Emple : Write n Eqution to Fit Dt Write n eqution to fit the dt in the tble of vlues. 0 4 4 Properties of the Eponentil Function =b () The

More information

TO: Next Year s AP Calculus Students

TO: Next Year s AP Calculus Students TO: Net Yer s AP Clculus Students As you probbly know, the students who tke AP Clculus AB nd pss the Advnced Plcement Test will plce out of one semester of college Clculus; those who tke AP Clculus BC

More information

A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H. Thomas Shores Department of Mathematics University of Nebraska Spring 2007

A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H. Thomas Shores Department of Mathematics University of Nebraska Spring 2007 A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H Thoms Shores Deprtment of Mthemtics University of Nebrsk Spring 2007 Contents Rtes of Chnge nd Derivtives 1 Dierentils 4 Are nd Integrls 5 Multivrite Clculus

More information

The semester B examination for Algebra 2 will consist of two parts. Part 1 will be selected response. Part 2 will be short answer.

The semester B examination for Algebra 2 will consist of two parts. Part 1 will be selected response. Part 2 will be short answer. ALGEBRA B Semester Em Review The semester B emintion for Algebr will consist of two prts. Prt will be selected response. Prt will be short nswer. Students m use clcultor. If clcultor is used to find points

More information

How can we approximate the area of a region in the plane? What is an interpretation of the area under the graph of a velocity function?

How can we approximate the area of a region in the plane? What is an interpretation of the area under the graph of a velocity function? Mth 125 Summry Here re some thoughts I ws hving while considering wht to put on the first midterm. The core of your studying should be the ssigned homework problems: mke sure you relly understnd those

More information

Interpreting Integrals and the Fundamental Theorem

Interpreting Integrals and the Fundamental Theorem Interpreting Integrls nd the Fundmentl Theorem Tody, we go further in interpreting the mening of the definite integrl. Using Units to Aid Interprettion We lredy know tht if f(t) is the rte of chnge of

More information

5.5 The Substitution Rule

5.5 The Substitution Rule 5.5 The Substitution Rule Given the usefulness of the Fundmentl Theorem, we wnt some helpful methods for finding ntiderivtives. At the moment, if n nti-derivtive is not esily recognizble, then we re in

More information

Sections 1.3, 7.1, and 9.2: Properties of Exponents and Radical Notation

Sections 1.3, 7.1, and 9.2: Properties of Exponents and Radical Notation Sections., 7., nd 9.: Properties of Eponents nd Rdicl Nottion Let p nd q be rtionl numbers. For ll rel numbers nd b for which the epressions re rel numbers, the following properties hold. i = + p q p q

More information

A. Limits - L Hopital s Rule. x c. x c. f x. g x. x c 0 6 = 1 6. D. -1 E. nonexistent. ln ( x 1 ) 1 x 2 1. ( x 2 1) 2. 2x x 1.

A. Limits - L Hopital s Rule. x c. x c. f x. g x. x c 0 6 = 1 6. D. -1 E. nonexistent. ln ( x 1 ) 1 x 2 1. ( x 2 1) 2. 2x x 1. A. Limits - L Hopitl s Rule Wht you re finding: L Hopitl s Rule is used to find limits of the form f ( ) lim where lim f or lim f limg. c g = c limg( ) = c = c = c How to find it: Try nd find limits by

More information

Answers for Ch. 5 Review: The Integral

Answers for Ch. 5 Review: The Integral Answers for Ch. 5 Review: The Integrl. So, I m to sketch grph of. I sketch things with softwre. Four subregions, Δ ½. The region R is onl in the first qudrnt, nd the -intercepts re nd, so the verticl seprtors

More information

We know that if f is a continuous nonnegative function on the interval [a, b], then b

We know that if f is a continuous nonnegative function on the interval [a, b], then b 1 Ares Between Curves c 22 Donld Kreider nd Dwight Lhr We know tht if f is continuous nonnegtive function on the intervl [, b], then f(x) dx is the re under the grph of f nd bove the intervl. We re going

More information

Unit #9 : Definite Integral Properties; Fundamental Theorem of Calculus

Unit #9 : Definite Integral Properties; Fundamental Theorem of Calculus Unit #9 : Definite Integrl Properties; Fundmentl Theorem of Clculus Gols: Identify properties of definite integrls Define odd nd even functions, nd reltionship to integrl vlues Introduce the Fundmentl

More information

x 2 1 dx x 3 dx = ln(x) + 2e u du = 2e u + C = 2e x + C 2x dx = arcsin x + 1 x 1 x du = 2 u + C (t + 2) 50 dt x 2 4 dx

x 2 1 dx x 3 dx = ln(x) + 2e u du = 2e u + C = 2e x + C 2x dx = arcsin x + 1 x 1 x du = 2 u + C (t + 2) 50 dt x 2 4 dx . Compute the following indefinite integrls: ) sin(5 + )d b) c) d e d d) + d Solutions: ) After substituting u 5 +, we get: sin(5 + )d sin(u)du cos(u) + C cos(5 + ) + C b) We hve: d d ln() + + C c) Substitute

More information

Improper Integrals. Type I Improper Integrals How do we evaluate an integral such as

Improper Integrals. Type I Improper Integrals How do we evaluate an integral such as Improper Integrls Two different types of integrls cn qulify s improper. The first type of improper integrl (which we will refer to s Type I) involves evluting n integrl over n infinite region. In the grph

More information

x ) dx dx x sec x over the interval (, ).

x ) dx dx x sec x over the interval (, ). Curve on 6 For -, () Evlute the integrl, n (b) check your nswer by ifferentiting. ( ). ( ). ( ).. 6. sin cos 7. sec csccot 8. sec (sec tn ) 9. sin csc. Evlute the integrl sin by multiplying the numertor

More information

Integration by Substitution. Pattern Recognition

Integration by Substitution. Pattern Recognition SECTION Integrtion b Substitution 9 Section Integrtion b Substitution Use pttern recognition to find n indefinite integrl Use chnge of vribles to find n indefinite integrl Use the Generl Power Rule for

More information

Obj: SWBAT Recall the many important types and properties of functions

Obj: SWBAT Recall the many important types and properties of functions Obj: SWBAT Recll the mny importnt types nd properties of functions Functions Domin nd Rnge Function Nottion Trnsformtion of Functions Combintions/Composition of Functions One-to-One nd Inverse Functions

More information

Unit 5. Integration techniques

Unit 5. Integration techniques 18.01 EXERCISES Unit 5. Integrtion techniques 5A. Inverse trigonometric functions; Hyperbolic functions 5A-1 Evlute ) tn 1 3 b) sin 1 ( 3/) c) If θ = tn 1 5, then evlute sin θ, cos θ, cot θ, csc θ, nd

More information

SCHOOL OF ENGINEERING & BUILT ENVIRONMENT. Mathematics. Basic Algebra

SCHOOL OF ENGINEERING & BUILT ENVIRONMENT. Mathematics. Basic Algebra SCHOOL OF ENGINEERING & BUILT ENVIRONMENT Mthemtics Bsic Algebr Opertions nd Epressions Common Mistkes Division of Algebric Epressions Eponentil Functions nd Logrithms Opertions nd their Inverses Mnipulting

More information

ES.182A Topic 30 Notes Jeremy Orloff

ES.182A Topic 30 Notes Jeremy Orloff ES82A opic 3 Notes Jerem Orloff 3 Non-independent vribles: chin rule Recll the chin rule: If w = f, ; nd = r, t, = r, t then = + r t r t r t = + t t t r nfortuntel, sometimes there re more complicted reltions

More information

Mathematics Number: Logarithms

Mathematics Number: Logarithms plce of mind F A C U L T Y O F E D U C A T I O N Deprtment of Curriculum nd Pedgogy Mthemtics Numer: Logrithms Science nd Mthemtics Eduction Reserch Group Supported y UBC Teching nd Lerning Enhncement

More information

4.6 Numerical Integration

4.6 Numerical Integration .6 Numericl Integrtion 5.6 Numericl Integrtion Approimte definite integrl using the Trpezoidl Rule. Approimte definite integrl using Simpson s Rule. Anlze the pproimte errors in the Trpezoidl Rule nd Simpson

More information

Quotient Rule: am a n = am n (a 0) Negative Exponents: a n = 1 (a 0) an Power Rules: (a m ) n = a m n (ab) m = a m b m

Quotient Rule: am a n = am n (a 0) Negative Exponents: a n = 1 (a 0) an Power Rules: (a m ) n = a m n (ab) m = a m b m Formuls nd Concepts MAT 099: Intermedite Algebr repring for Tests: The formuls nd concepts here m not be inclusive. You should first tke our prctice test with no notes or help to see wht mteril ou re comfortble

More information

Conservation Law. Chapter Goal. 5.2 Theory

Conservation Law. Chapter Goal. 5.2 Theory Chpter 5 Conservtion Lw 5.1 Gol Our long term gol is to understnd how mny mthemticl models re derived. We study how certin quntity chnges with time in given region (sptil domin). We first derive the very

More information

Improper Integrals, and Differential Equations

Improper Integrals, and Differential Equations Improper Integrls, nd Differentil Equtions October 22, 204 5.3 Improper Integrls Previously, we discussed how integrls correspond to res. More specificlly, we sid tht for function f(x), the region creted

More information

7. Indefinite Integrals

7. Indefinite Integrals 7. Indefinite Integrls These lecture notes present my interprettion of Ruth Lwrence s lecture notes (in Herew) 7. Prolem sttement By the fundmentl theorem of clculus, to clculte n integrl we need to find

More information

different methods (left endpoint, right endpoint, midpoint, trapezoid, Simpson s).

different methods (left endpoint, right endpoint, midpoint, trapezoid, Simpson s). Mth 1A with Professor Stnkov Worksheet, Discussion #41; Wednesdy, 12/6/217 GSI nme: Roy Zho Problems 1. Write the integrl 3 dx s limit of Riemnn sums. Write it using 2 intervls using the 1 x different

More information

Techniques of Integration

Techniques of Integration 7 Techniques of Integrtion Shown is photogrph of Omeg Centuri, which contins severl million strs nd is the lrgest globulr cluster in our gl. Astronomers use stellr stereogrph to determine the ctul densit

More information

Section - 2 MORE PROPERTIES

Section - 2 MORE PROPERTIES LOCUS Section - MORE PROPERTES n section -, we delt with some sic properties tht definite integrls stisf. This section continues with the development of some more properties tht re not so trivil, nd, when

More information