Antiderivatives and Indefinite Integration

Transcription

1 8 CHAPTER Integrtion Section EXPLORATION Finding Antiderivtives For ech derivtive, descrie the originl function F F F c F d F e F f F cos Wht strteg did ou use to find F? Antiderivtives nd Indefinite Integrtion Write the generl solution of differentil eqution Use indefinite integrl nottion for ntiderivtives Use sic integrtion rules to find ntiderivtives Find prticulr solution of differentil eqution Antiderivtives Suppose ou were sked to find function F whose derivtive is f From our knowledge of derivtives, ou would prol s tht F ecuse d d The function F is n ntiderivtive of f Definition of n Antiderivtive A function F is n ntiderivtive of f on n intervl I if F f for ll in I Video Note tht F is clled n ntiderivtive of f, rther thn the ntiderivtive of f To see wh, oserve tht F, F, nd F 97 re ll ntiderivtives of f In fct, for n constnt C, the function given F C is n ntiderivtive of f THEOREM Representtion of Antiderivtives If F is n ntiderivtive of f on n intervl I, then G is n ntiderivtive of f on the intervl I if nd onl if G is of the form G F C, for ll in I where C is constnt Proof The proof of Theorem in one direction is strightforwrd Tht is, if G F C, F f, nd C is constnt, then G d d F C F f To prove this theorem in the other direction, ssume tht G is n ntiderivtive of f Define function H such tht H G( F If H is not constnt on the intervl I, there must eist nd < in the intervl such tht H H Moreover, ecuse H is differentile on,, ou cn ppl the Men Vlue Theorem to conclude tht there eists some c in, such tht Hc H H Becuse H H, it follows tht Hc However, ecuse Gc Fc, ou know tht Hc Gc Fc, which contrdicts the fct tht Hc Consequentl, ou cn conclude tht H is constnt, C So, G F C nd it follows tht G F( C

2 SECTION Antiderivtives nd Indefinite Integrtion 9 Using Theorem, ou cn represent the entire fmil of ntiderivtives of function dding constnt to known ntiderivtive For emple, knowing tht D, ou cn represent the fmil of ll ntiderivtives of f G C Fmil of ll ntiderivtives of f ( where C is constnt The constnt C is clled the constnt of integrtion The fmil of functions represented G is the generl ntiderivtive of f, nd G( C is the generl solution of the differentil eqution G Differentil eqution A differentil eqution in nd is n eqution tht involves,, nd derivtives of For instnce, nd re emples of differentil equtions EXAMPLE Solving Differentil Eqution C = Find the generl solution of the differentil eqution C = Functions of the form Figure Editle Grph Video C = C Solution To egin, ou need to find function whose derivtive is One such function is is n ntiderivtive of Now, ou cn use Theorem to conclude tht the generl solution of the differentil eqution is C Generl solution The grphs of severl functions of the form C re shown in Figure Tr It Nottion for Antiderivtives When solving differentil eqution of the form d f d it is convenient to write it in the equivlent differentil form d f d Eplortion A The opertion of finding ll solutions of this eqution is clled ntidifferentition (or indefinite integrtion) nd is denoted n integrl sign The generl solution is denoted Vrile of integrtion Constnt of integrtion f d F C NOTE In this tet, the nottion f d F C mens tht F is n ntiderivtive of f on n intervl Integrnd The epression f d is red s the ntiderivtive of f with respect to So, the differentil d serves to identif s the vrile of integrtion The term indefinite integrl is snonm for ntiderivtive

3 CHAPTER Integrtion Bsic Integrtion Rules The inverse nture of integrtion nd differentition cn e verified sustituting F for f in the indefinite integrtion definition to otin F d F C Integrtion is the inverse of differentition Moreover, if f d F C, then d d f d f Differentition is the inverse of integrtion These two equtions llow ou to otin integrtion formuls directl from differentition formuls, s shown in the following summr Bsic Integrtion Rules Differentition Formul d C d d k k d d kf kf d d d f ± g f ± g d d n n n d sin cos d d cos sin d d d tn sec d sec sec tn d d d cot csc d csc csc cot d Integrtion Formul d C k d k C kf d kf d f ± g d f d ± g d n d n C, n cos d sin C sin d cos C sec d tn C sec tn d sec C csc d cot C n csc cot d csc C Power Rule NOTE Note tht the Power Rule for Integrtion hs the restriction tht n The evlution of d must wit until the introduction of the nturl logrithm function in Chpter

4 SECTION Antiderivtives nd Indefinite Integrtion EXAMPLE Appling the Bsic Integrtion Rules Descrie the ntiderivtives of Solution d d Constnt Multiple Rule d C C Rewrite s Power Rule n Simplif So, the ntiderivtives of re of the form C, where C is n constnt Tr It Eplortion A When indefinite integrls re evluted, strict ppliction of the sic integrtion rules tends to produce complicted constnts of integrtion For instnce, in Emple, ou could hve written d d C C However, ecuse C represents n constnt, it is oth cumersome nd unnecessr to write C s the constnt of integrtion So, is written in the simpler form, C C In Emple, note tht the generl pttern of integrtion is similr to tht of differentition Originl integrl Rewrite Integrte Simplif TECHNOLOGY Some softwre progrms, such s Derive, Mple, Mthcd, Mthemtic, nd the TI-89, re cple of performing integrtion smolicll If ou hve ccess to such smolic integrtion utilit, tr using it to evlute the indefinite integrls in Emple EXAMPLE Originl Integrl d d Rewriting Before Integrting Rewrite d d Integrte C C Simplif C C c sin d sin d cos C cos C Tr It Eplortion A Open Eplortion Video Rememer tht ou cn check our nswer to n ntidifferentition prolem differentiting For instnce, in Emple (), ou cn check tht C is the correct ntiderivtive differentiting the nswer to otin D C Use differentition to check ntiderivtive

5 CHAPTER Integrtion The sic integrtion rules listed erlier in this section llow ou to integrte n polnomil function, s shown in Emple EXAMPLE Integrting Polnomil Functions d d Integrnd is understood to e C Integrte d d d C C Integrte C The second line in the solution is usull omitted d c Integrte C C C C C Simplif Tr It Eplortion A EXAMPLE Rewriting Before Integrting d d d C C Rewrite s two frctions Rewrite with frctionl eponents Integrte Simplif C Tr It Eplortion A Eplortion B NOTE When integrting quotients, do not integrte the numertor nd denomintor seprtel This is no more vlid in integrtion thn it is in differentition For instnce, in Emple, e sure ou understnd tht d is not the sme s C d C d C EXAMPLE sin Rewriting Before Integrting cos d cos sin cos d sec tn d sec C Rewrite s product Rewrite using trigonometric identities Integrte Tr It Eplortion A

6 SECTION Antiderivtives nd Indefinite Integrtion Initil Conditions nd Prticulr Solutions The prticulr solution tht stisfies the initil condition F is F Figure Animtion C = C = C = C = C = C = C = C = F() = + C C = (, ) You hve lred seen tht the eqution f d hs mn solutions (ech differing from the others constnt) This mens tht the grphs of n two ntiderivtives of f re verticl trnsltions of ech other For emple, Figure shows the grphs of severl ntiderivtives of the form d C Generl solution for vrious integer vlues of C Ech of these ntiderivtives is solution of the differentil eqution d d In mn pplictions of integrtion, ou re given enough informtion to determine prticulr solution To do this, ou need onl know the vlue of F for one vlue of This informtion is clled n initil condition For emple, in Figure, onl one curve psses through the point (, To find this curve, ou cn use the following informtion F C F Generl solution Initil condition B using the initil condition in the generl solution, ou cn determine tht F 8 C, which implies tht C So, ou otin F EXAMPLE 7 Find the generl solution of Finding Prticulr Solution Prticulr solution F, > (, ) C = C = C = C = C = nd find the prticulr solution tht stisfies the initil condition F Solution To find the generl solution, integrte to otin F d d C F Fd Rewrite s power Integrte C = C, > Generl solution C = F() = + C C = Using the initil condition F, ou cn solve for C s follows F C C The prticulr solution tht stisfies the initil condition F is F, > Figure So, the prticulr solution, s shown in Figure, is F, > Prticulr solution Editle Grph Tr It Eplortion A

7 CHAPTER Integrtion So fr in this section ou hve een using s the vrile of integrtion In pplictions, it is often convenient to use different vrile For instnce, in the following emple involving time, the vrile of integrtion is t EXAMPLE 8 Solving Verticl Motion Prolem Height (in feet) Height of ll t time t Figure s t = t = Animtion s(t) = t + t + 8 t = t = t = t = Time (in seconds) NOTE In Emple 8, note tht the position function hs the form t A ll is thrown upwrd with n initil velocit of feet per second from n initil height of 8 feet Find the position function giving the height s s function of the time t When does the ll hit the ground? Solution Let t represent the initil time The two given initil conditions cn e written s follows s 8 s Initil height is 8 feet Initil velocit is feet per second Using feet per second per second s the ccelertion due to grvit, ou cn write st st st dt dt t C Using the initil velocit, ou otin s C, which implies tht C Net, integrting st, ou otin st stdt t dt t t C Using the initil height, ou otin s 8 C which implies tht C 8 So, the position function is st t t 8 See Figure Using the position function found in prt (), ou cn find the time tht the ll hits the ground solving the eqution st st t t 8 t t t, Becuse t must e positive, ou cn conclude tht the ll hits the ground seconds fter it ws thrown Tr It Eplortion A st gt v t s where g, v is the initil velocit, nd s is the initil height, s presented in Section Emple 8 shows how to use clculus to nlze verticl motion prolems in which the ccelertion is determined grvittionl force You cn use similr strteg to nlze other liner motion prolems (verticl or horizontl) in which the ccelertion (or decelertion) is the result of some other force, s ou will see in Eercises 77 8

8 SECTION Antiderivtives nd Indefinite Integrtion Before ou egin the eercise set, e sure ou relize tht one of the most importnt steps in integrtion is rewriting the integrnd in form tht fits the sic integrtion rules To illustrte this point further, here re some dditionl emples Originl Integrl Rewrite Integrte Simplif d d C C t dt d t t dt d t t t C C t t t C C d d 7 7 C 7 7

9 SECTION Antiderivtives nd Indefinite Integrtion Eercises for Section The smol Click on Click on indictes n eercise in which ou re instructed to use grphing technolog or smolic computer lger sstem to view the complete solution of the eercise to print n enlrged cop of the grph In Eercises, verif the sttement showing tht the derivtive of the right side equls the integrnd of the left side In Eercises, find the indefinite integrl nd check the result differentition 9 d C d d d C 7 d 8 d In Eercises 8, find the generl solution of the differentil eqution nd check the result differentition d 7 8 d In Eercises 9, complete the tle 9 d C d C d dr t dt Originl Integrl d d d d d d Rewrite d d d Integrte Simplif 9 d d d d d 7 d 8 9 d tt d dt d dt In Eercises, find the indefinite integrl nd check the result differentition sin cos d t sin t dt 7 csc t cot t dt 8 d d d d t dt sec d 9 sec sin d sec tn sec d tn cos d cos d

10 CHAPTER Integrtion In Eercises, the grph of the derivtive of function is given Sketch the grphs of two functions tht hve the given derivtive (There is more thn one correct nswer) To print n enlrged cop of the grph, select the MthGrph utton f f 9 d d,, d d,, In Eercises 7 nd 8, find the eqution for, given the derivtive nd the indicted point on the curve d 7 8 d f (, ) (, ) Slope Fields In Eercises 9, differentil eqution, point, nd slope field re given A slope field (or direction field) consists of line segments with slopes given the differentil eqution These line segments give visul perspective of the slopes of the solutions of the differentil eqution () Sketch two pproimte solutions of the differentil eqution on the slope field, one of which psses through the indicted point (To print n enlrged cop of the grph, select the MthGrph utton) () Use integrtion to find the prticulr solution of the differentil eqution nd use grphing utilit to grph the solution Compre the result with the sketches in prt () f d d d cos,, d Slope Fields In Eercises nd, () use grphing utilit to grph slope field for the differentil eqution, () use integrtion nd the given point to find the prticulr solution of the differentil eqution, nd (c) grph the solution nd the slope field in the sme viewing window d,, d In Eercises, solve the differentil eqution f, f g, g 7 ht 8t, h 8 fs s 8s, f 9 f, f, f f, f, f f, f, f f sin, f, f Tree Growth An evergreen nurser usull sells certin shru fter ers of growth nd shping The growth rte during those ers is pproimted dhdt t where t is the time in ers nd h is the height in centimeters The seedlings re centimeters tll when plnted t () Find the height fter t ers d d, >,, d,, d () How tll re the shrus when the re sold? Popultion Growth The rte of growth dpdt of popultion of cteri is proportionl to the squre root of t, where P is the popultion size nd t is the time in ds t Tht is dpdt kt The initil size of the popultion is After d the popultion hs grown to Estimte the popultion fter 7 ds 7

11 SECTION Antiderivtives nd Indefinite Integrtion 7 Writing Aout Concepts Use the grph of shown in the figure to nswer the following, given tht f () Approimte the slope of f t Eplin () Is it possile tht f? Eplin (c) Is f f >? Eplin (d) Approimte the vlue of where f is mimum Eplin (e) Approimte n intervls in which the grph of f is concve upwrd nd n intervls in which it is concve downwrd Approimte the -coordintes of n points of inflection (f) Approimte the -coordinte of the minimum of f (g) Sketch n pproimte grph of f To print n enlrged cop of the grph, select the MthGrph utton Figure for Figure for The grphs of f nd ech pss through the origin Use the grph of shown in the figure to sketch the grphs of f nd f To print n enlrged cop of the grph, select the MthGrph utton f f f 7 8 Verticl Motion In Eercises 7 7, use t feet per second per second s the ccelertion due to grvit (Neglect ir resistnce) 7 A ll is thrown verticll upwrd from height of feet with n initil velocit of feet per second How high will the ll go? 8 Show tht the height ove the ground of n oject thrown upwrd from point s feet ove the ground with n initil velocit of v feet per second is given the function f t t v t s 9 With wht initil velocit must n oject e thrown upwrd (from ground level) to rech the top of the Wshington Monument (pproimtel feet)? 7 A lloon, rising verticll with velocit of feet per second, releses sndg t the instnt it is feet ove the ground () How mn seconds fter its relese will the g strike the ground? () At wht velocit will it hit the ground? f f Verticl Motion In Eercises 7 7, use t 98 meters per second per second s the ccelertion due to grvit (Neglect ir resistnce) 7 Show tht the height ove the ground of n oject thrown upwrd from point s meters ove the ground with n initi velocit of v meters per second is given the function 7 The Grnd Cnon is 8 meters deep t its deepest point A rock is dropped from the rim ove this point Write the heigh of the rock s function of the time t in seconds How long wil it tke the rock to hit the cnon floor? 7 A sell is thrown upwrd from height of meters with n initil velocit of meters per second Determine its mimum height 7 With wht initil velocit must n oject e thrown upwrd (from height of meters) to rech mimum height of meters? 7 Lunr Grvit On the moon, the ccelertion due to grvit is meters per second per second A stone is dropped from cliff on the moon nd hits the surfce of the moon seconds lter How fr did it fll? Wht ws its velocit t impct? 7 Escpe Velocit The minimum velocit required for n ojec to escpe Erth s grvittionl pull is otined from the solution of the eqution where v is the velocit of the oject projected from Erth, is the distnce from the center of Erth, G is the grvittion constnt, nd M is the mss of Erth Show tht v nd re relted the eqution where v is the initil velocit of the oject nd R is the rdius of Erth Rectiliner Motion In Eercises 77 8, consider prticle moving long the -is where t is the position of the prticle t time t, t is its velocit, nd t is its ccelertion 77 f t 9t v t s v dv GM d v v GM R t t t 9t, t () Find the velocit nd ccelertion of the prticle () Find the open t-intervls on which the prticle is moving to the right (c) Find the velocit of the prticle when the ccelertion is 78 Repet Eercise 77 for the position function t t t, t 79 A prticle moves long the -is t velocit of vt t t > At time t, its position is Find the ccelertion nd position functions for the prticle

12 8 CHAPTER Integrtion 8 A prticle, initill t rest, moves long the -is such tht its ccelertion t time t > is given t cos t At the time t, its position is () Find the velocit nd position functions for the prticle () Find the vlues of t for which the prticle is t rest 8 Accelertion The mker of n utomoile dvertises tht it tkes seconds to ccelerte from kilometers per hour to 8 kilometers per hour Assuming constnt ccelertion, compute the following () The ccelertion in meters per second per second () The distnce the cr trvels during the seconds 8 Decelertion A cr trveling t miles per hour is rought to stop, t constnt decelertion, feet from where the rkes re pplied () How fr hs the cr moved when its speed hs een reduced to miles per hour? () How fr hs the cr moved when its speed hs een reduced to miles per hour? (c) Drw the rel numer line from to, nd plot the points found in prts () nd () Wht cn ou conclude? 8 Accelertion At the instnt the trffic light turns green, cr tht hs een witing t n intersection strts with constnt ccelertion of feet per second per second At the sme instnt, truck trveling with constnt velocit of feet per second psses the cr () How fr eond its strting point will the cr pss the truck? () How fst will the cr e trveling when it psses the truck? 8 Modeling Dt The tle shows the velocities (in miles per hour) of two crs on n entrnce rmp to n interstte highw The time t is in seconds () Assuming the decelertion of ech irplne is constnt, find the position functions s nd s for irplne A nd irplne B Let t represent the times when the irplnes re nd 7 miles from the irport () Use grphing utilit to grph the position functions (c) Find formul for the mgnitude of the distnce d etween the two irplnes s function of t Use grphing utilit to grph d Is d < for some time prior to the lnding of irplne A? If so, find tht time True or Flse? In Eercises 87 9, determine whether the sttement is true or flse If it is flse, eplin wh or give n emple tht shows it is flse 87 Ech ntiderivtive of n nth-degree polnomil function is n n th-degree polnomil function 88 If p is polnomil function, then p hs ectl one ntiderivtive whose grph contins the origin 89 If F nd G re ntiderivtives of f, then F G C 9 If f g, then gd f C 9 f gd f d gd 9 The ntiderivtive of f is unique 9 Find function f such tht the grph of f hs horizont tngent t, nd f 9 The grph of is shown Sketch the grph of f given tht f is continuous nd f f f t v v () Rewrite the tle converting miles per hour to feet per second () Use the regression cpilities of grphing utilit to find qudrtic models for the dt in prt () (c) Approimte the distnce trveled ech cr during the seconds Eplin the difference in the distnces 8 Accelertion Assume tht full loded plne strting from rest hs constnt ccelertion while moving down runw The plne requires 7 mile of runw nd speed of miles per hour in order to lift off Wht is the plne s ccelertion? 8 Airplne Seprtion Two irplnes re in stright-line lnding pttern nd, ccording to FAA regultions, must keep t lest three-mile seprtion Airplne A is miles from touchdown nd is grdull decresing its speed from miles per hour to lnding speed of miles per hour Airplne B is 7 miles from touchdown nd is grdull decresing its speed from miles per hour to lnding speed of miles per hour, < 9 If f f is continuous, nd f,, find f Is f differentile t? 9 Let s nd c e two functions stisfing s c nd c s for ll If s nd c, prove th s c Putnm Em Chllenge 97 Suppose f nd g re nonconstnt, differentile, rel-vlued functions on R Furthermore, suppose tht for ech pir o rel numers nd, f f f gg nd g f g gf If f, prove th f g for ll This prolem ws composed the Committee on the Putnm Prize Competition The Mthemticl Assocition of Americ All rights reserved

13 SECTION Are 9 Section Are Use sigm nottion to write nd evlute sum Understnd the concept of re Approimte the re of plne region Find the re of plne region using limits Sigm Nottion In the preceding section, ou studied ntidifferentition In this section, ou will look further into prolem introduced in Section tht of finding the re of region in the plne At first glnce, these two ides m seem unrelted, ut ou will discover in Section tht the re closel relted n etremel importnt theorem clled the Fundmentl Theorem of Clculus This section egins introducing concise nottion for sums This nottion is clled sigm nottion ecuse it uses the uppercse Greek letter sigm, written s Sigm Nottion The sum of n terms,,,, n is written s n i n i where i is the inde of summtion, i is the ith term of the sum, nd the upper nd lower ounds of summtion re n nd NOTE The upper nd lower ounds must e constnt with respect to the inde of summtion However, the lower ound doesn t hve to e An integer less thn or equl to the upper ound is legitimte EXAMPLE Emples of Sigm Nottion FOR FURTHER INFORMATION For geometric interprettion of summtion formuls, see the rticle, Looking t n k nd n k Geometricll Eric k k Heglom in Mthemtics Techer MthArticle c d e i i i i 7 j 7 j n k n n n n n k n f i f f f n i From prts () nd (), notice tht the sme sum cn e represented in different ws using sigm nottion Tr It Eplortion A Technolog Although n vrile cn e used s the inde of summtion i, j, nd k re often used Notice in Emple tht the inde of summtion does not pper in the terms of the epnded sum

14 CHAPTER Integrtion THE SUM OF THE FIRST INTEGERS Crl Friedrich Guss s (777 8) techer sked him to dd ll the integers from to When Guss returned with the correct nswer fter onl few moments, the techer could onl look t him in stounded silence This is wht Guss did: This is generlized Theorem, where i t The following properties of summtion cn e derived using the ssocitive nd commuttive properties of ddition nd the distriutive propert of ddition over multipliction (In the first propert, k is constnt) n k i k n i i i n i ± i n n i ± i i i i The net theorem lists some useful formuls for sums of powers A proof of this theorem is given in Appendi A THEOREM n i Summtion Formuls c cn n i n n n nn n i i n nn i i i EXAMPLE Evluting Sum i Evlute for n,,, nd, n i n n n i n i n n,, Solution Appling Theorem, ou cn write n i i i n n n i i n n n i i nnn n n n n n n Fctor constnt n out of sum Write s two sums Appl Theorem Simplif Simplif Now ou cn evlute the sum sustituting the pproprite vlues of n, s shown in the tle t the left Tr It Eplortion A Eplortion B In the tle, note tht the sum ppers to pproch limit s n increses Although the discussion of limits t infinit in Section pplies to vrile, where cn e n rel numer, mn of the sme results hold true for limits involving the vrile n, where n is restricted to positive integer vlues So, to find the limit of n n s n pproches infinit, ou cn write n lim n n

15 SECTION Are Rectngle: A h Figure h Are In Eucliden geometr, the simplest tpe of plne region is rectngle Although people often s tht the formul for the re of rectngle is A h, s shown in Figure, it is ctull more proper to s tht this is the definition of the re of rectngle From this definition, ou cn develop formuls for the res of mn other plne regions For emple, to determine the re of tringle, ou cn form rectngle whose re is twice tht of the tringle, s shown in Figure Once ou know how to find the re of tringle, ou cn determine the re of n polgon sudividing the polgon into tringulr regions, s shown in Figure 7 h Tringle: A h Figure Prllelogrm Hegon Polgon Figure 7 ARCHIMEDES (87 BC) Archimedes used the method of ehustion to derive formuls for the res of ellipses, prolic segments, nd sectors of spirl He is considered to hve een the gretest pplied mthemticin of ntiquit MthBio Finding the res of regions other thn polgons is more difficult The ncient Greeks were le to determine formuls for the res of some generl regions (principll those ounded conics) the ehustion method The clerest description of this method ws given Archimedes Essentill, the method is limiting process in which the re is squeezed etween two polgons one inscried in the region nd one circumscried out the region For instnce, in Figure 8 the re of circulr region is pproimted n n-sided inscried polgon nd n n-sided circumscried polgon For ech vlue of n the re of the inscried polgon is less thn the re of the circle, nd the re of the circumscried polgon is greter thn the re of the circle Moreover, s n increses, the res of oth polgons ecome etter nd etter pproimtions of the re of the circle FOR FURTHER INFORMATION For n lterntive development of the formul for the re of circle, see the rticle Proof Without Words: Are of Disk is R Russell J Hendel in Mthemtics Mgzine n = The ehustion method for finding the re of circulr region Figure 8 Animtion n = MthArticle A process tht is similr to tht used Archimedes to determine the re of plne region is used in the remining emples in this section

16 CHAPTER Integrtion The Are of Plne Region Recll from Section tht the origins of clculus re connected to two clssic prolems: the tngent line prolem nd the re prolem Emple egins the investigtion of the re prolem EXAMPLE Approimting the Are of Plne Region f() = + 8 () The re of the prolic region is greter thn the re of the rectngles f() = + 8 () The re of the prolic region is less thn the re of the rectngles Figure 9 Use the five rectngles in Figure 9() nd () to find two pproimtions of the re of the region ling etween the grph of f nd the -is etween nd Solution The right endpoints of the five intervls re i, where i,,,, The width of ech rectngle is, nd the height of ech rectngle cn e otined evluting f t the right endpoint of ech intervl,,,,,,, 8, 8, Evlute f t the right endpoints of these intervls The sum of the res of the five rectngles is Height Width f i i i i 8 Becuse ech of the five rectngles lies inside the prolic region, ou cn conclude tht the re of the prolic region is greter thn 8 The left endpoints of the five intervls re i, where i,,,, The width of ech rectngle is, nd the height of ech rectngle cn e otined evluting f t the left endpoint of ech intervl Height Width f i i i i 88 Becuse the prolic region lies within the union of the five rectngulr regions, ou cn conclude tht the re of the prolic region is less thn 88 B comining the results in prts () nd (), ou cn conclude tht 8 < Are of region < 88 Tr It Eplortion A Eplortion B Video NOTE B incresing the numer of rectngles used in Emple, ou cn otin closer nd closer pproimtions of the re of the region For instnce, using rectngles of width ech, ou cn conclude tht 77 < Are of region < 79

17 SECTION Are f Upper nd Lower Sums The procedure used in Emple cn e generlized s follows Consider plne region ounded ove the grph of nonnegtive, continuous function f, s shown in Figure The region is ounded elow the -is, nd the left nd right oundries of the region re the verticl lines nd To pproimte the re of the region, egin sudividing the intervl, into n suintervls, ech of width n, s shown in Figure The endpoints of the intervls re s follows n The region under curve Figure f(m i ) The intervl, is divided into n suintervls of width n Figure f f(m i ) < < < < n Becuse f is continuous, the Etreme Vlue Theorem gurntees the eistence of minimum nd mimum vlue of f in ech suintervl fm i Minimum vlue of f in ith suintervl fm i Mimum vlue of f in ith suintervl Net, define n inscried rectngle ling inside the ith suregion nd circumscried rectngle etending outside the ith suregion The height of the ith inscried rectngle is fm i nd the height of the ith circumscried rectngle is fm i For ech i, the re of the inscried rectngle is less thn or equl to the re of the circumscried rectngle Are of inscried rectngle The sum of the res of the inscried rectngles is clled lower sum, nd the sum of the res of the circumscried rectngles is clled n upper sum Lower sum sn n i Upper sum Sn n i fm i fm i Are of inscried rectngles Are of circumscried rectngles From Figure, ou cn see tht the lower sum sn is less thn or equl to the upper sum Sn Moreover, the ctul re of the region lies etween these two sums sn Are of region Sn fm i fm i Are of circumscried rectngle s(n) = f() = f() S(n) = f() Are of inscried rectngles Are of region Are of circumscried is less thn re of region rectngles is greter thn re of region Figure As n increses, oth the lower sum sn nd the upper sum Sn ecome closer to the ctul re of the region View the nimtion to see this Animtion

18 CHAPTER Integrtion EXAMPLE Finding Upper nd Lower Sums for Region f() = Inscried rectngles f() = Circumscried rectngles Figure Editle Grph Find the upper nd lower sums for the region ounded the grph of f nd the -is etween nd Solution To egin, prtition the intervl, into n suintervls, ech of width n Figure shows the endpoints of the suintervls nd severl inscried nd circumscried rectngles Becuse f is incresing on the intervl,, the minimum vlue on ech suintervl occurs t the left endpoint, nd the mimum vlue occurs t the right endpoint Left Endpoints m i i i n n Using the left endpoints, the lower sum is sn n i Using the right endpoints, the upper sum is Sn n i n fm i n f i i n n i n i n n fm i n i 8 n i i 8 n n i n 8 n nnn nn n n n n 8 n n i n n i f i n n n n i i i 8 n i i 8 n nnn n n n n 8 n n n n Right Endpoints M i i i n n i n i Lower sum Upper sum Tr It Eplortion A Eplortion B n

19 SECTION Are EXPLORATION For the region given in Emple, evlute the lower sum sn 8 n n nd the upper sum Sn 8 n n for n,, nd Use our results to determine the re of the region Emple illustrtes some importnt things out lower nd upper sums First, notice tht for n vlue of n, the lower sum is less thn (or equl to) the upper sum sn 8 n n < 8 n n Sn Second, the difference etween these two sums lessens s n increses In fct, if ou tke the limits s n 8, oth the upper sum nd the lower sum pproch lim sn lim n n 8 n n 8 lim Sn lim n n 8 n n 8 Lower sum limit Upper sum limit The net theorem shows tht the equivlence of the limits (s n ) of the upper nd lower sums is not mere coincidence It is true for ll functions tht re continuous nd nonnegtive on the closed intervl, The proof of this theorem is est left to course in dvnced clculus THEOREM Limits of the Lower nd Upper Sums Let f e continuous nd nonnegtive on the intervl, The limits s n of oth the lower nd upper sums eist nd re equl to ech other Tht is, lim sn lim n n n fm i i lim n n fm i i lim n Sn where n nd fm i nd fm i re the minimum nd mimum vlues of f on the suintervl f Becuse the sme limit is ttined for oth the minimum vlue fm i nd the mimum vlue fm i, it follows from the Squeeze Theorem (Theorem 8) tht the choice of in the ith suintervl does not ffect the limit This mens tht ou re free to choose n ritrr -vlue in the ith suintervl, s in the following definition of the re of region in the plne f(c i ) c i i i The width of the ith suintervl is i i Figure Definition of the Are of Region in the Plne Let f e continuous nd nonnegtive on the intervl, The re of the region ounded the grph of f, the -is, nd the verticl lines nd is Are lim n n fc i, i where n (see Figure ) i c i i Video

20 CHAPTER Integrtion EXAMPLE Finding Are the Limit Definition Find the re of the region ounded the grph f, the -is, nd the verticl lines nd, s shown in Figure f() = (, ) Solution Begin noting tht f is continuous nd nonnegtive on the intervl, Net, prtition the intervl, into n suintervls, ech of width n According to the definition of re, ou cn choose n -vlue in the ith suintervl For this emple, the right endpoints c i in re convenient (, ) The re of the region ounded the grph of f, the -is,, nd is Figure Editle Grph Are lim n n fc i lim i n n i i n n lim i n n n i n lim n n n lim n n n Right endpoints: c i i n The re of the region is Tr It EXAMPLE Eplortion A Finding Are the Limit Definition f() = Find the re of the region ounded the grph of f, the -is, nd the verticl lines nd, s shown in Figure Solution The function f is continuous nd nonnegtive on the intervl,, nd so egin prtitioning the intervl into n suintervls, ech of width n Choosing the right endpoint c i i i Right endpoints n The re of the region ounded the grph of f, the -is,, nd is Figure Editle Grph of ech suintervl, ou otin Are lim n n fc i lim i n n i n i n n n lim n n i i n i lim n n n i i i n n i n n i lim n n n n The re of the region is Tr It Eplortion A Open Eplortion

21 SECTION Are 7 The lst emple in this section looks t region tht is ounded the -is (rther thn the -is) EXAMPLE 7 A Region Bounded the -is Find the re of the region ounded the grph of, s shown in Figure 7 f nd the -is for (, ) f() = (, ) The re of the region ounded the grph of f nd the -is for is Figure 7 Editle Grph Solution When f is continuous, nonnegtive function of, ou still cn use the sme sic procedure shown in Emples nd Begin prtitioning the intervl, into n suintervls, ech of width n Then, using the upper endpoints c i in, ou otin Are lim n n fc i lim i n n i i n n The re of the region is lim i n n n i n lim n nnn lim n n n Upper endpoints: c i i n Tr It Eplortion A

22 SECTION Are 7 Eercises for Section The smol Click on Click on indictes n eercise in which ou re instructed to use grphing technolog or smolic computer lger sstem to view the complete solution of the eercise to print n enlrged cop of the grph In Eercises, find the sum Use the summtion cpilities of grphing utilit to verif our result i k c In Eercises 7, use sigm nottion to write the sum i k k 9 kk k j n n n n n n n n n n n n n j i i i In Eercises, use the properties of summtion nd Theorem to evlute the sum Use the summtion cpilities of grphing utilit to verif our result i 7 i 8 9 ii In Eercises nd, use the summtion cpilities of grphing utilit to evlute the sum Then use the properties of summtion nd Theorem to verif the sum n n n n n n n n n n i i i i i i i i i i i i ii i

23 8 CHAPTER Integrtion In Eercises, ound the re of the shded region pproimting the upper nd lower sums Use rectngles of width In Eercises 7, use upper nd lower sums to pproimte the re of the region using the given numer of suintervls (of equl width) In Eercises, find the limit of sn s n f sn 8 n n n sn n nnn sn 8 nnn f f sn nnn f In Eercises 8, use the summtion formuls to rewrite the epression without the summtion nottion Use the result to find the sum for n,,, nd, i i n kk 7 8 k n n In Eercises 9, find formul for the sum of n terms Use the formul to find the limit s n i 9 lim n n i n lim n n i n i lim n n n i i n n Numericl Resoning Consider tringle of re ounded the grphs of,, nd () Sketch the region () Divide the intervl, into n suintervls of equl width nd show tht the endpoints re < n < < n n < n n (c) Show tht sn n (d) Show tht Sn n (e) Complete the tle (f) Show tht lim sn lim Sn n n Numericl Resoning Consider trpezoid of re ounded the grphs of,,, nd () Sketch the region () Divide the intervl, into n suintervls of equl width nd show tht the endpoints re < n < < n n < n n (c) Show tht sn n (d) Show tht Sn n (e) Complete the tle n i i i n n i i n n n j j n i i n n i i n n (f) Show tht lim sn lim Sn n n n sn Sn n sn Sn i i n lim n n i i n n lim n n i i n n lim n n i i n n

24 SECTION Are 9 In Eercises 7, use the limit process to find the re of the region etween the grph of the function nd the -is over the given intervl Sketch the region 7,, 8, 9,,,,,,, [,,,,, In Eercises 7, use the limit process to find the re of the region etween the grph of the function nd the -is over the given -intervl Sketch the region 7 f, 8 g, 9 f, f, g, h, In Eercises, use the Midpoint Rule with n to pproimte the re of the region ounded the grph of the function nd the -is over the given intervl f,, f tn,, Progrmming Write progrm for grphing utilit to pproimte res using the Midpoint Rule Assume tht the function is positive over the given intervl nd the suintervls re of equl width In Eercises 7 7, use the progrm to pproimte the re of the region etween the grph of the function nd the -is over the given intervl, nd complete the tle n Are n i Approimte Are 7 f,, 8 f i i f,, f sin,, 8 f 8, 9 f tn, 7 f cos, 8, Writing Aout Concepts,,,,,,, Approimtion In Eercises 7 nd 7, determine which vlue est pproimtes the re of the region etween the -is nd the grph of the function over the given intervl (Mke our selection on the sis of sketch of the region nd not performing clcultions) 7 f,, () () (c) (d) (e) 8 Writing Aout Concepts (continued) 7 7 Grphicl Resoning Consider the region ounded the grphs of f,, nd, s shown in the figure To print n enlrged cop of the grph, select the MthGrph utton 8 f sin, () () (c) (d) 8 (e) 7 In our own words nd using pproprite figures, descrie the methods of upper sums nd lower sums in pproimting the re of region 7 Give the definition of the re of region in the plne 8, f () Redrw the figure, nd complete nd shde the rectngles representing the lower sum when n Find this lower sum () Redrw the figure, nd complete nd shde the rectngles representing the upper sum when n Find this upper sum (c) Redrw the figure, nd complete nd shde the rectngles whose heights re determined the functionl vlues the midpoint of ech suintervl when n Find this sum using the Midpoint Rule (d) Verif the following formuls for pproimting the re of the region using n suintervls of equl width Lower sum: Upper sum: Midpoint Rule: sn n i Sn n i Mn n f i n n f i n n i f i n n (e) Use grphing utilit nd the formuls in prt (d) to complete the tle n sn Sn Mn, 8

25 7 CHAPTER Integrtion (f) Eplin wh sn increses nd Sn decreses for incresing vlues of n, s shown in the tle in prt (e) 7 Monte Crlo Method The following computer progrm pproimtes the re of the region under the grph of monotonic function nd ove the -is etween nd Run the progrm for nd for severl vlues of N Eplin wh the Monte Crlo Method works [Adpttion of Monte Crlo Method progrm from Jmes M Sconers, Approimtion of Are Under Curve, MATHEMATICS TEACHER 77, no (Ferur 98) Copright 98 the Ntionl Council of Techers of Mthemtics Reprinted with permission] DEF FNF(X)=SIN(X) A= B=/ PRINT Input Numer of Rndom Points INPUT N N= 7 IF FNF(A)>FNF(B) THEN YMAX=FNF(A) ELSE YMAX=FNF(B) 8 FOR I= TO N 9 X=A+(B-A)*RND() Y=YMAX*RND() IF Y>=FNF(X) THEN GOTO N=N+ NEXT I AREA=(N/N)*(B-A)*YMAX PRINT Approimte Are: ; AREA END True or Flse? In Eercises 77 nd 78, determine whether the sttement is true or flse If it is flse, eplin wh or give n emple tht shows it is flse 77 The sum of the first n positive integers is nn 78 If f is continuous nd nonnegtive on,, then the limits s n of its lower sum sn nd upper sum Sn oth eist nd re equl 79 Writing Use the figure to write short prgrph eplining wh the formul n nn is vlid for ll positive integers n 8 Grphicl Resoning Consider n n-sided regulr polgon inscried in circle of rdius r Join the vertices of the polgon to the center of the circle, forming n congruent tringles (see figure) () Determine the centrl ngle in terms of n () Show tht the re of ech tringle is r sin (c) Let A n e the sum of the res of the n tringles Find lim A n n 8 Modeling Dt The tle lists the mesurements of lo ounded strem nd two stright rods tht meet t righ ngles, where nd re mesured in feet (see figure) () Use the regression cpilities of grphing utilit to find model of the form c d () Use grphing utilit to plot the dt nd grph the model (c) Use the model in prt () to estimte the re of the lot Figure for 8 Figure for 8 8 Building Blocks A child plces n cuic uilding locks in row to form the se of tringulr design (see figure) Ech successive row contins two fewer locks thn the preceding row Find formul for the numer of locks used in the design (Hint: The numer of uilding locks in the design depends on whether n is odd or even) 8 Prove ech formul mthemticl induction (You m need to review the method of proof induction from preclculus tet) () 8 Rod n i nn i () n i n n i Strem Rod n is even Putnm Em Chllenge Figure for 79 Figure for 8 8 A drt, thrown t rndom, hits squre trget Assuming th n two prts of the trget of equl re re equll likel to e hit, find the proilit tht the point hit is nerer to the center thn to n edge Write our nswer in the form cd where,, c, nd d re positive integers This prolem ws composed the Committee on the Putnm Prize Competition The Mthemticl Assocition of Americ All rights reserved

26 SECTION Riemnn Sums nd Definite Integrls 7 Section Riemnn Sums nd Definite Integrls Understnd the definition of Riemnn sum Evlute definite integrl using limits Evlute definite integrl using properties of definite integrls Riemnn Sums In the definition of re given in Section, the prtitions hve suintervls of equl width This ws done onl for computtionl convenience The following emple shows tht it is not necessr to hve suintervls of equl width EXAMPLE A Prtition with Suintervls of Unequl Widths n n n n f() = (n ) n n n The suintervls do not hve equl widths Figure 8 Editle Grph (, ) Are = (, ) = The re of the region ounded the grph of nd the -is for is Figure 9 Consider the region ounded the grph of f nd the -is for, s shown in Figure 8 Evlute the limit lim n n f c i i i where c is the right endpoint of the prtition given c i i n i nd i is the width of the ith intervl Solution So, the limit is The width of the ith intervl is given i i i n n i i i n i n lim n n f c i i lim i n n i i Tr It lim n lim n n lim n n n n i n n n n i i i n nn n Eplortion A nn From Emple 7 in Section, ou know tht the region shown in Figure 9 hs n re of Becuse the squre ounded nd hs n re of, ou cn conclude tht the re of the region shown in Figure 8 hs n re of This grees with the limit found in Emple, even though tht emple used prtition hving suintervls of unequl widths The reson this prticulr prtition gve the proper re is tht s n increses, the width of the lrgest suintervl pproches zero This is ke feture of the development of definite integrls

27 7 CHAPTER Integrtion Germn mthemticin Riemnn did his most fmous work in the res of non-eucliden geometr, differentil equtions, nd numer theor It ws Riemnn s results in phsics nd mthemtics tht formed the structure on which Einstein s theor of generl reltivit is sed GEORG FRIEDRICH BERNHARD RIEMANN (8 8) MthBio In the preceding section, the limit of sum ws used to define the re of region in the plne Finding re this mens is onl one of mn pplictions involving the limit of sum A similr pproch cn e used to determine quntities s diverse s rc lengths, verge vlues, centroids, volumes, work, nd surfce res The following definition is nmed fter Georg Friedrich Bernhrd Riemnn Although the definite integrl hd een defined nd used long efore the time of Riemnn, he generlized the concept to cover roder ctegor of functions In the following definition of Riemnn sum, note tht the function f hs no restrictions other thn eing defined on the intervl, (In the preceding section, the function f ws ssumed to e continuous nd nonnegtive ecuse we were deling with the re under curve) Definition of Riemnn Sum Let f e defined on the closed intervl,, nd let e prtition of, given < < < < n < n where i is the width of the ith suintervl If c i is n point in the ith suintervl, then the sum n f c i i, i i c i i is clled Riemnn sum of f for the prtition Video NOTE The sums in Section re emples of Riemnn sums, ut there re more generl Riemnn sums thn those covered there The width of the lrgest suintervl of prtition is the norm of the prtition nd is denoted If ever suintervl is of equl width, the prtition is regulr nd the norm is denoted n Regulr prtition n 8 = n does not impl tht Figure For generl prtition, the norm is relted to the numer of suintervls of, in the following w n Generl prtition So, the numer of suintervls in prtition pproches infinit s the norm of the prtition pproches Tht is, implies tht n The converse of this sttement is not true For emple, let n e the prtition of the intervl, given < n < n < < 8 < < < As shown in Figure, for n positive vlue of n, the norm of the prtition n is So, letting n pproch infinit does not force to pproch In regulr prtition, however, the sttements nd n re equivlent

28 SECTION Riemnn Sums nd Definite Integrls 7 Definite Integrls To define the definite integrl, consider the following limit lim n f c i i L i To s tht this limit eists mens tht for > there eists > such tht for ever prtition with < it follows tht L n i f c i < i (This must e true for n choice of in the ith suintervl of ) c i FOR FURTHER INFORMATION For insight into the histor of the definite integrl, see the rticle The Evolution of Integrtion A Shenitzer nd J Steprns in The Americn Mthemticl Monthl MthArticle Definition of Definite Integrl If f is defined on the closed intervl, nd the limit lim n f c i i i eists (s descried ove), then f is integrle on, nd the limit is denoted lim n f c i i f d i The limit is clled the definite integrl of f from to The numer is the lower limit of integrtion, nd the numer is the upper limit of integrtion It is not coincidence tht the nottion for definite integrls is similr to tht used for indefinite integrls You will see wh in the net section when the Fundmentl Theorem of Clculus is introduced For now it is importnt to see tht definite integrls nd indefinite integrls re different identities A definite integrl is numer, wheres n indefinite integrl is fmil of functions A sufficient condition for function f to e integrle on, is tht it is continuous on, A proof of this theorem is eond the scope of this tet THEOREM Continuit Implies Integrilit If function f is continuous on the closed intervl,, then f is integrle on, EXPLORATION The Converse of Theorem Is the converse of Theorem true? Tht is, if function is integrle, does it hve to e continuous? Eplin our resoning nd give emples Descrie the reltionships mong continuit, differentiilit, nd integrilit Which is the strongest condition? Which is the wekest? Which conditions impl other conditions?

29 7 CHAPTER Integrtion EXAMPLE Evluting Def inite Integrl s Limit Evlute the definite integrl d f() = Becuse the definite integrl is negtive, it does not represent the re of the region Figure Solution The function f is integrle on the intervl, ecuse it is continuous on, Moreover, the definition of integrilit implies tht n prtition whose norm pproches cn e used to determine the limit For computtionl convenience, define sudividing, into n suintervls of equl width i n Choosing c i s the right endpoint of ech suintervl produces c i i i n So, the definite integrl is given d lim n f c i i i lim n n f c i i lim n n i i n n lim n n n i i n lim n n n n lim n 9 9 n n nn Editle Grph Tr It Eplortion A Video Video f Becuse the definite integrl in Emple is negtive, it does not represent the re of the region shown in Figure Definite integrls cn e positive, negtive, or zero For definite integrl to e interpreted s n re (s defined in Section ), the function f must e continuous nd nonnegtive on,, s stted in the following theorem (The proof of this theorem is strightforwrd ou simpl use the definition of re given in Section ) You cn use definite integrl to find the re of the region ounded the grph of f, the -is,, nd Figure THEOREM The Definite Integrl s the Are of Region If f is continuous nd nonnegtive on the closed intervl,, then the re of the region ounded the grph of f, the -is, nd the verticl lines nd is given Are f d (See Figure )

30 SECTION Riemnn Sums nd Definite Integrls 7 f() = Are d Figure As n emple of Theorem, consider the region ounded the grph of nd the -is, s shown in Figure Becuse f is continuous nd nonnegtive on the closed intervl,, the re of the region is Are d A strightforwrd technique for evluting definite integrl such s this will e discussed in Section For now, however, ou cn evlute definite integrl in two ws ou cn use the limit definition or ou cn check to see whether the definite integrl represents the re of common geometric region such s rectngle, tringle, or semicircle EXAMPLE Ares of Common Geometric Figures Sketch the region corresponding to ech definite integrl Then evlute ech integrl using geometric formul f d d c d NOTE The vrile of integrtion in definite integrl is sometimes clled dumm vrile ecuse it cn e replced n other vrile without chnging the vlue of the integrl For instnce, the definite integrls d nd t dt hve the sme vlue Solution A sketch of ech region is shown in Figure This region is rectngle of height nd width d (Are of rectngle) 8 This region is trpezoid with n ltitude of nd prllel ses of lengths nd The formul for the re of trpezoid is h d (Are of trpezoid) c This region is semicircle of rdius The formul for the re of semicircle is r (Are of semicircle) d () f() = f() = + Editle Grph Editle Grph Editle Grph Figure () (c) f() = Tr It Eplortion A

31 7 CHAPTER Integrtion Properties of Definite Integrls The definition of the definite integrl of f on the intervl, specifies tht < Now, however, it is convenient to etend the definition to cover cses in which or > Geometricll, the following two definitions seem resonle For instnce, it mkes sense to define the re of region of zero width nd finite height to e Definitions of Two Specil Definite Integrls If f is defined t, then we define f d If f is integrle on,, then we define f d f d EXAMPLE Evluting Definite Integrls f f() d Becuse the sine function is defined t, nd the upper nd lower limits of integrtion re equl, ou cn write sin d The integrl d is the sme s tht given in Emple () ecept tht the upper nd lower limits re interchnged Becuse the integrl in Emple () hs vlue of, ou cn write d d The editle grph feture elow llows ou to edit the grph of function Editle Grph Tr It Eplortion A Open Eplortion c c f() d + Figure c f() d In Figure, the lrger region cn e divided t c into two suregions whose intersection is line segment Becuse the line segment hs zero re, it follows tht the re of the lrger region is equl to the sum of the res of the two smller regions THEOREM Additive Intervl Propert If f is integrle on the three closed intervls determined,, nd c, then c f d f d f d c EXAMPLE d Using the Additive Intervl Propert d d Theorem Tr It Eplortion A Are of tringle