C H A P T E R 5. Integrals. PDF Created with deskpdf PDF Writer - Trial :: A r e a s a n d D i s t a n c e s

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1 License o: PDF Cree wih eskpdf PDF Wrier - Tril :: hp:// C H A P T E R 5 To compue n re we pproime region recngles n le he numer of recngles ecome lrge. The precise re is he i of hese sums of res of recngles. Inegrls Coprigh 5 Thomson Lerning, Inc. All Righs Reserve. M no e copie, scnne, or uplice, in whole or in pr. License o: jsmuels@mcc.cun.eu Now is goo ime o re (or rere) A Preview of Clculus (see pge ). I iscusses he unifing ies of clculus n helps pu in perspecive where we hve een n where we re going. 5. A r e s n D i s n c e s FIGURE S=s(,), ƒ FIGURE In Chper we use he ngen n veloci prolems o inrouce he erivive, which is he cenrl ie in ifferenil clculus. In much he sme w, his chper srs wih he re n isnce prolems n uses hem o formule he ie of efinie inegrl, which is he sic concep of inegrl clculus. We will see in Chpers 6 n 8 how o use he inegrl o solve prolems concerning volumes, lenghs of curves, populion preicions, cric oupu, forces on m, work, consumer surplus, n sell, mong mn ohers. There is connecion eween inegrl clculus n ifferenil clculus. The Funmenl Theorem of Clculus reles he inegrl o he erivive, n we will see in his chper h i grel simplifies he soluion of mn prolems. In his secion we iscover h in ring o n he re uner curv e or he isnce rvele cr, we en up wih he sme specil pe of i. T h e A r e P r o l e m We egin emping o solve he re prolem: Fin he re of he region S h lies uner he curve f from o. This mens h S, illusre in Figure, is oune he grph of coninuous funcion f [where f ], he vericl lines n, n he -is. In ring o solve he re prolem we hve o sk ourselves: Wh is he mening of he wor re? This quesion is es o nswer for regions wih srigh sies. For recngle, he re is ene s he prouc of he lengh n he wih. The re of ringle is hlf he se imes he heigh. The re of polgon is foun iviing i ino ringles (s in Figure ) n ing he res of he ringles. w l A=lw = h S A= h Coprigh 5 Thomson Lerning, Inc. All Righs Reserve. M no e copie, scnne, or uplice, in whole or in pr. =ƒ = A A A A A=A +A +A +A 69

2 7 CHAPTER 5 INTEGRALS License o: jsmuels@mcc.cun.eu PDF Cree wih eskpdf PDF Wrier - Tril :: hp:// Tr plcing recngles o esime he re. Resources / Moule 6 / Wh Is Are? / Esiming Are uner Prol FIGURE 5 = (,) FIGURE FIGURE However, i isn so es o n he re of re gion wih curve sies. We ll hve n inuiive ie of wh he re of region is. Bu pr of he re prolem is o mke his inuiive ie precise giving n ec eniion of re. Recll h in ening ngen we rs pproime he slope of he ngen line slopes of secn lines n hen we ook he i of hese pproimions. We pursue similr ie for res. We rs pproime he re gion S recngles n hen we ke he i of he res of hese recngles s we increse he numer of recngles. The following emple illusres he proceure. EXAMPLE Use recngles o esime he re uner he prol from o (he prolic region S illusre in Figure ). SOLUTION We rs noice h he re of S mus e somewhere eween n ecuse S is conine in squre wih sie lengh, u we cn cerinl o eer hn h. Suppose we ivie S ino four srips S, S, S, n S rwing he vericl lines,, n s in Figure (). S S = S () We cn pproime ech srip recngle whose se is he sme s he srip n whose heigh is he sme s he righ ege of he srip [see Figure ()]. In oher wors, he heighs of hese recngles re he vlues of he funcion f he righ en poins of he suinervls [, ], [, ], [, ], n [, ]. Ech recngle hs wih n he heighs re (, ( ), ( ), n ). If we le R e he sum of he res of hese pproiming recngles, we ge R ( ) ( ) ( ) From Figure () we see h he re A of S is less hn S (,) = A.6875 Inse of using he recngles in Figure () we coul use he smller recngles in Figure 5 whose heighs re he vlues of f he lef enpoins of he suinervls. (The lefmos recngle hs collpse ecuse is heigh is.) The sum of he res of hese S (,) R, so () (,) License o: jsmuels@mcc.cun.eu pproiming recngles is SECTION 5. AREAS AND DISTANCES 7 We see h he re of S is lrger hn L, so we hve lower n upper esimes for A: We cn repe his proceure wih lrger numer of srips. Figure 6 shows wh hppens when we ivie he region S ino eigh srips of equl wih. FIGURE 6 Approiming S wih eigh recngles 8 () Using lef enpoins 8 () Using righ enpoins n n FIGURE 7 = Ln Rn (,) L ( ) ( ) ( ) =.875 A.6875 (,) B compuing he sum of he res of he smller recngles L8 n he sum of he res of he lrger recngles R8, we oin eer lower n upper esimes for A:.775 A.9875 So one possile nswer o he quesion is o s h he rue re of S lies somewhere eween.775 n We coul oin eer esimes incresing he numer of srips. The le he lef shows he resuls of similr clculions (wih compuer) using n recngles whose heighs re foun wih lef enpoins Ln or righ enpoins Rn. In priculr, we see using 5 srips h he re lies eween. n.. Wih srips we nrrow i own even more: A lies eween.85 n.85. A goo esime is oine verging hese numers: A.5. From he vlues in he le in Emple, i looks s if Rn is pproching s n increses. We conrm his in he ne emple. EXAMPLE For he region S in Emple, show h he sum of he res of he upper pproiming recngles pproches, h is, nl Rn SOLUTION Rn is he sum of he res of he n recngles in Figure 7. Ech recngle hs wih n n he heighs re he vlues of he funcion f he poins n, n, n,..., nn; h is, he heighs re n, n, n,..., nn. (,) Coprigh 5 Thomson Lerning, Inc. All Righs Reserve. M no e copie, scnne, or uplice, in whole or in pr. Coprigh 5 Thomson Lerning, Inc. All Righs Reserve. M no e copie, scnne, or uplice, in whole or in pr.

3 7 CHAPTER 5 INTEGRALS License o: jsmuels@mcc.cun.eu PDF Cree wih eskpdf PDF Wrier - Tril :: hp:// FIGURE 8 The ies in Emples n re eplore in Moule 5./5./7.7 for vrie of funcions. Here we re compuing he i of he sequence Rn. Sequences were iscusse in A Preview of Clculus n will e suie in eil in Chper. Their is re clcule in he sme w s is infini (Secion.6). In priculr, we know h nl n n= R =.85 Thus Here we nee he formul for he sum of he squres of he rs n posiive inegers: Perhps ou hve seen his formul efore. I is prove in Emple 5 in Appeni E. Puing Formul ino our epression for Rn, we ge Thus, we hve I cn e shown h he lower pproiming sums lso pproch, h is, From Figures 8 n 9 i ppers h, s n increses, oh Ln n Rn ecome eer n eer pproimions o he re of S. Therefore, we ene he re A o e he i of he Rn n n n n n n n n n n n n n n n Rn n nl n= R Å.5 nn n 6 n n Rn nl 6n nl 6 n n n 6 n n n nl 6 nl Ln nn n 6 n n 6n n=5 R =. License o: jsmuels@mcc.cun.eu n= L =.85 FIGURE 9 The re is he numer h is smller hn ll upper sums n lrger hn ll lower sums FIGURE FIGURE n= L Å.69 sums of he res of he pproiming recngles, h is, SECTION 5. AREAS AND DISTANCES 7 Les ppl he ie of Emples n o he more generl region S of Figure. We sr suiviing S ino n srips S, S,..., Sn of equl wih s in Figure. The wih of he inervl, is, so he wih of ech of he n srips is These srips ivie he inervl [, ] ino n suinervls where n n. The righ enpoins of he suinervls re, Les pproime he ih srip Si recngle wih wih n heigh f i, which is he vlue of f he righ enpoin (see Figure ). Then he re of he ih recngle A nl Rn nl Ln n,,,,,,..., n, n, =ƒ S S S S i S n... i- i... n- Î i- i n=5 L =.,... f( i ) Coprigh 5 Thomson Lerning, Inc. All Righs Reserve. M no e copie, scnne, or uplice, in whole or in pr. Coprigh 5 Thomson Lerning, Inc. All Righs Reserve. M no e copie, scnne, or uplice, in whole or in pr.

4 7 CHAPTER 5 INTEGRALS SECTION 5. AREAS AND DISTANCES 75 License o: jsmuels@mcc.cun.eu PDF Cree wih eskpdf PDF Wrier - Tril :: hp:// FIGURE () n= () n= (c) n=8 () n= This ells us o en wih i=n. This ells us o. This ells us o sr wih i=m. n µ f( i )Î i=m is f i. Wh we hink of inuiivel s he re of S is pproime he sum of he res of hese recngles, which is Figure shows his pproimion for n,, 8, n. Noice h his pproimion ppers o ecome eer n eer s he numer of srips increses, h is, s nl. Therefore, we ene he re A of he region S in he following w. f Definiion The re A of he region S h lies uner he grph of he coninuous funcion f is he i of he sum of he res of pproiming recngles: I cn e prove h he i in Deniion l ws eiss, since we re ssuming h is coninuous. I cn lso e shown h we ge he sme vlue if we use lef enpoins: In fc, inse of using lef enpoins or righ enpoins, we coul ke he heigh of he ih recngle o e he vlue of f n numer i* in he ih suinervl i, i. We cll he numers *, *,..., n* he smple poins. Figure shows pproiming recngles when he smple poins re no chosen o e enpoins. So more generl epression for he re of S is FIGURE A nl A nl Rn f f f n Rn f f f n nl Ln f f f n nl A nl f * f * f n* i- i n- * * * i * n * We ofen use sigm noion o wrie sums wih mn erms more compcl. For insnce, n f i f f f n i Î f( i *) License o: jsmuels@mcc.cun.eu If ou nee prcice wih sigm noion, look he emples n r some of he eercises in Appeni E. So he epressions for re in Equions,, n cn e wrien s follows: We coul lso rewrie Formul in he following w: EXAMPLE Le A e he re of he region h lies uner he grph of f e eween n. () Using righ enpoins, n n e pression for A s i. Do no evlue he i. () Esime he re king he smple poins o e mipoins n using four suinervls n hen en suinervls. SOLUTION () Since n, he wih of suinervl is So n, n, 6n, i in, n n nn. The sum of he res of he pproiming recngles is Accoring o Deniion, he re is A Rn nl nl n en e n e 6n e nn Using sigm noion we coul wrie A nl n f i i A nl n f i i A nl n f i* i n nn n i i 6 n Rn f f f n e n e e n n e e n n nn e n A nl n n e in i I is ifcul o e vlue his i irecl hn, u wih he i of compuer lger ssem i isn hr (see Eercise ). In Secion 5. we will e le o n A more esil using ifferen meho. () Wih n he suinervls of equl wih.5 re,.5,.5,,,.5, n.5,. The mipoins of hese suinervls re *.5, *.75, *.5, n Coprigh 5 Thomson Lerning, Inc. All Righs Reserve. M no e copie, scnne, or uplice, in whole or in pr. Coprigh 5 Thomson Lerning, Inc. All Righs Reserve. M no e copie, scnne, or uplice, in whole or in pr.

5 76 CHAPTER 5 INTEGRALS SECTION 5. AREAS AND DISTANCES 77 License o: jsmuels@mcc.cun.eu PDF Cree wih eskpdf PDF Wrier - Tril :: hp:// =e FIGURE =e FIGURE 5 n *.75, n he sum of he res of he four pproiming recngles (see Figure ) is M f i * i So n esime for he re is Wih n he suinervls re,.,.,.,...,.8, n he mipoins re *., *., *.5,..., *.9. Thus A M f. f. f.5 f.9.e. e. e.5 e.9.86 From Figure 5 i ppers h his esime is eer hn he esime wih n. T h e D i s n c e P r o l e m f.5 f.75 f.5 f.75 e.5.5 e.75.5 e.5.5 e.75.5 e.5 e.75 e.5 e A.8557 Now les consier he isnce prolem: Fin he isnce rvele n ojec uring cerin ime perio if he veloci of he ojec is known ll imes. (In sense his is he inverse prolem of he veloci prolem h we iscusse in Secion..) If he veloci remins consn, hen he isnce prolem is es o solve mens of he formul isnce veloci ime Bu if he veloci vries, is no so es o n he isnce r vele. We invesige he prolem in he following emple. EXAMPLE Suppose he oomeer on our cr is roken n we wn o esime he isnce riven over -secon ime inervl. We ke speeomeer reings ever v e secons n recor hem in he following le: Time (s) Veloci (mih) In orer o hve he ime n he veloci in consisen unis, les conver he veloci reings o fee per secon ( mih 586 fs): Time (s) Veloci (fs) During he rs v e secons he veloci oesn chnge ver much, so we cn esime he isnce rvele uring h ime ssuming h he veloci is consn. If we ke he veloci uring h ime inervl o e he iniil veloci (5 fs), hen we License o: jsmuels@mcc.cun.eu FIGURE 6 oin he pproime isnce rvele uring he rs v e secons: Similrl, uring he secon ime inervl he veloci is pproimel consn n we ke i o e he veloci when 5 s. So our esime for he isnce rvele from 5 s o s is fs 5 s 55 f If we similr esimes for he oher ime inervls, we oin n esime for he ol isnce rvele: We coul jus s well hve use he veloci he en of ech ime perio inse of he veloci he eginning s our ssume consn veloci. Then our esime ecomes If we h wne more ccure esime, we coul hve ken veloci reings ever wo secons, or even ever secon. Perhps he clculions in Emple remin ou of he sums we use erlier o esime res. The similri is epline when we skech grph of he veloci funcion of he cr in Figure 6 n rw recngles whose heighs re he iniil velociies for ech ime inervl. The re of he rs recngle is 5 5 5, which is lso our esime for he isnce rvele in he rs v e secons. In fc, he re of ech recngle cn e inerpree s isnce ecuse he heigh represens veloci n he wih represens ime. The sum of he res of he recngles in Figure 6 is L6 5, which is our iniil esime for he ol isnce rvele. In generl, suppose n ojec moves wih veloci v f, where n f (so he ojec lws moves in he posiive irecion). We ke veloci reings imes,,,..., n so h he veloci is pproimel consn on ech suinervl. If hese imes re equll spce, hen he ime eween consecuive reings is n. During he rs ime inerv l he veloci is pproimel f n so he isnce rvele is pproimel f. Similrl, he isnce rvele uring he secon ime inervl is ou f n he ol isnce rvele uring he ime inervl, is pproimel If we use he veloci righ enpoins inse of lef enpoins, our esime for he ol isnce ecomes The more frequenl we mesure he veloci, he more ccure our esimes ecome, so i seems plusile h he ec isnce rvele is he i of such epressions: f f f f f n n f i f f f n n f i nl n i 5 fs 5 s 5 f f i nl n We will see in Secion 5. h his is inee rue. i i i f i Coprigh 5 Thomson Lerning, Inc. All Righs Reserve. M no e copie, scnne, or uplice, in whole or in pr. Coprigh 5 Thomson Lerning, Inc. All Righs Reserve. M no e copie, scnne, or uplice, in whole or in pr.

6 78 CHAPTER 5 INTEGRALS License o: jsmuels@mcc.cun.eu PDF Cree wih eskpdf PDF Wrier - Tril :: hp:// 5. Eercises. () B reing vlues from he given grph of f, use v e recngles o n lo wer esime n n upper esime for he re uner he given grph of f from o. In ech cse skech he recngles h ou use. () Fin new esimes using recngles in ech cse.. () Use si recngles o n esimes of ech pe for he re uner he given grph of f from o. (i) L6 (smple poins re lef enpoins) (ii) R6 (smple poins re righ enpoins) (iii) M6 (smple poins re mipoins) () Is L6 n uneresime or overesime of he rue re? (c) Is R6 n uneresime or overesime of he rue re? () Which of he numers L6, R6, or M6 gives he es esime? Eplin =ƒ. () Esime he re uner he grph of f from o 5 using four pproiming recngles n righ enpoins. Skech he grph n he recngles. Is our esime n uneresime or n overesime? () Repe pr () using lef enpoins.. () Esime he re uner he grph of f 5 from o 5 using v e pproiming recngles n Becuse Equion 5 hs he sme form s our epressions for re in Equions n, i follows h he isnce rvele is equl o he re uner he grph of he veloci funcion. In Chpers 6 n 8 we will see h oher quniies of ineres in he nurl n socil sciencessuch s he w ork one vrile force or he cric oupu of he her cn lso e inerpree s he re uner curve. So when we compue res in his chper, er in min h he cn e inerpree in vrie of prcicl ws. =ƒ 8 righ enpoins. Skech he grph n he recngles. Is our esime n uneresime or n overesime? () Repe pr () using lef enpoins. 5. () Esime he re uner he grph of f from o using hree recngles n righ enpoins. Then improve our esime using si recngles. Skech he curve n he pproiming recngles. () Repe pr () using lef enpoins. (c) Repe pr () using mipoins. () From our skeches in prs () (c),which ppers o e he es esime? ; 6. () Grph he funcion f e,. () Esime he re uner he grph of f using four pproiming recngles n king he smple poins o e (i) righ enpoins (ii) mipoins In ech cse skech he curve n he recngles. (c) Improve our esimes in pr () using eigh recngles. 7 8 Wih progrmmle clculor (or compuer), i is possile o evlue he epressions for he sums of res of pproiming recngles, even for lrge vlues of n, using looping. (On TI use he Is commn or For-EnFor loop, on Csio use Isz, on n HP or in BASIC use FOR-NEXT loop.) Compue he sum of he res of pproiming recngles using equl suinervls n righ enpoins for n,, n 5. Then guess he vlue of he ec re. 7. The region uner sin from o 8. The region uner from o CAS 9. Some compuer lger ssems hve commns h will rw pproiming recngles n evlue he sums of heir res, les if i* is lef or righ enpoin. (For insnce, in Mple use lefo, righo, lefsum, n righsum.) () If f s,, n he lef n righ sums for n,, n 5. () Illusre grphing he recngles in pr (). (c) Show h he ec re uner f lies eween.6 n.7. CAS. () If f sinsin,, use he commns iscusse in Eercise 9 o n he lef n righ sums for n,, n 5. License o: jsmuels@mcc.cun.eu () Illusre grphing he recngles in pr (). (c) Show h he ec re uner f lies eween.87 n.9.. The spee of runner increse seil uring he rs hree secons of rce. Her spee hlf-secon inervls is given in he le. Fin lower n upper esimes for he isnce h she rvele uring hese hree secons. (s) v (fs) Speeomeer reings for moorccle -secon inervls re given in he le. () Esime he isnce rvele he moorccle uring his ime perio using he velociies he eginning of he ime inervls. () Give noher esime using he velociies he en of he ime perios. (c) Are our esimes in prs () n () upper n lower esimes? Eplin. (s) v (fs) Oil leke from nk re of r liers per hour. The re ecrese s ime psse n vlues of he re -hour ime inervls re shown in he le. Fin lower n upper esimes for he ol moun of oil h leke ou. h 6 8 r (Lh) When we esime isnces from veloci, i is someimes necessr o use imes,,,,... h re no equll spce. We cn sill esime isnces using he ime perios i i i. For emple, on M 7, 99, he spce shule Enevour ws lunche on mission STS-9, he purpose of which ws o insll new perigee kick moor in n Inels communicions sellie. The le, provie NASA, gives he veloci for he shule eween lifoff n he jeisoning of he soli rocke oosers. Even Time (s) Veloci (fs) Lunch Begin roll mneuver 85 En roll mneuver 5 9 Throle o 89% 7 Throle o 67% 7 Throle o % 59 5 Mimum nmic pressure 6 5 Soli rocke ooser seprion SECTION 5. AREAS AND DISTANCES 79 Use hese o esime he heigh ove Erhs surfce of he spce shule Enevour, 6 secons fer lifoff. The veloci grph of rking cr is shown. Use i o esime he isnce rvele he cr while he rkes re pplie. 6. The veloci grph of cr ccelering from res o spee of kmh over perio of secons is shown. Esime he isnce rvele uring his perio. 7 9 Use Deniion o n n e pression for he re uner he grph of f s i. Do no evlue he i. 7. f s, 6 8. f ln, 9. f cos, Deermine region whose re is equl o he given i. Do no evlue he i... nl n 5 i i n nl n i (f/s) 6 (km/h) n 8 (secons) n i n n 6 (secons). () Use Deniion o n n e pression for he re uner he curve from o s i. () The following formul for he sum of he cues of he rs n inegers is prove in Appeni E. Use i o evlue he i in pr (). nn n Coprigh 5 Thomson Lerning, Inc. All Righs Reserve. M no e copie, scnne, or uplice, in whole or in pr. Coprigh 5 Thomson Lerning, Inc. All Righs Reserve. M no e copie, scnne, or uplice, in whole or in pr.

7 8 CHAPTER 5 INTEGRALS License o: jsmuels@mcc.cun.eu CAS. () Epress he re uner he curve 5 from o s i. () Use compuer lger ssem o n he sum in our epression from pr (). (c) Evlue he i in pr (). PDF Cree wih eskpdf PDF Wrier - Tril :: hp:// CAS. Fin he ec re of he region uner he grph of e from o using compuer lger ssem o evlue he sum n hen he i in Emple (). Compre our nswer wih he esime oine in Emple (). 5. T h e D e f i n i e I n e g r l We sw in Secion 5. h i of he form CAS 5. Fin he ec re uner he cosine curve cos from o, where. (Use compuer lger ssem oh o evlue he sum n compue he i.) In priculr, wh is he re if? 6. () Le An e he re of polgon wih n equl sies inscrie in circle wih rius r. B iviing he polgon ino n congruen ringles wih cenrl ngle n, show h An nr sinn. () Show h nl An r. [Hin: Use Equion...] rises when we compue n re. We lso sw h i rises when we r o n he isnce rvele n ojec. I urns ou h his sme pe of i occurs in wie vrie of siuions even when f is no necessril posiive funcion. In Chpers 6 n 8 we will see h is of he form () lso rise in ning lenghs of curv es, volumes of solis, ceners of mss, force ue o wer pressure, n work, s well s oher quniies. We herefore give his pe of i specil nme n noion. Definiion of Definie Inegrl If f is coninuous funcion ene for, we ivie he inervl, ino n suinervls of equl wih n. We le,,,..., n ( ) e he enpoins of hese suinervls n we le *, *,..., n* e n smple poins in hese suinervls, so i* lies in he ih suinervl i, i. Then he enie inegrl of f from o is Becuse we hve ssume h f is coninuous, i cn e prove h he i in Deniion l ws eiss n gives he sme vlue no mer how we choose he smple poins i*. (See Noe for precise eniion of his pe of i.) If we k e he smple poins o e righ enpoins, hen i* i n he eniion of n ine grl ecomes nl n f i* f * i nl f * f n* f nl n f nl n i i f i* f i License o: jsmuels@mcc.cun.eu Bernhr Riemnn receive his Ph.D. uner he irecion of he legenr Guss he Universi of Göingen n remine here o ech. Guss, who ws no in he hi of prising oher mhemicins, spoke of Riemnn s creive, cive, rul mhemicl min n gloriousl ferile originli. The efiniion () of n inegrl h we use is ue o Riemnn. He lso me mjor conriuions o he heor of funcions of comple vrile, mhemicl phsics, numer heor, n he founions of geomer. Riemnn s ro concep of spce n geomer urne ou o e he righ seing, 5 ers ler, for Einsein s generl relivi heor. Riemnn s helh ws poor hroughou his life, n he ie of uerculosis he ge of 9. SECTION 5. THE DEFINITE INTEGRAL 8 If we choose he smple poins o e lef enpoins, hen i* i n he eniion ecomes Alernivel, we coul choose i* o e he mipoin of he suinervl or n oher numer eween i n i. Alhough mos of he funcions h we encouner re coninuous, he i in Deniion lso eiss if f hs nie numer of remo vle or jump isconinuiies (u no innie isconinuiies). (See Secion.5.) So we cn lso ene he enie ine grl for such funcions. NOTE The smol ws inrouce Leiniz n is clle n inegrl sign. I is n elonge S n ws chosen ecuse n inegrl is i of sums. In he noion f, f is clle he inegrn n n re clle he is of inegrion; is he lower i n is he upper i. The smol hs no ofcil mening iself; f is ll one smol. The proceure of clculing n inegrl is clle inegrion. NOTE The enie ine grl f is numer; i oes no epen on. In fc, we coul use n leer in plce of wihou chnging he vlue of he inegrl: NOTE The sum h occurs in Deniion is clle Riemnn sum fer he Germn mhemicin Bernhr Riemnn (86 866). We know h if f hppens o e posiive, hen he Riemnn sum cn e inerpree s sum of res of pproiming recngles (see Figure ). B compring Deniion wih he eniion of re in Secion 5., we see h he enie ine grl f cn e inerpree s he re uner he curve f from o. (See Figure.) Î i * FIGURE If ƒ, he Riemnn sum µf( i *)Î is he sum of res of recngles. f f f r r f n l n n f i* i i f i =ƒ FIGURE If ƒ, he inegrl j ƒ is he re uner he curve =ƒ from o. Coprigh 5 Thomson Lerning, Inc. All Righs Reserve. M no e copie, scnne, or uplice, in whole or in pr. Coprigh 5 Thomson Lerning, Inc. All Righs Reserve. M no e copie, scnne, or uplice, in whole or in pr.

8 8 CHAPTER 5 INTEGRALS SECTION 5. THE DEFINITE INTEGRAL 8 License o: jsmuels@mcc.cun.eu =ƒ PDF Cree wih eskpdf PDF Wrier - Tril :: hp:// + + _ FIGURE µf( i *)Î is n pproimion o he ne re + =ƒ _ FIGURE j ƒ is he ne re + If f kes on oh posiive n negive vlues, s in Figure, hen he Riemnn sum is he sum of he res of he recngles h lie ove he -is n he negives of he res of he recngles h lie elow he -is (he res of he gol recngles minus he res of he lue recngles). When we ke he i of such Riemnn sums, we ge he siuion illusre in Figure. A enie ine grl cn e inerpree s ne re, h is, ifference of res: where A is he re of he region ove he -is n elow he grph of f, n A is he re of he region elow he -is n ove he grph of f. NOTE In he spiri of he precise eniion of he i of funcion in Secion., we cn wrie he precise mening of he i h enes he ine grl in Deniion s follows: For ever numer here is n ineger N such h for ever ineger n N n for ever choice of i* in i, i. This mens h enie ine grl cn e pproime o wihin n esire egree of ccurc Riemnn sum. NOTE 5 Alhough we hve ene f iviing, ino suinervls of equl wih, here re siuions in which i is vngeous o work wih suinervls of unequl wih. For insnce, in Eercise in Secion 5. NASA provie veloci imes h were no equll spce, u we were sill le o esime he isnce rvele. An here re mehos for numericl inegrion h ke vnge of unequl suinervls. If he suinervl wihs re,,..., n, we hve o ensure h ll hese wihs pproch in he iing process. This hppens if he lrges wih, m i, pproches. So in his cse he eniion of enie ine grl ecomes EXAMPLE Epress s n inegrl on he inervl,. SOLUTION Compring he given i wih he i in Deniion, we see h he will e ienicl if we choose (So he smple poins re righ enpoins n he given i is of he form of Equion.) We re given h n. Therefore, Deniion or Equion, we hve i i sin i sin nl n i f n f i* i f nl n i i sin i i f sin f A A m n f i* i il i n i* i License o: jsmuels@mcc.cun.eu Formuls 7 re prove wriing ou ech sie in epne form. The lef sie of Equion 8 is c c c n The righ sie is c n These re equl he isriuive proper. The oher formuls re iscusse in Appeni E. Tr more prolems like his one. Resources / Moule 6 / Wh Is Are? / Prolems n Tess Ler, when we ppl he enie ine grl o phsicl siuions, i will e imporn o recognize is of sums s inegrls, s we i in Emple. When Leiniz chose he noion for n inegrl, he chose he ingreiens s reminers of he iing process. In generl, when we wrie we replce, i*, n. E v l u i n g I n e g r l s When we use he eniion o e vlue enie ine grl, we nee o know how o work wih sums. The following hree equions give formuls for sums of powers of posiive inegers. Equion m e fmilir o ou from course in lger. Equions 5 n 6 were iscusse in Secion 5. n re prove in Appeni E. 5 6 The remining formuls re simple rules for working wih sigm noion: nl n i n i i EXAMPLE () Evlue he Riemnn sum for f 6 king he smple poins o e righ enpoins n,, n n 6. () Evlue 6. SOLUTION () Wih n 6 he inervl wih is n nn i i n nn n i i 6 n ci c n i i n i i n i i n i i n i i n f i* f nn n c nc i i i n i n 6 i i i i Coprigh 5 Thomson Lerning, Inc. All Righs Reserve. M no e copie, scnne, or uplice, in whole or in pr. Coprigh 5 Thomson Lerning, Inc. All Righs Reserve. M no e copie, scnne, or uplice, in whole or in pr.

9 8 CHAPTER 5 INTEGRALS License o: jsmuels@mcc.cun.eu PDF Cree wih eskpdf PDF Wrier - Tril :: hp:// 5 = -6 FIGURE 5 5 = -6 A A FIGURE 6 j ( -6)=A -A =_6.75 In he sum, n is consn (unlike i), so we cn move n in fron of he sign. n he righ enpoins re.5,.,.5,., 5.5, n 6.. So he Riemnn sum is R6 6 f i i f.5 f. f.5 f. f.5 f Noice h f is no posiive funcion n so he Riemnn sum oes no represen sum of res of recngles. Bu i oes represen he sum of he res of he gol recngles (ove he -is) minus he sum of he res of he lue recngles (elow he -is) in Figure 5. () Wih n suinervls we hve Thus, n, 6n, 9n, n, in generl, i in. Since we re using righ enpoins, we cn use Equion : 6 nl n f i i nl f n n i i n nl n i i n 6 n i nl n i 7 n n i 8 n i nl 8 n n i nl 8 n i 5 n n nn nl 8 n n n i 5 nn n 7 n (Equion 8 wih c n) (Equions n 8) This inegrl cn e inerpree s n re ecuse f kes on oh posiive n negive vlues. Bu i cn e inerpree s he ifference of res A A, where A n re shown in Figure 6. Figure 7 illusres he clculion showing he posiive n negive erms in he righ Riemnn sum Rn for n. The vlues in he le show he Riemnn sums pproching he ec vlue of he inegrl, 6.75, s nl. i n A License o: jsmuels@mcc.cun.eu FIGURE 7 R Å_6.998 Becuse f e is posiive, he inegrl in Emple represens he re shown in Figure 8. FIGURE 8 = A compuer lger ssem is le o fin n eplici epression for his sum ecuse i is geomeric series. The i coul e foun using l Hospil s Rule. 5 = -6 SECTION 5. THE DEFINITE INTEGRAL 85 A much simpler meho for evluing he inegrl in Emple will e given in Secion 5.. EXAMPLE () Se up n epression for e s i of sums. () Use compuer lger ssem o evlue he epression. SOLUTION () Here we hve f e,,, n So, n, n, 6n, n From Equion, we ge () If we sk compuer lger ssem o evlue he sum n simplif, we oin Now we sk he compuer lger ssem o evlue he i: n i i n e f i i nl n f n i i n nl n nl n e in e nn e nn i e n e nl n e nn e nn e e e n We will lern much esier meho for he evluion of inegrls in he ne secion. n n i n e in n Rn Coprigh 5 Thomson Lerning, Inc. All Righs Reserve. M no e copie, scnne, or uplice, in whole or in pr. Coprigh 5 Thomson Lerning, Inc. All Righs Reserve. M no e copie, scnne, or uplice, in whole or in pr.

10 86 CHAPTER 5 INTEGRALS License o: jsmuels@mcc.cun.eu PDF Cree wih eskpdf PDF Wrier - Tril :: hp:// FIGURE 9 = œ - or + = FIGURE Moule 5./5./7.7 shows how he Mipoin Rule esimes improve s n increses. EXAMPLE Evlue he following inegrls inerpreing ech in erms of res. () () SOLUTION () Since f s, we cn inerpre his inegrl s he re uner he curve s from o. Bu, since, we ge, which shows h he grph of f is he qurer-circle wih rius in Figure 9. Therefore (In Secion 7. we will e le o prove h he re of circle of rius r is r.) () The grph of is he line wih slope shown in Figure. We compue he inegrl s he ifference of he res of he wo ringles: T h e M i p o i n R u l e We ofen choose he smple poin i* o e he righ enpoin of he ih suinervl ecuse i is convenien for compuing he i. Bu if he purpose is o n n pproimion o n inegrl, i is usull eer o choose i* o e he mipoin of he inervl, which we enoe i. An Riemnn sum is n pproimion o n inegrl, u if we use mipoins we ge he following pproimion. Mipoin Rule where n s A A.5 _ f n f i f f n i A n s =- i i i mipoin of i, i EXAMPLE 5 Use he Mipoin Rule wih n 5 o pproime. A (,) License o: jsmuels@mcc.cun.eu FIGURE = FIGURE M Å_6.756 SECTION 5. THE DEFINITE INTEGRAL 87 SOLUTION The enpoins of he v e suinervls re,.,.,.6,.8, n., so he mipoins re.,.,.5,.7, n.9. The wih of he suinervls is 5 5, so he Mipoin Rule gives.6998 Since f for, he inegrl represens n re, n he pproimion given he Mipoin Rule is he sum of he res of he recngles shown in Figure. A he momen we on know how ccure he pproimion in Emple 5 is, u in Secion 7.7 we will lern meho for esiming he error involve in using he Mipoin Rule. A h ime we will iscuss oher mehos for pproiming enie ine grls. If we ppl he Mipoin Rule o he inegrl in Emple, we ge he picure in Figure. The pproimion M is much closer o he rue vlue 6.75 hn he righ enpoin pproimion, R 6.998, shown in Figure 7. P r o p e r i e s o f h e D e f i n i e I n e g r l When we ene he enie ine grl f, we implicil ssume h. Bu he eniion s i of Riemnn sums mk es sense even if. Noice h if we reverse n, hen chnges from n o n. Therefore If, hen n so f. f. f.5 f.7 f = -6 f f We now evelop some sic properies of inegrls h will help us o evlue inegrls in simple mnner. We ssume h f n re coninuous funcions. f Coprigh 5 Thomson Lerning, Inc. All Righs Reserve. M no e copie, scnne, or uplice, in whole or in pr. Coprigh 5 Thomson Lerning, Inc. All Righs Reserve. M no e copie, scnne, or uplice, in whole or in pr.

11 88 CHAPTER 5 INTEGRALS SECTION 5. THE DEFINITE INTEGRAL 89 License o: jsmuels@mcc.cun.eu PDF Cree wih eskpdf PDF Wrier - Tril :: hp:// c =c FIGURE j c=c(-) re=c(-) g f+g FIGURE j [ƒ+ ]= j ƒ+j Proper seems inuiivel resonle ecuse we know h mulipling funcion posiive numer c sreches or shrinks is grph vericll fcor of c. So i sreches or shrinks ech pproiming recngle fcor c n herefore i hs he effec of mulipling he re c. f Properies of he Inegrl. c c, where c is n consn.. f, where c is n consn. f f cf c f f Proper ss h he inegrl of consn funcion f c is he consn imes he lengh of he inervl. If c n, his is o e epece ecuse c is he re of he she recngle in Figure. Proper ss h he inegrl of sum is he sum of he inegrls. For posiive funcions i ss h he re uner f is he re uner f plus he re uner. Figure helps us unersn wh his is rue: In view of how grphicl iion works, he corresponing vericl line segmens hve equl heigh. In generl, Proper follows from Equion n he fc h he i of sum is he sum of he is: Proper cn e prove in similr mnner n ss h he inegrl of consn imes funcion is he consn imes he inegrl of he funcion. In oher wors, consn (u onl consn) cn e ken in fron of n inegrl sign. Proper is prove wriing f f n using Properies n wih c. EXAMPLE 6 Use he properies of inegrls o evlue. SOLUTION Using Properies n of inegrls, we hve We know from Proper h f nl n nl n f i n i i i nl n i i f f i i f i nl n i i License o: jsmuels@mcc.cun.eu =ƒ c FIGURE 5 M =ƒ m FIGURE 6 n we foun in Emple in Secion 5. h. So The ne proper ells us how o comine inegrls of he sme funcion over jcen inervls: 5. This is no es o prove in generl, u for he cse where f n c Proper 5 cn e seen from he geomeric inerpreion in Figure 5: The re uner f from o c plus he re from c o is equl o he ol re from o. EXAMPLE 7 If i is known h f 7 n 8 f, n f. SOLUTION B Proper 5, we hve so Noice h Properies 5 re rue wheher,, or. The following properies, in which we compre sizes of funcions n sizes of inegrls, re rue onl if. Comprison Properies of he Inegrl 6. If f for, hen f. 7. If f for, hen f. 8. If m f M for, hen If f, hen f represens he re uner he grph of f, so he geomeric inerpreion of Proper 6 is simpl h res re posiive. Bu he proper cn e prove from he eniion of n ine grl (Eercise 6). Proper 7 ss h igger funcion hs igger inegrl. I follows from Properies 6 n ecuse f. Proper 8 is illusre Figure 6 for he cse where f. If f is coninuous we coul ke m n M o e he solue minimum n mimum vlues of f on he iner- 8 c 8 f f f f f f 8 f f 8 f 7 5 c m f M 5 8 Coprigh 5 Thomson Lerning, Inc. All Righs Reserve. M no e copie, scnne, or uplice, in whole or in pr. Coprigh 5 Thomson Lerning, Inc. All Righs Reserve. M no e copie, scnne, or uplice, in whole or in pr.

12 9 CHAPTER 5 INTEGRALS License o: jsmuels@mcc.cun.eu PDF Cree wih eskpdf PDF Wrier - Tril :: hp:// FIGURE 7 = =/e =e 5. Eercises vl,. In his cse Proper 8 ss h he re uner he grph of f is greer hn he re of he recngle wih heigh m n less hn he re of he recngle wih heigh M. Proof of Proper 8 Since m f M, Proper 7 gives Using Proper o evlue he inegrls on he lef n righ sies, we oin Proper 8 is useful when ll we wn is rough esime of he size of n inegrl wihou going o he oher of using he Mipoin Rule. EXAMPLE 8 Use Proper 8 o esime e. SOLUTION Becuse f e is ecresing funcion on,, is solue mimum vlue is M f n is solue minimum vlue is m f e. Thus, Proper 8, or Since e.679, we cn wrie. Evlue he Riemnn sum for f,, wih four suinervls, king he smple poins o e righ enpoins. Eplin, wih he i of igrm, wh he Riemnn sum represens.. If f ln,, evlue he Riemnn sum wih n 6, king he smple poins o e lef enpoins. (Give our nswer correc o si eciml plces.) Wh oes he Riemnn sum represen? Illusre wih igrm.. If f s, 6, n he Riemnn sum wih n 5 correc o si eciml plces, king he smple poins o e mipoins. Wh oes he Riemnn sum represen? Illusre wih igrm.. () Fin he Riemnn sum for f sin,, wih si erms, king he smple poins o e righ en- m f M m f M e e e.67 e e The resul of Emple 8 is illusre in Figure 7. The inegrl is greer hn he re of he lower recngle n less hn he re of he squre. poins. (Give our nswer correc o si eciml plces.) Eplin wh he Riemnn sum represens wih he i of skech. () Repe pr () wih mipoins s he smple poins. 5. The grph of funcion f is given. Esime f using four suinervls wih () righ enpoins, () lef enpoins, n (c) mipoins. f 8 License o: jsmuels@mcc.cun.eu 6. The grph of is shown. Esime wih si suinervls using () righ enpoins, () lef enpoins, n (c) mipoins. CAS 7. A le of vlues of n incresing funcion f is shown. Use he le o n lo wer n upper esimes for f. 8. The le gives he vlues of funcion oine from n eperimen. Use hem o esime 6 f using hree equl suinervls wih () righ enpoins, () lef enpoins, n (c) mipoins. If he funcion is known o e ecresing funcion, cn ou s wheher our esimes re less hn or greer hn he ec vlue of he inegrl? 9 Use he Mipoin Rule wih he given vlue of n o pproime he inegrl. Roun he nswer o four eciml plces. 9. s, n. If ou hve CAS h evlues mipoin pproimions n grphs he corresponing recngles (use milesum n mileo commns in Mple), check he nswer o Eercise n illusre wih grph. Then repe wih n n n.. Wih progrmmle clculor or compuer (see he insrucions for Eercise 7 in Secion 5.), compue he lef n righ Riemnn sums for he funcion f sin on he inervl, wih n. Eplin wh hese esimes show h Deuce h he pproimion using he Mipoin Rule wih n 5 in Eercise is ccure o wo eciml plces... sin, n sin f f g sec, n 6 e, n SECTION 5. THE DEFINITE INTEGRAL 9 5. Use clculor or compuer o mke le of vlues of righ Riemnn sums Rn for he inegrl sin wih n 5,, 5, n. Wh vlue o hese numers pper o e pproching? 6. Use clculor or compuer o mke le of vlues of lef n righ Riemnn sums Ln n Rn for he inegrl wih n 5,, 5, n. Beween wh wo e numers mus he vlue of he inegrl lie? Cn ou mke similr semen for he inegrl? Eplin. e 7 Epress he i s enie ine grl on he given inervl ,, nl n i* 6 i* 5 i 5 Use he form of he eniion of he ine grl given in Equion o evlue he inegrl n l n i n l n i nl n i i sin i,, e i,, 5 i si* i*,, 8] 6. () Fin n pproimion o he inegrl using Riemnn sum wih righ enpoins n n 8. () Drw igrm like Figure o illusre he pproimion in pr (). (c) Use Equion o evlue. () Inerpre he inegrl in pr (c) s ifference of res n illusre wih igrm like Figure. 7. Prove h.. 8. Prove h. 9 Epress he inegrl s i of Riemnn sums. Do no evlue he i ln Coprigh 5 Thomson Lerning, Inc. All Righs Reserve. M no e copie, scnne, or uplice, in whole or in pr. Coprigh 5 Thomson Lerning, Inc. All Righs Reserve. M no e copie, scnne, or uplice, in whole or in pr.

13 License o: CAS Epress he inegrl s i of sums. Then evlue, using compuer lger ssem o n oh he sum n he i. PDF Cree wih eskpdf PDF Wrier - Tril :: hp:// 9 CHAPTER 5 INTEGRALS. sin 5.. The grph of f is shown. Evlue ech inegrl inerpreing i in erms of res. () (c). The grph of consiss of wo srigh lines n semicircle. Use i o evlue ech inegrl. () () () () (c) 5 Evlue he inegrl inerpreing i in erms of res. 5. ( 6. s ( s9 ) f f Given h s, wh is s? 9. Evlue cos. 6 =ƒ 6 8 = 7. In Emple in Secion 5. we showe h. Use his fc n he properies of inegrls o evlue f f 7 5. Use he properies of inegrls n he resul of Emple o evlue e. 5. Use he resul of Emple o evlue e. 6. Use he resul of Eercise 7 n he fc h cos (from Eercise 5 in Secion 5.), ogeher wih he properies of inegrls, o evlue cos 5. Wrie s single inegrl in he form f : If n, n 5 f 5 f.6 f. 9. If n 9 f 7 9 6, n 9 f. 5. Fin f if 5 5 Use he properies of inegrls o verif he inequli wihou evluing he inegrls s s 5. 6 sin Use Proper 8 o esime he vlue of he inegrl n e Use properies of inegrls, ogeher wih Eercises 7 n 8, o prove he inequli s5 s 5 sin sin s 6 sin f 5 f f f for for 8 s sin License o: jsmuels@mcc.cun.eu 6. Prove Proper of inegrls. 6. Prove Proper 6 of inegrls. 65. If f is coninuous on,, show h f f f f f [Hin:.] 66. Use he resul of Eercise 65 o show h f sin f DISCOVERY PROJECT Are Funcions DISCOVERY PROJECT AREA FUNCTIONS Epress he i s enie ine grl. 67. [Hin: Consier f.] 68. nl n i nl n n i i n 5 in 69. Fin. Hin: Choose o e he geomeric men of i* i n i (h is, i* sii ) n use he ieni mm m m. () Drw he line n use geomer o n he re uner his line, ove he -is, n eween he vericl lines n. () If, le A e he re of he region h lies uner he line eween n. Skech his region n use geomer o n n e pression for A. (c) Differenie he re funcion A. Wh o ou noice?. () If, le A A represens he re of region. Skech h region. () Use he resul of Eercise 8 in Secion 5. o n n e pression for A. (c) Fin A. Wh o ou noice? () If n h is smll posiive numer, hen A h A represens he re of region. Descrie n skech he region. (e) Drw recngle h pproimes he region in pr (). B compring he res of hese wo regions, show h A h A h (f) Use pr (e) o give n inuiive eplnion for he resul of pr (c). ;. () Drw he grph of he funcion f cos in he viewing recngle,.5,.5. () If we ene ne w funcion cos hen is he re uner he grph of f from o [unil f ecomes negive, which poin ecomes ifference of res]. Use pr () o eermine he vlue of which srs o ecrese. [Unlike he inegrl in Prolem, i is impossile o evlue he inegrl ening o oin n eplici epression for.] (c) Use he inegrion commn on our clculor or compuer o esime (.), (.), (.6),...,(.8),(). Then use hese vlues o skech grph of. () Use our grph of from pr (c) o skech he grph of using he inerpreion of s he slope of ngen line. How oes he grph of compre wih he grph of f? Coprigh 5 Thomson Lerning, Inc. All Righs Reserve. M no e copie, scnne, or uplice, in whole or in pr. Coprigh 5 Thomson Lerning, Inc. All Righs Reserve. M no e copie, scnne, or uplice, in whole or in pr.

14 9 CHAPTER 5 INTEGRALS License o: jsmuels@mcc.cun.eu PDF Cree wih eskpdf PDF Wrier - Tril :: hp:// 5. T h e F u n m e n l T h e o r e m o f C l c u l u s Invesige he re funcion inercivel. Resources / Moule 6 / Ares n Derivives / Are s Funcion FIGURE FIGURE re= =f() =f(). Suppose f is coninuous funcion on he inervl, n we ene ne w funcion he equion f Bse on our resuls in Prolems, conjecure n epression for. The Funmenl Theorem of Clculus is ppropriel nme ecuse i eslishes connecion eween he wo rnches of clculus: ifferenil clculus n inegrl clculus. Differenil clculus rose from he ngen prolem, wheres inegrl clculus rose from seemingl unrele prolem, he re prolem. Newons echer Cmrige, Isc Brrow (6 677),iscovere h hese wo prolems re cull closel rele. In fc, he relize h iffereniion n inegrion re inverse processes. The Funmenl Theorem of Clculus gives he precise inverse relionship eween he erivive n he inegrl. I ws Newon n Leiniz who eploie his relionship n use i o evelop clculus ino ssemic mhemicl meho. In priculr, he sw h he Funmenl Theorem enle hem o compue res n inegrls ver esil wihou hving o compue hem s is of sums s we i in Secions 5. n 5.. The rs pr of he Funmenl Theorem els wih funcions ene n equion of he form where f is coninuous funcion on, n vries eween n. Oserve h epens onl on, which ppers s he vrile upper i in he inegrl. If is e numer, hen he inegrl f is enie numer. If we hen le vr, he numer f lso vries n enes funcion of enoe. If f hppens o e posiive funcion, hen cn e inerpree s he re uner he grph of f from o, where cn vr from o. (Think of s he re so f r funcion; see Figure.) EXAMPLE If f is he funcion whose grph is shown in Figure n f, n he v lues of,,,,, n 5. Then skech rough grph of. SOLUTION Firs we noice h f. From Figure we see h is he re of ringle: To n we o he re of recngle: f f f f f License o: jsmuels@mcc.cun.eu FIGURE g()= FIGURE =j f() g()= 5 +h FIGURE 5 ƒ h g We esime h he re uner f from o is ou., so For, f is negive n so we sr surcing res: We use hese vlues o skech he grph of in Figure. Noice h, ecuse f is posiive for, we keep ing re for n so is incresing up o, where i ins mimum vlue. For, ecreses ecuse f is negive. If we ke f n, hen, using Eercise 7 in Secion 5., we hve Noice h, h is, f. In oher wors, if is ene s he ine grl of f Equion, hen urns ou o e n nierivive of f, les in his cse. An if we skech he erivive of he funcion shown in Figure esiming slopes of ngens, we ge grph like h of f in Figure. So we suspec h f in Emple oo. To see wh his migh e generll rue we consier n coninuous funcion f wih f. Then f cn e inerpree s he re uner he grph of f from o, s in Figure. In orer o compue from he eniion of eri vive we rs oserv e h, for h, h is oine surcing res, so i is he re uner he grph of f from o h (he gol re in Figure 5). For smll h ou cn see from he gure h his re is pproimel equl o he re of he recngle wih heigh f n wih h: so g()å. f.. f f..7 g()å h hf h h SECTION 5. THE FUNDAMENTAL THEOREM OF CALCULUS 95 f g(5)å.7 Coprigh 5 Thomson Lerning, Inc. All Righs Reserve. M no e copie, scnne, or uplice, in whole or in pr. Coprigh 5 Thomson Lerning, Inc. All Righs Reserve. M no e copie, scnne, or uplice, in whole or in pr.

15 96 CHAPTER 5 INTEGRALS SECTION 5. THE FUNDAMENTAL THEOREM OF CALCULUS 97 License o: jsmuels@mcc.cun.eu PDF Cree wih eskpdf PDF Wrier - Tril :: hp:// We revie he nme of his heorem s FTC. In wors, i ss h he erivive of efinie inegrl wih respec o is upper i is he inegrn evlue he upper i. =ƒ FIGURE 6 m M u =+h Inuiivel, we herefore epec h The fc h his is rue, even when f is no necessril posiive, is he rs pr of he Funmenl Theorem of Clculus. The Funmenl Theorem of Clculus, Pr If f is coninuous on,, hen he funcion ene is coninuous on, n ifferenile on,, n f. Proof If n h re in,, hen n so, for h, ( Proper 5) For now le us ssume h h. Since f is coninuous on, h, he Ereme Vlue Theorem ss h here re numers u n v in, h such h f u m n f v M, where m n M re he solue minimum n mimum vlues of f on, h. (See Figure 6.) B Proper 8 of inegrls, we hve h is, Since h, we cn ivie his inequli h: Now we use Equion o replce he mile pr of his inequli: h f h l h h h f f h f f h f f f h h mh h f Mh f uh h f f vh f u h f f v h f u h h f v Inequli cn e prove in similr mnner for he cse h. (See Eercise 6.) h f h License o: jsmuels@mcc.cun.eu Moule 5. provies visul evience for FTC. Now we le hl. Then ul n vl, since u n v lie eween n h. Therefore n ecuse f is coninuous. We conclue, from () n he Squeeze Theorem, h If or, hen Equion cn e inerpree s one-sie i. Then Theorem.9. (moie for one-sie is) sho ws h is coninuous on,. Using Leiniz noion for erivives, we cn wrie FTC s 5 hl hl f u f u f ul f v f v f vl h f hl h f f when f is coninuous. Roughl speking, Equion 5 ss h if we rs ine gre f n hen ifferenie he resul, we ge ck o he originl funcion f. EXAMPLE Fin he erivive of he funcion s. SOLUTION Since f s is coninuous, Pr of he Funmenl Theorem of Clculus gives s EXAMPLE Alhough formul of he form f m seem like srnge w of ening funcion, ooks on phsics, chemisr, n sisics re full of such funcions. For insnce, he Fresnel funcion S sin is nme fer he French phsicis Augusin Fresnel (788 87),who is fmous for his works in opics. This funcion rs ppere in Fresnel s heor of he iffrcion of ligh wves, u more recenl i hs een pplie o he esign of highws. Pr of he Funmenl Theorem ells us how o ifferenie he Fresnel funcion: S sin This mens h we cn ppl ll he mehos of ifferenil clculus o nlze S (see Eercise 57). Coprigh 5 Thomson Lerning, Inc. All Righs Reserve. M no e copie, scnne, or uplice, in whole or in pr. Coprigh 5 Thomson Lerning, Inc. All Righs Reserve. M no e copie, scnne, or uplice, in whole or in pr.

16 98 CHAPTER 5 INTEGRALS SECTION 5. THE FUNDAMENTAL THEOREM OF CALCULUS 99 License o: jsmuels@mcc.cun.eu PDF Cree wih eskpdf PDF Wrier - Tril :: hp:// We revie his heorem s FTC. Figure 7 shows he grphs of f sin n he Fresnel funcion S f. A compuer ws use o grph S compuing he vlue of his inegrl for mn vlues of. I oes inee look s if S is he re uner he grph of f from o [unil. when S ecomes ifference of res]. Figure 8 shows lrger pr of he grph of S. f FIGURE 7 ƒ=sin(π /) S()=j sin(π@/) S If we now sr wih he grph of S in Figure 7 n hink ou wh is erivive shoul look like, i seems resonle h S f. [For insnce, S is incresing when f n ecresing when f.] So his gives visul conrmion of Pr of he Funmenl Theorem of Clculus. EXAMPLE Fin sec. SOLUTION Here we hve o e creful o use he Chin Rule in conjuncion wih FTC. Le u. Then sec u sec u u sec u u sec ( he Chin Rule) ( FTC) In Secion 5. we compue inegrls from he eniion s i of Riemnn sums n we sw h his proceure is someimes long n ifcul. The secon pr of he Funmenl Theorem of Clculus, which follows esil from he rs pr, pro vies us wih much simpler meho for he evluion of inegrls. The Funmenl Theorem of Clculus, Pr If f is coninuous on,, hen f F F where F is n nierivive of f, h is, funcion such h F f. FIGURE 8 The Fresnel funcion S()=j sin(π@/) sec u.5 License o: jsmuels@mcc.cun.eu Compre he clculion in Emple 5 wih he much hrer one in Emple in Secion 5.. Proof Le f. We know from Pr h f ; h is, is n nierivive of f. If F is n oher nierivive of f on,, hen we know from Corollr..7 h F n iffer consn: 6 for. Bu oh F n re coninuous on, n so, king is of oh sies of Equion 6 (s l n l ), we see h i lso hols when n. If we pu in he formul for, we ge So, using Equion 6 wih n, we hve Pr of he Funmenl Theorem ses h if we know n nierivive F of f, hen we cn evlue f simpl surcing he vlues of F he enpoins of he inervl,. Is ver surprising h f, which ws ene complice proceure involving ll of he vlues of f for, cn e foun knowing he vl- ues of F onl wo poins, n. Alhough he heorem m e surprising rs glnce, i ecomes plusile if we inerpre i in phsicl erms. If v is he veloci of n ojec n s is is posiion ime, hen v s, so s is n nierivive of v. In Secion 5. we consiere n ojec h lws moves in he posiive irecion n me he guess h he re uner he veloci curve is equl o he isnce rvele. In smols: Th is ecl wh FTC ss in his cone. EXAMPLE 5 Evlue he inegrl e. SOLUTION The funcion f e is coninuous everwhere n we know h n nierivive is F e, so Pr of he Funmenl Theorem gives Noice h FTC ss we cn use n nierivive F of f. So we m s well use he simples one, nmel F e, inse of e 7 or e C. We ofen use he noion F F C C f f F C v s s e F F e e F] F F Coprigh 5 Thomson Lerning, Inc. All Righs Reserve. M no e copie, scnne, or uplice, in whole or in pr. Coprigh 5 Thomson Lerning, Inc. All Righs Reserve. M no e copie, scnne, or uplice, in whole or in pr.

17 CHAPTER 5 INTEGRALS SECTION 5. THE FUNDAMENTAL THEOREM OF CALCULUS License o: jsmuels@mcc.cun.eu PDF Cree wih eskpdf PDF Wrier - Tril :: hp:// In ppling he Funmenl Theorem we use priculr nierivive F of f. I is no necessr o use he mos generl nierivive. re= FIGURE 9 =cos π So he equion of FTC cn e wrien s Oher common noions re F n F. EXAMPLE 6 Fin he re uner he prol from o. SOLUTION An nierivive of f is F. The require re A is foun using Pr of he Funmenl Theorem: If ou compre he clculion in Emple 6 wih he one in Emple in Secion 5., ou will see h he Funmenl Theorem gives much shorer meho. EXAMPLE 7 Evlue 6. A SOLUTION The given inegrl is n reviion for An nierivive of f is n, ecuse 6, we cn wrie F ln. So EXAMPLE 8 Fin he re uner he cosine curve from o, where. SOLUTION Since n nierivive of f cos is F sin, we hve A cos sin ] sin sin sin f F] 6 6 F ln where ln ] 6 ln 6 ln ln 6 ln In priculr, king, we hve prove h he re uner he cosine curve from o is sin. (See Figure 9.) When he French mhemicin Gilles e Roervl rs foun he re uner he sine n cosine curves in 65, his ws ver chllenging prolem h require gre el of ingenui. If we in hve he ene of he Funmenl Theorem, we woul hve o compue ifcul i of sums using oscure rigonomeric ieniies (or compuer lger ssem s in Eercise 5 in Secion 5.). I ws even more ifcul for Roerv l ecuse he pprus of is h no een invene in 65. Bu in he 66s n 67s, when he Funmenl Theorem ws iscovere Brrow n eploie Newon n Leiniz, such prolems ecme ver es, s ou cn see from Emple 8. F f License o: jsmuels@mcc.cun.eu EXAMPLE 9 Wh is wrong wih he following clculion? SOLUTION To sr, we noice h his clculion mus e wrong ecuse he nswer is negive u f n Proper 6 of inegrls ss h f when f. The Funmenl Theorem of Clculus pplies o coninuous funcions. I cn e pplie here ecuse f is no coninuous on,. In fc, f hs n innie isconinui, so D i f f e r e n i i o n n I n e g r i o n s I n v e r s e P r o c e s s e s We en his secion ringing ogeher he wo prs of he Funmenl Theorem. The Funmenl Theorem of Clculus Suppose f is coninuous on,.. If f, hen f.. f F F, where F is n nierivive of f, h is, F f. We noe h Pr cn e rewrien s which ss h if f is inegre n hen he resul is ifferenie, we rrive ck he originl funcion f. Since F f, Pr cn e rewrien s oes no eis f f F F F This version ss h if we ke funcion F, rs if ferenie i, n hen inegre he resul, we rrive ck he originl funcion F, u in he form F F. Tken ogeher, he wo prs of he Funmenl Theorem of Clculus s h iffereniion n inegrion re inverse processes. Ech unoes wh he oher oes. The Funmenl Theorem of Clculus is unquesionl he mos imporn heorem in clculus n, inee, i rnks s one of he gre ccomplishmens of he humn min. Before i ws iscovere, from he ime of Euous n Archimees o he ime of Glileo n Ferm, prolems of ning res, v olumes, n lenghs of curves were so ifcul h onl genius coul mee he chllenge. Bu now, rme wih he ssemic meho h Newon n Leiniz fshione ou of he Funmenl Theorem, we will see in he chpers o come h hese chllenging prolems re ccessile o ll of us. Coprigh 5 Thomson Lerning, Inc. All Righs Reserve. M no e copie, scnne, or uplice, in whole or in pr. Coprigh 5 Thomson Lerning, Inc. All Righs Reserve. M no e copie, scnne, or uplice, in whole or in pr.