5.3 The Fundamental Theorem of Calculus, Part I

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1 59 C HAPTER 5 THE INTEGRAL θ b 87. Suppose tht f nd g re continuous functions such tht, for ll, Z Z f./d g./d Give n intuitive rgument showing tht f./ g./.eplin our ide with grph. Let R f./d R g./d.considerwhthppenss decreses in size, becoming ver close to zero. Intuitivel, the res of the functions become.. //.f.//.f.// nd.. //.g.//.g.//.becuseweknow these res must be the sme, we hve.f.//.g.// nd therefore f./ g./. 88. Theorem 4 remins true without the ssumption b c. Verifthisforthecsesb<<cnd c<<b. The dditivit propert of definite integrls sttes for b c,wehve R c f./d R b f./dc R c b f./d. Suppose tht we hve b<<c.bthedditivitpropert,wehve R c b f./d R b f./dc R c R f./d.therefore, c f./d R c b f./d R b f./d R b f./dc R c b f./d. Now suppose tht we hve c < < b.bthedditivitpropert,wehve R b c f./d R c f./d C R b f./d. Therefore, R c f./d R c f./d R b f./d R b c f./d R b f./dc R c b f./d. Hence the dditivit propert holds for ll rel numbers, b,ndc, regrdlessoftheirreltionshipmongstechother. 5. The Fundmentl Theorem of Clculus, Prt I Preliminr Questions. Suppose tht F./ f./nd F./, F./ 7. () Wht is the re under f./over Œ; if f./? (b) Wht is the grphicl interprettion of F./ F./if f./tkes on both positive nd negtive vlues? () If f./ over Œ;, thenthereunder f./is F./ F./ 7 4. (b) If f./tkes on both positive nd negtive vlues, then F./ F./gives the signed re between f./nd the -is.. Suppose tht f./is negtive function with ntiderivtive F such tht F./ 7 nd F./ 4. Whtisthere(positive number) between the -is nd the grph of f./over Œ;? Z f./d represents the signed re bounded b the curve nd the intervl Œ;. Sincef./is negtive on Œ;, Z f./d is the negtive of the re. Therefore, if A is the re between the -is nd the grph of f./,wehve: Z A f./d.f./ F.//.4 7/. / :. Are the following sttements true or flse? Eplin. () FTC I is vlid onl for positive functions. (b) To use FTC I, ou hve to choose the right ntiderivtive. (c) If ou cnnot find n ntiderivtive of f./,thenthedefiniteintegrldoesnoteist. () Flse. The FTC I is vlid for continuous functions. (b) Flse. The FTC I works for n ntiderivtive of the integrnd. (c) Flse. If ou cnnot find n ntiderivtive of the integrnd, ou cnnot use the FTC I to evlute the definite integrl, but the definite integrl m still eist.

2 Z 9 4. Evlute f./ d where f./is differentible nd f./ f.9/ 4. Z 9 Becuse f is differentible, f./ d f.9/ f./ 4 4. SECTION 5. The Fundmentl Theorem of Clculus, Prt I 59 Eercises InEercises 4,sketchtheregionunderthegrphofthefunctionndfinditsreusingFTCI.. f./, Œ; We hve the re. f./, Œ; Z A d ˇˇˇˇ : Let A be thereindicted.then: Z Z Z A. /d d d ˇˇˇˇ ˇˇˇˇ. f./, Œ; 8.4 / 4 : We hve the re 4. f./ cos, ; Z A d ˇ C :

3 59 C HAPTER 5 THE INTEGRAL Let A be theshdedre.then Z = = A cos d sin ˇ : In Eercises 5 4, evlute the integrl using FTC I. Z 5. d Z d ˇˇˇˇ././ 7. Z 9. d Z 9 9 d ˇ.9/./ 8. Z /d Z.4 9 /d. / ˇ. /. /. Z 8. u du Z u du uˇˇˇˇ./. / 5. Z C 4/ d Z. 5 C 4/ d. C / ˇ.8 C 8 8/. C / 8. Z.. 9 C 5 /d Z. 9 C 5 /d C ˇˇˇˇ C C : Z..t t /dt Z.t t /dt t4 t ˇˇˇˇ 8. / Z..5u 4 C u u/ du Z.5u 4 C u u/ du u 5 C u ˇˇˇˇ u C 8. Z 4 p. d Z 4 Z p 4 d = d 4 =ˇˇˇˇ.4/=./=.

4 Z = d Z 8 4= d 8 7 7=ˇˇˇˇ 8.8 / 7 7. Z 5. t =4 dt = Z t =4 dt 4 = 5 t5=4ˇˇˇˇ 4 = Z. t 5= dt 4 Z t 5= dt 4 7 t7=ˇˇˇˇ. 8/ Z dt 7. t Z Z dt t t dt t ˇˇˇˇ C. Z d Z 4 4 d 4 ˇˇˇˇ.4/ C 4. Z 8 9. = d Z Z 8 = d 8 d 4 ˇˇˇˇ 4C. = = Z. d Z d ˇ. / C. / 8. Z.. /d Z. /d C ˇˇˇˇ 8 C C : Z 9. t = dt Z 9 t = dt t =ˇˇˇˇ 9.9/ =./ = 4. Z 7 t C. p dt t SECTION 5. The Fundmentl Theorem of Clculus, Prt I 59 Z t 4= 8t = 4. 8=7 t dt Z 7 t C p t Z 7 dt.t = C t = /dt t= C t ˇˇˇˇ = 7.8p / C p C p 8 : Z 8=7 t 4= 8t = t Z dt.t = 8t 5= /dt 8=7.t = C t = / ˇ. C /. C 7/ 5: 8=7

5 594 C HAPTER 5 THE INTEGRAL Z =4 5. sin d =4 Z =4 =4 p p sin d cos ˇ =4 =4 C p : Z 4. sin d Z 4 4 sin d cos ˇ. / : Z! = 7. cos d Z! = = cos d sin ˇˇˇˇ. =8 8. cos d =4 =8 cos d 5=8 sin =4 ˇˇˇˇ =4 sin 5 4 sin p 4. Z = 9. sec t dt Z = sec t dt tn t = ˇˇˇˇ p C p 4 p. Z =. sec tn d Z = = sec tn d sec ˇ sec sec p. Z =. csc 5 cot 5 d = Z = csc 5 cot 5 d = csc = 5 5ˇˇˇˇ p = 5 5.p /. Z =4. csc 7 d =8 Z =4 csc 7 d =4 cot =8 7 7ˇˇˇˇ =8 7 cot C 7 cot 4 7. Z. e d Z e d e ˇˇˇˇ e. 4. e 4 d e 4 d 5 4 e 4ˇˇˇˇ 4 e C 4 e. Z 5. e t dt Z e t dt ˇˇˇˇ e t e 7 C e.e e 7 /. Z. e 4t dt

6 SECTION 5. The Fundmentl Theorem of Clculus, Prt I 595 Z e 4t dt 4 e4t ˇˇˇˇ 4 e9 4 e5. Z d 7. Z d ln jj ˇ ln ln ln 5. Z 4 d 8. Z 4 d 4 ln jj ˇ ln j 4j ln j j ln ln. Z dt 9. t C Z dt ln jt C j t C ˇ ln ln ln. Z 4 dt 4. 5t C 4 Z 4 dt 5t C 4 4 ln j5t C 4j 5 ˇ 5 ln 4 5 ln 9 5 ln 4 9. Z 4.. 9e /d Z. 9e /d e ˇˇˇˇ. /. e / e 9. Z 4. C d Z C d C ln jjˇˇˇˇ.8 C ln /. C ln / C ln. In Eercises 4 48, write the integrl s sum of integrls without bsolute vlues nd evlute. Z 4. jj d 44. j j d 45. Z j j d Z Z jj d Z. /d C d ˇˇˇˇ C ˇˇˇˇ Z j j d. /d C. / d ˇˇˇˇ C 5 9 :.4/ C 5 : C Z Z Z j j d. /dc d 4 4ˇˇˇˇ C 4 4ˇˇˇˇ C 4. /4 C : 5 ˇˇˇˇ

7 59 C HAPTER 5 THE INTEGRAL 4. Z j j d 47. Z jcos j d 48. Z Z Z j j d. /dc. / d ˇˇˇˇ C.9 / : Z Z = jcos j d j 4 C j d Z cos dc = C ˇˇˇˇ =. cos /d sin ˇ sin ˇ. / : = j 4 C j d j. /. /j d Z Z. 4 C / d C. 4 C / d C. 4 C / d C ˇˇˇˇ C ˇˇˇˇ C 5 C ˇˇˇˇ 5 C.9 8 C 9/ C C C 5 C C 9/ 8 : InEercises49 54,evlutetheintegrlintermsoftheconstnts. 49. d d b 4 4ˇˇˇˇ 4 b4 4./4 b 4 for n number b. 4 Z 5. 4 d b Z 4 d b 5 5ˇˇˇˇ b b5 for n numbers, b d 5 d b ˇˇˇˇ b./.b / for n number b. Z 5..t C t/dt 5. d d Z.t C t/dt 4 t4 C ˇˇˇˇ t 5 ln jj ˇ ln j5j ln jj ln C 4 4 C :

8 SECTION 5. The Fundmentl Theorem of Clculus, Prt I d b d b ln jj b ˇ b Z 55. Clculte f./d,where ln jb j ln jbj ln jbj. f./ ( for for > Z Z Z Z Z f./d f./dc f./d. /dc d ˇˇˇˇ C 4 4ˇˇˇˇ./. /. / C Z 5. Clculte f./d,where 8 C : f./ ( cos cos sin for for > Z Z Z Z Z f./d f./dc f./d cos dc.cos sin / d sin ˇ C sin C ˇˇˇˇ cos. / C C C : Z 57. Use FTC I to show tht n d if n is n odd whole number. Eplin grphicll. We hve Z n d nc n C ˇ./nC n C Becuse n is odd, n C is even, which mens tht. / nc./ nc. Hence. /nc n C :./ nc n C. /nc n C n C n C : Grphicll speking, for n odd function such s shown here, the positivel signed re from to cncels the negtivel signed re from to

9 598 C HAPTER 5 THE INTEGRAL 58. Plot the function f./ sin.findthepositiverootoff./to three plces nd use it to find the re under the grph of f./in the first qudrnt. The grph of f./ sin is shown below t the left. In the figure below t the right, we zoom in on the positive root of f./nd find tht, to three deciml plces, this root is pproimtel :7. Thereunderthegrphoff./in the first qudrnt is then Z :7.sin /d cos :7 ˇˇˇˇ cos.:8/.:7/ C : Clculte F.4/given tht F./ nd F./. Hint: Epress F.4/ F./s definite integrl. B FTC I, Z 4 F.4/ F./ d 4 Therefore F.4/ F./C C 4.. Clculte G./, wheredg=dt t = nd G.9/ 5. B FTC I, Z G./ G.9/ t = dt. = /.9 = / 9 Therefore G./ 5C. Z. oes n d get lrger or smller s n increses? Eplin grphicll. Let n nd consider R n d. (Note:forn < the integrnd n! s! C, soweecludethis possibilit.) Now Z n d n C ncˇˇˇˇ n C./nC n C./nC n C ; which decreses s n increses. Recll tht R n d represents the re between the positive curve f./ n nd the -is over the intervl Œ;. Accordingl,thisregetssmllersn gets lrger. This is redil evident in the following grph, which shows curves for severl vlues of n. /4 / 4 8. Show tht the re of the shded prbolic rch in Figure is equl to four-thirds the re of the tringle shown. + b FIGURE Grph of. /.b /. b

10 SECTION 5. The Fundmentl Theorem of Clculus, Prt I 599 We first clculte the re of the prbolic rch:. /.b /d. /. b/d. b C b/ d b b C bˇˇˇˇ b ˇˇˇb C b.b b b C b /. b C b/. b C b /. C b/ C b b b.b / : The indicted tringle hs bse of length b nd height of C b b C b b : Thus, the re of the tringle is Finll, we note tht s required..b / b 8.b / :.b / 4 8.b / ; Further Insights nd Chllenges. Prove fmous result of Archimedes (generlizing Eercise ): For r<s,thereoftheshdedregioninfigureisequl to four-thirds the re of tringle 4ACE, wherec is the point on the prbol t which the tngent line is prllel to secnt line AE. () Show tht C hs -coordinte.r C s/=. (b) Show tht ABE hs re.s r/ =4 b viewing it s prllelogrm of height s r nd bse of length CF. (c) Show tht 4ACE hs re.s r/ =8 b observing tht it hs the sme bse nd height s the prllelogrm. (d) Compute the shded re s the re under the grph minus the re of trpezoid, nd prove Archimedes result. B C A r F r + s FIGURE Grph of f./. /.b /. E s () The slope of the secnt line AE is f.s/ f.r/ s r nd the slope of the tngent line long the prbol is.s /.b s/.r /.b r/ s r f./ C b : C b.r C s/ If C is the point on the prbol t which the tngent line is prllel to the secnt line AE, then its -coordinte must stisf C b C b.r C s/ or r C s :

11 C HAPTER 5 THE INTEGRAL (b) Prllelogrm ABE hs height s r nd bse of length CF. Since the eqution of the secnt line AE is the length of the segment CF is r C s b r C s Œ C b.r C s/. r/ C.r /.b r/; r C s Œ C b.r C s/ r.r /.b r/.s r/ : 4 Thus, the re of ABE is.s r/ 4. (c) Tringle ACE is comprised of ACF nd CEF.Echofthesesmllertringleshsheight s r Thus, the re of ACE is s r.s r/ C s r.s r/ 4 4 (d) The re under the grph of the prbol between r nd s is Z s. /.b /d b C r. C b/ s ˇˇˇˇ r.s r/ : 8 nd bse of length.s r/ 4. bs C. C b/s s C br. C b/r C r while the re of the trpezoid under the shded region is b.r s/ C. C b/.s r/.s C r/ C.r s/.r C rs C s /;.s r/œ.s /.b s/ C.r /.b r/ h.s r/ b C. C b/.r C s/ r s i Thus, the re of the shded region is.r s/ r C rs C s r s which is four-thirds the re of the tringle ACE. b.r s/ C. C b/.s r/.r C s/ C.r s/.r C s /:.s r/ r rs C s.s r/ ; 4. () Appl the Comprison Theorem (Theorem 5 in Section 5.) to the inequlit sin (vlid for ) toprovetht (b) Appl it gin to prove tht cos (c) Verif these inequlities for :. sin.for / Z Z () We hve sin tdt cos t ˇ cos C nd tdt tˇˇˇˇ. Hence cos C : Solving, this gives cos. cos follows utomticll. (b) The previous prt gives us t cos t, fort>.theorem5givesus,fterintegrtingovertheintervlœ;, sin : (c) Substituting : into the inequlities obtined in () nd (b) ields respectivel. :955 : nd :955 :9559 :;

12 SECTION 5. The Fundmentl Theorem of Clculus, Prt I 5. Use the method of Eercise 4 to prove tht cos C 4 4 sin C 5 (for ) Verif these inequlities for :. Whhvewespecified for sin but not for cos? B Eercise 4, t t sin t t for t>.integrtingthisinequlitovertheintervlœ;,ndthensolvingfor cos,ields: 4 4 cos cos C 4 4 : These inequlities ppl for.sincecos,, nd C 4 4 re ll even functions, the lso ppl for. Hving estblished tht for ll t,weintegrteovertheintervlœ;,toobtin: t cos t t C t4 4 ; sin C 5 : The functions sin, nd C 5 re ll odd functions, so the inequlities re reversed for <. Evluting these inequlities t : ields both of which re true. :995 :99545 :99547 :998 :9984 :99847;. Clculte the net pir of inequlities for sin nd cos b integrting the results of Eercise 5. Cn ou guess the generl pttern? Integrting over the intervl Œ; ields t t sin t t t C t5 (for t ) Solving for cos ields 4 4 cos 4 4 C 7 : C 4 cos 4 7 C 4 4 : Replcing ech b t nd integrting over the intervl Œ; produces C 5 7 sin 54 C 5 : To see the pttern,it is best to compre consecutive inequlities for sin nd those for cos : sin sin sin C 5 :

13 C HAPTER 5 THE INTEGRAL Ech itertion dds n dditionl term. Looking t the highest order terms, we get the following pttern: 5 5Š We guessthttheledingtermofthepolnomilsreoftheform Š. / n nc.n C /Š : Similrl, for cos,theledingtermsofthepolnomilsintheinequlitreoftheform. / n n.n/š : 7. Use FTC I to prove tht if jf./j K for Œ; b, thenjf./ f./jkj j for Œ; b. Let >bbe rel numbers, nd let f./be such tht jf./j K for Œ; b.bftc, Z f.t/ dt f./ f./: Since f./ K for ll Œ; b, weget: Z f./ f./ f.t/ dt K. /: Since f./ K for ll Œ; b, weget: Z f./ f./ f.t/ dt K. /: Combining these two inequlities ields so tht, b definition, K. / f./ f./ K. /; jf./ f./jkj j: 8. () Use Eercise 7 to prove tht j sin sin bj j bj for ll ; b. (b) Let f./ sin. C / sin.useprt()toshowthtthegrphoff./lies between the horizontl lines. (c) Plot f./ nd the lines to verif (b) for :5 nd :. () Let f./ sin,sothtf./ cos,nd for ll.fromeercise7,weget: jf./j jsin sin bj j bj: (b) Let f./ sin. C / sin./.appling(),wegettheinequlit: This is equivlent, b definition, to the two inequlities: jf./jjsin. C / sin./j j. C /jjj: sin. C / sin./ : (c) The plots of sin. C :5/ sin./ nd of sin. C :/ sin./ re shown below. The inequlit is stisfied in both plots