4.5 THE FUNDAMENTAL THEOREM OF CALCULUS

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1 4.5 The Funmentl Theorem of Clculus Contemporry Clculus 4.5 THE FUNDAMENTAL THEOREM OF CALCULUS This section contins the most importnt n most use theorem of clculus, THE Funmentl Theorem of Clculus. Discovere inepenently by Newton n Leibniz in the lte 6s, it estblishes the connection between erivtives n integrls, provies wy of esily clculting mny integrls, n ws key step in the evelopment of moern mthemtics to support the rise of science n technology. Clculus is one of the most significnt intellectul structures in the history of humn thought, n the Funmentl Theorem of Clculus is most importnt brick in tht beutiful structure. The previous sections emphsize the mening of the efinite integrl, efine it, n begn to eplore some of its pplictions n properties. In this section, the emphsis is on the Funmentl Theorem of Clculus. You will use this theorem often in lter sections. There re two prts of the Funmentl Theorem. They re similr to results in the lst section but more generl. Prt of the Funmentl Theorem of Clculus sys tht every continuous function hs n ntierivtive n shows how to ifferentite function efine s n integrl. Prt shows how to evlute the efinite integrl of ny function if we know n ntierivtive of tht function. Prt : Antierivtives Every continuous function hs n ntierivtive, even those nonifferentible functions with "corners" such s bsolute vlue. The Funmentl Theorem of Clculus (Prt ) If f is continuous n A() = f(t) t the n ( f(t) t ) = A() = f(). A() is n ntierivtive of f(). Proof: Assume f is continuous function n let A() = f(t) t. By the efinition of erivtive of A, A() = A( + h) # A() lim = lim h" h h" h %' +h & $ f(t)t # $ f(t)t (' By Property 6 of efinite integrls (Section 4.), for h > )' * + ' = lim h" h +h # f(t) t.

2 4.5 The Funmentl Theorem of Clculus Contemporry Clculus +h { min of f on [, +h] }. h f(t) t { m of f on [, +h] }. h. (Fig. ) Diviing ech prt of the inequlity by h, we hve tht +h h f(t) t is between the minimum n the mimum of f on the intervl [, +h]. The function f is continuous (by the hypothesis) n the intervl [,+h] is shrinking (since h pproches ), so lim h" { min of f on [, +h] } = f() n lim { m of f on [, +h] } = f(). Therefore, +h h f(t) t is stuck between two h" { min of f on [, +h] }!! { m of f on [, +h] } f() s h "! h +h Fig. f(t) t s h "!?! f() s h " quntities (Fig. ) which both pproch f(). +h Then h f(t) t must lso pproch f(), n A() = lim h" h +h # f(t) t = f(). Emple : A() = f(t) t for f in Fig.. Evlute A() n A'() for =,, n 4. f(t) t = /, Solution: A() = f(t) t = /, A() = f(t) t =, A() = A(4) = 4 f(t) t = /. Since f is continuous, A '() = f() so A'() = f() =, A'() = f() =, A'() = f() =, A'(4) = f(4) =.

3 4.5 The Funmentl Theorem of Clculus Contemporry Clculus Prctice : A() = f(t) t for f in Fig. 4. Evlute A() n A'() for =,, n 4. Emple : A() = f(t)t for the function f shown in Fig. 5. For which vlue of is A() mimum? For which is the rte of chnge of A mimum? Solution: Since A is ifferentible, the only criticl points re where A'() = or t enpoints. A'() = f() = t =, n A hs mimum t =. Notice tht the vlues of A() increse s goes from to n then the A vlues ecrese. The rte of chnge of A() is A'() = f(), n f() ppers to hve mimum t = so the rte of chnge of A() is mimum when =. Ner =, slight increse in the vlue of yiels the mimum increse in the vlue of A(). Prt : Evluting Definite Integrls If we know n cn evlute some ntierivtive of function, then we cn evlute ny efinite integrl of tht function. The Funmentl Theorem of Clculus (Prt ) If f() is continuous n F() is ny ntierivtive of f ( F '() = f() ), then b f() = F() b = F(b) F(). Proof: If F is n ntierivtive of f, then F() n A() = f(t) t re both ntierivtives of f, F'() = f() = A'(), so F n A iffer by constnt: A() F() = C for ll. At =, we hve C = A() F() = F() = F() so C = F() n the eqution A() F() = C becomes A() F() = F(). Then A() = F() F() for ll so A(b) = F(b) F() n b f() = A(b) = F(b) F(), the formul we wnte.

4 4.5 The Funmentl Theorem of Clculus Contemporry Clculus 4 The efinite integrl of continuous function f cn be foun by fining n ntierivtive of f (ny ntierivtive of f will work) n then oing some rithmetic with this ntierivtive. The theorem oes not tell us how to fin n ntierivtive of f, n it oes not tell us how to fin the efinite integrl of iscontinuous function. It is possible to evlute efinite integrls of some iscontinuous functions (Section 4.), but the Funmentl Theorem of Clculus cn not be use to o so. Emple : Evlute ( ). Solution: F() = is n ntierivtive of f() = (check tht D( ) = ), so ( ) = = { } { } = / = /. If friens h picke ifferent ntierivtive of, sy F() = + 4, then their clcultions woul be slightly ifferent but the result woul be the sme: 4 = /. ( ) = + 4 = ( Prctice : Evlute ( ). + 4) ( + 4) = 4/ Emple 4: Evlute.7 INT(). (INT() is the lrgest integer less thn or equl to. Fig. 6).5 Solution: f() = INT() is not continuous t = in the intervl [.5,.7] so the Funmentl Theorem of Clculus cn not be use. We cn, however, use our unerstning of the mening of n integrl to get.7 INT() = (re for between.5 n ) + (re for between n.7).5 Prctice : Evlute. = (bse)(height) + (bse)(height) = (.5)() + (.7)() =.9..4 INT().

5 4.5 The Funmentl Theorem of Clculus Contemporry Clculus 5 Clculus is the stuy of erivtives n integrls, their menings n their pplictions. The Funmentl Theorem of Clculus shows tht ifferentition n integrtion re closely relte n tht integrtion is relly ntiifferentition, the inverse of ifferentition. Applictions The Future Clculus is importnt for mny resons, but stuents re usully require to stuy clculus becuse it is neee for unerstning concepts n oing pplictions in vriety of fiels. The Funmentl Theorem of Clculus is very importnt to both pursuits. Most pplie problems in integrl clculus require the following steps to get from the problem to numericl nswer: Applie Riemnn efinite number problem sum (or re) integrl In some cses, the pth from the problem to the nswer my be bbrevite, but the three steps re commonly use. Step is bsolutely vitl. If we cn not trnslte the ies of n pplie problem into n re or Riemnn sum or efinite integrl, then we cn not use integrl clculus to solve the problem. For few types of pplie problems, we will be ble to go irectly from the problem to n integrl, but usully it will be esier to first brek the problem into smller pieces n to buil Riemnn sum. Section 4.8 n ll of chpter 5 will focus on trnslting ifferent types of pplie problems into Riemnn sums n efinite integrls. Computers n clcultors re selom of ny help with Step. n Step is usully esy. If we hve Riemnn sum f(ck ) k on the intervl [,b], then the limit of the k= b f(). sum is simply the efinite integrl Step cn be hnle in severl wys. If the function f is reltively simple, there re severl wys to fin n ntierivtive of f (sections 4.6, prts of chpter 6 n others), n then Prt of the Funmentl Theorem of Clculus cn be use to get numericl nswer. If the function f is more complicte, then integrl tbles (section 4.8) or computers (symbolic mnipultors such s Mple or Mthemtic) cn be use to fin n ntierivtive of f. Then Prt of the Funmentl Theorem of Clculus cn be use to get numericl nswer.

6 4.5 The Funmentl Theorem of Clculus Contemporry Clculus 6 If n ntierivtive of f cnnot be foun, pproimte numericl nswers for the efinite integrl cn be foun by vrious summtion methos (section 4.9). These summtion methos re typiclly one on computers, n progrm listings re inclue in n Appeni. Usully the ifficulties in solving n pplie problem come with the st n r steps, n the most time will be spent working with them. There re techniques n etils to mster n unerstn, but it is lso importnt to keep in min where these techniques n etils fit into the bigger picture. The net Emple illustrtes these steps for the problem of fining volume of soli. Problems of fining volumes of solis will be emine in more etil in Section 5.. Emple 5: Fin the volume of the soli in Fig. 7 for. (Ech perpeniculr "slice" through the soli is squre.) Solution: Step : Going from the figure to Riemnn sum. If we brek the soli into n "slices" with cuts perpeniculr to the is, t,,,..., n (like cutting lof of bre), then the volume of the originl soli is the sum of the volumes of the "slices" (Fig. 8): n Totl Volume = (volume of the i th slice ). i= The volume of the i th slice is pproimtely equl to the volume of bo: (height of the slice). (bse of the slice). (thickness) ( c i + ). ( c i + ). i where c i is ny vlue between i n i. Therefore, n Totl Volume ( c i + ). ( c i + ). i i= which is Riemnn sum.

7 4.5 The Funmentl Theorem of Clculus Contemporry Clculus 7 Step : Going from the Riemnn sum to efinite integrl. The Riemnn sum pproimtion of the totl volume in Step is improve by tking thinner slices (mking ll of the i smll), n Totl volume = = lim mesh" & N ( *( '%(c i +) # (c i +) # $ i + )( i=,( ( + )( + ) = ( + + ). Step : Going from the efinite integrl to numericl nswer. We cn use Prt of the Funmentl Theorem of Clculus to evlute the integrl. F() = + + is n ntierivtive of + + (check by ifferentiting F() ), so ( + + ) = F() F() = { + + } { + + } = { 6 } { } = 6 = 8. The volume of the soli shpe in Fig. 7 is ectly 8 cubic inches. Prctice 4: Fin the volume of the soli shpe in Fig. 9 for. (Ech "slice" through the soli perpeniculr to the is is squre.) Leibniz' Rule For Differentiting Integrls If the enpoint of n integrl is function of rther thn simply, then we nee to use the Chin Rule together with prt of the Funmentl Theorem of Clculus to clculte the erivtive of the integrl. Accoring to the Chin Rule, if A() = f(), then A( ) = f( ). n, pplying the Chin Rule to the erivtive of the integrl, g() ( f(t)t ) = A( g() ) = f( g() ). g'().

8 4.5 The Funmentl Theorem of Clculus Contemporry Clculus 8 If f is continuous function n A() = f(t)t then ( f(t)t ) = A() = f() (Funmentl Theorem, Prt I) n, if g is ifferentible, ( g() f(t)t ) = A( g() ) = f( g() ). g'() (Leibniz' Rule) Emple 6: Clculte 5 ( t t ), ( cos(u) u ), w ( sin(w) z z ). Solution: ( 5 t t ) = (5). 5 = 5. ( cos(u) u ) = cos( ). =. cos( ) ( w sin(w) z z ) = (sin(w)). cos(w) = sin (w)cos(w). Prctice 5: Fin ( sin(t) t ). PROBLEMS:. A() = t t () Use prt of the Funmentl Theorem to fin formul for A() n then ifferentite A() to obtin formul for A'(). Evlute A'() t =,, n. (b) Use prt of the Funmentl Theorem to evlute A'() t =,, n.. A() = ( + t ) t () Use prt of the Funmentl Theorem to fin formul for A() n then ifferentite A() to obtin formul for A'(). Evlute A'() t =,, n. (b) Use prt of the Funmentl Theorem to evlute A'() t =,, n.

9 4.5 The Funmentl Theorem of Clculus Contemporry Clculus 9 In problems 8, evlute A'() t =,, n. t t. A() = t t 4. A() = t t 5. A() = 6. A() = t t ( t ) t 7. A() = sin(t) t 8. A() = In problems 9, A() = f(t) t for the functions in Figures 4. Evlute A'(), A'(), A'(). 9. f in Fig.. f in Fig.. f in Fig.. f in Fig. In problems, verify tht F() is n ntierivtive of the integrn f() n use Prt of the Funmentl Theorem to evlute the efinite integrls. 4, F() =., F() = , F() = ( + 4 ), F() = , F() = ln( ) 8.. 5, F() = ln( ) , F() = ln( )., F() = ln( ) + / π/ π, F() = / cos(), F() = sin( ). sin(), F() = cos(). 4 4., F() = / 5. 7, F() = / 6. 4, F() =

10 4.5 The Funmentl Theorem of Clculus Contemporry Clculus , F() = 9. e, F() = e, F() = 8. +, F() = ln( + ). π/4 sec (), F() = tn(). e ln(), F() =. ln(). +, F() = ( + ) / For problems 4 48, fin n ntierivtive of the integrn n use Prt of the Funmentl Theorem to evlute the efinite integrl. 5 e ( + 4 ) 7. π/ 8. sin() 9. π/ " e " e 47. sin(). ln() π/ π/6 ( ) sec () In problems 49 54, fin the re of ech she region. 49. Region in Fig Region in Fig Region in Fig. 6.

11 4.5 The Funmentl Theorem of Clculus Contemporry Clculus 5. Region in Fig Region in Fig Region in Fig. 9. Leibniz' Rule 55. If D( A() ) = tn(), then fin D( A() ), D( A( ) ), n D( A( sin() ) ). 56. If D( B() ) = sec(), then fin D( B() ), D( B( ) ), n D( B( sin() ) ) ( + t t ) 58. ( + t t ) 59. sin() ( + t t ) 6. + ( t + 5 t ) 6. ( t + t ) 6. ( 9 t + t ) 6. ( π cos(t) t ) 64. ( 7 π cos(t) t ) 65. ( tn(t) t ) 66. ( π cos(t) t ) 67. ( ln() 5t. cos(t) t) ) 68. ( π tn(7t) t ) Very Optionl Problems ice. Wht clculus stuent puts in rink. b. cbin Where Abe Lincoln ws born. cerely jm c. cos() How the clculus stuent ene letter.. Wht forester puts on tost.

12 4.5 The Funmentl Theorem of Clculus Contemporry Clculus Section 4.5 PRACTICE Answers Prctice : A() =, A() =.5, A() =, A(4) =.5 Prctice : A '() = f() so A '() = f() =, A '() = f() =, A '() =, A '(4) =. F() = is one ntierivtive of f() = ( F ' = f ) so = = ( ) ( ) = 4. F() = + 7 is nother ntierivtive of f() = so = + 7 = ( + 7) ( + 7) = 4. No mtter which ntierivtive of f() = you use, the vlue of the efinite integrl is 4. Prctice :..4 INT() =.9. Since f() = INT() is not continuous on the intervl [.,.4] so we cn not use the Funmentl Theorem of Clculus. Inste, we cn think of the efinite integrl s n re (Fig. ). Prctice 4: Totl volume = lim mesh" = = ' N ) + ) (&(# c i ) $ (# c i ) $ % i, *) i= -) ( )( ) (9 6 + ) = 9 + = (8 + 8 ) ( +) = 8. Prctice 5: ( sin(t) t ) = sin( ) =. sin( ).