4.5 The Fundamental Theorem of Calculus

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1 4 the integrl 4.5 The Funmentl Theorem of Clculus This section contins the most importnt n most frequently use theorem of clculus, THE Funmentl Theorem of Clculus. Discovere inepenently by Newton n Leibniz uring the lte 6s, it estblishes connection between erivtives n integrls, provies wy to esily clculte mny efinite 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 the most importnt brick in tht beutiful structure. Prior sections hve emphsize the mening of the efinite integrl, efine it, n begn to eplore some of its pplictions n properties. In this section, the emphsis shifts to the Funmentl Theorem of Clculus. You will use this theorem often in lter sections. The Funmentl Theorem hs two prts. They resemble results in the previous section but pply to more generl situtions. The first prt (FTC ) sys tht every continuous function hs n ntierivtive n shows how to ifferentite function efine s n integrl. The secon prt (FTC ) shows how to evlute the efinite integrl of ny function if we know formul for n ntierivtive of tht function. Prt : Antierivtives Every continuous function hs n ntierivtive, even functions with corners, such s the bsolute vlue function f (), tht fil to be ifferentible t one or more points. The Funmentl Theorem of Clculus Prt (FTC ) If f is continuous n A() f (t) t then A () f (t) t f () so A() is n ntierivtive of f (). Proof. For continuous function f, let A() efinition of erivtive, A A( + h) A() () lim lim h h h +h f (t) t. f (t) t h f (t) t By the Using one of the integrl properties from Section 4., we know tht: +h +h f (t) t f (t) t f (t) t + f (t) t +h +h f (t) t f (t) t

2 4.5 the funmentl theorem of clculus 4 Assume for the moment tht h >. Becuse f is continuous on, + h we know tht f ttins mimum n minimum on tht intervl, so there re vlues m h n M h with < m h < + h n < M h < + h so tht f (m h ) f (t) f (M h ) when t + h. Hence: +h f (m h ) h f (m h ) f (m h ) t +h +h +h f (t) t +h f (t) t f (M h ) h f (t) t f (M h h ) Becuse < m h < + h, we know lim h + m h becuse f (t) is continuous we lso know tht lim Likewise, lim h + f (M h ) t ; consequently h + f (m h) f (). f (M h) f (), so the Squeezing Theorem tells us tht: lim h + +h f (t) t f () h Repeting this rgument for h < is reltively strightforwr. Emple. Define A() f (t) t for f in the mrgin figure. Evlute A() n A () for,, n 4. Solution. A() f (t) t, A() f (t) t, A() f (t) t 4 n A(4) f (t) t. Becuse f is continuous, FTC tells us tht A () f (), so A () f (), A () f (), A () f () n A (4) f (4). Prctice. Define A() g(t) t for g in the mrgin figure. Evlute A() n A () for,, n 4. Emple. Define A() f (t) t for f in the mrgin figure. For which vlue of is A() mimum? For which is the rte of chnge of A mimum? Solution. Becuse 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 vlues of A 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().

3 4 the integrl Prt : Evluting Definite Integrls If we know formul for n ntierivtive of function, then we cn compute ny efinite integrl of tht function. The Funmentl Theorem of Clculus Prt (FTC ) If then f () is continuous n F() is ny ntierivtive of f (so tht F () f ()) b b f () F() F(b) F(). Proof. Define A() f (t) t. If F is n ntierivtive of f, then F () f () n by FTC we know tht A () f () so F () A (), hence F() n A() iffer by constnt: A() F() C for ll n some constnt C. 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 setting b yiels A(b) F(b) F(), hence formul we wnte. b f () F(b) F(), the We cn evlute the efinite integrl of continuous function f by fining n ntierivtive of f (ny ntierivtive of f will work) n then oing some rithmetic with this ntierivtive. FTC 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 (s we sw in Section 4.) but not by using FTC irectly. ( ) Emple. Evlute. Solution. F() ( is ) n ntierivtive of f () (you shoul check tht D ), so: ( ) If your frien h picke ifferent ntierivtive of, sy G() + 4, then her clcultions woul be slightly ifferent : ( ) but the result woul be the sme

4 4.5 the funmentl theorem of clculus 4 Prctice. Evlute Emple 4. Evlute ( )..7.5 (where INT() is the lrgest integer less thn or equl to, s in the mrgin figure). Solution. f () is not continuous t in the intervl.5,.7, so we cnnot employ the Funmentl Theorem of Clculus irectly. We cn, however, use our unerstning of the geometric mening of efinite integrl to compute:.7.5 (re below y for.5 ) + (re below y for.7) (first bse) (first height) + (secon bse) (secon height) (.5)() + (.7)().9 We coul lso split the integrl into two pieces: (.7) (.) using the fct tht for.5 <. n the fct tht for..7. (We lso nee to reefine the first integrn to equl t its right enpoint n the secon integrn to equl t its right enpoint so tht ech integrn is continuous on close intervl). Problem 5 in Section 4. inictes tht this reefinition is perfectly legl. Prctice. Evlute.4.. Clculus is the stuy of erivtives n integrls, their menings n their pplictions. The Funmentl Theorem of Clculus emonstrtes how ifferentition n integrtion re closely relte processes: integrtion is relly nti-ifferentition, the inverse of ifferentition. Applictions: The Future Clculus is importnt for mny resons, but stuents re usully require to stuy clculus becuse they will nee to pply clculus concepts in vriety of fiels. Most pplie problems in integrl clculus require the following steps to get from rel-life problem to numericl nswer: pplie problem Riemnn sum efinite integrl number

5 44 the integrl 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 specil types of pplie problems, we will be ble to move 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 4.7 n ll of Chpter 5 focus on trnslting ifferent types of pplie problems into Riemnn sums n efinite integrls. Computers n clcultors re selom of ny help with Step. Step is usully esy. If we hve Riemnn sum n k f (c k ) k on n intervl, b, then the limit of the sum (s n ) is simply the efinite integrl b f (). Step cn be hnle in severl wys. If the function f is reltively simple, we my be ble to fin n ntierivtive for f (using techniques from Section 4.6 n Chpter 8) n then pply FTC to get numericl nswer. If the function f is more complicte, then integrl tbles or computers (Section 4.8) my help us fin n ntierivtive for f, in which cse we cn pply FTC to get numericl nswer. If we cnnot fin n ntierivtive for f, we cn compute pproimte numericl nswers for the efinite integrl using vrious pproimtion methos (Sections 4.9 n 8.7); we typiclly employ computers to crry out the hevy-uty rithmetic. Usully ny ifficulties in solving n pplie problem rise in the first n thir steps. 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. We will eplore techniques for fining volumes of solis in greter etil in Chpter 5. Emple 5. Fin the volume of the soli shown in the mrgin for. (Ech slice perpeniculr to the y-plne 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 (n the y-plne) using prtition P with cuts t,,,..., n (like slicing block of cheese or lof of bre), then the volume of the originl soli is equl to the sum of the volumes of the slices. The volume of the k-th slice is pproimtely equl to the volume of thin, rectngulr bo: (height) (bse) (thickness) (c k + ) (c k + ) k

6 4.5 the funmentl theorem of clculus 45 where c k is ny chosen vlue between k n k. Therefore: totl volume n n (volume of the k-th slice) (c k + ) k k k which is Riemnn sum. Step : Going from the Riemnn sum to efinite integrl. We cn improve the Riemnn sum pproimtion of the totl volume from Step by tking thinner slices (mking ll of the k smller n smller) so tht the mesh of the prtition P pproches : lim P n (c k + ) k k ( + ) + + Step : Going from the efinite integrl to numericl nswer. We cn now use FTC to evlute the integrl: F() + + is n ntierivtive of + + (check this by ifferentiting F()), so: The volume of the soli shpe is ectly 6 cubic inches. Prctice 4. Fin the volume of the soli shpe in the mrgin figure for. (Ech slice perpeniculr to the y-plne is squre.) Leibniz s 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 FTC to clculte the erivtive of the integrl. For emple: A () f () A ( ) A () f ( ) We cn generlize this result by pplying the Chin Rule to the erivtive of the integrl: g() f (t) t A (g()) f (g()) g () n combine this with some integrl properties to further eten FTC. Leibniz s Rule If f is continuous function, A() then f (t) t n g () n g () re both ifferentible functions g () f (t) t f (g ()) g () f (g ()) g () g ()

7 46 the integrl Proof. Assume for simplicity tht f, g n g re continuous on (, ) n let c be ny number. Then: g () g () f (t) t g () c g () c c f (t) t + f (t) t g () g () c f (t) t f (t) t Now pply the preceing result. Emple 6. If is ny constnt, compute the erivtives 5 cos(u) u n sin w z z. w πw Solution. Applying Leibniz s Rule: 5 t t (5) 5 5 cos(u) u cos( ) cos( ) w sin(w) πw z z (sin(w)) cos(w) (πw) π t t, The lst quntity simplifies to sin (w) cos(w) π 4 w. Prctice 5. Compute sin(t)t. 4.5 Problems In Problems, () Use FTC to fin formul for A(), ifferentite A() to obtin formul for A (), n evlute A () t, n. (b) Use FTC to evlute A () t, n.. A() t t. A() ( + t) t In Problems 8, compute A (), A () n A ().. A() 5. A() 7. A() t t 4. A() t t 6. A() sin(t) t 8. A() t t ( t ) t t t In 9, A() f (t) t, with f (t) given grphiclly. Evlute A (), A () n A ()

8 4.5 the funmentl theorem of clculus 47 In, verify tht F() is n ntierivtive of the integrn n use FTC to evlute the integrl π π π 4 e, F() + 5, F() +, F() + 4, F() +, F() ln(), F() ln() + 4, F() ln(), F() ln() + cos(), sin(), F() sin() F() cos(), F(), F(), F(), F(), F(), e, F() F() e +, F() ln( + ) sec (), ln(), F() tn() F() ln() +, F() ( + ) For 4 48, fin n ntierivtive of the integrn n use FTC to evlute the efinite integrl π π 4 sin() e e π e 47. ( ) π 6 sec () sin() ln() In 49 54, fin the re of the she region

9 48 the integrl 55. Given tht A () tn(), fin D (A()), D ( A( ) ) n D (A(sin())). 56. Given tht B () sec(), fin D (B()), D ( B( ) ) n D (B(sin())) In 57 68, pply Leibniz s Rule. 5 + t t 58. sin() + t t 6. + t t + ( ) t + 5 t π 7 ( ) t + t ( ) t + t cos(t) t π cos(t) t π tn(7t) t π y ln() cos(t) t tn(t) t 5t cos(t) t y tn(θ) θ 4.5 Prctice Answers. A(), A().5, A(), A(4).5; A () f g) so A () g(), A () g(), A (), A (4).. F() is n ntierivtive of f () so: 4 F() + 7 is nother ntierivtive of f () so: No mtter which ntierivtive of f () you use, the vlue of the efinite integrl is 4.. Becuse f () is not continuous on.,.4 we cnnot use the Funmentl Theorem of Clculus. Inste, we cn think of the efinite integrl s n re (see mrgin figure) n compute: First brek the soli into slices n pproimte the volume of the k-th slice by ( c k ) k where c k is ny point in the k-th subintervl. Net up these pproimte volumes to get Riemnn Sum: n ( c k ) k k

10 4.5 the funmentl theorem of clculus n then tke the limit of these Riemnn sums s the mesh of the prtitions pproches (n n, where n is the number of subintervls in the prtition): lim P n k ( c k ) k ( ) ( ) sin(t) t sin( ) sin( )