5.3 The Fundamental Theorem of Calculus

Transcription

1 CHAPTER 5. THE DEFINITE INTEGRAL The Funmentl Theorem of Clculus Emple. Let f(t) t +. () Fin the re of the region below f(t), bove the t-is, n between t n t. (You my wnt to look up the re formul for trpezoi.) (b) Repet prt () for the re between t n t. Repet gin for the re between t n t 3. (c) For >, let A() be the re uner the line f(t), bove the t-is, n between t n t. Fin formul for A(). () Tke the erivtive of the formul A() tht you foun in prt (c). Wht o you notice? (e) Wht o prts (c) n () hve to o with inegrls? Solution. () The re is shown in gry below: 3 This forms trpezoi with bse n heights h n h f(). The re formul for trpezoi is A b(h + h )sowehve A ()( + ) () 3. (b) This time we hve two bigger trpezois 3 3 The clcultions re bout the sme s in prt (). A(t tot ) ()( + ) A(t tot ) (3)( + ) 5 (c) This time we hve trpezoi, but the position of the right hn sie is vrible:

2 CHAPTER 5. THE DEFINITE INTEGRAL 3 Still, we o things just s in prt (b) 3 A() A(t tot ) ()( + + ) ( + ) ( +) + () Tke the erivtive of the formul A() tht you foun in prt (c). Wht o you notice? A() ( +) + We notic tht the formul for A () is the sme s the formul for the line f(t) t +. (e) We cn epress A() using n integrl: A() Z t +t In this lnguge, we foun tht tking the integrl of + gives n ntierivtive of +. This is where we ene on Mony, November 5 Theorem (Funmentl Theorem of Clculus I). Let f() be continuous function efine on [, b] n efine F () Z f(t) t Then F () is n nti-erivtive of f(). In other wors, F () f(). Comments. One thing tht stuents often fin confusing in the previous theorem is the use of two vribles: t the top of the integrl sign Z, n t in f(t) t. In some sense, this is technicl point tht you on t nee to worry bout too much, but here s the issue. The vrible sttes how fr to the right we re tking the re. The vrible t is use to efine the curve. They re plying i erent roles,

3 CHAPTER 5. THE DEFINITE INTEGRAL 37 s shown by the previous emple where mrke the right ege of the trpezoi n t efine the curve on top. On the other hn, this technicl point is often something stuents on t unerstn, n ignore, n it ens up cusing no problem! So my vice is to think bout, try to unerstn it, n then move on without worries. Proof. By efinition of the erivtive we hve Applying the efinition of F () wehve F () lim h! h F () lim h! F ( + h) F () h Z +h f(t) t Z f(t) t lim h! h Z +h f(t) t Since the limit involves infinitely smll vlues of h, we hve tht the re in the integrl is given by single rectngle of height f() n with h f() Z +h f(t) t she re! re of rectngle f() h f() + h h so F () becomes F () lim re of rectngle lim h f() f() h! h h! h Comments. The bove proof justifies the correctness of the Funmentl Theorem of Clculus I, but it oesn t eplin it much. Here s n intuitive eplntion. Suppose you efine function F (), tht mesures the mount of re, up to some vlue, unercurveefinebyf(t). The rte of chnge of F () ishowmuchthe re is chnging t the point ; this is given by the height of the function f(t) t t, i.e.f(). The bigger f() is, the more the re is chnging, n the smller f() is, the less the re is chnging. Finlly, this mkes it cler why the vlue oesn t pper in the theorem: whtever the height f() is, it oesn t tell you how much the re is chnging t. This is where we ene on Tuesy, November Definition. An nti-erivtive of f() is function F () such tht F () f(). Theorem.. If F () is n nti-erivtive of f(), then F ()+C is lso n nti-erivtive for every constnt C.

4 CHAPTER 5. THE DEFINITE INTEGRAL 3. If F () n G() re ny two nti-erivtives of f(), then G() F ()+C for some constnt C. Proof. We ssume tht F () f(). Then F ()+C F ()+f(). Conversely, if G() is nother nti-erivtive of f() thenf () f() n G () f(). So F () G (). By Corollry. of the Men Vlue Theorem this shows tht F () G()+C. We cn rewrite this s G() F ()+C where C C. As result of the previous theorem, we lwys write nti-erivtives using +C t the en of the formul. Theorem 3 (Funmentl Theorem of Clculus II). Let f() be continuous function efine on [, b]. Then f() F (b) F () where F () is ny nti-erivtive of f(). Comments. Nottion: we usully write F () b s n bbrevition for F (b) F (). Proof. Cse : We prove the theorem for one prticulr nti-erivtive of f(). Let F () clculte: Z F (b) F () f(t) t. By FTCI we know F () is n nti-erivtive of f(). Now we f(t) t f(t) t Z f(t) t f() (t this point there is no i erence between n t) Cse : We prove the theorem for ny nti-erivtive. Suppose G() is ny nti-erivtive of f(). Then, by Theorem, we hve G() F ()+C, for some constnt C. Now we clculte: G(b) G() F (b)+c F ()+C F (b) F () f() by cse Emple. Fin This is where we ene on Mony, December Z sin().

5 CHAPTER 5. THE DEFINITE INTEGRAL 39 Solution. FTC II sys we shoul strt by fining n nti-erivtive: we o this through guessing n checking. F ()? sin() F () cos() No. F ()? cos() F () F ()? sin() No. But close. cos() F () ( sin()) sin()x Now we use the nti-erivtive we ve foun. We write it (n you shoul two) in two steps: one with the nti-erivtive s function, n one with the numbers plugge in: Emple 3. Fin Z Z sin() cos() cos( ) ( cos()). ( ) ( ) Solution. We strt by guessing n nti-erivtive. In other wors, guess n check:?? No. 3 No. p p No. 5 5 No, but closer. ln() No, but closer. 5 X ln() X Now we combine these n finish the integrl: Z ln() 5 ln() ( 5 ) ln() + ln() ln() ( 5 )