Linear Approximation and the Fundamental Theorem of Calculus

Transcription

1 Mth 3A Discussion Session Week 9 Notes Mrch nd 3, 26 Liner Approimtion nd the Fundmentl Theorem of Clculus We hve three primry ols in tody s discussion of the fundmentl theorem of clculus. By the end of our discussion, we should:. know why the FTC mtters tht is, we should know tht it ctully sys somethin; 2. know wht the FTC sys we should know which two concepts it connects, nd how it connects them; 3. know how to use the FTC we should be ble to pply the FTC to mke certin computtions esier. A fourth, more mbitious ol is tht we miht understnd why the FTC is true. Our discussion will hopefully convince us tht the FTC is true, but deeper pprecition of the vlidity of the FTC my tke some time to ferment. In tody s discussion we ll need to recll the followin definitions: Definition. Let f : [, b] R be continuous. Then the definite interl of f over [, b] is iven by b f()d := f ( + i ), () where = (b )/n. Definition. If f nd F re continuous functions on [, b], then we sy tht F is n ntiderivtive of f if F = f. Note tht unlike derivtives, ntiderivtives re not unique. Note. Remember tht priori, definite interls nd ntiderivtives hve nothin to do with one nother. The ltter comes from our study of clculus, while the former is simple ide we developed usin shrinkin rectnles. We wnt to eplore the interction between these two importnt ides, nd we bein with motivtin emple. Recll the ide of locl liner pproimtion: iven differentible function f nd fied point, we pproimte the vlue of f() by f() f() + f ()( ) when is sufficiently close to. For emple, suppose you drive for 4 hours in one direction nd I m hopin to estimte the distnce you hve trveled. Sy tht t time t = you were

2 trvelin t rte of 5 mph; then my estimte for the distnce you ve covered in 4 hours would be loction(4 hours) loction( hours) speed( hours)(4 ) = 2 miles. Of course, this (hopefully) isn t ret estimte. In ll likelihood, the 5 mph mesurement cme while you were pullin out of the re, nd you didn t trvel t this speed for the full 4 hours. I cn improve my estimte if I know your speed t some other point lon the journey. Suppose I now know tht fter 2 hours, you were drivin 7 mph. Then I cn refine my estimte by ssumin tht you drove 5 mph for the first 2 hours, then drove 7 mph for the lst 2 hours. For nottionl purposes, let s write f(t) for your loction t time t nd f (t) for your speed t time t. Then f(4) f() = [f(4) f(2)] + [f(2) f()] f (2)(4 2) + f ()(2 ) = 4 miles + miles = 5 miles. So my new estimte sys tht you ve trveled 5 miles over 4 hours, which is hopefully more relistic. Let s refine the estimte once more, ssumin I now know your speeds t hours nd 3 to hve been 6 nd 45 mph, respectively. Then f(4) f() = [f(4) f(3)] + [f(3) f(2)] + [f(2) f()] + [f() f()] f (3)(4 3) + f (2)(3 2) + f ()(2 ) + f ()( ) = 45 miles + 7 miles + 6 miles + 5 miles = 8 miles. Notice tht we cn use our summtion nottion to write this estimte s f(4) f() = f((i + ) t) f(i t) f (i t) t, where n = 4 nd t = (4 )/4 =. It should be the cse tht s I lern your speed t more nd more instnces durin your journey, I cn et more ccurte estimte of the distnce you ve trveled. So we should hve f(4) f() = f (i t) t = 4 f (t)dt, where t = (4 )/n. This emple is ctully quite typicl, nd motivtes the followin result. Theorem (The Fundmentl Theorem of Clculus). If f : [, b] R is continuous nd F is n ntiderivtive of f on [, b], then b f()d = F (b) F (). (2) 2

3 Proof. As with our emple, we bein with the usul locl liner pproimtion F (b) F () + F ()(b ) nd refine it. First, suppose we slice [, b] into n intervls, ech of width = (b )/n. Then the difference between F () nd F (b) is the sum of the differences between the endpoints of these intervls: F (b) F () = [F (b) F ( + (n ) )] + [F ( + (n ) ) F ( + (n 2) )] + + F ( + ) F () = F ( + (i + ) ) F ( + i ), where = (b )/n. Also, on ech of our intervls we cn mke the liner pproimtion so we hve F ( + (i + ) ) F ( + i ) + F ( + i )[( + (i + ) ) ( + i )] = F ( + i ) + F ( + i ), F ( + (i + ) ) F ( + i ) F ( + i ). Substitutin this into our sum, we obtin the followin estimte for the difference between F () nd F (b): F (b) F () F ( + i ), where = b n. As usul, incresin n decreses. This mens we re mkin our liner pproimtions on smller nd smller intervls, so these pproimtions re becomin more ccurte. So Since F = f, we cn rewrite this s which is our desired result. F (b) F () = F ( + i ). F (b) F () = f( + i ) = b f()d, 3

4 Note. Notice tht it doesn t mtter which ntiderivtive of f we hve, it just mtters tht F is some ntiderivtive of F. This mens tht if F nd G re both ntiderivtives of f on [, b], then G(b) G() = F (b) F (). In fct, it s not difficult to show tht F nd G must differ by constnt. Also notice tht this theorem justifies our use of the nottion f()d for n ntiderivtive of F. Of course, the rel usefulness of this result is its computtionl power. The quntities F (b) nd F () re usully quite esy to determine, but b f()d, the sined re between the curve y = f() nd the -is, cn be quite difficult to compute. This theorem ives us n esy formul, ssumin we cn find n ntiderivtive of f. Now tht we ve proven the fundmentl theorem (we did it!) let s check bck in on our ols. We sid tht we wnt to know why the FTC mtters. For this, it s importnt to keep in mind tht eqution (2) is not the definition of the definite interl. The definition of the definite interl is the much ulier eqution (). Just lookin t the riht-hnd side of these two equtions is n indictor of why the FTC mtters: we cn now epress res under curves s the difference of two function vlues. Gol 2 ws to know wht the FTC sys, which relly mens understndin the riht side of eqution (2). The riht side of (2) is iven by findin n ntiderivtive of f nd evlutin tht ntiderivtive t the two endpoints of our intervl. The FTC sys tht the re under f between nd b is equl to the chne in the vlue of ny of its ntiderivtives. Fully pprecitin wht the FTC sys will require workin some emples; chievin ol 3 will require workin severl. The net two emples re intended to illustrte the power of the fundmentl theorem (nd thus help us pprecite its eistence). Unlike most emples, you shouldn t feel too uilty if you d prefer to just red the solution to the first one rther thn findin it yourself it s desined to be nsty. Emple. Suppose bll is thrown striht up with initil velocity v >. Then its velocity t time t is iven by v(t) = v + t, where 9.8 m/s is the ccelertion due to rvity. Assumin tht the bll reches its mimum heiht t time t = v /, use Riemnn sum to compute the difference between the initil heiht of the bll nd the mimum heiht of the bll. (Solution) From our discussion, it should be the cse tht n heiht( v /) heiht() = v(i t) t, where t = ( v / )/n = v /(n). Then the difference in heihts is iven by ( ) i v v v(i t) t = v n n 4

5 [ = v i ] ( n v v n = v 2 = v2 = v2 = v2 2. n n(n + ) n 2 2 n i n = v2 = v2 n 2 n(n + ) 2n 2 ( i n n i ) ) n Emple. Confirm the bove result by findin trjectory whose velocity function is v(t) = v + t (tht is, find n ntiderivtive of v(t)) nd evlutin this trjectory t t = nd t = v /. (Solution) Consider the trjectory where is constnt. Then r(t) = + v t + 2 t2, r (t) = v + t = v(t), so r is n ntiderivtive of v. We hve r() = nd ( ) ( ) v v r = + v so ( ) v r + 2 ( v r() = v2 2, ) 2 = v 2 2, s desired. Notice tht the constnt doesn t mtter we cn choose ny ntiderivtive of v. This emple serves to illustrte the power of the fundmentl theorem; findin n ntiderivtive of v(t) nd evlutin t the endpoints ws much simpler thn computin the Riemnn sum. Hopefully these emples helped to convince us tht the FTC is useful tool; to ccomplish ol 3, one should work mny emple problems. Hvin mde ood proress towrds our ols, we ll finish with n ppliction of Theorem tht is very importnt, nd is often referred to s the second fundmentl theorem of clculus. Theorem 2. For nice enouh functions f : [, b] R, ( d ) f(t)dt = f(), d for every (, b). 5

6 Note. There re two typicl objections to this theorem:. If you re little too fmilir with the FTC, this miht look like we re syin the derivtive of the ntiderivtive of function is the function you strted with, which seems trivil. This is, in effect, how we justify the sttement, but it s little deeper thn tht. 2. Even more likely, you miht look t this eqution nd think it s completely useless. Both of these concerns cn (hopefully) be clered up by thinkin of n re function A() = f(t)dt. This function represents the (sined) re under f between nd, nd on its fce hs nothin to do with ntiderivtives. So the theorem is syin somethin: it s syin tht the derivtive of our re function t point is precisely f(). So if f() >, the re under f is incresin t, if f() <, the re is decresin t, nd if f() =, we hve criticl point for the re function. This mkes sense; when our function is positive, we re ddin re to the totl, when our function is netive, we re tkin re wy, nd when our function is the re isn t chnin. Hopefully we cn see tht the theorem is mkin useful sttement. Proof. By nice enouh function, we men tht f hs n ntiderivtive, sy F. Then, usin Theorem, ( d ) f(t)dt = d (F () F ()) d d = d d (F ()) (F ()) d d = F () = f(). We hve d d (F ()) = becuse F () is constnt. We hve F () = f() becuse F is n ntiderivtive of f. We cn ctully prove the second FTC usin the sme locl pproimtion ides we used to prove the first FTC, s the followin, lterntive proof demonstrtes. So the second FTC isn t just corollry to the first FTC; it comes out of the sme ides. Proof. Let A() = f(t)dt. We wnt to compute A A( + h) A() () =. h h 6

7 Notice tht A( + h) = +h so, usin properties of the interl, A( + h) A() = f(t)dt nd A() = +h f(t)dt f(t)dt = +h f(t)dt, f(t)dt. For smll h we cn pproimte the vlue of this lst interl by h f(). This is becuse the interl is bein tken over n intervl of width h, nd the heiht of the function will sty close to f() when h is smll. In prticulr, we should hve +h But this just mens nd this is our desired result. f(t)dt hf(), so h h A( + h) A() h h = h h +h +h f(t)dt = f(). f(t)dt = f(), Emple. Use the second fundmentl theorem of clculus nd intertion by substitution to find [ ] d 2 f(t)dt. d (Solution) First, let s split our interl into two interls, ech of which hs constnt s it of intertion: [ ] [ d 2 f(t)dt = d ] [ 2 f(t)dt + f(t)dt = d 2 ] f(t)dt f(t)dt. d d d Now the second FTC epects liner term s it of intertion, not qudrtic term. To fi this, we introduce vrible u for which u 2 = t. Then dt = 2udu nd our its of intertion become nd, becuse u = when t = nd u = when t = 2. Then d d [ ] 2 f(t)dt = d [ 2uf(u)du d = 2f( 2 ) f(), where the second step comes from the second FTC. ] f(t)dt 7