Fundamental Theorem of Calculus

Transcription

1 Chater 6 Fundamental Theorem of Calulus 6. Definition (Nie funtions.) I will say that a real valued funtion f defined on an interval [a, b] is a nie funtion on [a, b], if f is ontinuous on [a, b] and integrable on every subinterval of [a, b]. Remark: We know that ieewise monotoni ontinuous funtions on [a, b] are nie. It turns out that every ontinuous funtion on [a, b] is nie, but we are not going to rove this. The theorems stated in this hater for nie funtions are usually stated for ontinuous funtions. You an find a roof that every ontinuous funtion on an interval [a, b] is integrable on [a, b] (and hene that every ontinuous funtion on [a, b] is nie on [a, b]) in [44, age 246] or in [, age 53]. However both of these soures use a slightly different definition of ontinuity and of integral than we do, so you will need to do some work to translate the roofs in these referenes into roofs in our terms. You might try to rove the result yourself, but the roof is rather triky. For all the aliations we will make in this ourse, the funtions examined will be ontinuous and ieewise monotoni so the theorems as we rove them are good enough. 6.2 Exerise. Can you give an examle of a ontinuous funtion on a losed interval that is not ieewise monotoni? You may desribe your examle rather loosely, and you do not need to rove that it is ontinuous. 6.3 Theorem (Fundamental theorem of alulus I.) Let g be a nie funtion on [a, b]. Suose G is an antiderivative for g on [a, b]. Then G is 32

2 32 an indefinite integral for g on [a, b]; i.e., g = G(q) G() for all, q [a, b]. (6.4) Proof: By the definition of antiderivative, G is ontinuous on [a, b] and G = g on (a, b). Let, q be arbitrary oints in [a, b]. I will suose < q. (Note that if (6.4) holds when < q, then it holds when q <, sine both sides of the equation hange sign when and q are interhanged. Also note that the theorem learly holds for = q.) Let P = {x,, x m } be any artition of [, q], and let i be an integer with i m. If x i < x i we an aly the mean value theorem to G on [x i, x i ] to find a number s i (x i, x i ) suh that g(s i )(x i x i ) = G (s i )(x i x i ) = G(x i ) G(x i ). If x i = x i, let s i = x i. Then S = {s i,, s m } is a samle for P suh that (g, P, S) = = m i= m i= g(s i )(x i x i ) G(x i ) G(x i ) = G(x m ) G(x ) = G(q) G(). We have shown that for every artition P of [, q] there is a samle S for P suh that (g, P, S) = G(q) G(). Let {P n } be a sequene of artitions for [, q] suh that {µ(p n )}, and for eah n Z + let S n be a samle for P n suh that (g, Pn, S n ) = G(q) G(). Then, sine g is integrable on [q, ], g = lim{ (g, P n, S n )} = lim{g(q) G()} = G(q) G().

3 322 CHAPTER 6. FUNDAMENTAL THEOREM OF CALCULUS 6.5 Examle. The fundamental theorem will allow us to evaluate many integrals easily. For examle, we know that d ( ) artan(x) = dx + x2. Hene, by the fundamental theorem, + x 2dx = artan(x) = artan() artan() = π 4 = π 4. y = x 2 y = +x 2 Two sets with the same area This says that the two sets {(x, y): x and y x 2 } and {(x, y): x and y + x } 2 have the same area a rather remarkable result. 6.6 Theorem (Mean value theorem for integrals.) Let f be a nie funtion on an interval [, q], where < q. Then there is a number (, q) suh that f = f()(q ) i.e., f() = f. q Proof: Sine f is ontinuous on [, q] we an find numbers r, s [, q] suh that f(r) f(x) f(s) for all x [, q].

4 323 By the inequality theorem for integrals f(r) f f(s); (here f(r) and f(s) denote onstant funtions) i.e., i.e., f(r)(q ) f f(s)(q ), f(r) f f(s). q We an now aly the intermediate value roerty to f on the interval whose endoints are r and s to get a number between r and s suh that f() = f. q The number is in the interval (, q), so we are done. 6.7 Corollary. Let f be a nie funtion on a losed interval whose endoints are and q where q. Then there is a number between and q suh that f() = f. q 6.8 Exerise. Exlain how orollary 6.7 follows from theorem 6.6. (There is nothing to show unless q < ) 6.9 Lemma. Let f be a funtion suh that f is integrable on every subinterval of [a, b]. Let [a, b] and let F(x) = Then F is ontinuous on [a, b]. f for all x [a, b]. Proof: Let t [a, b]. I will show that F is ontinuous at t. Sine f is integrable on [a, b] there is a number M suh that M f(x) M for all x [a, b].

5 324 CHAPTER 6. FUNDAMENTAL THEOREM OF CALCULUS By the orollary to the inequality theorem for integrals (8.7), it follows that t f M s t s for all s, t [a, b]. Thus, for all s, t [a, b], F(s) F(t) = s f t f s = f M s t. Now lim s t M s t =, so by the squeezing rule for limits of funtions, lim s t F(s) F(t) =. It follows that lim s t F(s) = F(t).. 6. Theorem (Fundamental theorem of alulus II.) Let f be a nie funtion on [a, b], and let [a, b]. Let G(x) = Then G is an antiderivative for f, i.e. f for all x [a, b]. d x f = d f(t) dt = f(x). (6.) dx dx In artiular, every nie funtion on [a, b] has an antiderivative on [a, b]. t Proof: Let G(x) = f for all x [a, b] and let t be a oint in (a, b). Let {x n } be any sequene in [a, b] \ {t} suh that {x n } t. Then G(x n ) G(t) x n t = = [ n f x n t xn f. x n t t t f ] By the mean value theorem for integrals, there is a number s n between x n and t suh that G(x n ) G(t) = f(s n ). x n t

6 325 Now s n t x n t for all n and sine { x n t }, we have { s n t }, by the squeezing rule for sequenes. Sine f is ontinuous, we onlude that {f(s n )} f(t); i.e., i.e., { G(x n ) G(t) } f(t); x n t G(x) G(t) lim x t x t = f(t). This roves that G (t) = f(t) for t (a, b). In addition G is ontinuous on [a, b] by lemma 6.9. Hene G is an antiderivative for f on [a, b]. Remark Leibnitz s statement of the fundamental rinile of the alulus was the following: Differenes and sums are the inverses of one another, that is to say, the sum of the differenes of a series is a term of the series, and the differene of the sums of a series is a term of the series; and I enuniate the former thus, dx = x, and the latter thus, d = x[34, age 42]. To see the relation between Leibnitz s formulas and ours, in the equation d = x, write x = ydt to get d ydt = ydt, or d ydt = y. This orresonds to equation (6.). Equation (6.4) an be written dt as dg dx = G(q) G(). dx If we anel the dx s (in the next hater we will show that this is atually justified!) we get dg = G(q) G(). This is not quite the same as dx = x. However if you hoose the origin of oordinates to be (, G()), then the two formulas oinide. To emhasize the inverse-like relation between differentiation and integration, I will restate our formulas for both arts of the the fundamental theorem, ignoring all hyotheses: d x f(t)dt = f(x) and dx d f(t)dt = f(x) f(). dt

7 326 CHAPTER 6. FUNDAMENTAL THEOREM OF CALCULUS By exloiting the ambiguous notation for indefinite integrals, we an get a form almost idential with Leibniz s: d d f(x)dx = f(x) and f(x)dx = f(x). dx dx 6.2 Examle. Let F(x) = G(x) = 3 e t2 dt, e t2 dt, H(x) = e x2 e t2 dt. We will alulate the derivatives of F, G, and H. By the fundamental theorem, F (x) = e x2. Now G(x) = F(x 3 ), so by the hain rule, G (x) = F (x 3 ) 3x 2 = e (x3 ) 2 3x 2 = 3x 2 e x6. We have H(x) = e x2 F(x), so by the rodut rule, H (x) = e x2 F (x) + e x2 ( 2x)F(x) = e x2 e x2 + e x2 ( 2x) = 2xe x2 e t2 dt. e t2 dt 6.3 Exerise. Calulate the derivatives of the following funtions. Simlify your answers as muh as you an. a) F(x) = b) G(x) = ) H(x) = 2 + t 2 dt t2 dt t2 dt

8 327 d) K(x) = e) L(x) = sinh(x) osh(x) 2 + t 2 dt t2 dt. (We defined osh and sinh in exerise 4.56.) Find simle formulas (not involving any integrals) for K and for L. 6.4 Exerise. Use the fundamental theorem of alulus to find a) b) ) π 4 2 se 2 x dx. x 2 dx. e x dx. 6.5 Exerise. Let F and F 2 be the funtions whose grahs are shown below: {y = F (x)} {y = F 2 (x)} Let G i (x) = F i (t) dt for t 8. Sketh the grahs of G and G 2. Inlude some disussion about why your answer is orret.