f(x)dx . Show that there 1, 0 < x 1 does not exist a differentiable function g : [ 1, 1] R such that g (x) = f(x) for all

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1 3 Definite Integrl 3.1 Introduction In school one comes cross the definition of the integrl of rel vlued function defined on closed nd bounded intervl [, b] between the limits nd b, i.e., f(x)dx s the number g(b) g(), where g is function whose derivtive is f t ll points in the open intervl (, b). One immedite question tht one would like to rise is the following: Given ny function f : [, b] R, does there exist differentible function g : [, b] R such tht g (x) = f(x) for ll x (, b)? Obviously, it is not necessry to hve such function g (See the exercise below). { 1, 1 x 0, Exercise 3.1 Consider the function f(x) =. Show tht there 1, 0 < x 1 does not exist differentible function g : [ 1, 1] R such tht g (x) = f(x) for ll x (0, 1). Another point one reclls from school is tht if g : [, b] R is differentible function, then g (x) hs geometric mening, nmely, it represents the slope of the tngent to the grph of g t the point x. Do we hve geometric mening to the integrl f(t)dt? We nswer both the bove questions ffirmtively for certin clss of functions by giving geometric definition of the concept of integrl. Suppose f : [, b] R is bounded function. Our ttempt is to ssocite number γ to such function such tht, in cse f(x) 0 for x [, b], then γ is the re of the region under the grph of f, i.e., the region bounded by the grph of f, x-xis, nd the ordintes t nd b. We my not succeed to do this for ll bounded functions f. 112

2 Lower nd Upper Sums 113 Suppose, for moment, the f(x) 0 for ll x [, b]. Let us gree tht we hve some ide bout wht the re A f of the region under the grph of f. Then it is cler tht m(b ) A f M(b ), (1) where m = inf f(x) nd M = sup f(x). x [,b] x [,b] Thus, we get n upper nd lower bound for the A f. To get better estimtes, let us consider point c such tht < c < b. Then we hve m 1 f(x) M 1 x [, c]; m 2 f(x) M 2 x [c, b], where m 1 = inf f(x), m 2 = inf f(x), x [,c] x [c,b] Then it is obvious tht M 1 = sup f(x), x [,c] M 2 = sup f(x). x [c,b] Since m 1 (c ) + m 2 (b c) A f M 1 (c ) + M 2 (b c). (2) m(b ) = m(c ) + m(b c) m 1 (c ) + m 2 (b c), M(b ) = M(c ) + M(b c) M 1 (c ) + M 2 (b c), we cn infer tht the estimtes in (2) re better thn those in (1). We my be ble to improve these bounds by tking more nd more points in [, b]. This is the bsic ide of Riemnn integrtion. 3.2 Upper nd Lower Sums Let f : [, b] R be bounded function, nd let P be prtition of [, b], i.e., finite set P = {x 0, x 1, x 2,..., x k } of points in [, b] such tht = x 0 < x 1 < x 2 <... < x k = b. We shll denote such prtitions lso by P = {x i } k i=0. Corresponding to the prtition P = {x i } k i=0 nd the function f, we ssocite two numbers: where L(P, f) := for i = 1,..., k. m i (x i x i 1 ), U(P, f) := m i = M i (x i x i 1 ), inf f(x), M i = sup f(x) x [x i 1,x i ] x [x i 1,x i ]

3 114 Definite Integrl M.T. Nir Remrk 3.1 If f is continuous, then L(P, f) bd U(P, f) cn be represented s L(P, f) := f(c i )(x i x i 1 ), U(P, f) := respectively, for some c i, d i, i {1,..., k} (Why?). f(d i )(x i x i 1 ), Definition 3.1 The quntities L(P, f) nd U(P, f) re clled lower sum nd upper sum, respectively, of the function f ssocited with the prtition P. Note tht if f(x) 0 for ll x [, b], then L(P, f) is the totl re of the rectngles with lengths m i nd widths x i, nd U(P, f) is the totl re of the rectngle with lengths M i nd widths x i x i 1 for i = 1,..., k. Thus, it is intuitively cler tht the required re A f under the grph of f must stisfy the reltion: for ll prtitions P of [, b]. L(P, f) A f U(P, f) Throughout this chpter, functions defined on [, b] re considered to be bounded functions, nd P denotes the set of ll prtitions of [, b]. Thus we hve L(P, f) U(P, f) P P. NOTATION: If P nd Q re prtitions of [, b], then we denote by P Q the prtition obtined by tking ll points in P nd Q with the condition tht common points would be tken only once. It cn be seen (Verify!) tht if P, Q P nd P = P Q, then L(P, f) L( P, f) nd U( P, f) U(Q, f). Hence Also, we observe tht L(P, f) U(Q, f) P, Q P. where m = m(b ) L(P, f) U(P, f) M(b ) P P, inf f(x) nd M = sup f(x). Hence, the set x [,b] x [,b] {L(P, f) : P P} is bounded bove by M(b ), nd the set {U(P, f) : P P} is bounded below by m(b ). Therefore, α f := sup{l(p, f) : P P}, β f := inf{u(p, f) : P P}

4 Integrbility nd Integrl 115 exist, nd hence L(P, f) α f β f U(Q, f) for ll P, Q P. Another sums corresponding to prtition P of [, b] re the Riemnn sum. Definition 3.2 Corresponding to prtition P : = x 0 < x 1 < x k = b of [, b] nd set T := {ξ i } k of tgs on P, the sum S(P, f, T ) := f(ξ i )(x i x i 1 ) is clled the Riemnn sum for f corresponding to (P, T ). We my observe tht for ny prtition P of [, b], for every tg T on P. L(P, f) S(P, f, T ) U(P, f) 3.3 Integrbility nd Integrl Definition 3.3 If there exists unique γ such tht for every P P, L(P, f) γ U(P, f), then we sy tht f is Riemnn integrble on [, b], nd this γ is clled the Riemnn integrl of f, nd it is denoted by f(x) dx. Remrk 3.2 In the due course, Riemnn integrl will be simply referred to s integrl Recll tht L(P, f) α f β f U(Q, f) for ll P, Q P, where α f := sup L(P, f), P P β f := inf U(P, f). P P The proof of the following theorem is left s n exercise.

5 116 Definite Integrl M.T. Nir Theorem 3.1 A bounded function f : [, b] R is integrble if nd only if α f = β f, nd in tht cse f(x)dx = α f = β f. NOTATION: In view of Theorem 3.1, the quntities α f nd β f re known s lower integrl nd upper integrl, respectively, nd they re denoted by f(x)dx nd f(x)dx, respectively. Remrk 3.3 Not ll functions re integrble! For exmple, consider f : [, b] R defined by { 0, x Q, f(x) = 1, x Q. For this function we hve L(P, f) = 0 nd U(P, f) = b for ny prtition P of [, b]. Thus, in this cse α f = 0, β f = b, nd hence, f is not integrble. Theorem 3.2 Suppose f is integrble on [, b], nd m, M re such tht m f(x) M x [, b]. Then m(b ) f(x) dx M(b ). Proof. We know tht for ny prtition P on [, b], m(b ) L(P, f) Hence the result. f(x) dx U(P, f) M(b ). Definition 3.4 Suppose f : [, b] R is integrble. Then we define b f(x) dx := f(x) dx. Also, for ny function function f : [, b] R, we define τ τ f(x) dx := 0 τ [, b].

6 Integrbility nd Integrl 117 Exmple 3.1 Let f(x) = c for ll x [, b], for some c R. Then we hve L(P, f) = c(b ), U(P, f) = c(b ) Hence c(b ) = L(P, f) α f β f U(P, f) = c(b ). Thus, f is integrble nd f(x)dx = c(b b). Exmple 3.2 Let f(x) = x for ll x [, b]. Let us consider n rbitrry prtition P : = x 0 < x 1 < < x k = b. Then L(P, f) = x i 1 (x i x i 1 ), U(P, f) = x i (x i x i 1 ). Since x i 1 (x i x i 1 ) 1 2 (x i + x i 1 )(x i x i 1 ) x i (x i x i 1 ) we hve L(P, f) 1 2 (x 2 i x 2 i 1) U(P, f). Thus, L(P, f) b2 2 2 for ll prtitions P. Also, we hve U(P, f) U(P, f) L(P, f) = (x i x i 1 ) 2. Thus, tking x i = + i (b ), i = 0, 1,..., k, k ( b ) 2 (b ) 2 U(P, f) L(P, f) = k =. k k Since L(P, f) α f β f U(P, f), we obtin, β f α f (b )2 k k N showing tht α f = β f nd hence f is integrble, nd the integrl is (b 2 2 )/2.

7 118 Definite Integrl M.T. Nir 3.4 Some Bsic Properties In Exmple 3.2, wht we hve showed is tht for sequence (P n ) of prtitions, the sequences {U(P n, f)} nd {L(P n, f)} converge to the sme point. We shll see (Corollry 3.4 below) tht this is true for every integrble function. Theorem 3.3 A bounded function f : [, b] R is Riemnn integrble if nd only if for every ε > 0, there exists prtition P of [, b] such tht U(P, f) L(P, f) < ε. Proof. Suppose f is integrble nd γ is its integrl. Then γ is the unique number such tht for every prtition P of [, b], L(P, f) γ U(P, f). (1) Hence, if there exists n ε 0 > 0 stisfying U(P, f) L(P, f) ε 0 for ll prtitions P of [, b], then we rrive t contrdiction to the fct tht γ is the unique number stisfying (1). This proves the necessry prt. For the sufficiency prt, suppose tht for every ε > 0, there exists prtition P ε of [, b] such tht U(P ε, f) L(P ε, f) < ε. Since L(P ε, f) α f β f U(P ε, f) we obtin tht β f α f < ε for every ε > 0. This shows tht α f = β f. Thus, f is integrble. The proof of the following corollry is immedite from Theorem 3.3. Corollry 3.4 A bounded function f : [, b] R is Riemnn integrble if nd only if there exists sequence (P n ) of prtitions of [, b] such tht nd in tht cse U(P n, f) L(P n, f) 0 s n L(P n, f) f(x) dx, U(P n, f) f(x) dx nd S(P n, f, T n ) f(x) dx, where T n is ny set of tgs on P n, n N. s n. In the ppendix (Section 3.8), we shll give chrcteriztion of Riemnn integrbility in terms of Riemnn sums (see Theorem 3.24). Exercise 3.2 Supply detils of the proof of Theorem 3.3.

8 Some Bsic Properties 119 Theorem 3.5 Let f nd g be integrble over [, b]. Then f + g is integrble nd [f(x) + g(x)]dx = g(x) dx + f(x) dx, Proof. For ny prtition P of [, b], it cn be seen tht L(P, f) + L(P, g) L(P, f + g) nd U(P, f + g) U(P, f) + U(P, g). Now, let P 1 nd P 2 be ny two rbitrry prtitions of [, b] nd P = P 1 P 2. Since L(P 1, f) + L(P 2, g) L(P, f) + L(P, g) L(P, f + g), U(P, f + g) U(P, f) + U(P, g) U(P 1, f) + U(P 2, g), nd since we obtin L(P, f + g) α f+g β f+g U(P, f + g), L(P 1, f) + L(P 2, g) α f+g β f+g U(P 1, f) + U(P 2, g). This is true for ny two rbitrry prtitions P 1, P 2 of [, b]. Therefore (How?), α f + α g α f+g β f+g β f + β g. But, Hence, α f + α g = β f + β g = α f+g = β f+g = f(x)dx + f(x)dx + g(x)dx. g(x)dx so tht f + g is integrble nd [f(x) + g(x)]dx = f(x)dx + g(x)dx. Exercise 3.3 Give n lternte proof of Theorem 3.5 using Corollry 3.4. Theorem 3.6 If f is integrble on [, b] nd c R, then cf is integrble on [, b], nd cf(x)dx = c f(x)dx. Proof. Let c 0. Then for ny given prtition P, we hve L(P, cf) = cl(p, f), U(P, cf) = cu(p, f) so tht sup L(P, cf) = c f(x)dx, inf U(P, cf) = c f(x)dx.

9 120 Definite Integrl M.T. Nir Hence, If c < 0, then we hve Thus, once we prove we obtin sup L(P, cf) = inf U(P, cf) = c cf(x)dx = ( c)( f)(x)dx = ( c) [ f(x)]dx = cf(x)dx = ( c)( 1) f(x)dx. ( f)(x)dx. f(x)dx, (1) (x)dx = c (x)dx. To prove (1), let P be ny prtition of [, b]. Now, we my observe tht, for ny bounded set S R, inf{ s : s S} = sup{s : s S}, sup{ s : s S} = inf{s : s S}. (2) Hence, for ny sub-intervl [, d] of [, b], we hve so tht inf [c,d] [ f(x)] = sup f(x), [c,d] L(P, f) = U(P, f), sup[ f(x)] = inf f(x) [c,d] [c,d] L(P, f) = U(P, f). Agin using (2), we hve sup L(P, f) = inf U(P, f), inf U(P, f) = sup L(P, f). Thus, so tht sup L(P, f) = f(x)dx, inf U(P, f) = sup L(P, f) = inf U(P, f) = f(x)dx. Thus, we hve proved tht cf(x)dx = c f(x)dx for ll c R. f(x)dx Theorem 3.7 Suppose f is integrble on [, c] nd [c, b]. Then f is integrble on [, b], nd f(x) dx = c f(x) dx + c f(x) dx.

10 Some Bsic Properties 121 Proof. Let f 1 = f [,c], f 2 = f [c,b]. Let ε > 0 be given. Since f 1 nd f 2 re integrble, there exist prtitions P 1 nd P 2 of [, c] nd [c, b] respectively such tht U(P 1, f 1 ) L(P 1, f 1 ) < ε 2, U(P 2, f 2 ) L(P 2, f 2 ) < ε 2. Suppose P = P 1 P 2. Then, it cn be seen tht Hence, L(P, f) = L(P 1, f 1 ) + L(P 2, f 2 ), U(P, f) = U(P 1, f 1 ) + U(P 2, f 2 ). U(P, f) L(P, f) = [U(P 1, f 1 ) L(P 1, f 1 )] + [U(P 2, f 2 ) L(P 2, f 2 )] < ε. Thus f is integrble. Since L(P 1, f 1 ) + L(P 2, f 2 ) it follows tht c L(P, f) c f(x) dx + f(x)dx + c f(x)dx U(P, f), c f(x) dx This is true for ll ε > 0. Hence the finl result. f(x)dx U(P 1, f 1 ) + U(P 2, f 2 ), f(x) dx < ε. Theorem 3.8 If f is integrble on [, b] such tht f(x) 0 for ll x [, b], then f(x) dx 0. More generlly, if f nd g re integrble on [, b], then f(x) g(x) x [, b] = f(x) dx g(x) dx. Proof. Since L(P, f) f(x)dx for every prtition P on [, b], nd since L(P, f) 0 by the ssumption on f, we obtin f(x)dx 0. The generl cse follows by pplying the bove result for the function g f. Corollry 3.9 If f is integrble on [, b] such tht f(x) 0 for ll x [, b], then for [c, d] [, b], f(x) dx d c f(x) dx. Corollry 3.10 Suppose f is integrble on [, b]. Then f(x)dx f(x) dx. Exercise 3.4 Prove Corollries 3.9 nd 3.10.

11 122 Definite Integrl M.T. Nir 3.5 Integrl of Continuous Functions Definition 3.5 Let P = {x i : i = 0, 1,..., k} be prtition of [, b]. Then the quntity µ(p ) := mx{x i x i 1 : i = 1,..., k} is clled the mesh of the prtition P. Theorem 3.11 Suppose f : [, b] R is continuous function. Then f is integrble. In fct, we hve the following: (i) For every ε > 0 there exists δ > 0 such tht for every prtition P of [, b] with µ(p ) < δ, we hve U(P, f) L(P, f) < ε. (ii) Suppose (P n ) is sequence of prtitions of [, b] such tht µ(p n ) 0 s n, nd for ech n N, let T n be set of tgs on P n. Then the sequences {U(P n, f)}, {L(P n, f)} {S(P n, f, T n )} converge to the sme limit f(x) dx. The min property of continuous function f : [, b] R tht is required to prove the first prt is its uniform continuity. Definition 3.6 A rel vlued function f defined on n intervl I is sid to be uniformly continuous on I if for every ε > 0, there exists δ > 0 such tht x, y I, x y < δ = f(x) f(y) < ε. Clerly, every uniformly continuous function is continuous. But, the converse is not true. To see this, consider the function f(x) = 1 x, 0 < x < 1. Clerly, f is continuous on I := (0, 1). Now, let ε > 0 be given. For x, y (0, 1), f(x) f(y) < ε x y xy < ε x y < εxy. Thus, for f to be uniformly continuous, there must exist δ > 0 such tht δ ε xy, which is not possible. In prticulr, we re not in position to choose δ independent of the points x, y. To see this fct, more clerly, consider x n = 1/n nd y n = 1/(n + 1). Then we know tht x n y n 0 s n nd f(x n ) f(y n ) = 1 for ll n N. Hence, the condition in the definition of uniform continuity is not stisfied if we tke ε < 1. The bove kind of sitution will not rise if the intervl I is closed nd bounded. More specificlly, we hve the following theorem.

12 Integrl of Continuous Functions 123 Theorem 3.12 Every rel vlued continuous function defined on closed nd bounded intervl is uniformly continuous. Proof. Suppose f : [, b] R is continuous. We hve to show tht for every ε > 0, there exists δ > 0 such tht x y < δ = f(x) f(y) < ε. Suppose this is not true. Then there exists ε 0 > 0 such tht for every δ > 0, there re x, y such tht x y < δ but f(x) f(y) ε 0. (3) Tking δ = 1/n, there exists x n, y n such tht x n y n < 1/n but f(x n ) f(y n ) ε 0. Since (x n ) is bounded sequence, it hs convergent subsequence, sy x kn c for some c R. Since x kn y kn 0, we lso hve the convergence y kn c. Now, by the continuity of f, we hve f(x kn ) f(c) nd f(y kn ) f(c). Thus, f(x kn ) f(y kn ) 0. This is contrdiction to (3). Proof of Theorem Let f : [, b] R be continuous function, nd let ε > 0 be given. Let P : = x 0 < x 1 < x 2... < x k = b be prtition of [, b]. Then U(P, f) L(P, f) = (M i m i )(x i x i 1 ). Since f is continuous on the closed intervl [, b], there exists ξ i, η i in [x i 1, x i ] such tht M i = f(ξ i ), m i = f(η i ) for i = 1,..., k. Hence, U(P, f) L(P, f) = [f(ξ i ) f(η i )](x i x i 1 ). Agin, since f is uniformly continuous on [, b], there exists δ > 0 such tht f(t) f(s) < ε/(b ) whenever t s < δ. Hence, if we tke P such tht µ(p ) < δ, then we hve U(P, f) L(P, f) = [f(ξ i ) f(η i )](x i x i 1 ) < ε. Therefore, by Theorem 3.3, f is integrble, nd the proof of (i) is lso over. Next, suppose (P n ) is sequence of prtitions such tht µ(p n ) 0 s n. Let N N be such tht µ(p n ) < δ for ll n N. Then, it follows by (i) bove tht

13 124 Definite Integrl M.T. Nir U(P n, f) L(P n, f) < ε for ll n N, showing tht U(P n, f) L(P n, f) 0 s n. Now, the conclusion in (ii) follows from the observtions for ll n N. L(P n, f) f(x)dx U(P n, f), L(P n, f) S(P n, f, T n ) U(P n, f) Remrk 3.4 It cn be lso shown tht if bounded function f : [, b] R is piecewise continuous, i.e., there re t most finite number of points in [, b] t which f is discontinuous, then f is integrble. For proof of this, see Theorem 3.27 in the ppendix (Section 3.8). Theorem 3.13 Suppose f : [, b] R is continuous, f(x) 0 for ll x [, b] nd f(x)dx = 0. Then f(x) = 0 for ll x [, b]. Proof. Suppose the conclusion is not true. Then there exists x 0 [, b] such tht f(x 0 ) 0. Without loss of generlity, ssume tht f(x 0 ) > 0. Then, by continuity of f t x 0, there exists closed intervl [c, d] [, b] contining x 0 such tht f(x) f(x 0 )/2 for ll x [c, d]. Therefore, 0 = which is contrdiction. f(x)dx d c f(x)dx α dx = α(d c) 0, Exercise 3.5 Suppose f : [, b] R is continuous function. Show tht for every ε > 0 there exists δ > 0 such tht for every prtition P of [, b] with µ(p ) < δ, we hve () f(x)dx L(P, f) < ε (b) U(P, f) f(x)dx < ε (c) S(P, f, T ) f(x)dx < ε for ny tg T on P. Remrk 3.5 The property (c) in the bove exercise is written s lim S(P, f, T ) µ(p ) 0 f(x)dx. Exercise 3.6 For continuous functions f nd g give (simpler) proof for Theorem 3.5 using Riemnn sums.

14 Integrl of Continuous Functions 125 Exmple 3.3 Let f(x) = e x for ll x [, b]. Then, f is continuous. Let h n = (b ), x i = + ih n, t i = x i 1, i = 1,..., n. n Then with P n = {x i } n nd T = {t i} n, we hve µ(p n) 0, nd S(P n, f, T n ) = h n e +(i 1)hn = h n e where α n = e hn. Since α n n = e b, we hve α (i 1) n = h n e αn n 1 α n 1, S(P n, f, T n ) = h n e αn n 1 α n 1 = e [e b 1] e hn 1 = [eb e ] e hn 1. Since lim n e hn 1 h n = 1, we hve S(P n, f, T n ) e [e b 1] = e b e. We close this section by test for convergence of series using integrls. Theorem 3.14 (Integrl test) Suppose f(x) is continuous, non-negtive nd decresing function for x [1, ). For ech n N, let n := n 1 f(x)dx. Then h n f(n) converges ( n ) converges. n=1 Proof. First we observe tht n = n 1 f(x) dx = n k k=2 k 1 f(x) dx. Now, since f(x) is decresing function for x [1, ), we hve for ech k N, Hence, for k = 2, 3,..., so tht Thus, Now, let s n := k 1 x k = f(k) f(x) f(k 1). f(k) f(k) k=2 k k=2 f(k) k=2 k 1 k n f(k) for n N. k=1 1 k 1 f(x) dx f(k 1) f(x) dx f(x) dx f(k 1). k=2 f(k 1). k=2 h n Then from the bove inequlities, together with the fct tht (s n ) is monotoniclly incresing sequence, it follows tht ( n ) converges if nd only if (s n ) converges.

15 126 Definite Integrl M.T. Nir 3.6 Men Vlue Theorems Theorem 3.15 (Men-vlue theorem) Suppose f is continuous on [, b]. Then there exists ξ [, b] such tht 1 b f(x) dx = f(ξ). Proof. Since f is continuous, we know tht there exist u, v [, b] such tht f(u) = m := min f(x) nd f(v) = M := mx f(x). Hence, by Theorem 3.2, f(u) 1 b f(x) dx f(v). Hence, by intermedite vlue theorem, there exists ξ [, b] such tht Hence the result. 1 b f(x) dx = f(ξ). Theorem 3.16 (Generlized men-vlue theorem) Suppose f nd g re continuous on [, b] where g(x 0 ) 0 for some x 0 [, b] nd g(x) 0 for ll x [, b]. Then there exists ξ [, b] such tht f(x)g(x) dx = f(ξ) g(x)dx. Proof. Let m := inf f(x) nd M = sup f(x). Then we hve m g(x)dx f(x)g(x) dx M g(x)dx. Since g(x 0 ) 0 for some x 0 [, b] nd g(x) 0 for ll x [, b], it follows (How?) tht g(x)dx > 0. Hence, m f(x)g(x) dx g(x)dx M. Therefore, by the intermedite vlue theorem, there exists ξ [, b] such tht f(ξ) = f(x)g(x) dx g(x)dx. Thus, f(x)g(x) dx = f(ξ) g(x)dx. Exercise 3.7 The conclusion of Theorem 3.16 holds if g(x) 0 is replced by g(x) 0. How?

16 Fundmentl Theorems 127 Let us illustrte the bove two men vlue theorems by n exmple. Exmple 3.4 Let f be continuous in [, b] where > 1. We show tht there exist ξ, η [, b] such tht f(x) x ( (b ) 1 dx = f(ξ) + b 1 ) ( η ) f(t)dt. b To see this, first we pply the product rule for integrtion: f(x) [ 1 x dx = x ] b f(t)dt ( 1x 2 ) ( ) f(t)dt dx. Now, by Men Vlue Theorem 3.15, there exists ξ [, b] such tht f(t)dt = (b )f(ξ), ( ) nd by Generlized Men Vlue Theorem 3.16, there exists η [, b] such tht ( 1 x ) ( η ) 1 x 2 f(t)dt dx = f(t)dt x 2 dx. ( 1 1 Hence, from ( ), using the fct tht x 2 dx = 1 ), we get b f(x) x ( (b ) 1 dx = f(ξ) + b 1 ) ( η f(t)dt b In prticulr, if = 1, b = 2, then 2 f(x) 1 x dx = 1 2 f(ξ) + 1 ( η ) f(t)dt. 2 ). 3.7 Fundmentl Theorems The first theorem tht we prove in this section justifies wht we do in school for the clcultion of integrls is correct. Theorem 3.17 (Fundmentl theorem-i) Suppose f is Riemnn integrble on [, b] such tht there exists continuous function g on [, b] which is differentible in (, b) nd g (x) = f(x) for ll x (, b). Then f(t)dt = g(b) g().

17 128 Definite Integrl M.T. Nir Proof. Let P : = x 0 < x 1 <... < x n = b be ny prtition of [, b]. Then by Lgrnge s men vlue theorem, there exists ξ i (x i 1, x i ) such tht Hence, Hence, we hve g(x i ) g(x i 1 ) = g (ξ i )(x i x i 1 ) = f(ξ i )(x i x i 1 ). g(b) g() = [g(x i ) g(x i 1 )] = f(ξ i )(x i x i 1 ). L(P, f) g(b) g() U(P, f) for ll prtition P of [, b]. Since f(x)dx is the unique number which lies between L(P, f) nd U(P, f) for ll prtition P of [, b], we obtin This completes the proof. f(x)dx = g(b) g(). Remrk 3.6 The conclusion of the bove theorem is lso known s Newton- Leibniz formul. The difference g(b) g() is usully written s [g(x)] b. Definition 3.7 A continuous function g on [, b] is clled n nti-derivtive or primitive of f if g is differentible in (, b) nd g (x) = f(x) for ll x (, b). The following exmples hve been worked out by knowing the ntiderivtives of the functions involved. Exmple 3.5 For k 0, Exmple 3.6 For α 0, [ x x k k+1 dx = k + 1 [ e e αx αx dx = α ] b ] b = bk+1 k+1. k + 1 = eαb e α k + 1. Theorem 3.18 (Fundmentl theorem-ii) Suppose f is Riemnn integrble on [, b], nd g(x) = f(t)dt, x [, b]. Then g is continuous on [, b]. If, in ddition, f is continuous on [, b], then g differentible nd g (x) = f(x) x (, b).

18 Fundmentl Theorems 129 Hence, Proof. Let x [, b] nd h R be such tht x + h [, b]. Then we hve g(x + h) g(x) = +h g(x + h) g(x) f(t)dt +h x f(t)dt = +h x f(t) dt M h, f(t)dt. where M > 0 is such tht f(x) M for ll x [, b]. Thus, g(x + h) g(x) 0 s h 0, showing tht g is continuous t x. Next ssume tht f is continuous on [, b] nd x (, b). h R be such tht x + h [, b]. Then, by men-vlue theorem, there exists ξ h between x nd x + h such tht 1 x+h f(t)dt = f(ξ h ). h Since f is continuous t x, we hve f(ξ h ) f(x) s h 0. Hence Thus, g (x) exists nd g (x) = f(x). x g(x + h) g(x) lim = lim f(ξ h ) = f(x). h 0 h h 0 Exercise 3.8 Derive Theorem 3.17 from Theorem The following theorem shows tht integrl of continuous function is Riemnn sum. Theorem 3.19 Suppose f is continuous function.then for every prtition P of [, b], there exists set T of tgs on P such tht S(P, f, T ) = f(x) dx. Proof. Let P = {x i : i = 1,..., k} be prtition of [, b]. Since f is continuous, by men vlue theorem (Theorem 3.15), there exists ξ i [x i 1, x i ] such tht i x i 1 f(x) dx = f(ξ i )(x i x i 1 ), i = 1,..., k. Hence, tking T = {ξ i : i = 1,..., k}, S(P, f, T ) = f(ξ i )(x i x i 1 ) = i x i 1 f(x) dx = f(x) dx. This completes the proof.

19 130 Definite Integrl M.T. Nir Applictions of fundmentl theorems Theorem 3.20 (Product formul) Suppose f nd g re continuous functions on [, b], nd let G be n nti-derivtive of g. If f is differentible on [, b], then f(x)g(x) dx = [f(x)g(x)] b f (x)g(x) dx. Hence, Proof. Recll tht if u nd v re differentible, then [u(x)v(x)] dx = Using fundmentl theorem, Thus, [u(x)v(x)] b = (uv) = u v + uv. u (x)v(x) dx + u (x)v(x) dx + u(x)v (x) dx = [u(x)v(x)] b u(x)v (x) dx. u(x)v (x) dx. u (x)v(x) dx. Now, tking f(x) = u(x) nd v(x) = G(x), we obtin the required formul. Theorem 3.21 (Chnge of vrible formul) Suppose ψ : [α, β] R is differentible function such tht ψ(α) = nd ψ(β) = b. Then, for ny continuous function f : [, b] R, f(x) dx = β α f(ψ(t))ψ (t)dt. Proof. Let F be n nti-derivtive of f, i.e., F (x) = f(x). Then tking G(t) = F (ψ(t)) for t [α, β], we hve G (t) = F (ψ(t))ψ (t) = f(ψ(t))ψ (t), t [α, β]. Hence, by fundmentl theorem, Hence, β α f(ψ(t))ψ (t)dt = β α This completes the proof. β α G (t)dt = G(β) G(α) = F (ψ(β)) F (ψ(α)). f(ψ(t))ψ (t)dt = F (b) F () = f(x) dx.

20 Fundmentl Theorems 131 Theorem 3.22 (The Cuchy integrl reminder formul) Let I be n open intervl, x 0 I nd f hs n + 1 continuous derivtives. Then for every x I, f (k) (x 0 ) f(x) = (x x 0 ) k + 1 f (n+1) (t)(x t) n dt. k! n! x 0 k=0 Proof. Let x I nd x x 0. By Fundmentl theorem-i, we hve f(x) = f(x 0 ) + Hence, by integrtion by prts, we hve x 0 f (t)dt. f (t)dt = f (t) d (x t)dt x 0 x 0 dt [ ] x = f (t)(x t) + f (t)(x t)dt x 0 x 0 = f (x 0 )(x x 0 ) + f (t)(x t)dt. x 0 Thus, theorem holds for n = 1. Now, let m < n nd ssume tht theorem holds for n = m 1, i.e., m 1 f (k) (x 0 ) f(x) = (x x 0 ) k 1 x + f (m) (t)(x t) m 1 dt. k! (m 1)! x 0 Then we hve k=0 x 0 f (m) (t)(x t) m 1 dt = 1 m x 0 f (m) (t) d dt (x t)m dt = 1 m f (m) (x 0 )(x x 0 ) m + 1 m Thus, the theorem holds for n = m s well. x 0 f (m+1) (t)(x t) m dt. Corollry 3.23 (Tylor s formul) Let I be n open intervl, x 0 I nd f hs n + 1 continuous derivtives. Then for every x I, there exists ξ x strictly lying between x nd x 0 such tht f(x) = k=0 f (k) (x 0 ) k! Proof. By Theorem 3.22, f (k) (x 0 ) f(x) = (x x 0 ) k + 1 k! n! k=0 (x x 0 ) k + f (n+1) (ξ x ) (x x 0 ) n+1. (n + 1)! x 0 f (n+1) (t)(x t) n dt nd by Theorem 3.16, there exists ξ x between x nd x 0 such tht Hence, the required formul follows. f (n+1) (t)(x t) n dt = f (n+1) (ξ x ) (x x 0) n+1 x 0 n + 1.