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


 Justin Marsh
 9 months ago
 Views:
Transcription
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, xxis, 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 subintervl [, 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, nonnegtive 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 (Menvlue 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 menvlue 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 theoremi) 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 ntiderivtive 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 theoremii) 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 menvlue 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 ntiderivtive 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 ntiderivtive 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 theoremi, 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.
Review of Riemann Integral
1 Review of Riemnn Integrl In this chpter we review the definition of Riemnn integrl of bounded function f : [, b] R, nd point out its limittions so s to be convinced of the necessity of more generl integrl.
More informationThe Regulated and Riemann Integrals
Chpter 1 The Regulted nd Riemnn Integrls 1.1 Introduction We will consider severl different pproches to defining the definite integrl f(x) dx of function f(x). These definitions will ll ssign the sme vlue
More informationAdvanced Calculus: MATH 410 Notes on Integrals and Integrability Professor David Levermore 17 October 2004
Advnced Clculus: MATH 410 Notes on Integrls nd Integrbility Professor Dvid Levermore 17 October 2004 1. Definite Integrls In this section we revisit the definite integrl tht you were introduced to when
More informationCalculus in R. Chapter Di erentiation
Chpter 3 Clculus in R 3.1 Di erentition Definition 3.1. Suppose U R is open. A function f : U! R is di erentible t x 2 U if there exists number m such tht lim y!0 pple f(x + y) f(x) my y =0. If f is di
More informationMath 554 Integration
Mth 554 Integrtion Hndout #9 4/12/96 Defn. A collection of n + 1 distinct points of the intervl [, b] P := {x 0 = < x 1 < < x i 1 < x i < < b =: x n } is clled prtition of the intervl. In this cse, we
More information1 The Riemann Integral
The Riemnn Integrl. An exmple leding to the notion of integrl (res) We know how to find (i.e. define) the re of rectngle (bse height), tringle ( (sum of res of tringles). But how do we find/define n re
More informationHomework 4. (1) If f R[a, b], show that f 3 R[a, b]. If f + (x) = max{f(x), 0}, is f + R[a, b]? Justify your answer.
Homework 4 (1) If f R[, b], show tht f 3 R[, b]. If f + (x) = mx{f(x), 0}, is f + R[, b]? Justify your nswer. (2) Let f be continuous function on [, b] tht is strictly positive except finitely mny points
More informationChapter 6. Riemann Integral
Introduction to Riemnn integrl Chpter 6. Riemnn Integrl WonKwng Prk Deprtment of Mthemtics, The College of Nturl Sciences Kookmin University Second semester, 2015 1 / 41 Introduction to Riemnn integrl
More informationUNIFORM CONVERGENCE MA 403: REAL ANALYSIS, INSTRUCTOR: B. V. LIMAYE
UNIFORM CONVERGENCE MA 403: REAL ANALYSIS, INSTRUCTOR: B. V. LIMAYE 1. Pointwise Convergence of Sequence Let E be set nd Y be metric spce. Consider functions f n : E Y for n = 1, 2,.... We sy tht the sequence
More informationThe Riemann Integral
Deprtment of Mthemtics King Sud University 20172018 Tble of contents 1 Antiderivtive Function nd Indefinite Integrls 2 3 4 5 Indefinite Integrls & Antiderivtive Function Definition Let f : I R be function
More information1 i n x i x i 1. Note that kqk kp k. In addition, if P and Q are partition of [a, b], P Q is finer than both P and Q.
Chpter 6 Integrtion In this chpter we define the integrl. Intuitively, it should be the re under curve. Not surprisingly, fter mny exmples, counter exmples, exceptions, generliztions, the concept of the
More informationAdvanced Calculus I (Math 4209) Martin Bohner
Advnced Clculus I (Mth 4209) Spring 2018 Lecture Notes Mrtin Bohner Version from My 4, 2018 Author ddress: Deprtment of Mthemtics nd Sttistics, Missouri University of Science nd Technology, Roll, Missouri
More informationWeek 10: Riemann integral and its properties
Clculus nd Liner Algebr for Biomedicl Engineering Week 10: Riemnn integrl nd its properties H. Führ, Lehrstuhl A für Mthemtik, RWTH Achen, WS 07 Motivtion: Computing flow from flow rtes 1 We observe the
More informationDefinite integral. Mathematics FRDIS MENDELU. Simona Fišnarová (Mendel University) Definite integral MENDELU 1 / 30
Definite integrl Mthemtics FRDIS MENDELU Simon Fišnrová (Mendel University) Definite integrl MENDELU / Motivtion  re under curve Suppose, for simplicity, tht y = f(x) is nonnegtive nd continuous function
More information7.2 Riemann Integrable Functions
7.2 Riemnn Integrble Functions Theorem 1. If f : [, b] R is step function, then f R[, b]. Theorem 2. If f : [, b] R is continuous on [, b], then f R[, b]. Theorem 3. If f : [, b] R is bounded nd continuous
More informationDefinite integral. Mathematics FRDIS MENDELU
Definite integrl Mthemtics FRDIS MENDELU Simon Fišnrová Brno 1 Motivtion  re under curve Suppose, for simplicity, tht y = f(x) is nonnegtive nd continuous function defined on [, b]. Wht is the re of the
More informationReview of Calculus, cont d
Jim Lmbers MAT 460 Fll Semester 200910 Lecture 3 Notes These notes correspond to Section 1.1 in the text. Review of Clculus, cont d Riemnn Sums nd the Definite Integrl There re mny cses in which some
More informationReview on Integration (Secs ) Review: Sec Origins of Calculus. Riemann Sums. New functions from old ones.
Mth 20B Integrl Clculus Lecture Review on Integrtion (Secs. 5.  5.3) Remrks on the course. Slide Review: Sec. 5.5.3 Origins of Clculus. Riemnn Sums. New functions from old ones. A mthemticl description
More informationAdvanced Calculus: MATH 410 Uniform Convergence of Functions Professor David Levermore 11 December 2015
Advnced Clculus: MATH 410 Uniform Convergence of Functions Professor Dvid Levermore 11 December 2015 12. Sequences of Functions We now explore two notions of wht it mens for sequence of functions {f n
More informationBig idea in Calculus: approximation
Big ide in Clculus: pproximtion Derivtive: f (x) = df dx f f(x +h) f(x) =, x h rte of chnge is pproximtely the rtio of chnges in the function vlue nd in the vrible in very short time Liner pproximtion:
More informationLecture 1. Functional series. Pointwise and uniform convergence.
1 Introduction. Lecture 1. Functionl series. Pointwise nd uniform convergence. In this course we study mongst other things Fourier series. The Fourier series for periodic function f(x) with period 2π is
More informationThe final exam will take place on Friday May 11th from 8am 11am in Evans room 60.
Mth 104: finl informtion The finl exm will tke plce on Fridy My 11th from 8m 11m in Evns room 60. The exm will cover ll prts of the course with equl weighting. It will cover Chpters 1 5, 7 15, 17 21, 23
More informationMATH 409 Advanced Calculus I Lecture 19: Riemann sums. Properties of integrals.
MATH 409 Advnced Clculus I Lecture 19: Riemnn sums. Properties of integrls. Drboux sums Let P = {x 0,x 1,...,x n } be prtition of n intervl [,b], where x 0 = < x 1 < < x n = b. Let f : [,b] R be bounded
More information1. On some properties of definite integrals. We prove
This short collection of notes is intended to complement the textbook Anlisi Mtemtic 2 by Crl Mdern, published by Città Studi Editore, [M]. We refer to [M] for nottion nd the logicl stremline of the rguments.
More informationProperties of the Riemann Integral
Properties of the Riemnn Integrl Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University Februry 15, 2018 Outline 1 Some Infimum nd Supremum Properties 2
More informationLecture 1: Introduction to integration theory and bounded variation
Lecture 1: Introduction to integrtion theory nd bounded vrition Wht is this course bout? Integrtion theory. The first question you might hve is why there is nything you need to lern bout integrtion. You
More informationMAA 4212 Improper Integrals
Notes by Dvid Groisser, Copyright c 1995; revised 2002, 2009, 2014 MAA 4212 Improper Integrls The Riemnn integrl, while perfectly welldefined, is too restrictive for mny purposes; there re functions which
More informationWeek 7 Riemann Stieltjes Integration: Lectures 1921
Week 7 Riemnn Stieltjes Integrtion: Lectures 1921 Lecture 19 Throughout this section α will denote monotoniclly incresing function on n intervl [, b]. Let f be bounded function on [, b]. Let P = { = 0
More informationPrinciples of Real Analysis I Fall VI. Riemann Integration
21355 Principles of Rel Anlysis I Fll 2004 A. Definitions VI. Riemnn Integrtion Let, b R with < b be given. By prtition of [, b] we men finite set P [, b] with, b P. The set of ll prtitions of [, b] will
More informationx = b a n x 2 e x dx. cdx = c(b a), where c is any constant. a b
CHAPTER 5. INTEGRALS 61 where nd x = b n x i = 1 (x i 1 + x i ) = midpoint of [x i 1, x i ]. Problem 168 (Exercise 1, pge 377). Use the Midpoint Rule with the n = 4 to pproximte 5 1 x e x dx. Some quick
More informationMain topics for the First Midterm
Min topics for the First Midterm The Midterm will cover Section 1.8, Chpters 23, Sections 4.14.8, nd Sections 5.15.3 (essentilly ll of the mteril covered in clss). Be sure to know the results of the
More informationCalculus and linear algebra for biomedical engineering Week 11: The Riemann integral and its properties
Clculus nd liner lgebr for biomedicl engineering Week 11: The Riemnn integrl nd its properties Hrtmut Führ fuehr@mth.rwthchen.de Lehrstuhl A für Mthemtik, RWTH Achen Jnury 9, 2009 Overview 1 Motivtion:
More informationChapter 6. Infinite series
Chpter 6 Infinite series We briefly review this chpter in order to study series of functions in chpter 7. We cover from the beginning to Theorem 6.7 in the text excluding Theorem 6.6 nd Rbbe s test (Theorem
More informationLecture 3 ( ) (translated and slightly adapted from lecture notes by Martin Klazar)
Lecture 3 (5.3.2018) (trnslted nd slightly dpted from lecture notes by Mrtin Klzr) Riemnn integrl Now we define precisely the concept of the re, in prticulr, the re of figure U(, b, f) under the grph of
More informationCalculus III Review Sheet
Clculus III Review Sheet 1 Definitions 1.1 Functions A function is f is incresing on n intervl if x y implies f(x) f(y), nd decresing if x y implies f(x) f(y). It is clled monotonic if it is either incresing
More informationMath 360: A primitive integral and elementary functions
Mth 360: A primitive integrl nd elementry functions D. DeTurck University of Pennsylvni October 16, 2017 D. DeTurck Mth 360 001 2017C: Integrl/functions 1 / 32 Setup for the integrl prtitions Definition:
More informationRiemann Sums and Riemann Integrals
Riemnn Sums nd Riemnn Integrls Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University August 26, 2013 Outline 1 Riemnn Sums 2 Riemnn Integrls 3 Properties
More informationMAT612REAL ANALYSIS RIEMANN STIELTJES INTEGRAL
MAT612REAL ANALYSIS RIEMANN STIELTJES INTEGRAL DR. RITU AGARWAL MALVIYA NATIONAL INSTITUTE OF TECHNOLOGY, JAIPUR, INDIA302017 Tble of Contents Contents Tble of Contents 1 1. Introduction 1 2. Prtition
More informationIntegration Techniques
Integrtion Techniques. Integrtion of Trigonometric Functions Exmple. Evlute cos x. Recll tht cos x = cos x. Hence, cos x Exmple. Evlute = ( + cos x) = (x + sin x) + C = x + 4 sin x + C. cos 3 x. Let u
More informationA REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H. Thomas Shores Department of Mathematics University of Nebraska Spring 2007
A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H Thoms Shores Deprtment of Mthemtics University of Nebrsk Spring 2007 Contents Rtes of Chnge nd Derivtives 1 Dierentils 4 Are nd Integrls 5 Multivrite Clculus
More informationUNIFORM CONVERGENCE. Contents 1. Uniform Convergence 1 2. Properties of uniform convergence 3
UNIFORM CONVERGENCE Contents 1. Uniform Convergence 1 2. Properties of uniform convergence 3 Suppose f n : Ω R or f n : Ω C is sequence of rel or complex functions, nd f n f s n in some sense. Furthermore,
More informationRiemann is the Mann! (But Lebesgue may besgue to differ.)
Riemnn is the Mnn! (But Lebesgue my besgue to differ.) Leo Livshits My 2, 2008 1 For finite intervls in R We hve seen in clss tht every continuous function f : [, b] R hs the property tht for every ɛ >
More informationRiemann Sums and Riemann Integrals
Riemnn Sums nd Riemnn Integrls Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University August 26, 203 Outline Riemnn Sums Riemnn Integrls Properties Abstrct
More informationProperties of the Riemann Stieltjes Integral
Properties of the Riemnn Stieltjes Integrl Theorem (Linerity Properties) Let < c < d < b nd A,B IR nd f,g,α,β : [,b] IR. () If f,g R(α) on [,b], then Af +Bg R(α) on [,b] nd [ ] b Af +Bg dα A +B (b) If
More informationW. We shall do so one by one, starting with I 1, and we shall do it greedily, trying
Vitli covers 1 Definition. A Vitli cover of set E R is set V of closed intervls with positive length so tht, for every δ > 0 nd every x E, there is some I V with λ(i ) < δ nd x I. 2 Lemm (Vitli covering)
More informationProperties of Integrals, Indefinite Integrals. Goals: Definition of the Definite Integral Integral Calculations using Antiderivatives
Block #6: Properties of Integrls, Indefinite Integrls Gols: Definition of the Definite Integrl Integrl Clcultions using Antiderivtives Properties of Integrls The Indefinite Integrl 1 Riemnn Sums  1 Riemnn
More informationNumerical Integration
Chpter 5 Numericl Integrtion Numericl integrtion is the study of how the numericl vlue of n integrl cn be found. Methods of function pproximtion discussed in Chpter??, i.e., function pproximtion vi the
More informationIMPORTANT THEOREMS CHEAT SHEET
IMPORTANT THEOREMS CHEAT SHEET BY DOUGLAS DANE Howdy, I m Bronson s dog Dougls. Bronson is still complining bout the textbook so I thought if I kept list of the importnt results for you, he might stop.
More informationSections 5.2: The Definite Integral
Sections 5.2: The Definite Integrl In this section we shll formlize the ides from the lst section to functions in generl. We strt with forml definition.. The Definite Integrl Definition.. Suppose f(x)
More information38 Riemann sums and existence of the definite integral.
38 Riemnn sums nd existence of the definite integrl. In the clcultion of the re of the region X bounded by the grph of g(x) = x 2, the xxis nd 0 x b, two sums ppered: ( n (k 1) 2) b 3 n 3 re(x) ( n These
More informationUnit #9 : Definite Integral Properties; Fundamental Theorem of Calculus
Unit #9 : Definite Integrl Properties; Fundmentl Theorem of Clculus Gols: Identify properties of definite integrls Define odd nd even functions, nd reltionship to integrl vlues Introduce the Fundmentl
More informationThe Banach algebra of functions of bounded variation and the pointwise Helly selection theorem
The Bnch lgebr of functions of bounded vrition nd the pointwise Helly selection theorem Jordn Bell jordn.bell@gmil.com Deprtment of Mthemtics, University of Toronto Jnury, 015 1 BV [, b] Let < b. For f
More informationFUNDAMENTALS OF REAL ANALYSIS by. III.1. Measurable functions. f 1 (
FUNDAMNTALS OF RAL ANALYSIS by Doğn Çömez III. MASURABL FUNCTIONS AND LBSGU INTGRAL III.. Mesurble functions Hving the Lebesgue mesure define, in this chpter, we will identify the collection of functions
More informationINTRODUCTION TO INTEGRATION
INTRODUCTION TO INTEGRATION 5.1 Ares nd Distnces Assume f(x) 0 on the intervl [, b]. Let A be the re under the grph of f(x). b We will obtin n pproximtion of A in the following three steps. STEP 1: Divide
More informationA BRIEF INTRODUCTION TO UNIFORM CONVERGENCE. In the study of Fourier series, several questions arise naturally, such as: c n e int
A BRIEF INTRODUCTION TO UNIFORM CONVERGENCE HANS RINGSTRÖM. Questions nd exmples In the study of Fourier series, severl questions rise nturlly, such s: () (2) re there conditions on c n, n Z, which ensure
More informationIndefinite Integral. Chapter Integration  reverse of differentiation
Chpter Indefinite Integrl Most of the mthemticl opertions hve inverse opertions. The inverse opertion of differentition is clled integrtion. For exmple, describing process t the given moment knowing the
More informationMath 118: Honours Calculus II Winter, 2005 List of Theorems. L(P, f) U(Q, f). f exists for each ǫ > 0 there exists a partition P of [a, b] such that
Mth 118: Honours Clculus II Winter, 2005 List of Theorems Lemm 5.1 (Prtition Refinement): If P nd Q re prtitions of [, b] such tht Q P, then L(P, f) L(Q, f) U(Q, f) U(P, f). Lemm 5.2 (Upper Sums Bound
More information11 An introduction to Riemann Integration
11 An introduction to Riemnn Integrtion The PROOFS of the stndrd lemms nd theorems concerning the Riemnn Integrl re NEB, nd you will not be sked to reproduce proofs of these in full in the exmintion in
More informationEuler, Ioachimescu and the trapezium rule. G.J.O. Jameson (Math. Gazette 96 (2012), )
Euler, Iochimescu nd the trpezium rule G.J.O. Jmeson (Mth. Gzette 96 (0), 36 4) The following results were estblished in recent Gzette rticle [, Theorems, 3, 4]. Given > 0 nd 0 < s
More informationMA123, Chapter 10: Formulas for integrals: integrals, antiderivatives, and the Fundamental Theorem of Calculus (pp.
MA123, Chpter 1: Formuls for integrls: integrls, ntiderivtives, nd the Fundmentl Theorem of Clculus (pp. 27233, Gootmn) Chpter Gols: Assignments: Understnd the sttement of the Fundmentl Theorem of Clculus.
More informationTHE EXISTENCEUNIQUENESS THEOREM FOR FIRSTORDER DIFFERENTIAL EQUATIONS.
THE EXISTENCEUNIQUENESS THEOREM FOR FIRSTORDER DIFFERENTIAL EQUATIONS RADON ROSBOROUGH https://intuitiveexplntionscom/picrdlindeloftheorem/ This document is proof of the existenceuniqueness theorem
More informationII. Integration and Cauchy s Theorem
MTH6111 Complex Anlysis 200910 Lecture Notes c Shun Bullett QMUL 2009 II. Integrtion nd Cuchy s Theorem 1. Pths nd integrtion Wrning Different uthors hve different definitions for terms like pth nd curve.
More informationIntegrals along Curves.
Integrls long Curves. 1. Pth integrls. Let : [, b] R n be continuous function nd let be the imge ([, b]) of. We refer to both nd s curve. If we need to distinguish between the two we cll the function the
More informationn f(x i ) x. i=1 In section 4.2, we defined the definite integral of f from x = a to x = b as n f(x i ) x; f(x) dx = lim i=1
The Fundmentl Theorem of Clculus As we continue to study the re problem, let s think bck to wht we know bout computing res of regions enclosed by curves. If we wnt to find the re of the region below the
More informationAppendix to Notes 8 (a)
Appendix to Notes 8 () 13 Comprison of the Riemnn nd Lebesgue integrls. Recll Let f : [, b] R be bounded. Let D be prtition of [, b] such tht Let D = { = x 0 < x 1
More informationPresentation Problems 5
Presenttion Problems 5 21355 A For these problems, ssume ll sets re subsets of R unless otherwise specified. 1. Let P nd Q be prtitions of [, b] such tht P Q. Then U(f, P ) U(f, Q) nd L(f, P ) L(f, Q).
More informationCalculus II: Integrations and Series
Clculus II: Integrtions nd Series August 7, 200 Integrls Suppose we hve generl function y = f(x) For simplicity, let f(x) > 0 nd f(x) continuous Denote F (x) = re under the grph of f in the intervl [,x]
More informationAbstract inner product spaces
WEEK 4 Abstrct inner product spces Definition An inner product spce is vector spce V over the rel field R equipped with rule for multiplying vectors, such tht the product of two vectors is sclr, nd the
More informationDEFINITE INTEGRALS. f(x)dx exists. Note that, together with the definition of definite integrals, definitions (2) and (3) define b
DEFINITE INTEGRALS JOHN D. MCCARTHY Astrct. These re lecture notes for Sections 5.3 nd 5.4. 1. Section 5.3 Definition 1. f is integrle on [, ] if f(x)dx exists. Definition 2. If f() is defined, then f(x)dx.
More informationMath 116 Calculus II
Mth 6 Clculus II Contents 5 Exponentil nd Logrithmic functions 5. Review........................................... 5.. Exponentil functions............................... 5.. Logrithmic functions...............................
More informationON THE CINTEGRAL BENEDETTO BONGIORNO
ON THE CINTEGRAL BENEDETTO BONGIORNO Let F : [, b] R be differentible function nd let f be its derivtive. The problem of recovering F from f is clled problem of primitives. In 1912, the problem of primitives
More informationThe First Fundamental Theorem of Calculus. If f(x) is continuous on [a, b] and F (x) is any antiderivative. f(x) dx = F (b) F (a).
The Fundmentl Theorems of Clculus Mth 4, Section 0, Spring 009 We now know enough bout definite integrls to give precise formultions of the Fundmentl Theorems of Clculus. We will lso look t some bsic emples
More informationNotes on Calculus. Dinakar Ramakrishnan Caltech Pasadena, CA Fall 2001
Notes on Clculus by Dinkr Rmkrishnn 25337 Cltech Psden, CA 91125 Fll 21 1 Contents Logicl Bckground 2.1 Sets... 2.2 Functions... 3.3 Crdinlity... 3.4 EquivlenceReltions... 4 1 Rel nd Complex Numbers 6
More informationExam 2, Mathematics 4701, Section ETY6 6:05 pm 7:40 pm, March 31, 2016, IH1105 Instructor: Attila Máté 1
Exm, Mthemtics 471, Section ETY6 6:5 pm 7:4 pm, Mrch 1, 16, IH115 Instructor: Attil Máté 1 17 copies 1. ) Stte the usul sufficient condition for the fixedpoint itertion to converge when solving the eqution
More informationThe Fundamental Theorem of Calculus
The Fundmentl Theorem of Clculus MATH 151 Clculus for Mngement J. Robert Buchnn Deprtment of Mthemtics Fll 2018 Objectives Define nd evlute definite integrls using the concept of re. Evlute definite integrls
More informationMath 324 Course Notes: Brief description
Brief description These re notes for Mth 324, n introductory course in Mesure nd Integrtion. Students re dvised to go through ll sections in detil nd ttempt ll problems. These notes will be modified nd
More informationAnalysis III. Ben Green. Mathematical Institute, Oxford address:
Anlysis III Ben Green Mthemticl Institute, Oxford Emil ddress: ben.green@mths.ox.c.uk 2000 Mthemtics Subject Clssifiction. Primry Contents Prefce 1 Chpter 1. Step functions nd the Riemnn integrl 3 1.1.
More informationMath Advanced Calculus II
Mth 452  Advnced Clculus II Line Integrls nd Green s Theorem The min gol of this chpter is to prove Stoke s theorem, which is the multivrible version of the fundmentl theorem of clculus. We will be focused
More informationSection Areas and Distances. Example 1: Suppose a car travels at a constant 50 miles per hour for 2 hours. What is the total distance traveled?
Section 5.  Ares nd Distnces Exmple : Suppose cr trvels t constnt 5 miles per hour for 2 hours. Wht is the totl distnce trveled? Exmple 2: Suppose cr trvels 75 miles per hour for the first hour, 7 miles
More informationa n+2 a n+1 M n a 2 a 1. (2)
Rel Anlysis Fll 004 Tke Home Finl Key 1. Suppose tht f is uniformly continuous on set S R nd {x n } is Cuchy sequence in S. Prove tht {f(x n )} is Cuchy sequence. (f is not ssumed to be continuous outside
More informationThe Fundamental Theorem of Calculus. The Total Change Theorem and the Area Under a Curve.
Clculus Li Vs The Fundmentl Theorem of Clculus. The Totl Chnge Theorem nd the Are Under Curve. Recll the following fct from Clculus course. If continuous function f(x) represents the rte of chnge of F
More informationStuff You Need to Know From Calculus
Stuff You Need to Know From Clculus For the first time in the semester, the stuff we re doing is finlly going to look like clculus (with vector slnt, of course). This mens tht in order to succeed, you
More information1 The fundamental theorems of calculus.
The fundmentl theorems of clculus. The fundmentl theorems of clculus. Evluting definite integrls. The indefinite integrl new nme for ntiderivtive. Differentiting integrls. Tody we provide the connection
More information1.3 The Lemma of DuBoisReymond
28 CHAPTER 1. INDIRECT METHODS 1.3 The Lemm of DuBoisReymond We needed extr regulrity to integrte by prts nd obtin the Euler Lgrnge eqution. The following result shows tht, t lest sometimes, the extr
More informationMath 61CM  Solutions to homework 9
Mth 61CM  Solutions to homework 9 Cédric De Groote November 30 th, 2018 Problem 1: Recll tht the left limit of function f t point c is defined s follows: lim f(x) = l x c if for ny > 0 there exists δ
More informationLine Integrals. Partitioning the Curve. Estimating the Mass
Line Integrls Suppose we hve curve in the xy plne nd ssocite density δ(p ) = δ(x, y) t ech point on the curve. urves, of course, do not hve density or mss, but it my sometimes be convenient or useful to
More informationIntegrals  Motivation
Integrls  Motivtion When we looked t function s rte of chnge If f(x) is liner, the nswer is esy slope If f(x) is nonliner, we hd to work hrd limits derivtive A relted question is the re under f(x) (but
More informationChapter 6 Notes, Larson/Hostetler 3e
Contents 6. Antiderivtives nd the Rules of Integrtion.......................... 6. Are nd the Definite Integrl.................................. 6.. Are............................................ 6. Reimnn
More informationarxiv: v1 [math.ca] 11 Jul 2011
rxiv:1107.1996v1 [mth.ca] 11 Jul 2011 Existence nd computtion of Riemnn Stieltjes integrls through Riemnn integrls July, 2011 Rodrigo López Pouso Deprtmento de Análise Mtemátic Fcultde de Mtemátics, Universidde
More informationMath 113 Exam 1Review
Mth 113 Exm 1Review September 26, 2016 Exm 1 covers 6.17.3 in the textbook. It is dvisble to lso review the mteril from 5.3 nd 5.5 s this will be helpful in solving some of the problems. 6.1 Are Between
More information1.9 C 2 inner variations
46 CHAPTER 1. INDIRECT METHODS 1.9 C 2 inner vritions So fr, we hve restricted ttention to liner vritions. These re vritions of the form vx; ǫ = ux + ǫφx where φ is in some liner perturbtion clss P, for
More informationLECTURE. INTEGRATION AND ANTIDERIVATIVE.
ANALYSIS FOR HIGH SCHOOL TEACHERS LECTURE. INTEGRATION AND ANTIDERIVATIVE. ROTHSCHILD CAESARIA COURSE, 2015/6 1. Integrtion Historiclly, it ws the problem of computing res nd volumes, tht triggered development
More informationTest 3 Review. Jiwen He. I will replace your lowest test score with the percentage grade from the final exam (provided it is higher).
Test 3 Review Jiwen He Test 3 Test 3: Dec. 46 in CASA Mteril  Through 6.3. No Homework (Thnksgiving) No homework this week! Hve GREAT Thnksgiving! Finl Exm Finl Exm: Dec. 1417 in CASA You Might Be Interested
More informationGeometric and Mechanical Applications of Integrals
5 Geometric nd Mechnicl Applictions of Integrls 5.1 Computing Are 5.1.1 Using Crtesin Coordintes Suppose curve is given by n eqution y = f(x), x b, where f : [, b] R is continuous function such tht f(x)
More informationMath Calculus with Analytic Geometry II
orem of definite Mth 5.0 with Anlytic Geometry II Jnury 4, 0 orem of definite If < b then b f (x) dx = ( under f bove xxis) ( bove f under xxis) Exmple 8 0 3 9 x dx = π 3 4 = 9π 4 orem of definite Problem
More informationACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS 1 MATH00030 SEMESTER /2019
ACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS MATH00030 SEMESTER 208/209 DR. ANTHONY BROWN 7.. Introduction to Integrtion. 7. Integrl Clculus As ws the cse with the chpter on differentil
More informationDefinition of Continuity: The function f(x) is continuous at x = a if f(a) exists and lim
Mth 9 Course Summry/Study Guide Fll, 2005 [1] Limits Definition of Limit: We sy tht L is the limit of f(x) s x pproches if f(x) gets closer nd closer to L s x gets closer nd closer to. We write lim f(x)
More information7.2 The Definite Integral
7.2 The Definite Integrl the definite integrl In the previous section, it ws found tht if function f is continuous nd nonnegtive, then the re under the grph of f on [, b] is given by F (b) F (), where
More informationCzechoslovak Mathematical Journal, 55 (130) (2005), , Abbotsford. 1. Introduction
Czechoslovk Mthemticl Journl, 55 (130) (2005), 933 940 ESTIMATES OF THE REMAINDER IN TAYLOR S THEOREM USING THE HENSTOCKKURZWEIL INTEGRAL, Abbotsford (Received Jnury 22, 2003) Abstrct. When relvlued
More informationReview. April 12, Definition 1.2 (Closed Set). A set S is closed if it contains all of its limit points. S := S S
Review April 12, 2017 1 Definitions nd Some Theorems 1.1 Topology Definition 1.1 (Limit Point). Let S R nd x R. Then x is limit point of S if, for ll ɛ > 0, V ɛ (x) = (x ɛ, x + ɛ) contins n element s S
More information