The Riemann Integral


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1 Deprtment of Mthemtics King Sud University
2 Tble of contents 1 Antiderivtive Function nd Indefinite Integrls
3 Indefinite Integrls & Antiderivtive Function Definition Let f : I R be function defined on n intervl I. A function F : I R is clled n ntiderivtive of f on I if F is differentible on I nd F (x) = f (x), forllx I. There re mny ntiderivtives of the function f (x) = 2x on R such s F 1 (x) = x 2 +1, F 2 (x) = x 2 7, F 3 (x) = x , F 4(x) = x 2 +c.
4 Proposition Let F nd G be two ntiderivtives of function f on n intervl I, then there is constnt c R such tht F (x) = G(x) + c, x I.
5 Definition Let F be n ntiderivtive of function f on n intervl I, we denote f (x)dx ny ntiderivtive i.e. f (x)dx = F (x) + c, x I (1) f (x)dx is clled the indefinite integrl of f on I. In the eqution (1), the constnt c is clled the constnt of integrtion, x is clled the vrible of integrtion, f (x) is clled the integrnd. The mpping f f (x)dx is clled s indefinite integrl or evluting the integrl or integrting f.
6 Bsic tble of indefinite integrls f (x) f (x)dx x r ; r 1, R Q cos(x) sin(x) sec 2 (x) csc 2 (x) sec(x) tn(x) x r+1 r c sin(x) + c cos(x) + c tn(x) + c cot(x) + c sec(x) + c
7 Proposition (Some importnt formuls) If f is differentible on n intervl I, then d f (x)dx = f (x) + c. dx If f hs n ntiderivtive on n intervl I, then d f (x)dx = f (x). dx αf (x)dx = α f (x)dx. f (x) + g(x)dx = f (x)dx + g(x)dx.
8 Substitution Method Theorem (Substitution) If F is n ntiderivtive of f, then f (g(x))g (x) hs ntiderivtive F (g(x)). Or, f (g(x))g (x)dx = F (g(x)) + c. This is obvious. It is clled substitution since it cn be obtined by substituting u = g(x) nd du = g (x)dx into f (u)du = F (u) + c.
9 Remrk Substitution method is lso clled chnging vrible method. Exmples 1 2 (x 2 + 1) n 2xdx u=x2 +1 = sin(2x + 3)dx u=2x+3 = 1 2 u n du = un+1 n + 1 = (x 2 + 1) n+1 n + 1 sin udu = 1 2 cos u + c = + c. 3 1 cos(2x + 3) + c. 2 1 u=πx cos 2 dx = 1 (πx) π 1 cos 2 (u) du = 1 tn(πx) + c. π
10 Theorem Let I be n intervl, r Q \ { 1} nd f : I R differentible function. Assume tht f r (x) is defined for every x I. Then f r (x)f (x)dx = f r+1 (x) r c.
11 Summtion Nottions Definition Given set of numbers { 1, 2,..., n }, the symbol represents their sum s follows n k = n. k=1 n k=1 k
12 Theorem For every c R nd n N, we hve n c = nc. k=1 Theorem Let α, β R nd n N. For every 1, 2,..., n, b 1, b 2,..., b n R we hve n n n (α k + βb k ) = α k + β b k. k=1 k=1 k=1
13 Theorem For ll n N, we hve n k = k=1 n k 2 = k=1 n k 3 = k=1 n(n + 1) 2 n(n + 1)(2n + 1) 6 [ n(n + 1) ] 2. 2
14 Exmple 1 : n (3k 2 2k + 1) = 3 k=1 n k 2 2 k=1 n k + k=1 n 1, k=1 n(n + 1)(2n + 1) = = n 2 n(n + 1) 2 [ (n + 1)(2n + 1) 2(n + 1) + 2 = n 2 (2n2 + 3n + 1 2n 2 + 2), = n 2 (2n2 + n + 1). + 1.n, ],
15 The pproch of the integrl of function by res gives the geometricl sense of integrtion. The second pproch consists in introducing priori the ntiderivtive of function. The ide of the first pproch is to cut the intervl [, b] by subdivision in subintervls [ j, j+1 ], then to dd the res of rectngles bsed on the intervls [ j, j+1 ].
16 = + k b. Definitions 1 A prtition P of the closed intervl [, b] is finite set of points P = { 0, 1, 2,..., n } such tht = 0 < 1 < 2 <... < n 1 < n = b. Ech [ j 1, j ] is clled subintervl of the prtition nd the number h j = j j 1 is clled the mplitude of this intervl. 2 The norm of prtition is defined to be the length of the longest subintervl [ j, j+1 ], tht is, it is P = mx{ j j 1, j = 1,..., n}. 3 A prtition P = { 0, 1, 2,..., n } of the closed intervl [, b] is clled uniform if k+1 k = b. Then in this n cse
17 Definitions 1 A mrk on the prtition P = { 0, 1, 2,..., n } is set of points w = {x 1,..., x n } such tht x j [ j 1, j ]. 2 A pointed prtition of n intervl is prtition of n intervl together with finite sequence of numbers x 1, x 2,..., x n such tht j = 1,..., n, x j [ j 1, j ]. This pointed prtition will be denoted by P = {([ j 1, j ], x j )} 1 j n.
18 Definition Let P = {([ j 1, j ], x j )} 1 j n be pointed prtition of the intervl [, b]. The Riemnn sum of f with respect to the pointed prtition P is the number R(f, P) = n f (x j )( j j 1 ) = j=1 n f (x j ) j (2) Ech term in the sum is the product of the vlue of the function t given point nd the length of n intervl. Consequently, ech term represents the re of rectngle with height f (x j ) nd length j j 1. The Riemnn sum is the signed re under ll the rectngles. j=1
19 The Riemnn sum s D (f ) is the lgebric re of the union of the rectngles of width j nd height f (x j ). This is n lgebric re since f (x j ) j is counted positively if f (x j ) > 0 nd negtively if f (x j ) < 0. Intuitively the lgebric re A under the grph of f is the limit of s D (f ) when the j tend to 0. One possible choice is uniform prtition of the intervl [, b] j = + j b n, 0 j n where j = h = b n, which could be combined with ny choice of x j.
20 y f(x i) = 0 i 1 i xi i+1 b = n x
21 Definition For ny bounded function f defined on the bounded nd closed intervl [, b], the definite integrl of f from to b is b f (x)dx = lim n k=1 n f (w k ) x k, ( P 0) whenever the limit exists. (The limit is over ll pointed prtitions P = {([x j 1, x j ], w j )} 1 j n ). When the limit exists, we sy tht f is Riemnn integrble (or integrble) on [, b].
22 Exmple 2 : Let f : [0, 1] R defined by f (x) = 1 if x Q [0, 1] nd f (x) = 0 if x Q [0, 1]. If P = {x 0, x 1, x 2,..., x n } prtition of the closed intervl [0, 1], we tke the mrks t = {t 0, t 1, t 2,..., t n }, t = {t 0, t 1, t 2,..., t n} such tht t k [x k, x k+1 ] Q nd t k [x k, x k+1 ] (R \ Q), for ll k = 0,..., n 1. Then R(f, P, t) = 1 nd R(f, P, t ) = 0. Then f is not Riemnn integrble.
23 Theorem Any continuous function f : [, b] R is Riemnn integrble nd b f (x)dx = x k [ + (k 1) b n, + k b n ] b lim n + n n f (x k ). k=1
24 Definition A function f : [, b] R is clled piecewise continuous on closed intervl [, b] if there exists prtition P = {x 0, x 1, x 2,..., x n } of the closed intervl [, b] such tht f is continuous on every intervl ]x k, x k+1 [, lim x x + k f (x) exist in R, for ll k = 0,..., n 1. lim x x k+1 f (x) nd
25 Theorem Any piecewise continuous function f : [, b] R is Riemnn integrble nd b f (x)dx = x k [ + (k 1) b n, + k b n ] b lim n + n n f (x k ). k=1
26 Fundmentl Properties. 1 Linerity. If f, g : [, b] R re two functions nd α,β two rels numbers, then R(αf + βg, P) = αr(f, P) + βr(g, P). 2 Monotony. If f, g : [, b] R re two functions, then f g R(f, P) R(g, P). In prticulr, if f 0, then R(f, P) 0. 3 Chsles s Formul. If < c < b re three rels numbers nd f be function defined on [, b]. If P 1 is pointed prtition of [, c] nd P 2 be pointed prtition of [b, c], then P 1 P 2 is pointed prtition of [, b] nd R(f, P 1 P 2 ) = R(f, P 1 ) + R(f, P 2 ).
27 Theorem If f : [, b] R is integrble nd f (x) 0, x [, b], then the re A of the region under the grph of f from to b is A = b f (x)dx.
28 Theorem We hve the following properties of the the definite integrls: (P 1 ) If α is rel number, then b αdx = α(b ). (P 2 ) If α is rel number nd f : [, b] R is n integrble function, then αf is integrble on [, b] nd b b αf (x)dx = α f (x)dx.
29 Theorem (P 3 )If f nd g re two integrble functions on [, b], then f + g is integrble on [, b] nd b f (x) + g(x)dx = b b f (x)dx + g(x)dx. (P 4 )If f nd g re two integrble functions on [, b], then f g is integrble on [, b] nd b f (x) g(x)dx = b b f (x)dx g(x)dx.
30 Theorem (P 5 ) If < c < b nd if f is n integrble function on [, b], then f is integrble on [, c] nd on [c, b], moreover b f (x)dx = c b f (x)dx + f (x)dx. c (P 6 )If f is integrble on closed intervl I nd if, b nd c three numbers in I, then b c b f (x)dx = f (x)dx + f (x)dx. c
31 Theorem (P 7 ) If f is integrble on [, b] nd x [, b], f (x) 0. Then b f (x)dx 0. (P 8 ) If f nd g re integrble on [, b] nd x [, b], f (x) g(x). Then b f (x)dx b f (x)dx.
32 Exmple 3 : Verify the following inequlity 4 1 (2x + 2)dx 4 1 (3x + 1)dx. Solution. Here [, b] = [1, 4] nd f (x) = 3x + 1, g(x) = 2x + 2. We hve f (x) g(x) = (3x + 1) (2x + 2) = x 1 0, x [1, 4], then f (x) g(x), x [1, 4].
33 Using the property (P 8 ), we obtin 4 1 g(x)dx or 4 (2x + 2)dx f (x)dx (3x + 1)dx.
34 Exmple 4 : Compute the following integrl 3 3 x 2 x 2 dx. Solution. x 2 x 2 = (x 2)(x + 1), then 3 3 x 2 x 2 dx = = x 2 x 2dx + x 2 x 2dx 1 2 x 2 + xdx
35 Theorem (Men Vlue Theorem for the definite integrls) If f is continuous on [, b], then there is number c [, b] such tht b Proof Let m = inf f (x) nd M = sup x [,b] f (x)dx = (b )f (c). x [,b] b f (x). Since m f M, m 1 f (x)dx M. By the b Intermedite Vlue Theorem, there exists c [, b] such tht 1 b f (x)dx = f (c). b
36 Definition Let f be continuous on [, b]. Then the verge vlue f v of f is given by f v = 1 b f (x)dx. b Exmple 5 : Let f (x) = 3x + 7 on the intervl [0, 1]. We know tht 1 0 (3x + 7)dx = 17. Then the point c where f ssumed its 2 verge vlue verify 3c + 7 = 172, then c = 1 2.
37 Theorem () prt I If f is continuous function on [, b], then F (x) = x f (t)dt is continuous on [, b] nd differentible on [, b] nd its derivtive is F (x) = f (x). Proof Let x [, b] nd h 0 such tht x + h [, b]. Then it results fron the The Men Vlue Theorem for Definite Integrls tht there exists c [x, x + h] or c [x + h, x] such tht f (c) = F (x + h) F (x) h lim f (c) = f (x) = F (x). h 0 = 1 h x+h x f (t)dt. As f is continuous
38 Theorem () prt II If f is continuous function on [, b] nd F is n ntiderivtive of f on [, b], then Proof Let G(x) = x b f (t)dt = F (b) F () f (t)dt. We know tht G (x) = f (x), then there exists c R such tht F (x) = G(x) + C for some constnt C for x b. Since G() = 0, then C = F (), nd G(x) = F (x) F (), for ll x [, b].
39 Theorem Let f be continuous on the closed intervl [, b]. Let c [, b] nd Then G(x) = x c f (t)dt; x [, b]. G (x) = f (x); x [, b]. Proof G(x) = x c f (t)dt f (t)dt, then G (x) = f (x).
40 Theorem Let f be continuous on n intervl I. If v nd u be two differentible functions on n intervl J such tht v(j) I nd u(j) I, then the function x v(x) u(x) f (t)dt is differentible on the intervl J. Moreover d dx ( v(x) u(x) ) f (t)dt = v (x)f (v(x)) u (x)f (u(x)); x J.
41 Proof Let F (x) = v(x) u(x) x f (t)dt, where I. f (t)dt = F (u(x)) F (v(x)). Since F (x) = f (x), the Chin Rule Formul yields d dx ( v(x) u(x) ) f (t)dt = v (x)f (v(x)) u (x)f (u(x)); x J.
42 Definition 1 A function f : [, ] R is odd if f ( x) = f (x) for ll x [, ]. 2 A function f : (, ) R is even if f ( x) = f (x) for ll x [, ]. 3 A function f : R R is T periodic if f (x + T ) = f (x) for ll x R
43 Theorem 1 If f is n odd function on [, ], then f (x) dx = 0. 2 If f is n even function on [, ], then 3 If f is T periodic, then R. f (x) dx = 2 +T 0 f (x) dx. f (x) dx = T 0 f (x) dx, for ll
44 Exmple 1 : 1 1 x 2 dx = [ 1 x 3 dx = 1 4 x 4 5+π sin(x) dx = 5 π [ 1 x 2 dx = 2 3 x 3 ] 1 1 π π = 0. ] 1 0 sin(x) dx = 0. = 2 3.
45 Very often definite integrtions cnnot be done in closed form. When this hppens we need some simple nd useful techniques for pproximting definite integrls. In this section we discuss three such simple nd useful methods.
46 The Trpezoidl Rule Let f : [, b] R be non negtive continuous function. In this method, to pproximte the re under the grph of f, we join the point (x j, f (x j )) with the point (x j+1, f (x j+1 )) for ech subintervl [x j, x j+1 ], by stright line nd find the re under this line. Which mens tht we replce f on [x j, x j+1 ] by the polynomil P of degree 1 such tht P(x j ) = f (x j ) nd P(x j+1 ) = f (x j+1 ). We sy tht P interpoltes f t the points x j nd x j+1. Then P(x) = f (x j ) x j+1 x + f (x j+1 ) x x j. x j+1 x j x j+1 x j The re under the grph of P on the intervl [x j, x j+1 ] is the re of trpezoid with vlue equl to 1 2 (x j+1 x j )(f (x j+1 ) + f (x j+1 )).
47 The re under the grph of f is pproximted by n j=1 1 2 (x j+1 x j )(f (x j+1 ) + f (x j )). In the cse where x j+1 x j = b, this re is pproximted by n b 2n n j=1 b (f (x j+1 )+f (x j+1 )) = b 2n f (x)dx b 2n n 1 f () + 2 f (x j ) n 1 f () + 2 f (x j ) j=1 j=1 + f (b). + f (b). (3) This formul (3) is clled the Trpezoidl Rule. This formul is exct for polynomils of degree t most 1.
48 If the function f is C 2 on the intervl [, b]. The reminder for this method is pproximted s follows R n (b )3 M 2 12n 2, M 2 = sup f (2) (x). x [,b]
49 The Simpson Method In this method, we replce f on [x j, x j+1 ] by the polynomil P of degree 2 which interpoltes f t the points x j, x j+1 nd the middle point m j = x j +x j+1 2. xj+1 x j f (x)dx xj+1 x j P j (x)dx = x j+1 x j (f (x j )+f (x j+1 )+4f (m j )). 6 P j (x) = f (x j ) (x j+1 x)(x m j ) (x j+1 x j )(x j m j ) + f (m j) (x j+1 x)(x x j ) (x j+1 m j )(m j x j ) (x x j )(x m j ) + f (x j+1 ) (x j+1 x j )(x j+1 m j ). xj+1 x j f (x)dx xj+1 x j P 2 (x)dx = x j+1 x j (f (x j )+f (x j+1 )+4f (m j )). 6
50 If the prtition is uniform, h = x j+1 x j = b n, then S n (f ) = n 1 j=0 b x 6n (f (x j) + f (x j+1 ) + 4f (m j )) = h ( n 1 f () + f (b) j=1 n 1 f (x j ) + 4 j=0 ) f (m j ). b f (x)dx (b ) ( f ()+f (b)+2 6n n 1 j=1 n 1 f (x j )+4 j=0 ) f (m j ). (4)
51 In other words if we tke n = 2m, the Simpson formul is given by the following: b f (x)dx (b ) ( m 1 f ()+4 3n j=0 m 1 f (x 2j+1 )+2 j=1 ) f (x 2j )+f (b). (5) x j = + j b 2m. This formul (5) is clled The Simpson Formul nd it is exct for polynomils of degree t most 3.
52 If the function f is C 4 on the intervl [, b], The reminder for this method is pproximted s follows R n (b )5 M n 4, M 4 = sup f (4) (x). x [,b]
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