Lecture The definite Integral

Transcription

1 Lecture The definite Integrl Qingling Zeng Deprtment of Mthemtics nd Sttistics York University 1

2 We study the problem of defining re, lis the definite integrl. The re of more complicted region is defined s the limit of the res of unions of rectngles which pproximte this region. (Note: the re of rectngle is the product of its length nd width.) 1: Prtitions Definition A Prtition P = {x 0, x 1,..., x n 1, x n } of n intervl [, b] consists of n incresing sequence of points: = x 0 x 1... x n 1 x n = b The mesh of the prtition is the lrgest of the number x k x k 1 for 1 k n. 2

3 Remrk: We cn put Definition in the other wy: A Prtition P = {x 0, x 1,..., x n 1, x n } of closed intervl [, b] is subdivision of the intervl [, b] into n subintervls. The numbers of P re rrnged in incresing order: The n subintervls re: = x 0 x 1... x n 1 x n = b [x 0, x 1 ], [x 1, x 2 ],... [x n 2, x n 1 ], [x n 1, x n ]. The mesh m(p ) of the prtition P is the length of the longest of these subintervls:. m(p ) = mx{x 1 x 0, x 2 x 1,... x n 1 x n 2, x n x n 1 } Exmple 1: The intervl [0, 1] cn be prtitioned into 4 subintervls in mny wys. 3

4 Regulr Prtition: We cn prtition the intervl [, b] into n subintervls of equl length. This prtition is clled the regulr prtition of the intervl [, b] into n subintervls. The intervl [, b] hs length b, ech subintervl will hve length. Thus P = {x 0, x 1,..., x n } with b n for 0 k n. x k = + k ( b n Exmple 2: Find the regulr prtition of the intervl [3, 10] into five subintervls. Exmple 3: List three prtitions (one of them is regulr prtition) of the intervl [ 2, 3] into 5 subintervls. Find the mesh of ech prtition. ) 4

5 2. Riemnn Sum Definition Let f be function with domin [, b], nd let P = {x 0, x 1,..., x n 1, x n } be prtition of the intervl [, b]. Let T = {t 1,..., t n } be given with t k [x k 1, x k ] for ech 1 k n. The Riemnn sum determined by P, T nd f is defined s:. R(P, T, f) = f(t 1 )(x 1 x 0 ) f(t n )(x n x n 1 ) Exmple 4: P = { 4, 2, 0, 3, 5}, T = { 3, 1, 2, 4}, f(x) = x Evlute the Riemnn sum R(P, T, f). Exmple 5: Approximte the re A bounded by the prbol f(x) = x 2, the x-xis, x = 0 nd x = 1 using the prtition P = {x 0 = 0, x 1 = 1 4, x 2 = 1 2, x 3 = 3 4, x 4 = 1}. Let t k be the midpoint of [x k 1, x k ] for 1 k 4. 5

6 3. Lower nd Upper Riemnn Sums Definition Let f be continuous function with domin [, b]. Let P be prtition of the intervl [, b]. () Choose T = {t 1,..., t n} so tht minimum vlue of y = f(x) on the intervl [x k 1, x k ] occurs t x = t k for 1 k n. Then L(P, f) = R(P, T, f) is clled the lower Riemnn sum of f for the prtition P. () Choose T = {t 1,..., t n } so tht mximum vlue of y = f(x) on the intervl [x k 1, x k ] occurs t x = t k for 1 k n. Then U(P, f) = R(P, T, f) is clled the upper Riemnn sum of f for the prtition P. Exmple 6: Find the lower nd upper Riemnn sums determined by f(x) = x 2, [, b] = [0, 1] nd P = {0, 1 4, 1 2, 3 4, 1}. 6

7 Proposition Let f be continuous function with domin [, b]. If R(P, T, f) is ny Riemnn sum then L(P, f) R(P, T, f) U(P, f). 7

8 4. Definite Integrl Definition Let P nd P be prtition of the intervl [, b]. If P P, we cll P refinement of P. Proposition [, b]. Then Let P be refinement of the prtition P of mesh P mesh P. If f is continuous function with domin [, b], then L(P, f) L(P, f) U(P, f) U(P, f). 8

9 Definition [, b] where b. Let f be continuous function with domin () If there is unique number A such tht L(P, f) A U(P, f) for ll prtition P of [, b], the we sy tht the definite integrl of f on the intervl [, b] exists nd equls A. The usul nottion is (b) Define A = f(x)dx = f(x)dx. b f(x)dx. Theorem Let f be continuous function with domin [, b]. Then the definite integr f(x)dx exists. 9

10 Exmple 7: (1) The definite integrl of function f(x) = 6 x on the intervl [0, 6], intervl [6, 8] nd intervl [0, 8]. (2) (3) f(x)dx = 0 where f is ny function with Domin f. cdx = c(b ). (4) If f is n odd function, f( x) = f(x) for x [, ], then f(x)dx = 0. Remrk: There re two importnt difference between the definite integrl f(x)dx nd re A bounded by the grph of y = f(x), the x-xis nd the lines x = nd x = b. (1) If < b, then b f(x)dx is defined s f(x)dx. (2) Riemnn sums R(P, T, f) nd hence under the x-xis s negtive. f(x)dx computes re 10

11 Properties of the definite integrl: Proposition Let f nd g be two continuous functions with domin [, b] such tht both f(x)dx nd g(x)dx exist. () If f(x) 0 for x [, b], then f(x)dx 0. (b) If f(x) g(x) for x [, b], then (c) f(x)dx f(x)dx g(x)dx. f(x) dx. 11

12 Theorem (Additivity of Definite Integrls) Let, b, c be three distinct numbers. Let f be continuous function whose domin is the intervl [A, B] where A is the smllest of, b, c nd B is the lrgest of these three numbers. Assume tht f(x)dx nd c b f(x)dx exist. Then c f(x)dx exist nd c f(x)dx = f(x)dx + c b f(x)dx. Exmple 8: Evlute 2 f(x)dx where 5 f(x) = 1 for 5 x < 1 tn πx 4 for 1 x 1 2 x for 1 < x 2 12

13 Theorem (Men Vlue Theorem for Definite Integrls) Let f be continuous function with domin [, b] where < b. Then there is t lest one number c [, b] such tht f(c) = f(x)dx. b Corollry Let f be continuous function with domin [, b]. Let the minimum vlue of f occur t x = m, nd let the mximum vlue of f occur t x = M. Then f(m)(b ) f(x)dx f(m)(b ). Exmple 9: Verify the men vlue theorem for the definite integrls when f(x) = 3x on the intervl [0, 6]. 13