Definite Integrals. The area under a curve can be approximated by adding up the areas of rectangles = 1 1 +


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1 Definite Integrls 5 The re under curve cn e pproximted y dding up the res of rectngles. Exmple. Approximte the re under y = from x = to x = using equl suintervls nd + x evluting the function t the lefthnd endpoints. 9 n= f ( n ) 9 = n= ( n + ) 9 = You cn use clcultor to pproximte this sum; it s round n= 8 + n. Exmple. Note tht the suintervls don t hve to e the sme size, nd I don t hve to choose the evlution points systemticlly. These re oth conveniences for the prolems. For exmple, here is n pproximtion to the re under y = x from x = to x = 6. intervl x f(x) f(x) x [, ]... [, ] [,.5] [.5, 6] sum 5.8 This gives n pproximte re of 5.8. The ctul re is 77. In generl, suppose I m trying to find the re under y = f(x) from x = to x =. I rek the intervl [, ] up into n suintervls of lengths x, x,..., x n. In the kth suintervl, I pick some point x k, nd use f(x k ) s the height of the kth rectngle. f( x k ) x k Dx k The sum of the res of the rectngles pproximtes the re under the curve: Are f(x k ) x k. k=
2 The more rectngles I tke, the etter the pproximtion. So it s resonle to suppose tht the exct re would e given y the limit of such sums, s n goes to infinity: Are = lim n k= f(x k ) x k. The expression on the right limit of sum, or Riemnn sum is clled the definite integrl of f(x) from to nd is denoted s follows: f(x)dx = lim n k= f(x k ) x k. It is possile to compute res using the formul ove, though it s not esy. Exmple. Use the limit of sum to compute the re under y = x + from x = to x =. I will need the following summtion formuls: k = k= n(n ) nd c = nc. k= Divide up the intervl [, ] into n equl suintervls. Ech hs length x =. I ll evlute the n function t the lefthnd endpoints. These re lim n n n, n, n,..., (n ) n. The kth point is k, so the height of the kth rectngle is n ( ) k f = k n n +. I get the following expression for the re: lim n k= ( ) k n + n = lim n n Apply the formuls from the eginning of the prolem: k= k= ( ) n k +. ( ) ( ) n k + n(n ) n = lim + n = lim ((n ) + n) = lim =. n n n n n n n While this pproch works, it s horrendously complicted. I ll discuss etter wys to compute definite integrls shortly. Since I m tking limit in computing the definite integrl, it s possile for the limit (nd hence, the definite integrl) to e undefined. f(x) is integrle on the intervl x if following fct sys tht mny of the functions you ll use in clculus re integrle: f(x)dx is defined. The
3 A ounded function with finitely mny discontinuities is integrle. (A function on n intervl x is ounded if there is numer M such tht f(x) M for ll x in the intervl.) For exmple, { if x f(x) = x if x > is integrle on ny intervl In prticulr, continuous function is integrle. On the other hnd, f(x) = is not integrle on ny intervl contining. x In some cses, you cn use the fct tht the definite integrl represents the re under curve to evlute the integrl geometriclly. Exmple. Compute ( x)dx. y =  x ( x)dx = = 8.
4 Exmple. Compute 6 ( x)dx. y =  x 6 The re consists of the piece in the lst prolem, together with piece of re. But the second piece is elow the xxis, so it is tken s negtive in the definite integrl: 6 ( x)dx = 8 = 6. Exmple. Compute x dx This is hlf the re of circle of rdius : x dx = π.578. Here re some properties of definite integrls. I ll present them without proofs.. If k is numer, then k dx = k( ). This is nother wy of sying tht the re of rectngle is the se times the height. k
5 Exmple. 5 7 dx = 7 ( 5) = 56.. If f nd g re integrle, then so is f + g, nd (f(x) + g(x)) dx = f(x)dx + This sys tht the integrl of sum is the sum of the integrls.. If f nd g re integrle nd f(x) g(x) for x, then f(x)dx This sys tht igger functions hve igger integrls. g(x) dx. g(x) dx. Exmple. You cn use the lst rule to get estimtes for integrls. For exmple, Therefore, 7 dx 7 x x + for ll x. x 7 x dx, or 5 x + x + dx.. If f is integrle, then f(x)dx = f(x) dx. Tht is, switching the limits of integrtion multiplies the integrl y. 5. If f is integrle on the intervl x c nd c, then f(x)dx + c f(x)dx = c f(x) dx. Tht is, integrting from to nd then from to c is the sme s integrting ll the wy from to c: f(x) f(x) dx c f(x) dx c 5
6 Exmple. f(x)dx f(x)dx f(x)dx = f(x)dx + f(x)dx + f(x)dx = f(x) dx. 6. (The Men Vlue Theorem for Integrls) There is numer c, c, such tht f(x)dx = f(c) ( ). f(c) represents the height of rectngle on the integrl [, ] which hs the sme re s the re under the grph of f(x). f(x) f(c) Exmple. for some c etween nd. Now Therefore, So c + 5, dx x + = c ( ) = + c + c gives c. c + 5, c + 5. dx x + 5. I ve gotten rough estimte for the vlue of the integrl. c 5 y Bruce Ikeng 6
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