38 Riemann sums and existence of the definite integral.

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1 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 x-xis nd 0 x b, two sums ppered: ( n (k 1) 2) b 3 n 3 re(x) ( n These sums re exmples of wht re clled Riemnn sums. k 2) b 3 n 3 Equl length subintervls Riemnn sum. Suppose f is function with domin x b. We crete (equl length subintervls Riemnn sum s follow: Tke positive integer nd divide the intervl [,b] in n equl length subintervls of length x n = ( ) b n. Clerly the endpoints of the vrious subintervls re: x 0 =,, x k = x 0 + k x n,, x n = x 0 + n x n = b. = x 0 x 1 x2 x k-1 x k x = b n

2 For ech subintervl I k = [x k 1,x k ] we choose point x k in I k. Imge rectngle with bse the intervl I k nd height f(x k ). The sum f(x k) x n, (where x n = ( b ) ) n is clled Riemnn sum. f(x * k) = x x x x = b 0 k-1 k n

3 We re free to choose the points x k to be ny point we wnt in the intervl J k, so there re mny Riemnn sums. Exmples Left endpoint Riemnn sum. Here, we lwys tke the point x k to be the left endpoint x k 1 = + (k 1) x n in the subintervl I k = [x k 1,x k ]. The Riemnn sum is f(x k 1 ) x n = f( + (k 1) x n ) x n, (where x n = ( b ) ) n Right endpoint Riemnn sum. Here, we tke x k to lwys be the right endpoint of I k. The Riemnn sum is f(x k ) x n = f( + k x n ) x n Midpoint Riemnn sum. We tke x k to lwys be the midpoint x k 1+x k 2 of I k. The midpoint is +(k 1 2 ) x n), nd the Riemnn sum is f( x k 1 +x k ) x n = 2 f( + (k 1 2 ) x n) x n

4 Nice property. Suppose the function f hs domin [, b], nd on ny subintervl [c,d], the function chieves minimum nd mximum vlue. Exmples of functions which stisfy the nice property re: Any continuous function. An incresing function mening x 1 < x 2 = f(x 1 ) f(x 2 ). A decresing function mening x 1 < x 2 = f(x 1 ) f(x 2 ). Ifthefunctionf stisfiestheniceproperty,forechpositiveintegern,wecncretetworiemnn sums by tking x min k to be n input which provides the minimum vlue of f on I k, nd x mx k to be n input which provides the mximum vlue of f on I k. We cll the sums: f(x min k ) x n n under-estimte Riemnn sum f(x mx k ) x n n over-estimte Riemnn sum

5 For ny other x k in I k, we hve f(x min k ) f(x k ) f(xmx k ), so f(x min k ) x n f(x k) x n f(x mx k ) x n which sys ny Riemnn sum for the prtition into n equl length subintervls of length x n = ( b n ) is between the under nd the over estimte Riemnn sums. A function f stisfying the nice property is integrble if the under nd over estimte Riemnn sums hve equl limits s n. lim n f(x min k ) x n equls lim n f(x mx k ) x n. The common limit vlue is clled the definite integrl of the function, nd denoted s: f(x) dx

6 Note. For the exmple g(x) = x 2 with domin [0,b], the left endpoints of the intervls I k s give the under-estimte Riemnn sum: f(x min k ) x n = n 1 l=0 l ( 2 b 3 ) b 3 n 3 3 s n The right endpoints of the intervls I k give the over-estimte Riemnn sum: f(x mx k ) x n = k ( 2 b 3 ) b 3 s n n 3 3

7 Existence of definite integrl for incresing functions Suppose function f with domin [, b] is incresing (or decresing) on the intervl. Then, the under nd over estimte Riemnn sums hve limits which re the sme. So, the function is integrble. These pictures give the reson why the ssertion is true when f is n incresing function. Since f is n incresing function on ech subintervl I k, the the function f hs minimum vlue t the left endpoint nd mximum vlue t the right endpoint. This mens: The under estimte Riemnn sum is the left endpoint Riemnn sum n f(x k 1) x n. The over estimte Riemnn sum is the right endpoint Riemnn sum n f(x k) x n.

8 The difference of the over estimte nd under estimte Riemnn sums (for n equl length subintervls) re the shded rectngles in the left figure. These rectngles ll hve bse length x n. They cn be slided to produce verticl column of bse x n nd height f(b) f(). The difference is: ( ) f(b) f() xn = ( f(b) f() ) b n As n, the fctor b n in the right hnd side hs limit zero, so the difference of the over nd under estimte hs limit zero. This mens the limits of the under nd over estimtes must exist nd hve the sme vlue. The logic in chnging nottion from Riemnn sum to definite integrl is: f(x k ) x n is converted to f(x) dx

9 Bsic properties of definite integrls. If the function f is hs domin [,c], nd the function is integrble on the subintervls [,b] nd [b,c], then the function is integrble on the entire intervl [,c], nd f(x) dx + c b f(x) dx = c f(x) dx This property is quite useful. A function which is incresing (or decresing) on n intervl is integrble. If the domin [r,s] of function f cn be prtitioned into ( finite number of) subintervls on which the function is either incresing or decresing, then the function is integrble on ech subintervl, nd therefore integrble on the entire intervl [r, s]. In the figure, the intervl hs been prtition into 6 subintervls on which the function is decresing or incresing. On ech subintervl, the function is integrble, so the function is integrble on the entire intervl.

10 Most common functions such s polynomils, exponentil, trignometric, rtionl, power nd their combintions nd composites stisfy the bove property, nd re therefore integrble on ny intevrl [,b] on which they re defined (mening no undefined points such s n infinite limit). If f nd g re two functions integrble on the domin [,b], then: Their sum f + g is integrble, nd (f(x) + g(x)) dx = If we multiply by constnt C, we hve Cf(x) dx = C f(x) dx + f(x) dx. g(x) dx. If f nd g re two functions integrble on the domin [,b], nd f(x) g(x) throughout the intervl, then f(x) dx = b f(x) dx, nd If the function vlues re 0, then f(x) dx f(x) dx = 0. g(x) dx. f(x) dx = minus re between grph of f nd x-xis (on intervl [,b]).

11 If function f, with domin [,b], is integrble, then it is lso integrble on ny inside subintervl [c, d]. If function f, with domin [,b], is continuous, then it is integrble. For common continuous functions, with domin [, b], we cn decompose the intervl into finite sumber of subintervls on which the function is incresing or decresing nd the 1st property pplies to show f is integrble. But there re continuous function on n intervl [,b] for which it is not possible to divide the intervl in finite number of subintervls of increse nd decrese. An exmple is the function f(0) = 0, nd f(x) = x sin(1/x) for 0 < x < 1. It is continuous on [0,1], but [0,1] cn be divided into finite number of subintervls where the function is only incresing or decresing x*sin(1/x) is continuous

12 Exmple of function which is not integrble. Tke the following function with domin [0, 1]: { 0 when x is n irrtionl number f(x) = 1 when x is n rtionl number On ny subintervl, the minimum vlue of the function is 0 nd the mximum vlue is 1. This mens ny over-estimte Riemnn sum equls 1 nd ny under-estimte Riemnn sum equls 0. For equl length subintervls, the over-estimte Riemnn sums hs limit 1 nd the under-estimte Riemnn sums hs limit 0. So, the function is not integrble.

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