MATH 149 Spring 2008 Riemann Sum, Right Endpoints April, 2008Suppose that J =Rbaf(x) dx, that h = (b − a)/n, and that x0= a, xj= a + jh for j = 1, 2, . . . ,n. Let the sampling points for the Riemann sum approximating the definite integral J be the rightendpoints x∗j= xjfor j = 1, 2, . . . , n. Then the Riemann sum approximation for J is simply thearithmetic mean of the right endpoints multiplied by the length of the interval [ a , b ] from a to b,namely b − a,RRS(n) =nXj=1f(x∗j)∆xj=nXj=1f(x∗j) · h = hnXj=1f(xj) .We claim that if | f0(x) | ≤ M for a ≤ x ≤ b, then| J − RRS(n) | ≤ M(b − a)22n.Let Jj=Zxjxj−1f(x) dx for j = 1, 2, . . . , n, denote the pieces of the definite integral over eachof the sub-intervals. Then we may write the definite integral as a sum J = J1+ J2+ · · · + Jn.Therefore, the difference between the definite integral and its’ approximation by a Riemann sumwith sampling points being right endpoints, for an equally spaced partition, isJ − RSS(n) = J1− hf(x1) + J2− hf(x2) + · · · + Jn− hf(xn) =nXj=1(Jj− hf(xj)) .Hence, | J − RSS(n) | ≤nXj=1| Jj− hf(xj) | .To bound the individual terms in the sum, consider the situation for a more general interval,namely,Zdd−hf(x) dx − hf(d) =Zdd−hf(x) dx −Zdd−hf(d) dx =Zdd−h(f(x) − f(d)) dx .By applying the Mean Value Theorem to the difference in the last integrand, f(x) − f(d) = f0(cx) ·(x − d) for some cxbetween x and d, for each x in the closed interval [ d − h , d ]. If we assume thatthe derivative of f(x) is bounded, that is, if we assume that there is a positive real number M suchthat for x in the interval [a, b] the derivative satisfies | f0(x) | ≤ M , then we get a bound on theabsolute value of the integral becauseZdd−h(f(x) − f(d)) dx≤Zdd−h| f(x) − f(d) | dx =Zdd−h| f0(cx) · (x − d) | dx .Hence, applying the bound on the derivative,Zdd−h| f0(cx) · (x − d) | dx ≤Zdd−hM · | x − d | dx =Zdd−hM · (d − x) dx ,because d − h ≤ x ≤ d .1MATH 149 Spring 2008 Riemann Sum, Right Endpoints April, 2008But,Zdd−hM · (d − x) dx = M ·−(d − x)22dd−h= M ·−(d − d)2+ (d − (d − h))22=Mh22.Therefore,Zdd−h(f(x) − f(d)) dx≤Zdd−h| f0(cx) · (x − d) | dx ≤Mh22.This result can be applied to each of the sub-intervals of the partition of [ a , b ] by setting d = xjfor the right hand endpoint and noting that then xj−1= xj− h = d − h for the left hand endpoint.Therefore,| Jj− hf(xj) | ≤Zxjxj−1f(x) dx − hf(xj)≤Mh22.Adding these terms together yields the following bound on the error in approximating the definiteintegral with the arithmetic mean of equally spaced function values in terms of an upper bound onthe derivative of the integrand:| J − RRS(n) | ≤nXj=1| Jj− hf(xj) | ≤nXj=1Mh22= nMh22= M(b − a)h2= M(b − a)22n,since n h = b − a. This is the result we wished to show.Note 1. The preceding argument was inspired by Courant and John. See Chapter 6, Numerical Methods, Section6.1, Computation of Integrals, Sub-section a., Approximation by Rectangles, pp. 482–483, and 6.1 Problem 4, p. 507,below.Note 2. For a recent treatment of error bounds for other numerical approximations for the definite integral, seethe paper by Talman.referencesRichard Courant and Fritz John, Introduction to Analysis, vol. 1, Wiley-Interscience, 1965.ISBN 0 471- 17860-4Louis A. Talman, Simpson’s rule is exact for quintics, American Mathematical Monthly, 113(2006)
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