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<math> E \le \frac{M(b-a)^2}{N}. </math> | <math> E \le \frac{M(b-a)^2}{N}. </math> | ||
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Latest revision as of 09:49, 14 October 2008
Suppose that $ f(x) $ is continuously differentiable on the interval [a,b].
- Let N be a positive integer
- Let $ M = Max \{ |f'(x)| : a \leq x \leq b \} $
- Let $ h = \frac{(b-a)}{N} $
- Let $ R_N $ denote the "right endpoint"
Riemann Sum for the integral
$ I = \int_a^b f(x) dx . $
In other words,
$ R_N = \sum_{n=1}^N f(a + n h) h . $
Explain why the error, $ E = | R_N - I | $, satisfies
$ E \le \frac{M(b-a)^2}{N}. $
(I moved all the discussion that used to be here to the discussion tab. I still think it would be nice to put a polished proof of the estimate here. --Bell 14:49, 14 October 2008 (UTC))
