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Let $f\in \mathcal{C}$ and $(0,\frac{1}{n},...,1)$ be an equidistant partiton of the Intervall $[0,1]$. $\int_{0}^{1}f(x)dx=\sum_{i=0}^{n}f(\zeta_{i})\frac{1}{n}+Error$. And $\zeta_{i}\in[\frac{i-1}{n},\frac{i}{n}]$ How can I explicitly represent the Error Term? Or can I give an fine upper bound?

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What is $\zeta_i$? And why are you multiplying $f(\zeta_i)$ by $\frac{i}{n}$? Shouldn't that be $\frac{1}{n}$ instead? –  Alex Becker Sep 19 '12 at 23:07
    
Corrected and added. –  Peter Moor Sep 19 '12 at 23:11
    
You should also be indexing from $1$ rather than $0$. –  Alex Becker Sep 19 '12 at 23:28
    
It depends of the specific $f$. And if $f$ is integrable, for any positive number you can find an $n$ so that the error is bounded by that number. –  leo Sep 19 '12 at 23:29
    
It would be enough for me to know how the error depends on $n$.Something like $Constant \cdot \alpha(n)$ would be enough. –  Peter Moor Sep 19 '12 at 23:32
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The error term is $$\mathrm{Error}=\sum\limits_{i=1}^n \int_{\frac{i-1}{n}}^{\frac{i}{n}}(f(x)-f(\zeta_i))\mathrm{d}x$$ however this is usually less useful than the upper bound for the error (assuming $f$ is differentiable) given by $$\mathrm{Error}\leq \sum\limits_{i=1}^n \int_{\frac{i-1}{n}}^{\frac{i}{n}}|f(x)-f(\zeta_i)|\mathrm{d}x\leq \sum\limits_{i=1}^n \int_{\frac{i-1}{n}}^{\frac{i}{n}}\frac{1}{n}\sup_{x\in [\frac{i-1}{n},\frac{i}{n}]}|f'(x)|\mathrm{d}x\leq\frac{1}{n}\sup_{x\in [0,1]}|f'(x)|.$$ If $f$ is continuous or at least Riemann-integrable but not differentiable, then the error term will still go to $0$ as $n$ goes to $\infty$, but we in general cannot bound the error term for any particular $n$.

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