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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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    It would be enough for me to know how the error depends on $n$.Something like $Constant \cdot \alpha(n)$ would be enough.2012-09-19

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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 $n$.