0
$\begingroup$

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?

  • 0
    What is $\zeta_i$? And why are you multiplying $f(\zeta_i)$ by $\frac{i}{n}$? Shouldn't that be $\frac{1}{n}$ instead?2012-09-19
  • 0
    Corrected and added.2012-09-19
  • 0
    You should also be indexing from $1$ rather than $0$.2012-09-19
  • 0
    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.2012-09-19
  • 0
    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

1 Answers 1