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Mathematics-Online course: Prepcourse Mathematics - Analysis - Integral Calculus

Riemann Integral

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The definite integral of a piecewise continuous function $ f$ is defined by

$\displaystyle \int_a^b f(x)\,dx = \lim_{\vert\Delta\vert\to0} \int_a^b f_\Delta = \lim_{\vert\Delta\vert\to0} \sum_{k} f(\xi_k)\,\Delta x_k \quad . $

Here $ \Delta:\,a=x_0<x_1<\cdots<x_n=b$ is a decomposition of $ [a,b]$;

$\displaystyle \vert\Delta\vert=\max_k \Delta x_k\,, \qquad \Delta x_k=x_k-x_{k-1}\,, $

denotes the maximal length of the interval and $ \xi_k$ is an arbitrary point in the $ k$-th interval. The sums on the righthand side of the integral's definition are called Riemann sums.

\includegraphics[width=0.6\linewidth]{riemann_bild}

For a positive function $ f$, $ \int_a^b f$ corresponds to the area below the graph of $ f$.

(Authors: Höllig/Hörner/Wipper/Abele)
In order to determine $ \int_0^1 x^2\,dx$ as a Riemann integral, use the sequence of partitions

$\displaystyle \Delta_n: x_i= i/n\,,\, i=0,\ldots ,n\, , $

with evaluation points

$\displaystyle \xi_i=(2i-1)/(2n)\,,\, i=1,\ldots,n\, , $

for the $ n$-th partition.

\includegraphics[width=0.6\linewidth]{bsp_riemann_x2_bild}

According to definition this yields

$\displaystyle \int f_{\Delta_n}$ $\displaystyle =$ $\displaystyle \sum_{i=1}^n \frac{1}{n}\left(\frac{2i-1}{2n}\right)^2$  
  $\displaystyle =$ $\displaystyle \frac{1}{4n^3} \left(4\sum_{i=1}^n i^2-4 \sum_{i=1}^n i + \sum_{i=1}^n 1\right)$  
  $\displaystyle =$ $\displaystyle \frac{1}{4n^3} \left( \frac{4n(n+1)(2n+1)}{6} - \frac{4n(n+1)}{2} +n \right)$  
  $\displaystyle =$ $\displaystyle \frac{1}{3} -\frac{1}{12n^2}$  

and thus $ {\displaystyle{\lim_{n\to\infty} \int f_{\Delta_n} = \frac{1}{3}}}\,$.

(Authors: Höllig/Hörner/Abele)
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  automatically generated 9/18/2007