Mo logo [home] [lexicon] [problems] [tests] [courses] [auxiliaries] [notes] [staff] german flag
Mathematics-Online course: Preparatory Course Mathematics - Analysis - Integral Calculus

Riemann Integral

[previous page] [next page] [table of contents][page overview]
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 partition 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 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$.

To compute $ \int_0^1 x^2\,dx$ with Riemann sums, we choose a sequence of partitions

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

with points

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

The Riemann sums are

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

and

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

in accordance with the exact value of the integral.
[previous page] [next page] [table of contents][page overview]
  automatically generated 1/9/2017