Mathematics-Online lexicon: Annotation toRiemann Integral

	[home] [lexicon] [problems] [tests] [courses] [auxiliaries] [notes] [staff]
	Mathematics-Online lexicon: Annotation to
	Riemann Integral

A B C D E F G H I J K L M N O P Q R S T U V W X Y Z

overview

The definite integral of a piecewise continuous function

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

;

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

denotes the maximal length of the interval and $\xi_k$ is an arbitrary point in the

-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 , $\int_a^b f$ corresponds to the area below the graph of .

Slightly strengthening the hypotheses, we assume that

is continuously differentiable, and prove the convergence of the Riemann sums using Cauchy's criterion.

For a sequence $(\Delta_i)$ of partitions with $\vert\Delta_i\vert\to 0$ , we compare the Riemann sums corresponding to $\Delta_m$ and $\Delta_n$ with the aid of a partition $\Delta$ , which consists of the union of the points of $\Delta_m$ and $\Delta_n$ :

$\displaystyle \Delta_m: x_0<\cdots<x_{k_m},\, \Delta_n: y_0<\cdots<y_{k_n} \quad\leadsto\quad \Delta: z_0<\cdots<z_k \,.$

The difference of the Riemann sum for $\Delta$ ,

$\displaystyle \sum_{j=1}^k f(\zeta_j)\,\Delta z_j,\quad \zeta_i\in[z_{i-1},z_i] \,,$

to the Riemann sum for $\Delta_m$ can be estimated via the mean value theorem. From

$\displaystyle \vert f(t_1)-f(t_2)\vert \le (t_2-t_1) \max_{t\in [t_1,t_2]} \vert f^\prime(t)\vert$

we obtain

$\displaystyle \left\vert\int f_\Delta- \int f_{\Delta_m}\right\vert$	$\displaystyle =$	$\displaystyle \left\vert \sum_{j=1}^{k} f(\zeta_j)\,\Delta z_j - \sum_{i=1}^{k_m} f(\xi_i)\,\Delta x_i\right\vert$
	$\displaystyle =$	$\displaystyle \left\vert \sum_{i=1}^{k_m} \sum_{x_{i-1}\le z_{j-1}<z_{j}\le x_{i}} (f(\zeta_j)-f(\xi_i))\Delta z_j\right\vert$
	$\displaystyle \leq$	$\displaystyle \vert\Delta_m\vert\, \underbrace{\max_{t\in[a,b]} \vert f^\prime(... ...um_{i=1}^{k_m} \sum_{x_{i-1}\le z_{j-1}<z_{j}\le x_{i}} \Delta z_j }_{=b-a} \,,$

i.e.,

$\displaystyle \left\vert\int f_\Delta- \int f_{\Delta_m}\right\vert \le c\,(b-a)\,\vert\Delta_m\vert \,.$

With an analogous estimate for $\vert\int f_\Delta-\int f_{\Delta_n}\vert$ , Cauchy's criterion follows:

$\displaystyle \left\vert\int f_{\Delta_m}-\int f_{\Delta_n}\right\vert \le c\,(b-a)\,\left(\vert\Delta_m\vert+\vert\Delta_n\vert\right)\to 0$ for $\displaystyle \quad m,n\to\infty \,.$

The convergence of two sequences to the same limit can be established with an identical argument.

For piecewise continuous , the proof is technically more complicated. It relies on the uniform continuity of on the closed interval of integration:

$\displaystyle \left\vert f(x_1)-f(x_2)\right\vert\leq\varepsilon$ für $\displaystyle \quad \vert x_1-x_2\vert<\delta \,.$

[Back]

automatisch erstellt am 22. 9. 2016