=for timestamp
Sa Okt  8 17:22:36 CEST 2005

=head2 8. Hausaufgabe

=head3 Analysis-Buch Seite 36, Aufgabe 17a

Berechne mit Hilfe der Streifenmetnode den Flächeninhalt von

M<F := \left\{ (x,y) \,|\, 0 \leq x \leq 1 \wedge 0 \leq y \leq 1 - x^2 \right\};>

[M<\sum\limits_{i = 1}^n i = \frac{n\left(n + 1\right)}{2}; \quad>
M<\sum\limits_{i = 1}^n i^2 = \frac{n\left(n + 1\right)\left(2n +
1\right)}{6};>]

M<\mathrm{f}(x) = 1 - x^2;>

M<<
{} \renewcommand{\arraystretch}{3.0} \begin{array}{rcl}
{}   S_n &=& \sum\limits_{i = 1}^n \frac{1}{n} \mathrm{f}\!\left(\frac{i - 1}{n}\right) =
{}           \frac{1}{n} \sum\limits_{i = 1}^n \left[1 - \frac{i^2 - 2i + 1}{n^2}\right] =
{}           \frac{n \cdot 1}{n} + \frac{1}{n}\sum\limits_{i = 1}^n - \frac{i^2 - 2i + 1}{n^2} = \\
{}       &=& 1 - \frac{1}{n^3}\sum\limits_{i = 1}^n \left[i^2 - 2i + 1\right] =
{}           1 - \frac{1}{n^3}\left(\,\sum\limits_{i = 1}^n i^2 + 2 \sum\limits_{i = 1}^n i + n\right) = \\
{}       &=& 1 - \frac{1}{n^3}\left[\frac{n\left(n + 1\right)\left(2n + 1\right)}{6} + n\left(n + 1\right) + n\right] = 1 - \frac{2n^2 + 9n + 13}{6n^2};
{} \end{array}
>>

M<<
{} \renewcommand{\arraystretch}{3.0} \begin{array}{rcl}
{}   s_n &=& \sum\limits_{i = 1}^n \frac{1}{n} \mathrm{f}\!\left(\frac{i}{n}\right) =
{}           \frac{1}{n} \sum\limits_{i = 1}^n \left[1 - \frac{i^2}{n^2}\right] =
{}           \frac{n \cdot 1}{n} - \frac{1}{n^3} \sum\limits_{i = 1}^n i^2 = \\
{}       &=& 1 - \frac{n\left(n + 1\right)\left(2n + 1\right)}{6n^3} =
{}           1 - \frac{2n^2 + n + 2n + 1}{6n^2} =
{}           1 - \frac{2n^2 + 3n + 1}{6n^2};
{} \end{array}
>>

M<< \left.\begin{array}{l}
{} \lim\limits_{n \to \infty} S_n = 1 - \frac{1}{3} = \frac{2}{3}; \\
{} \lim\limits_{n \to \infty} s_n = 1 - \frac{1}{3} = \frac{2}{3};
\end{array}\right\} \Rightarrow \lim\limits_{n \to \infty} S_n = \lim\limits_{n \to \infty} s_n = \frac{2}{3} =: A; >>