=for timestamp
Mo Nov 27 13:04:51 CET 2006

=head2 116. Hausaufgabe

=head3 Analysis-Buch Seite 257, Aufgabe 19

Berechne:

=over

=item a)

M<\int\limits_0^1 2x \left(2 + x^2\right)^2 \mathrm{d}x =
{}\int\limits_0^1 \left(2 + x^2\right)^2 \left(2 + x^2\right)' \mathrm{d}x =
{}\left[\int t^2 \,\mathrm{d}t\right]_0^1 =
{}\left[\frac{1}{3} \left(2 + x^2\right)^3\right]_0^1 =
{}\frac{19}{3};>

=item b)

M<\int\limits_0^{\pi} \sin x \cos^3 x \,\mathrm{d}x =
{}\int\limits_0^{\pi} \cos^3 x \cdot \left(\cos x\right)' \cdot \left(-1\right) \,\mathrm{d}x =
{}\left[-\int t^3 \,\mathrm{d}t\right]_0^{\pi} =
{}\left[-\frac{1}{4} \cos^4 x\right]_0^{\pi} = 0;>

=item c)

M<\int\limits_{-1}^1 x e^{-x^2} \,\mathrm{d}x =
{}\int\limits_{-1}^1 e^{-x^2} \cdot \underbrace{\left(-x^2\right)'}_{-2x} \cdot \left(-\frac{1}{2}\right) \,\mathrm{d}x =
{}\left[-\frac{1}{2} \int e^t \,\mathrm{d}t\right]_{-1}^1 =
{}\left[-\frac{1}{2} e^{-x^2}\right]_{-1}^1 = 0;>

=item d)

M<\int\limits_1^e \frac{\ln x}{x} \mathrm{d}x =
{}\int\limits_1^e I(\ln x) \cdot \underbrace{\left(\ln x\right)'}_{\frac{1}{x}} \mathrm{d}x =
{}\left[\int \underbrace{I(t)}_t \,\mathrm{d}t\right]_1^e =
{}\left[\frac{1}{2} \ln^2 x\right]_1^e = \frac{1}{2}> mit M<I(t) = t;>

=back