«K12/K13» 115. Hausaufgabe «PDF», «POD»

0.0.1 ↑ 115. Hausaufgabe

0.0.1.1 ↑ Analysis-Buch Seite 256, Aufgabe 15h

\displaystyle\int\limits_0^{\infty} x^2 e^{-x} \,\mathrm{d}x = {}\int\limits_0^{\infty} x^2 \left(-e^{-x}\right)' \,\mathrm{d}x = {}\lim\limits_{\alpha \to \infty} \left[-x^2 e^{-x} - \int 2x \left(-e^{-x}\right) \mathrm{d}x\right]_0^{\alpha} = \\ {}\lim\limits_{\alpha \to \infty} \left[-x^2 e^{-x} + 2 \int x \left(-e^{-x}\right)' \mathrm{d}x\right]_0^{\alpha} = {}\lim\limits_{\alpha \to \infty} \left[-x^2 e^{-x} + 2 \left(-x e^{-x} - \int -e^{-x} \,\mathrm{d}x\right)\right]_0^{\alpha} = {}\lim\limits_{\alpha \to \infty} \left[e^{-x} \left(-x^2 - 2 x - 2\right)\right]_0^{\alpha} = 2;$\underset{0}{\overset{\infty }{\int \; <!--nolimits-->}}{x}^2{e}^{-x}\text{d}x=\underset{0}{\overset{\infty }{\int \; <!--nolimits-->}}{x}^2{\left(-e^{-x}\right)}^{\prime }\text{d}x=\underset{\alpha \to \infty }{lim}{\left[-x^2{e}^{-x}-\int \; <!--nolimits-->2x\left(-e^{-x}\right)\text{d}x\right]}_0^{\alpha }=\underset{\alpha \to \infty }{lim}{\left[-x^2{e}^{-x}+2\int \; <!--nolimits-->x{\left(-e^{-x}\right)}^{\prime }\text{d}x\right]}_0^{\alpha }=\underset{\alpha \to \infty }{lim}{\left[-x^2{e}^{-x}+2\left(-x{e}^{-x}-\int \; <!--nolimits-->-e^{-x}\text{d}x\right)\right]}_0^{\alpha }=\underset{\alpha \to \infty }{lim}{\left[e^{-x}\left(-x^2-2x-2\right)\right]}_0^{\alpha }=2;$

0.0.1.2 ↑ Analysis-Buch Seite 256, Aufgabe 16

a)

Für n \neq 1$n\ne 1$:

\int\limits_1^a \frac{\ln x}{x^n} \mathrm{d}x = {}\int\limits_1^a \left(\frac{x^{1-n}}{1-n}\right)' \ln x \,\mathrm{d}x = {}\left[\frac{x^{1-n}}{1 - n} \ln x - \int \underbrace{\frac{x^{1-n}}{1 - n} \cdot \frac{1}{x}}_{\frac{x^{-n}}{1-n}} \mathrm{d}x\right]_1^a = {}\left[\frac{x^{1-n}}{1-n} \left(\ln x - \frac{1}{1-n}\right)\right]_1^a = {}\frac{a^{1-n}}{1-n} \left(\ln a - \frac{1}{1-n}\right) + \left(\frac{1}{1-n}\right)^2;$\underset{1}{\overset{a}{\int \; <!--nolimits-->}}\frac{lnx}{x^n}\text{d}x=\underset{1}{\overset{a}{\int \; <!--nolimits-->}}{\left(\frac{x^{1-n}}{1-n}\right)}^{\prime }lnx\text{d}x={\left[\frac{x^{1-n}}{1-n}lnx-\int \; <!--nolimits-->\underset{\frac{x^{-n}}{1-n}}{\underbrace{\frac{x^{1-n}}{1-n}\cdot \frac{1}{x}}}\text{d}x\right]}_1^a={\left[\frac{x^{1-n}}{1-n}\left(lnx-\frac{1}{1-n}\right)\right]}_1^a=\frac{a^{1-n}}{1-n}\left(lna-\frac{1}{1-n}\right)+{\left(\frac{1}{1-n}\right)}^2;$

Für n = 1$n=1$: \int \frac{\ln x}{x} \,\mathrm{d}x = \int I(\ln x) \left(\ln x\right)' \,\mathrm{d}x = \int I(t) \,\mathrm{d}t = \frac{1}{2} \ln^2 x$\int \; <!--nolimits-->\frac{lnx}{x}\text{d}x=\int \; <!--nolimits-->I\left(lnx\right){\left(lnx\right)}^{\prime }\text{d}x=\int \; <!--nolimits-->I\left(t\right)\text{d}t=\frac{1}{2}{ln}^{2}x$ mit I(t) = t;$I\left(t\right)=t;$

c)

\int\limits_0^a x^n \ln x \,\mathrm{d}x = {}\left[\int \frac{\ln x}{x^{-n}} \mathrm{d}x\right]_0^a = {}\left[\frac{x^{1+n}}{1+n} \left(\ln x - \frac{1}{1+n}\right)\right]_0^a = {}\frac{a^{1+n}}{1+n} \left(\ln a - \frac{1}{1+n}\right);$\underset{0}{\overset{a}{\int \; <!--nolimits-->}}{x}^{n}lnx\text{d}x={\left[\int \; <!--nolimits-->\frac{lnx}{x^{-n}}\text{d}x\right]}_0^a={\left[\frac{x^{1+n}}{1+n}\left(lnx-\frac{1}{1+n}\right)\right]}_0^a=\frac{a^{1+n}}{1+n}\left(lna-\frac{1}{1+n}\right);$

Speziell für a = 1$a=1$:

\int\limits_0^1 x^n \ln x \,\mathrm{d}x = -\left(\frac{1}{1 + n}\right)^2;$\underset{0}{\overset{1}{\int \; <!--nolimits-->}}{x}^{n}lnx\text{d}x=-{\left(\frac{1}{1+n}\right)}^2;$