«K12/K13» 114. Hausaufgabe «PDF», «POD»

0.0.1 ↑ 114. Hausaufgabe

0.0.1.1 ↑ Analysis-Buch Seite 256, Aufgabe 14a

\int e^x \sin x \,\mathrm{d}x = {}e^x \sin x - \int e^x \cos x \,\mathrm{d}x = {}e^x \sin x - e^x \cos x - \int e^x \sin x \,\mathrm{d}x;$\int \; <!--nolimits-->e^{x}sinx\text{d}x=e^{x}sinx-\int \; <!--nolimits-->e^{x}cosx\text{d}x=e^{x}sinx-e^{x}cosx-\int \; <!--nolimits-->e^{x}sinx\text{d}x;$ ⇔ \int e^x \sin x \,\mathrm{d}x = \frac{e^x}{2}\left(\sin x - \cos x\right)\!;$\int \; <!--nolimits-->e^{x}sinx\text{d}x=\frac{e^x}{2}\left(sinx-cosx\right);$

\int\limits_0^{\pi/2} e^x \sin x \,\mathrm{d}x = {}\frac{1}{2}\left(e^{\pi/2} + 1\right)\!;$\underset{0}{\overset{\pi ∕2}{\int \; <!--nolimits-->}}{e}^{x}sinx\text{d}x=\frac{1}{2}\left(e^{\pi ∕2}+1\right);$

0.0.1.2 ↑ Analysis-Buch Seite 256, Aufgabe 15

e)

\displaystyle\int\limits_{-1}^1 \ln x^2 \,\mathrm{d}x = {}2 \int\limits_0^1 \ln x^2 \,\mathrm{d}x = {}2 \int\limits_0^1 x' \cdot \ln x^2 \,\mathrm{d}x = {}\lim\limits_{\alpha \to 0+} 2 \left[x \ln x^2 - \int \underbrace{x \cdot \frac{1}{x^2} \cdot 2x}_{2} \,\mathrm{d}x\right]_{\alpha}^1 = {}\lim\limits_{\alpha \to 0+} 2 \left[x \ln x^2 - 2x\right]_{\alpha}^1 = -4;$\underset{-1}{\overset{1}{\int \; <!--nolimits-->}}ln{x}^2\text{d}x=2\underset{0}{\overset{1}{\int \; <!--nolimits-->}}ln{x}^2\text{d}x=2\underset{0}{\overset{1}{\int \; <!--nolimits-->}}{x}^{\prime }\cdot ln{x}^2\text{d}x=\underset{\alpha \to 0+}{lim}2{\left[xln{x}^2-\int \; <!--nolimits-->\underset{2}{\underbrace{x\cdot \frac{1}{x^2}\cdot 2x}}\text{d}x\right]}_{\alpha }^1=\underset{\alpha \to 0+}{lim}2{\left[xln{x}^2-2x\right]}_{\alpha }^1=-4;$

f)

\displaystyle\int\limits_1^{\sqrt{2}} x \ln\!\left(1+x^2\right) \,\mathrm{d}x = {}\int\limits_1^{\sqrt{2}} \left(\frac{1}{2} x^2\right)' \ln\!\left(1+x^2\right) \,\mathrm{d}x = {}\left[\frac{1}{2} x^2 \ln\!\left(1 + x^2\right) - \int \underbrace{\frac{1}{2} x^2 \cdot \frac{1}{1 + x^2} \cdot 2x}_{\frac{x^3}{1+x^2}} \,\mathrm{d}x\right]_1^{\sqrt{2}} = {}\left[\frac{1}{2} x^2 \ln\!\left(1 + x^2\right) - \int x \,\mathrm{d}x - \underbrace{\frac{x}{1 + x^2}}_{\frac{\left(1 + x^2\right)'}{1 + x^2} \cdot \frac{1}{2}} \mathrm{d}x\right]_1^{\sqrt{2}} = {}\left[\frac{1}{2} x^2 \ln\!\left(1 + x^2\right) - \frac{1}{2} x^2 + \frac{1}{2} \ln \left|1 + x^2\right|\right]_1^{\sqrt{2}} = {}\frac{1}{2} \left[\left(1 + x^2\right) \ln\!\left(1 + x^2\right) - x^2\right]_1^{\sqrt{2}} = {}\frac{1}{2}\left(3 \ln 3 - 2 \ln 2 - 1\right);$\underset{1}{\overset{\sqrt{2}}{\int \; <!--nolimits-->}}xln\left(1+x^2\right)\text{d}x=\underset{1}{\overset{\sqrt{2}}{\int \; <!--nolimits-->}}{\left(\frac{1}{2}{x}^2\right)}^{\prime }ln\left(1+x^2\right)\text{d}x={\left[\frac{1}{2}{x}^{2}ln\left(1+x^2\right)-\int \; <!--nolimits-->\underset{\frac{x^3}{1+x^2}}{\underbrace{\frac{1}{2}{x}^2\cdot \frac{1}{1+x^2}\cdot 2x}}\text{d}x\right]}_1^{\sqrt{2}}={\left[\frac{1}{2}{x}^{2}ln\left(1+x^2\right)-\int \; <!--nolimits-->x\text{d}x-\underset{\frac{{\left(1+x^2\right)}^{\prime }}{1+x^2}\cdot \frac{1}{2}}{\underbrace{\frac{x}{1+x^2}}}\text{d}x\right]}_1^{\sqrt{2}}={\left[\frac{1}{2}{x}^{2}ln\left(1+x^2\right)-\frac{1}{2}{x}^2+\frac{1}{2}ln\left|1+x^2\right|\right]}_1^{\sqrt{2}}=\frac{1}{2}{\left[\left(1+x^2\right)ln\left(1+x^2\right)-x^2\right]}_1^{\sqrt{2}}=\frac{1}{2}\left(3ln3-2ln2-1\right);$

g)

\displaystyle\int\limits_0^1 x^{-1/2} \ln x \,\mathrm{d}x = {}\int\limits_0^1 \left(2 x^{1/2}\right)' \ln x \,\mathrm{d}x = {}\lim\limits_{\alpha \to 0+} \left[2 \sqrt{x} \ln x - \int 2 \underbrace{x^{1/2} x^{-1}}_{x^{-1/2}} \,\mathrm{d}x\right]_{\alpha}^1 = {}\lim\limits_{\alpha \to 0+} \left[2 \sqrt{x} \left(\ln x - 2\right)\right]_{\alpha}^1 = {}-4;$\underset{0}{\overset{1}{\int \; <!--nolimits-->}}{x}^{-1∕2}lnx\text{d}x=\underset{0}{\overset{1}{\int \; <!--nolimits-->}}{\left(2x^{1∕2}\right)}^{\prime }lnx\text{d}x=\underset{\alpha \to 0+}{lim}{\left[2\sqrt{x}lnx-\int \; <!--nolimits-->2\underset{x^{-1∕2}}{\underbrace{x^{1∕2}{x}^{-1}}}\text{d}x\right]}_{\alpha }^1=\underset{\alpha \to 0+}{lim}{\left[2\sqrt{x}\left(lnx-2\right)\right]}_{\alpha }^1=-4;$

0.0.1.3 ↑ Analysis-Buch Seite 256, Aufgabe 17a

Zeige, dass gilt:

\displaystyle\int \sin^n x \,\mathrm{d}x = {}-\frac{1}{n} \left(\sin x\right)^{n-1} \cos x + \frac{n - 1}{n} \int \left(\sin x\right)^{n-2} \mathrm{d}x;$\int \; <!--nolimits-->{sin}^{n}x\text{d}x=-\frac{1}{n}{\left(sinx\right)}^{n-1}cosx+\frac{n-1}{n}\int \; <!--nolimits-->{\left(sinx\right)}^{n-2}\text{d}x;$

\displaystyle\renewcommand{\arraystretch}{2.2}\begin{array}{@{}cl} {} & \left[-\frac{1}{n} \left(\sin x\right)^{n-1} \cos x + \frac{n - 1}{n} \int \left(\sin x\right)^{n-2} \mathrm{d}x\right]' = \\ {} =& \frac{1}{n} \left[\left(\sin x\right)^{n-1} \sin x - \left(n-1\right) \left(\sin x\right)^{n-2} \underbrace{\cos^2 x}_{1 - \sin^2 x} + \left(n-1\right) \left(\sin x\right)^{n-2}\right] = \\ {} =& \frac{1}{n} \sin^n x \left[1 - \left(n-1\right) \left(\sin x\right)^{-2} \left(1 - \sin^2 x - 1\right)\right] = \\ {} =& \frac{1}{n} \sin^n x \cdot \left(1 + n - 1\right) = \sin^n x; \end{array}$\begin{array}{cc}\hfill \hfill & {\left[-\frac{1}{n}{\left(sinx\right)}^{n-1}cosx+\frac{n-1}{n}\int \; <!--nolimits-->{\left(sinx\right)}^{n-2}\text{d}x\right]}^{\prime }=\hfill \\ \hfill =\hfill & \frac{1}{n}\left[{\left(sinx\right)}^{n-1}sinx-\left(n-1\right){\left(sinx\right)}^{n-2}\underset{1-{sin}^{2}x}{\underbrace{{cos}^{2}x}}+\left(n-1\right){\left(sinx\right)}^{n-2}\right]=\hfill \\ \hfill =\hfill & \frac{1}{n}{sin}^{n}x\left[1-\left(n-1\right){\left(sinx\right)}^{-2}\left(1-{sin}^{2}x-1\right)\right]=\hfill \\ \hfill =\hfill & \frac{1}{n}{sin}^{n}x\cdot \left(1+n-1\right)={sin}^{n}x;\hfill \end{array}$