«K12/K13» 113. Hausaufgabe «PDF», «POD»

0.0.1 ↑ 113. Hausaufgabe

0.0.1.1 ↑ Analysis-Buch Seite 256, Aufgabe 15

a)

\int \sin^2 x \,\mathrm{d}x = {}\int \sin x \left(-\cos x\right)' \mathrm{d}x = {}-\sin x \cos x + \int \underbrace{\cos^2 x}_{1 - \sin^2 x} \,\mathrm{d}x = {}-\sin x \cos x + x - \int \sin^2 \,\mathrm{d}x;$\int \; <!--nolimits-->{sin}^{2}x\text{d}x=\int \; <!--nolimits-->sinx{\left(-cosx\right)}^{\prime }\text{d}x=-sinxcosx+\int \; <!--nolimits-->\underset{1-{sin}^{2}x}{\underbrace{{cos}^{2}x}}\text{d}x=-sinxcosx+x-\int \; <!--nolimits-->{sin}^2\text{d}x;$

⇔ \int \sin^2 x \,\mathrm{d}x = \frac{1}{2} \left(x - \sin x \cos x\right);$\int \; <!--nolimits-->{sin}^{2}x\text{d}x=\frac{1}{2}\left(x-sinxcosx\right);$

\int\limits_0^{\pi} \sin^2 x \,\mathrm{d}x = {}\frac{1}{2} \left[x - \sin x \cos x\right]_0^{\pi} = {}\frac{\pi}{2};$\underset{0}{\overset{\pi }{\int \; <!--nolimits-->}}{sin}^{2}x\text{d}x=\frac{1}{2}{\left[x-sinxcosx\right]}_0^{\pi }=\frac{\pi }{2};$

b)

\int\limits_1^e x \ln x \,\mathrm{d}x = {}\int\limits_1^e \left(\frac{1}{2} x^2\right)' \cdot \ln x \,\mathrm{d}x = {}\left[\frac{1}{2} x^2 \cdot \ln x - \int \frac{1}{x} \cdot \frac{1}{2} x^2 \,\mathrm{d}x\right]_1^e = {}\left[\frac{1}{2} x^2 \cdot \ln x - \frac{1}{4} x^2\right]_1^e = {}\frac{e^2}{4} + \frac{1}{4};$\underset{1}{\overset{e}{\int \; <!--nolimits-->}}xlnx\text{d}x=\underset{1}{\overset{e}{\int \; <!--nolimits-->}}{\left(\frac{1}{2}{x}^2\right)}^{\prime }\cdot lnx\text{d}x={\left[\frac{1}{2}{x}^2\cdot lnx-\int <!--nolimits-->\frac{1}{x}\cdot \frac{1}{2}{x}^2\text{d}x\right]}_1^e={\left[\frac{1}{2}{x}^2\cdot lnx-\frac{1}{4}{x}^2\right]}_1^e=\frac{e^2}{4}+\frac{1}{4};$

c)

\int\limits_1^{e^2} \sqrt{x} \ln x \,\mathrm{d}x = {}\int\limits_1^{e^2} \left(\frac{2}{3} x^{3/2}\right)' \cdot \ln x \,\mathrm{d}x = {}\left[\frac{2}{3} x^{3/2} \cdot \ln x - \int \frac{1}{x} \cdot \frac{2}{3} x^{3/2} \,\mathrm{d}x\right]_1^{e^2} = {}\frac{2}{3} \left[x^{3/2} \cdot \ln x - \frac{2}{3} x^{3/2}\right]_1^{e^2} = {}\frac{8}{9} e^3 + \frac{4}{9};$\underset{1}{\overset{e^2}{\int \; <!--nolimits-->}}\sqrt{x}lnx\text{d}x=\underset{1}{\overset{e^2}{\int <!--nolimits-->}}{\left(\frac{2}{3}{x}^{3∕2}\right)}^{\prime }\cdot lnx\text{d}x={\left[\frac{2}{3}{x}^{3∕2}\cdot lnx-\int <!--nolimits-->\frac{1}{x}\cdot \frac{2}{3}{x}^{3∕2}\text{d}x\right]}_1^{e^2}=\frac{2}{3}{\left[x^{3∕2}\cdot lnx-\frac{2}{3}{x}^{3∕2}\right]}_1^{e^2}=\frac{8}{9}{e}^3+\frac{4}{9};$

d)

\int\limits_{\sqrt{e}}^e \ln^2 x \,\mathrm{d}x = {}\int\limits_{\sqrt{e}}^e x' \cdot \ln^2 x \,\mathrm{d}x = {}\left[x \cdot \ln^2 x - \int x \cdot 2 \ln x \cdot \frac{1}{x} \,\mathrm{d}x\right]_{\sqrt{e}}^e = {}\left[x \cdot \ln^2 x - 2 \int x \ln x \,\mathrm{d}x\right]_{\sqrt{e}}^e = {}\left[x \cdot \ln^2 x - 2 \left(x \cdot \ln x - x\right)\right]_{\sqrt{e}}^e = {}\left[x \cdot \left(\ln^2 x - 2 \ln x + 2\right)\right]_{\sqrt{e}}^e = {}e - \frac{5}{4} \sqrt{e};$\underset{\sqrt{e}}{\overset{e}{\int \; <!--nolimits-->}}{ln}^{2}x\text{d}x=\underset{\sqrt{e}}{\overset{e}{\int \; <!--nolimits-->}}{x}^{\prime }\cdot {ln}^{2}x\text{d}x={\left[x\cdot {ln}^{2}x-\int \; <!--nolimits-->x\cdot 2lnx\cdot \frac{1}{x}\text{d}x\right]}_{\sqrt{e}}^e={\left[x\cdot {ln}^{2}x-2\int \; <!--nolimits-->xlnx\text{d}x\right]}_{\sqrt{e}}^e={\left[x\cdot {ln}^{2}x-2\left(x\cdot lnx-x\right)\right]}_{\sqrt{e}}^e={\left[x\cdot \left({ln}^{2}x-2lnx+2\right)\right]}_{\sqrt{e}}^e=e-\frac{5}{4}\sqrt{e};$