0.0.1 ↑ 4. Hausaufgabe
0.0.1.1 ↑ Analysis-Buch Seite 15, Aufgabe 9
Zeige die Richtigkeit von
- a)
\int \left(x^2 - x\right) \mathrm{d}x = \frac{1}{3}x^3 - \frac{1}{2}x^2 + C;
\left(\frac{1}{3}x^3 - \frac{1}{2}x^2 + C\right)' = x^2 - x;
- b)
\int \sqrt{x} \,\mathrm{d}x = \frac{2}{3}x\sqrt{x} + C;
\left(\frac{2}{3}x\sqrt{x} + C\right)' = \left(\frac{2}{3}x^{\frac{3}{2}} + C\right)' = x^{\frac{1}{2}} = \sqrt{x};
- c)
\int \frac{1}{x^2} \,\mathrm{d}x = -\frac{1}{x} + C;
\left(-\frac{1}{x} + C\right)' = \frac{1}{2}x^2;
- d)
\int \sin^2 x \,\mathrm{d}x = \frac{1}{2}\left(x - \sin x \cos x\right) + C;
\left[\frac{1}{2}\left(x - \sin x \cos x\right) + C\right]' = \frac{1}{2}\left(1 - \cos x \cos x + \sin x \sin x\right) = \frac{1}{2}\left(1 - 1 + \sin^2 x + \sin^2 x\right) = \sin^2 x;
- e)
\int \cos^2 x \,\mathrm{d}x = \frac{1}{2}\left(x + \sin x \cos x\right) + C;
\left[\frac{1}{2}\left(x + \sin x \cos x\right) + C\right]' = \frac{1}{2}\left(1 + \cos x \cos x - \sin x \sin x\right) = \frac{1}{2}\left(1 - 1 + \cos^2 x + \cos^2 x\right) = \cos^2 x;
- f)
\int \frac{x}{\sqrt{a^2 - x^2}} \,\mathrm{d}x = -\sqrt{a^2 - x^2} + C;
\left(-\sqrt{a^2 - x^2} + C\right)' = -\dfrac{1}{2\sqrt{a^2 - x^2}}\left(0 - 2x\right) = \dfrac{x}{\sqrt{a^2 - x^2}};