Gemischte Integrale

Aufgabenstellung

Berechnen Sie die folgenden bestimmten Integrale. Wählen Sie jeweils eine geeignete Methode.

$\displaystyle\int_0^1 x \cdot e^{-x}\,dx$

Wahl: $u = x$ , $v' = e^{-x}$ → $u' = 1$ , $v = -e^{-x}$ .

$= \big[-x\,e^{-x}\big]_0^1 + \int_0^1 e^{-x}\,dx$

$= -e^{-1} - 0 + \big[-e^{-x}\big]_0^1 = -e^{-1} + (-e^{-1} + 1) = 1 - 2e^{-1}$

$\boxed{\int_0^1 x\,e^{-x}\,dx = 1 - \frac{2}{e} \approx 0{,}264}$

$\displaystyle\int_0^2 x\sqrt{x^2+1}\,dx$

Substitution: $u = x^2 + 1$ , $du = 2x\,dx$ → $x\,dx = \frac{1}{2}\,du$ .

Grenzen: $x = 0 \Rightarrow u = 1$ ; $x = 2 \Rightarrow u = 5$ .

$\frac{1}{2}\int_1^5 \sqrt{u}\,du = \frac{1}{2} \cdot \left[\frac{2}{3}u^{3/2}\right]_1^5 = \frac{1}{3}\left(5\sqrt{5} - 1\right)$

$\boxed{\int_0^2 x\sqrt{x^2+1}\,dx = \frac{5\sqrt{5}-1}{3} \approx 3{,}393}$

$\displaystyle\int_0^{\pi/2} \sin^2(x)\,dx$

Mit der Identität $\sin^2(x) = \frac{1 - \cos(2x)}{2}$ :

$\int_0^{\pi/2} \frac{1 - \cos(2x)}{2}\,dx = \frac{1}{2}\left[x - \frac{\sin(2x)}{2}\right]_0^{\pi/2}$

$= \frac{1}{2}\left(\frac{\pi}{2} - \frac{\sin(\pi)}{2} - 0\right) = \frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4}$

$\boxed{\int_0^{\pi/2} \sin^2(x)\,dx = \frac{\pi}{4} \approx 0{,}785}$

$\displaystyle\int_1^e x \cdot \ln(x)\,dx$

Wahl: $u = \ln(x)$ , $v' = x$ → $u' = \frac{1}{x}$ , $v = \frac{x^2}{2}$ .

$= \left[\frac{x^2}{2}\ln(x)\right]_1^e - \int_1^e \frac{x^2}{2} \cdot \frac{1}{x}\,dx$

$= \frac{e^2}{2} \cdot 1 - 0 - \frac{1}{2}\int_1^e x\,dx = \frac{e^2}{2} - \frac{1}{2}\left[\frac{x^2}{2}\right]_1^e$

$= \frac{e^2}{2} - \frac{1}{4}(e^2 - 1) = \frac{e^2}{4} + \frac{1}{4} = \frac{e^2 + 1}{4}$

$\boxed{\int_1^e x\,\ln(x)\,dx = \frac{e^2+1}{4} \approx 2{,}097}$