= ${\left[2e^{3x-5}\right]}_0^2$

$=$ $2e^{3\cdot 2-5}-2e^{3\cdot 0-5}$

= $2e^{6-5}-2e^{0-5}$

= $2e^1-2e^{-5}$

= $2e-2e^{-5}$

= ${\left[\frac{2}{3}{\left(3x-6\right)}^{\frac{3}{2}}\right]}_{\frac{22}{3}}^{\frac{31}{3}}$

= ${\left[\frac{2}{3}{\left(\sqrt{3x-6}\right)}^3\right]}_{\frac{22}{3}}^{\frac{31}{3}}$

$=$ $\frac{2}{3}{\left(\sqrt{3\cdot \left(\frac{31}{3}\right)-6}\right)}^3-\frac{2}{3}{\left(\sqrt{3\cdot \left(\frac{22}{3}\right)-6}\right)}^3$

= $\frac{2}{3}{\left(\sqrt{31-6}\right)}^3-\frac{2}{3}{\left(\sqrt{22-6}\right)}^3$

= $\frac{2}{3}{\left(\sqrt{25}\right)}^3-\frac{2}{3}{\left(\sqrt{16}\right)}^3$

= $\frac{2}{3}\cdot 5^3-\frac{2}{3}\cdot 4^3$

= $\frac{2}{3}\cdot 125-\frac{2}{3}\cdot 64$

= $\frac{250}{3}-\frac{128}{3}$

= $\frac{122}{3}$

= ${\left[-6e^{-\mathrm{0,7}x+\mathrm{0,2}}\right]}_0^u$

$=$ $-6e^{-\mathrm{0,7}u+\mathrm{0,2}}+6e^{-\mathrm{0,7}\cdot 0+\mathrm{0,2}}$

= $-6e^{-\mathrm{0,7}u+\mathrm{0,2}}+6e^{0+\mathrm{0,2}}$

= $-6e^{-\mathrm{0,7}u+\mathrm{0,2}}+6e^{\mathrm{0,2}}$

Diese Integralfunktion soll ja den Wert $5$ annehmen, deswegen setzen wir sie gleich :

Wir berechnen den Mittelwert mit der üblichen Formel:

= $\frac{1}{\pi }$ ${\left[4\cdot \sin \left(x+\frac{3}{2}\pi \right)\right]}_{\frac{1}{2}\pi }^{\frac{3}{2}\pi }$

$=$ $\frac{1}{\pi }·\left(4\cdot \sin \left(\frac{3}{2}\pi +\frac{3}{2}\pi \right)-4\cdot \sin \left(\frac{1}{2}\pi +\frac{3}{2}\pi \right)\right)$

= $\frac{1}{\pi }·\left(4\cdot \sin \left(3\pi \right)-4\cdot \sin \left(2\pi \right)\right)$

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

= $\frac{1}{\pi }·\left(0+0\right)$

= $\frac{1}{\pi }·0$

= 0

= ${\left[x^{-2}\right]}_u^5$

= ${\left[\frac{1}{x^2}\right]}_u^5$

$=$ $\frac{1}{5^2}-\frac{1}{u^2}$

= $\frac{1}{25}-\frac{1}{u^2}$

Für u → 0 (u>0, also von rechts) gilt: A(u) = $\frac{1}{25}-\frac{1}{u^2}$ → $-\infty$