= ${\left[-3{\left(x-3\right)}^{-2}\right]}_5^7$

= ${\left[-\frac{3}{{\left(x-3\right)}^2}\right]}_5^7$

$=$ $-\frac{3}{{\left(7-3\right)}^2}+\frac{3}{{\left(5-3\right)}^2}$

= $-\frac{3}{4^2}+\frac{3}{2^2}$

= $-3\cdot \left(\frac{1}{16}\right)+3\cdot \left(\frac{1}{4}\right)$

= $-\frac{3}{16}+\frac{3}{4}$

= $-\frac{3}{16}+\frac{12}{16}$

= $\frac{9}{16}$

= ${\left[\frac{2}{3}{\left(3x-5\right)}^{\frac{3}{2}}\right]}_7^{10}$

= ${\left[\frac{2}{3}{\left(\sqrt{3x-5}\right)}^3\right]}_7^{10}$

$=$ $\frac{2}{3}{\left(\sqrt{3\cdot 10-5}\right)}^3-\frac{2}{3}{\left(\sqrt{3\cdot 7-5}\right)}^3$

= $\frac{2}{3}{\left(\sqrt{30-5}\right)}^3-\frac{2}{3}{\left(\sqrt{21-5}\right)}^3$

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

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

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

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

= $\frac{122}{3}$

= ${\left[7e^{\mathrm{0,2}x-\mathrm{0,3}}\right]}_0^u$

$=$ $7e^{\mathrm{0,2}u-\mathrm{0,3}}-7e^{\mathrm{0,2}\cdot 0-\mathrm{0,3}}$

= $7e^{\mathrm{0,2}u-\mathrm{0,3}}-7e^{0-\mathrm{0,3}}$

= $7e^{\mathrm{0,2}u-\mathrm{0,3}}-7e^{-\mathrm{0,3}}$

Diese Integralfunktion soll ja den Wert $13$ annehmen, deswegen setzen wir sie gleich :

Wir berechnen den Mittelwert mit der üblichen Formel:

= $\frac{1}{4}$ ${\left[6e^{x-1}\right]}_0^4$

$=$ $\frac{1}{4}\left(6e^{4-1}-6e^{0-1}\right)$

= $\frac{1}{4}\left(6e^3-6e^{-1}\right)$

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

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

$=$ $\frac{2}{2}-\frac{2}{u}$

= $2\cdot \left(\frac{1}{2}\right)-\frac{2}{u}$

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

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