= ${\left[2{\left(2x-2\right)}^{\frac{3}{2}}\right]}_9^{\frac{27}{2}}$

= ${\left[2{\left(\sqrt{2x-2}\right)}^3\right]}_9^{\frac{27}{2}}$

$=$ $2{\left(\sqrt{2\cdot \left(\frac{27}{2}\right)-2}\right)}^3-2{\left(\sqrt{2\cdot 9-2}\right)}^3$

= $2{\left(\sqrt{27-2}\right)}^3-2{\left(\sqrt{18-2}\right)}^3$

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

= $2\cdot 5^3-2\cdot 4^3$

= $2\cdot 125-2\cdot 64$

= $250-128$

= $122$

= ${\left[e^{2x-4}\right]}_2^3$

$=$ $e^{2\cdot 3-4}-e^{2\cdot 2-4}$

= $e^{6-4}-e^{4-4}$

= $e^2-e^0$

= $e^2-1$

= ${\left[-4e^{-\mathrm{0,7}x+\mathrm{0,7}}\right]}_0^u$

$=$ $-4e^{-\mathrm{0,7}u+\mathrm{0,7}}+4e^{-\mathrm{0,7}\cdot 0+\mathrm{0,7}}$

= $-4e^{-\mathrm{0,7}u+\mathrm{0,7}}+4e^{0+\mathrm{0,7}}$

= $-4e^{-\mathrm{0,7}u+\mathrm{0,7}}+4e^{\mathrm{0,7}}$

Diese Integralfunktion soll ja den Wert $4$ annehmen, deswegen setzen wir sie gleich :

Wir berechnen den Mittelwert mit der üblichen Formel:

= $\frac{2}{3\pi }$ ${\left[-3\cdot \cos (2x+\pi )\right]}_0^{\frac{3}{2}\pi }$

$=$ $\frac{2}{3\pi }·\left(-3\cdot \cos (2\cdot \left(\frac{3}{2}\pi \right)+\pi )+3\cdot \cos (2\cdot \left(0\right)+\pi )\right)$

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

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

= $\frac{2}{3\pi }·\left(-3-3\right)$

= $\frac{2}{3\pi }·\left(-6\right)$

= $-\frac{4}{\pi }$

= ${\left[\frac{1}{4}{\left(2x-2\right)}^{-2}\right]}_u^2$

= ${\left[\frac{1}{4{\left(2x-2\right)}^2}\right]}_u^2$

$=$ $\frac{1}{4{\left(2\cdot 2-2\right)}^2}-\frac{1}{4{\left(2u-2\right)}^2}$

= $\frac{1}{4{\left(4-2\right)}^2}-\frac{1}{4{\left(2u-2\right)}^2}$

= $\frac{1}{4\cdot 2^2}-\frac{1}{4{\left(2u-2\right)}^2}$

= $\frac{1}{4}\cdot \left(\frac{1}{4}\right)-\frac{1}{4{\left(2u-2\right)}^2}$

= $\frac{1}{16}-\frac{1}{4{\left(2u-2\right)}^2}$

Für u → 1 (u>1, also von rechts) gilt: A(u) = $-\frac{1}{4{\left(2u-2\right)}^2}+\frac{1}{16}$ → $-\infty$