= ${\left[-2{\left(2x-4\right)}^{-1}\right]}_4^5$

= ${\left[-\frac{2}{2x-4}\right]}_4^5$

$=$ $-\frac{2}{2\cdot 5-4}+\frac{2}{2\cdot 4-4}$

= $-\frac{2}{10-4}+\frac{2}{8-4}$

= $-\frac{2}{6}+\frac{2}{4}$

= $-2\cdot \left(\frac{1}{6}\right)+2\cdot \left(\frac{1}{4}\right)$

= $-\frac{1}{3}+\frac{1}{2}$

= $-\frac{2}{6}+\frac{3}{6}$

= $\frac{1}{6}$

= ${\left[\frac{4}{3}{\left(x-2\right)}^{\frac{3}{2}}\right]}_{11}^{27}$

= ${\left[\frac{4}{3}{\left(\sqrt{x-2}\right)}^3\right]}_{11}^{27}$

$=$ $\frac{4}{3}{\left(\sqrt{27-2}\right)}^3-\frac{4}{3}{\left(\sqrt{11-2}\right)}^3$

= $\frac{4}{3}{\left(\sqrt{25}\right)}^3-\frac{4}{3}{\left(\sqrt{9}\right)}^3$

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

= $\frac{4}{3}\cdot 125-\frac{4}{3}\cdot 27$

= $\frac{500}{3}-36$

= $\frac{500}{3}-\frac{108}{3}$

= $\frac{392}{3}$

= ${\left[4e^{\mathrm{0,8}x-\mathrm{0,8}}\right]}_0^u$

$=$ $4e^{\mathrm{0,8}u-\mathrm{0,8}}-4e^{\mathrm{0,8}\cdot 0-\mathrm{0,8}}$

= $4e^{\mathrm{0,8}u-\mathrm{0,8}}-4e^{0-\mathrm{0,8}}$

= $4e^{\mathrm{0,8}u-\mathrm{0,8}}-4e^{-\mathrm{0,8}}$

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

Wir berechnen den Mittelwert mit der üblichen Formel:

= $\frac{1}{3}$ ${\left[\frac{2}{3}{\left(2x-2\right)}^3\right]}_2^5$

$=$ $\frac{1}{3}\left(\frac{2}{3}\cdot {\left(2\cdot 5-2\right)}^3-\frac{2}{3}\cdot {\left(2\cdot 2-2\right)}^3\right)$

= $\frac{1}{3}\left(\frac{2}{3}\cdot {\left(10-2\right)}^3-\frac{2}{3}\cdot {\left(4-2\right)}^3\right)$

= $\frac{1}{3}\left(\frac{2}{3}\cdot 8^3-\frac{2}{3}\cdot 2^3\right)$

= $\frac{1}{3}\left(\frac{2}{3}\cdot 512-\frac{2}{3}\cdot 8\right)$

= $\frac{1}{3}\left(\frac{1024}{3}-\frac{16}{3}\right)$

= $\frac{1}{3}·336$

= $112$

= ${\left[-e^{-x+3}\right]}_1^u$

$=$ $-e^{-u+3}+e^{-1+3}$

= $-e^{-u+3}+e^2$

Für u → ∞ gilt: A(u) = $-e^{-u+3}+e^2$ → $0+e^2$ = $e^2$ ≈ 7.389

Für den Flächeninhalt (immer positiv) gilt also I = 7.389