= ${\left[-\frac{4}{3}{\left(x-3\right)}^{-3}\right]}_5^7$

= ${\left[-\frac{4}{3{\left(x-3\right)}^3}\right]}_5^7$

$=$ $-\frac{4}{3{\left(7-3\right)}^3}+\frac{4}{3{\left(5-3\right)}^3}$

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

= $-\frac{4}{3}\cdot \left(\frac{1}{64}\right)+\frac{4}{3}\cdot \left(\frac{1}{8}\right)$

= $-\frac{1}{48}+\frac{1}{6}$

= $-\frac{1}{48}+\frac{8}{48}$

= $\frac{7}{48}$

= ${\left[\frac{2}{3}{e}^{3x-5}\right]}_1^3$

$=$ $\frac{2}{3}{e}^{3\cdot 3-5}-\frac{2}{3}{e}^{3\cdot 1-5}$

= $\frac{2}{3}{e}^{9-5}-\frac{2}{3}{e}^{3-5}$

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

= ${\left[e^{\mathrm{0,6}x-\mathrm{0,8}}\right]}_0^u$

$=$ $e^{\mathrm{0,6}u-\mathrm{0,8}}-e^{\mathrm{0,6}\cdot 0-\mathrm{0,8}}$

= $e^{\mathrm{0,6}u-\mathrm{0,8}}-e^{0-\mathrm{0,8}}$

= $e^{\mathrm{0,6}u-\mathrm{0,8}}-e^{-\mathrm{0,8}}$

Diese Integralfunktion soll ja den Wert $16$ annehmen, deswegen setzen wir sie gleich :

Wir berechnen den Mittelwert mit der üblichen Formel:

= $\frac{1}{3}$ ${\left[\frac{2}{9}{\left(3x-7\right)}^3\right]}_0^3$

$=$ $\frac{1}{3}\left(\frac{2}{9}\cdot {\left(3\cdot 3-7\right)}^3-\frac{2}{9}\cdot {\left(3\cdot 0-7\right)}^3\right)$

= $\frac{1}{3}\left(\frac{2}{9}\cdot {\left(9-7\right)}^3-\frac{2}{9}\cdot {\left(0-7\right)}^3\right)$

= $\frac{1}{3}\left(\frac{2}{9}\cdot 2^3-\frac{2}{9}\cdot {\left(-7\right)}^3\right)$

= $\frac{1}{3}\left(\frac{2}{9}\cdot 8-\frac{2}{9}\cdot \left(-343\right)\right)$

= $\frac{1}{3}\left(\frac{16}{9}+\frac{686}{9}\right)$

= $\frac{1}{3}·78$

= $26$

= ${\left[-{\left(2x-6\right)}^{\frac{1}{2}}\right]}_u^{\mathrm{7,5}}$

= ${\left[-\sqrt{2x-6}\right]}_u^{\mathrm{7,5}}$

$=$ $-\sqrt{2\cdot \mathrm{7,5}-6}+\sqrt{2u-6}$

= $-\sqrt{15-6}+\sqrt{2u-6}$

= $-\sqrt{9}+\sqrt{2u-6}$

= $-3+\sqrt{2u-6}$

= $\sqrt{2u-6}-3$

Für u → 3 (u>3, also von rechts) gilt: A(u) = $\sqrt{2u-6}-3$ → $0-3$ = $-3$ ≈ -3

Für den Flächeninhalt (immer positiv) gilt also I = 3