= ${\left[-\frac{2}{9}{\left(3x-7\right)}^{-3}\right]}_4^6$

= ${\left[-\frac{2}{9{\left(3x-7\right)}^3}\right]}_4^6$

$=$ $-\frac{2}{9{\left(3\cdot 6-7\right)}^3}+\frac{2}{9{\left(3\cdot 4-7\right)}^3}$

= $-\frac{2}{9{\left(18-7\right)}^3}+\frac{2}{9{\left(12-7\right)}^3}$

= $-\frac{2}{9\cdot {11}^3}+\frac{2}{9\cdot 5^3}$

= $-\frac{2}{9}\cdot \left(\frac{1}{1331}\right)+\frac{2}{9}\cdot \left(\frac{1}{125}\right)$

= $-\frac{2}{11979}+\frac{2}{1125}$

= $-\frac{250}{1497375}+\frac{2662}{1497375}$

= $\frac{268}{166375}$

= ${\left[-{\left(x-2\right)}^{-1}\right]}_3^4$

= ${\left[-\frac{1}{x-2}\right]}_3^4$

$=$ $-\frac{1}{4-2}+\frac{1}{3-2}$

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

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

= $\frac{1}{2}$

= ${\left[8e^{\mathrm{0,6}x-\mathrm{0,9}}\right]}_0^u$

$=$ $8e^{\mathrm{0,6}u-\mathrm{0,9}}-8e^{\mathrm{0,6}\cdot 0-\mathrm{0,9}}$

= $8e^{\mathrm{0,6}u-\mathrm{0,9}}-8e^{0-\mathrm{0,9}}$

= $8e^{\mathrm{0,6}u-\mathrm{0,9}}-8e^{-\mathrm{0,9}}$

Diese Integralfunktion soll ja den Wert $11$ annehmen, deswegen setzen wir sie gleich :

Wir berechnen den Mittelwert mit der üblichen Formel:

= $\frac{1}{3}$ ${\left[\frac{3}{4}{\left(2x-3\right)}^4+5x\right]}_0^3$

$=$ $\frac{1}{3}(\frac{3}{4}\cdot {\left(2\cdot 3-3\right)}^4+5\cdot 3-\left(\frac{3}{4}\cdot {\left(2\cdot 0-3\right)}^4+5\cdot 0\right))$

= $\frac{1}{3}(\frac{3}{4}\cdot {\left(6-3\right)}^4+15-\left(\frac{3}{4}\cdot {\left(0-3\right)}^4+0\right))$

= $\frac{1}{3}(\frac{3}{4}\cdot 3^4+15-\left(\frac{3}{4}\cdot {\left(-3\right)}^4+0\right))$

= $\frac{1}{3}(\frac{3}{4}\cdot 81+15-\left(\frac{3}{4}\cdot 81+0\right))$

= $\frac{1}{3}(\frac{243}{4}+15-\left(\frac{243}{4}+0\right))$

= $\frac{1}{3}(\frac{243}{4}+\frac{60}{4}-\left(\frac{243}{4}+0\right))$

= $\frac{1}{3}\left(\frac{303}{4}-\frac{243}{4}\right)$

= $\frac{1}{3}·15$

= $5$

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

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

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

Für u → ∞ gilt: A(u) = $-3e^{u-3}+3e^{-2}$ → $-\infty$