= ${\left[\frac{4}{9}{\left(3x-4\right)}^{\frac{3}{2}}\right]}_{\frac{8}{3}}^{\frac{13}{3}}$

= ${\left[\frac{4}{9}{\left(\sqrt{3x-4}\right)}^3\right]}_{\frac{8}{3}}^{\frac{13}{3}}$

$=$ $\frac{4}{9}{\left(\sqrt{3\cdot \left(\frac{13}{3}\right)-4}\right)}^3-\frac{4}{9}{\left(\sqrt{3\cdot \left(\frac{8}{3}\right)-4}\right)}^3$

= $\frac{4}{9}{\left(\sqrt{13-4}\right)}^3-\frac{4}{9}{\left(\sqrt{8-4}\right)}^3$

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

= $\frac{4}{9}\cdot 3^3-\frac{4}{9}\cdot 2^3$

= $\frac{4}{9}\cdot 27-\frac{4}{9}\cdot 8$

= $12-\frac{32}{9}$

= $\frac{108}{9}-\frac{32}{9}$

= $\frac{76}{9}$

= ${\left[2e^{3x-6}\right]}_1^2$

$=$ $2e^{3\cdot 2-6}-2e^{3\cdot 1-6}$

= $2e^{6-6}-2e^{3-6}$

= $2e^0-2e^{-3}$

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

= ${\left[-9e^{-\mathrm{0,4}x+\mathrm{0,8}}\right]}_0^u$

$=$ $-9e^{-\mathrm{0,4}u+\mathrm{0,8}}+9e^{-\mathrm{0,4}\cdot 0+\mathrm{0,8}}$

= $-9e^{-\mathrm{0,4}u+\mathrm{0,8}}+9e^{0+\mathrm{0,8}}$

= $-9e^{-\mathrm{0,4}u+\mathrm{0,8}}+9e^{\mathrm{0,8}}$

Diese Integralfunktion soll ja den Wert $1$ annehmen, deswegen setzen wir sie gleich :

Wir berechnen den Mittelwert mit der üblichen Formel:

= $\frac{1}{21}$ ${\left[4{\left(x-1\right)}^{\frac{3}{2}}\right]}_5^{26}$

= $\frac{1}{21}$ ${\left[4{\left(\sqrt{x-1}\right)}^3\right]}_5^{26}$

$=$ $\frac{1}{21}\left(4{\left(\sqrt{26-1}\right)}^3-4{\left(\sqrt{5-1}\right)}^3\right)$

= $\frac{1}{21}\left(4{\left(\sqrt{25}\right)}^3-4{\left(\sqrt{4}\right)}^3\right)$

= $\frac{1}{21}\left(4\cdot 5^3-4\cdot 2^3\right)$

= $\frac{1}{21}\left(4\cdot 125-4\cdot 8\right)$

= $\frac{1}{21}\left(500-32\right)$

= $\frac{1}{21}·468$

= $\frac{156}{7}$

= ${\left[2{\left(x-3\right)}^{\frac{1}{2}}\right]}_{19}^u$

= ${\left[2\sqrt{x-3}\right]}_{19}^u$

$=$ $2\sqrt{u-3}-2\sqrt{19-3}$

= $2\sqrt{u-3}-2\sqrt{16}$

= $2\sqrt{u-3}-2\cdot 4$

= $2\sqrt{u-3}-8$

Für u → ∞ gilt: A(u) = $2\sqrt{u-3}-8$ → $\infty$