= ${\left[4{\left(x-3\right)}^{\frac{3}{2}}\right]}_{19}^{28}$

= ${\left[4{\left(\sqrt{x-3}\right)}^3\right]}_{19}^{28}$

$=$ $4{\left(\sqrt{28-3}\right)}^3-4{\left(\sqrt{19-3}\right)}^3$

= $4{\left(\sqrt{25}\right)}^3-4{\left(\sqrt{16}\right)}^3$

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

= $4\cdot 125-4\cdot 64$

= $500-256$

= $244$

= ${\left[4{\left(x-2\right)}^{\frac{3}{2}}\right]}_{18}^{27}$

= ${\left[4{\left(\sqrt{x-2}\right)}^3\right]}_{18}^{27}$

$=$ $4{\left(\sqrt{27-2}\right)}^3-4{\left(\sqrt{18-2}\right)}^3$

= $4{\left(\sqrt{25}\right)}^3-4{\left(\sqrt{16}\right)}^3$

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

= $4\cdot 125-4\cdot 64$

= $500-256$

= $244$

= ${\left[-6e^{-\mathrm{0,9}x+\mathrm{0,2}}\right]}_0^u$

$=$ $-6e^{-\mathrm{0,9}u+\mathrm{0,2}}+6e^{-\mathrm{0,9}\cdot 0+\mathrm{0,2}}$

= $-6e^{-\mathrm{0,9}u+\mathrm{0,2}}+6e^{0+\mathrm{0,2}}$

= $-6e^{-\mathrm{0,9}u+\mathrm{0,2}}+6e^{\mathrm{0,2}}$

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

Wir berechnen den Mittelwert mit der üblichen Formel:

= $\frac{1}{3}$ ${\left[\frac{3}{8}{\left(2x-4\right)}^4\right]}_0^3$

$=$ $\frac{1}{3}\left(\frac{3}{8}\cdot {\left(2\cdot 3-4\right)}^4-\frac{3}{8}\cdot {\left(2\cdot 0-4\right)}^4\right)$

= $\frac{1}{3}\left(\frac{3}{8}\cdot {\left(6-4\right)}^4-\frac{3}{8}\cdot {\left(0-4\right)}^4\right)$

= $\frac{1}{3}\left(\frac{3}{8}\cdot 2^4-\frac{3}{8}\cdot {\left(-4\right)}^4\right)$

= $\frac{1}{3}\left(\frac{3}{8}\cdot 16-\frac{3}{8}\cdot 256\right)$

= $\frac{1}{3}\left(6-96\right)$

= $\frac{1}{3}·\left(-90\right)$

= $-30$

= ${\left[\ln \left(\left|$ x-3$\right|\right)\right]}_5^u$

$=$ $\ln \left(\left|$ u-3$\right|\right)-\ln \left(\left|$ 5-3$\right|\right)$

= $\ln \left(\left|$ u-3$\right|\right)-\ln \left(2\right)$

Für u → ∞ gilt: A(u) = $-\ln \left(2\right)+\ln \left(\left|$ x-3$\right|\right)$ → $\infty$