= ${\left[\frac{10}{3}{\left(x-3\right)}^{\frac{3}{2}}\right]}_{19}^{28}$

= ${\left[\frac{10}{3}{\left(\sqrt{x-3}\right)}^3\right]}_{19}^{28}$

$=$ $\frac{10}{3}{\left(\sqrt{28-3}\right)}^3-\frac{10}{3}{\left(\sqrt{19-3}\right)}^3$

= $\frac{10}{3}{\left(\sqrt{25}\right)}^3-\frac{10}{3}{\left(\sqrt{16}\right)}^3$

= $\frac{10}{3}\cdot 5^3-\frac{10}{3}\cdot 4^3$

= $\frac{10}{3}\cdot 125-\frac{10}{3}\cdot 64$

= $\frac{1250}{3}-\frac{640}{3}$

= $\frac{610}{3}$

= ${\left[3e^{x-1}\right]}_0^2$

$=$ $3e^{2-1}-3e^{0-1}$

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

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

= ${\left[-5e^{-\mathrm{0,8}x+\mathrm{0,3}}\right]}_0^u$

$=$ $-5e^{-\mathrm{0,8}u+\mathrm{0,3}}+5e^{-\mathrm{0,8}\cdot 0+\mathrm{0,3}}$

= $-5e^{-\mathrm{0,8}u+\mathrm{0,3}}+5e^{0+\mathrm{0,3}}$

= $-5e^{-\mathrm{0,8}u+\mathrm{0,3}}+5e^{\mathrm{0,3}}$

Diese Integralfunktion soll ja den Wert $3$ annehmen, deswegen setzen wir sie gleich :

Wir berechnen den Mittelwert mit der üblichen Formel:

= $\frac{1}{3}$ ${\left[-2{\left(x-2\right)}^{-1}\right]}_4^7$

= $\frac{1}{3}$ ${\left[-\frac{2}{x-2}\right]}_4^7$

$=$ $\frac{1}{3}\left(-\frac{2}{7-2}+\frac{2}{4-2}\right)$

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

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

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

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

= $\frac{1}{3}·\frac{3}{5}$

= $\frac{1}{5}$

= ${\left[-2{\left(2x-4\right)}^{-\frac{1}{2}}\right]}_{10}^u$

= ${\left[-\frac{2}{\sqrt{2x-4}}\right]}_{10}^u$

$=$ $-\frac{2}{\sqrt{2u-4}}+\frac{2}{\sqrt{2\cdot 10-4}}$

= $-\frac{2}{\sqrt{2u-4}}+\frac{2}{\sqrt{20-4}}$

= $-\frac{2}{\sqrt{2u-4}}+\frac{2}{\sqrt{16}}$

= $-\frac{2}{\sqrt{2u-4}}+\frac{2}{4}$

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

= $-\frac{2}{\sqrt{2u-4}}+\frac{1}{2}$

Für u → ∞ gilt: A(u) = $-\frac{2}{\sqrt{2u-4}}+\frac{1}{2}$ → $0+\frac{1}{2}$ = $\frac{1}{2}$ ≈ 0.5

Für den Flächeninhalt (immer positiv) gilt also I = 0.5