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

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

= $e^{6-3}-e^{0-3}$

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

= ${\left[\frac{2}{3}{\left(x-2\right)}^{\frac{3}{2}}\right]}_{11}^{27}$

= ${\left[\frac{2}{3}{\left(\sqrt{x-2}\right)}^3\right]}_{11}^{27}$

$=$ $\frac{2}{3}{\left(\sqrt{27-2}\right)}^3-\frac{2}{3}{\left(\sqrt{11-2}\right)}^3$

= $\frac{2}{3}{\left(\sqrt{25}\right)}^3-\frac{2}{3}{\left(\sqrt{9}\right)}^3$

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

= $\frac{2}{3}\cdot 125-\frac{2}{3}\cdot 27$

= $\frac{250}{3}-18$

= $\frac{250}{3}-\frac{54}{3}$

= $\frac{196}{3}$

= ${\left[-3e^{-\mathrm{0,7}x+\mathrm{0,8}}\right]}_0^u$

$=$ $-3e^{-\mathrm{0,7}u+\mathrm{0,8}}+3e^{-\mathrm{0,7}\cdot 0+\mathrm{0,8}}$

= $-3e^{-\mathrm{0,7}u+\mathrm{0,8}}+3e^{0+\mathrm{0,8}}$

= $-3e^{-\mathrm{0,7}u+\mathrm{0,8}}+3e^{\mathrm{0,8}}$

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

Wir berechnen den Mittelwert mit der üblichen Formel:

= $\frac{1}{2}$ ${\left[-\frac{4}{3}{\left(3x-5\right)}^{-1}\right]}_2^4$

= $\frac{1}{2}$ ${\left[-\frac{4}{3\left(3x-5\right)}\right]}_2^4$

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

= $\frac{1}{2}\left(-\frac{4}{3\left(12-5\right)}+\frac{4}{3\left(6-5\right)}\right)$

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

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

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

= $\frac{1}{2}\left(-\frac{4}{21}+\frac{28}{21}\right)$

= $\frac{1}{2}·\frac{8}{7}$

= $\frac{4}{7}$

= ${\left[-\frac{2}{3}{e}^{-3x+4}\right]}_0^u$

$=$ $-\frac{2}{3}{e}^{-3u+4}+\frac{2}{3}{e}^{-3\cdot 0+4}$

= $-\frac{2}{3}{e}^{-3u+4}+\frac{2}{3}{e}^{0+4}$

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

Für u → ∞ gilt: A(u) = $-\frac{2}{3}{e}^{-3u+4}+\frac{2}{3}{e}^4$ → $0+\frac{2}{3}{e}^4$ = $\frac{2}{3}{e}^4$ ≈ 36.399

Für den Flächeninhalt (immer positiv) gilt also I = 36.399