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

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

= $\frac{2}{3}{e}^{9-3}-\frac{2}{3}{e}^{3-3}$

= $\frac{2}{3}{e}^6-\frac{2}{3}{e}^0$

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

= ${\left[\frac{4}{3}{\left(2x-4\right)}^{\frac{3}{2}}\right]}_{10}^{\frac{29}{2}}$

= ${\left[\frac{4}{3}{\left(\sqrt{2x-4}\right)}^3\right]}_{10}^{\frac{29}{2}}$

$=$ $\frac{4}{3}{\left(\sqrt{2\cdot \left(\frac{29}{2}\right)-4}\right)}^3-\frac{4}{3}{\left(\sqrt{2\cdot 10-4}\right)}^3$

= $\frac{4}{3}{\left(\sqrt{29-4}\right)}^3-\frac{4}{3}{\left(\sqrt{20-4}\right)}^3$

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

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

= $\frac{4}{3}\cdot 125-\frac{4}{3}\cdot 64$

= $\frac{500}{3}-\frac{256}{3}$

= $\frac{244}{3}$

= ${\left[-3e^{-\mathrm{0,9}x+\mathrm{0,1}}\right]}_0^u$

$=$ $-3e^{-\mathrm{0,9}u+\mathrm{0,1}}+3e^{-\mathrm{0,9}\cdot 0+\mathrm{0,1}}$

= $-3e^{-\mathrm{0,9}u+\mathrm{0,1}}+3e^{0+\mathrm{0,1}}$

= $-3e^{-\mathrm{0,9}u+\mathrm{0,1}}+3e^{\mathrm{0,1}}$

Diese Integralfunktion soll ja den Wert $2$ annehmen, deswegen setzen wir sie gleich :

Wir berechnen den Mittelwert mit der üblichen Formel:

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

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

= ${\left(6-4\right)}^3-{\left(4-4\right)}^3$

= $2^3-0^3$

= $8-0$

= $8$

= ${\left[2{\left(-x+3\right)}^{-1}\right]}_5^u$

= ${\left[\frac{2}{-x+3}\right]}_5^u$

$=$ $\frac{2}{-u+3}-\frac{2}{-5+3}$

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

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

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

Für u → ∞ gilt: A(u) = $\frac{2}{-u+3}+1$ → $0+1$ = $1$ ≈ 1

Für den Flächeninhalt (immer positiv) gilt also I = 1