= ${\left[-\frac{4}{3}{\left(3x-5\right)}^{-1}\right]}_3^5$

= ${\left[-\frac{4}{3\left(3x-5\right)}\right]}_3^5$

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

= $-\frac{4}{3\left(15-5\right)}+\frac{4}{3\left(9-5\right)}$

= $-\frac{4}{3\cdot 10}+\frac{4}{3\cdot 4}$

= $-\frac{4}{3}\cdot \left(\frac{1}{10}\right)+\frac{4}{3}\cdot \left(\frac{1}{4}\right)$

= $-\frac{2}{15}+\frac{1}{3}$

= $-\frac{2}{15}+\frac{5}{15}$

= $\frac{1}{5}$

= ${\left[6e^{x-3}\right]}_2^5$

$=$ $6e^{5-3}-6e^{2-3}$

= $6e^2-6e^{-1}$

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

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

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

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

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

Wir berechnen den Mittelwert mit der üblichen Formel:

= $\frac{2}{\pi }$ ${\left[\frac{1}{2}\cdot \sin \left(2x+\frac{1}{2}\pi \right)\right]}_{\frac{1}{2}\pi }^{\pi }$

$=$ $\frac{2}{\pi }·\left(\frac{1}{2}\cdot \sin \left(2\cdot \pi +\frac{1}{2}\pi \right)-\frac{1}{2}\cdot \sin \left(2\cdot \left(\frac{1}{2}\pi \right)+\frac{1}{2}\pi \right)\right)$

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

= $\frac{2}{\pi }·\left(\frac{1}{2}\cdot 1-\frac{1}{2}\cdot \left(-1\right)\right)$

= $\frac{2}{\pi }·\left(\frac{1}{2}+\frac{1}{2}\right)$

= $\frac{2}{\pi }·\left(\mathrm{0,5}+\mathrm{0,5}\right)$

= $\frac{2}{\pi }·1$

= $\frac{2}{\pi }$

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

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

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

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

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

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

Für u → ∞ gilt: A(u) = $-\frac{3}{-u+3}-\frac{3}{2}$ → $0-\frac{3}{2}$ = $-\frac{3}{2}$ ≈ -1.5

Für den Flächeninhalt (immer positiv) gilt also I = 1.5