= ${\left[e^{2x-5}\right]}_0^2$

$=$ $e^{2\cdot 2-5}-e^{2\cdot 0-5}$

= $e^{4-5}-e^{0-5}$

= $e^{-1}-e^{-5}$

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

= ${\left[-\frac{1}{x-1}\right]}_2^4$

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

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

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

= $\frac{2}{3}$

= ${\left[-2x^2\right]}_3^u$

$=$ $-2u^2+2\cdot 3^2$

= $-2u^2+2\cdot 9$

= $-2u^2+18$

Diese Integralfunktion soll ja den Wert $-\mathrm{94,5}$ annehmen, deswegen setzen wir sie gleich :

Da u= $-\mathrm{7,5}$ < 3 ist u= $\mathrm{7,5}$ die einzige Lösung.

Wir berechnen den Mittelwert mit der üblichen Formel:

= $\frac{2}{\pi }$ ${\left[-\frac{5}{2}\cdot \cos (2x+\pi )\right]}_{\frac{1}{2}\pi }^{\pi }$

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

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

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

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

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

= $\frac{2}{\pi }·5$

= $\frac{10}{\pi }$

= ${\left[-\frac{1}{6}{\left(3x-5\right)}^{-2}\right]}_3^u$

= ${\left[-\frac{1}{6{\left(3x-5\right)}^2}\right]}_3^u$

$=$ $-\frac{1}{6{\left(3u-5\right)}^2}+\frac{1}{6{\left(3\cdot 3-5\right)}^2}$

= $-\frac{1}{6{\left(3u-5\right)}^2}+\frac{1}{6{\left(9-5\right)}^2}$

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

= $-\frac{1}{6{\left(3u-5\right)}^2}+\frac{1}{6}\cdot \left(\frac{1}{16}\right)$

= $-\frac{1}{6{\left(3u-5\right)}^2}+\frac{1}{96}$

Für u → ∞ gilt: A(u) = $-\frac{1}{6{\left(3u-5\right)}^2}+\frac{1}{96}$ → $0+\frac{1}{96}$ = $\frac{1}{96}$ ≈ 0.01

Für den Flächeninhalt (immer positiv) gilt also I = 0.01