= ${\left[2e^{3x-5}\right]}_0^3$

$=$ $2e^{3\cdot 3-5}-2e^{3\cdot 0-5}$

= $2e^{9-5}-2e^{0-5}$

= $2e^4-2e^{-5}$

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

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

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

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

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

= $5^3-4^3$

= $125-64$

= $61$

= ${\left[-4x^2\right]}_3^u$

$=$ $-4u^2+4\cdot 3^2$

= $-4u^2+4\cdot 9$

= $-4u^2+36$

Diese Integralfunktion soll ja den Wert $-108$ annehmen, deswegen setzen wir sie gleich :

Da u= $-6$ < 3 ist u= $6$ die einzige Lösung.

Wir berechnen den Mittelwert mit der üblichen Formel:

= $\frac{1}{2}$ ${\left[\frac{1}{6}{\left(2x-3\right)}^3\right]}_2^4$

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

= $\frac{1}{2}\left(\frac{1}{6}\cdot {\left(8-3\right)}^3-\frac{1}{6}\cdot {\left(4-3\right)}^3\right)$

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

= $\frac{1}{2}\left(\frac{1}{6}\cdot 125-\frac{1}{6}\cdot 1\right)$

= $\frac{1}{2}\left(\frac{125}{6}-\frac{1}{6}\right)$

= $\frac{1}{2}·\frac{62}{3}$

= $\frac{31}{3}$

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

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

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

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

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

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

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

Für u → ∞ gilt: A(u) = $-\frac{3}{2\left(2u-2\right)}+\frac{3}{4}$ → $0+\frac{3}{4}$ = $\frac{3}{4}$ ≈ 0.75

Für den Flächeninhalt (immer positiv) gilt also I = 0.75