= ${\left[3e^{2x-3}\right]}_2^3$

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

= $3e^{6-3}-3e^{4-3}$

= $3e^3-3e^1$

= $3e^3-3e$

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

$=$ $4e^{3-2}-4e^{2-2}$

= $4e^1-4e^0$

= $4e-4$

= ${\left[5x^2\right]}_2^u$

$=$ $5u^2-5\cdot 2^2$

= $5u^2-5\cdot 4$

= $5u^2-20$

Diese Integralfunktion soll ja den Wert $25$ annehmen, deswegen setzen wir sie gleich :

Da u= $-3$ < 2 ist u= $3$ die einzige Lösung.

Wir berechnen den Mittelwert mit der üblichen Formel:

= $\frac{1}{3}$ ${\left[\frac{4}{3}{\left(x-2\right)}^3+\frac{5}{2}{x}^2\right]}_0^3$

$=$ $\frac{1}{3}(\frac{4}{3}\cdot {\left(3-2\right)}^3+\frac{5}{2}\cdot 3^2-\left(\frac{4}{3}\cdot {\left(0-2\right)}^3+\frac{5}{2}\cdot 0^2\right))$

= $\frac{1}{3}(\frac{4}{3}\cdot 1^3+\frac{5}{2}\cdot 9-\left(\frac{4}{3}\cdot {\left(-2\right)}^3+\frac{5}{2}\cdot 0\right))$

= $\frac{1}{3}(\frac{4}{3}\cdot 1+\frac{45}{2}-\left(\frac{4}{3}\cdot \left(-8\right)+0\right))$

= $\frac{1}{3}(\frac{4}{3}+\frac{45}{2}-\left(-\frac{32}{3}+0\right))$

= $\frac{1}{3}(\frac{8}{6}+\frac{135}{6}-\left(-\frac{32}{3}+0\right))$

= $\frac{1}{3}\left(\frac{143}{6}+\frac{32}{3}\right)$

= $\frac{1}{3}\left(\frac{143}{6}+\frac{64}{6}\right)$

= $\frac{1}{3}·\frac{69}{2}$

= $\frac{23}{2}$

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

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

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

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

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

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

= $-\frac{1}{3{\left(-3u+5\right)}^2}+\frac{1}{48}$

Für u → ∞ gilt: A(u) = $-\frac{1}{3{\left(-3u+5\right)}^2}+\frac{1}{48}$ → $0+\frac{1}{48}$ = $\frac{1}{48}$ ≈ 0.021

Für den Flächeninhalt (immer positiv) gilt also I = 0.021