= ${\left[-\frac{1}{6}{\left(3x-3\right)}^{-2}\right]}_3^4$

= ${\left[-\frac{1}{6{\left(3x-3\right)}^2}\right]}_3^4$

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

= $-\frac{1}{6{\left(12-3\right)}^2}+\frac{1}{6{\left(9-3\right)}^2}$

= $-\frac{1}{6\cdot 9^2}+\frac{1}{6\cdot 6^2}$

= $-\frac{1}{6}\cdot \left(\frac{1}{81}\right)+\frac{1}{6}\cdot \left(\frac{1}{36}\right)$

= $-\frac{1}{486}+\frac{1}{216}$

= $-\frac{4}{1944}+\frac{9}{1944}$

= $\frac{5}{1944}$

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

$=$ $2e^{4-1}-2e^{2-1}$

= $2e^3-2e^1$

= $2e^3-2e$

= ${\left[2x^2\right]}_1^u$

$=$ $2u^2-2\cdot 1^2$

= $2u^2-2\cdot 1$

= $2u^2-2$

Diese Integralfunktion soll ja den Wert $6$ annehmen, deswegen setzen wir sie gleich :

Da u= $-2$ < 1 ist u= $2$ die einzige Lösung.

Wir berechnen den Mittelwert mit der üblichen Formel:

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

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

= $\frac{1}{2}(\frac{2}{3}\cdot {\left(12-3\right)}^3+\frac{1}{2}\cdot 16-\left(\frac{2}{3}\cdot {\left(6-3\right)}^3+\frac{1}{2}\cdot 4\right))$

= $\frac{1}{2}(\frac{2}{3}\cdot 9^3+8-\left(\frac{2}{3}\cdot 3^3+2\right))$

= $\frac{1}{2}(\frac{2}{3}\cdot 729+8-\left(\frac{2}{3}\cdot 27+2\right))$

= $\frac{1}{2}(486+8-\left(18+2\right))$

= $\frac{1}{2}(494-1·20)$

= $\frac{1}{2}\left(494-20\right)$

= $\frac{1}{2}·474$

= $237$

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

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

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

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

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

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

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

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