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

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

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

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

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

= $-\left(\frac{1}{9}\right)+1$

= $\frac{8}{9}$

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

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

$=$ $-\frac{2}{4-2}+\frac{2}{3-2}$

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

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

= $-1+2$

= $1$

= ${\left[8e^{\mathrm{0,2}x-\mathrm{0,1}}\right]}_0^u$

$=$ $8e^{\mathrm{0,2}u-\mathrm{0,1}}-8e^{\mathrm{0,2}\cdot 0-\mathrm{0,1}}$

= $8e^{\mathrm{0,2}u-\mathrm{0,1}}-8e^{0-\mathrm{0,1}}$

= $8e^{\mathrm{0,2}u-\mathrm{0,1}}-8e^{-\mathrm{0,1}}$

Diese Integralfunktion soll ja den Wert $9$ annehmen, deswegen setzen wir sie gleich :

Wir berechnen den Mittelwert mit der üblichen Formel:

= $\frac{1}{6}$ ${\left[2{\left(2x-2\right)}^{\frac{3}{2}}\right]}_3^9$

= $\frac{1}{6}$ ${\left[2{\left(\sqrt{2x-2}\right)}^3\right]}_3^9$

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

= $\frac{1}{6}\left(2{\left(\sqrt{18-2}\right)}^3-2{\left(\sqrt{6-2}\right)}^3\right)$

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

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

= $\frac{1}{6}\left(2\cdot 64-2\cdot 8\right)$

= $\frac{1}{6}\left(128-16\right)$

= $\frac{1}{6}·112$

= $\frac{56}{3}$

= ${\left[-{\left(x-2\right)}^{-2}\right]}_4^u$

= ${\left[-\frac{1}{{\left(x-2\right)}^2}\right]}_4^u$

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

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

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

Für u → ∞ gilt: A(u) = $-\frac{1}{{\left(u-2\right)}^2}+\frac{1}{4}$ → $0+\frac{1}{4}$ = $\frac{1}{4}$ ≈ 0.25

Für den Flächeninhalt (immer positiv) gilt also I = 0.25