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

= ${\left[-\frac{2}{x-2}\right]}_3^6$

$=$ $-\frac{2}{6-2}+\frac{2}{3-2}$

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

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

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

= $-\mathrm{0,5}+2$

= $\mathrm{1,5}$

= ${\left[2{\left(2x-5\right)}^{\frac{3}{2}}\right]}_{\frac{21}{2}}^{15}$

= ${\left[2{\left(\sqrt{2x-5}\right)}^3\right]}_{\frac{21}{2}}^{15}$

$=$ $2{\left(\sqrt{2\cdot 15-5}\right)}^3-2{\left(\sqrt{2\cdot \left(\frac{21}{2}\right)-5}\right)}^3$

= $2{\left(\sqrt{30-5}\right)}^3-2{\left(\sqrt{21-5}\right)}^3$

= $2{\left(\sqrt{25}\right)}^3-2{\left(\sqrt{16}\right)}^3$

= $2\cdot 5^3-2\cdot 4^3$

= $2\cdot 125-2\cdot 64$

= $250-128$

= $122$

= ${\left[2e^{\mathrm{0,9}x-\mathrm{0,2}}\right]}_0^u$

$=$ $2e^{\mathrm{0,9}u-\mathrm{0,2}}-2e^{\mathrm{0,9}\cdot 0-\mathrm{0,2}}$

= $2e^{\mathrm{0,9}u-\mathrm{0,2}}-2e^{0-\mathrm{0,2}}$

= $2e^{\mathrm{0,9}u-\mathrm{0,2}}-2e^{-\mathrm{0,2}}$

Diese Integralfunktion soll ja den Wert $18$ annehmen, deswegen setzen wir sie gleich :

Wir berechnen den Mittelwert mit der üblichen Formel:

= $\frac{1}{2}$ ${\left[5e^{x-1}\right]}_0^2$

$=$ $\frac{1}{2}\left(5e^{2-1}-5e^{0-1}\right)$

= $\frac{1}{2}\left(5e^1-5e^{-1}\right)$

= $\frac{1}{2}\left(5e-5e^{-1}\right)$

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

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

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

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

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

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

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

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

Für u → ∞ gilt: A(u) = $\frac{2}{3\left(-3u+3\right)}+\frac{1}{9}$ → $0+\frac{1}{9}$ = $\frac{1}{9}$ ≈ 0.111

Für den Flächeninhalt (immer positiv) gilt also I = 0.111