= ${\left[\frac{4}{3}{\left(x-1\right)}^{\frac{3}{2}}\right]}_{10}^{26}$

= ${\left[\frac{4}{3}{\left(\sqrt{x-1}\right)}^3\right]}_{10}^{26}$

$=$ $\frac{4}{3}{\left(\sqrt{26-1}\right)}^3-\frac{4}{3}{\left(\sqrt{10-1}\right)}^3$

= $\frac{4}{3}{\left(\sqrt{25}\right)}^3-\frac{4}{3}{\left(\sqrt{9}\right)}^3$

= $\frac{4}{3}\cdot 5^3-\frac{4}{3}\cdot 3^3$

= $\frac{4}{3}\cdot 125-\frac{4}{3}\cdot 27$

= $\frac{500}{3}-36$

= $\frac{500}{3}-\frac{108}{3}$

= $\frac{392}{3}$

= ${\left[\frac{5}{3}{\left(2x-3\right)}^{\frac{3}{2}}\right]}_6^{14}$

= ${\left[\frac{5}{3}{\left(\sqrt{2x-3}\right)}^3\right]}_6^{14}$

$=$ $\frac{5}{3}{\left(\sqrt{2\cdot 14-3}\right)}^3-\frac{5}{3}{\left(\sqrt{2\cdot 6-3}\right)}^3$

= $\frac{5}{3}{\left(\sqrt{28-3}\right)}^3-\frac{5}{3}{\left(\sqrt{12-3}\right)}^3$

= $\frac{5}{3}{\left(\sqrt{25}\right)}^3-\frac{5}{3}{\left(\sqrt{9}\right)}^3$

= $\frac{5}{3}\cdot 5^3-\frac{5}{3}\cdot 3^3$

= $\frac{5}{3}\cdot 125-\frac{5}{3}\cdot 27$

= $\frac{625}{3}-45$

= $\frac{625}{3}-\frac{135}{3}$

= $\frac{490}{3}$

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

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

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

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

Diese Integralfunktion soll ja den Wert $10$ annehmen, deswegen setzen wir sie gleich :

Wir berechnen den Mittelwert mit der üblichen Formel:

= $\frac{2}{\pi }$ ${\left[5\cdot \sin \left(x+\frac{3}{2}\pi \right)\right]}_{\frac{1}{2}\pi }^{\pi }$

$=$ $\frac{2}{\pi }·\left(5\cdot \sin \left(\pi +\frac{3}{2}\pi \right)-5\cdot \sin \left(\frac{1}{2}\pi +\frac{3}{2}\pi \right)\right)$

= $\frac{2}{\pi }·\left(5\cdot \sin \left(\frac{5}{2}\pi \right)-5\cdot \sin \left(2\pi \right)\right)$

= $\frac{2}{\pi }·\left(5\cdot 1-5\cdot 0\right)$

= $\frac{2}{\pi }·\left(5+0\right)$

= $\frac{2}{\pi }·5$

= $\frac{10}{\pi }$

= ${\left[-\frac{1}{6}{\left(3x-5\right)}^{-2}\right]}_2^u$

= ${\left[-\frac{1}{6{\left(3x-5\right)}^2}\right]}_2^u$

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

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

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

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

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

Für u → ∞ gilt: A(u) = $-\frac{1}{6{\left(3u-5\right)}^2}+\frac{1}{6}$ → $0+\frac{1}{6}$ = $\frac{1}{6}$ ≈ 0.167

Für den Flächeninhalt (immer positiv) gilt also I = 0.167