= ${\left[\frac{8}{9}{\left(3x-5\right)}^{\frac{3}{2}}\right]}_7^{10}$

= ${\left[\frac{8}{9}{\left(\sqrt{3x-5}\right)}^3\right]}_7^{10}$

$=$ $\frac{8}{9}{\left(\sqrt{3\cdot 10-5}\right)}^3-\frac{8}{9}{\left(\sqrt{3\cdot 7-5}\right)}^3$

= $\frac{8}{9}{\left(\sqrt{30-5}\right)}^3-\frac{8}{9}{\left(\sqrt{21-5}\right)}^3$

= $\frac{8}{9}{\left(\sqrt{25}\right)}^3-\frac{8}{9}{\left(\sqrt{16}\right)}^3$

= $\frac{8}{9}\cdot 5^3-\frac{8}{9}\cdot 4^3$

= $\frac{8}{9}\cdot 125-\frac{8}{9}\cdot 64$

= $\frac{1000}{9}-\frac{512}{9}$

= $\frac{488}{9}$

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

= ${\left[-\frac{5}{x-1}\right]}_2^3$

$=$ $-\frac{5}{3-1}+\frac{5}{2-1}$

= $-\frac{5}{2}+\frac{5}{1}$

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

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

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

= $\mathrm{2,5}$

= ${\left[3e^{\mathrm{0,3}x-\mathrm{0,2}}\right]}_0^u$

$=$ $3e^{\mathrm{0,3}u-\mathrm{0,2}}-3e^{\mathrm{0,3}\cdot 0-\mathrm{0,2}}$

= $3e^{\mathrm{0,3}u-\mathrm{0,2}}-3e^{0-\mathrm{0,2}}$

= $3e^{\mathrm{0,3}u-\mathrm{0,2}}-3e^{-\mathrm{0,2}}$

Diese Integralfunktion soll ja den Wert $20$ annehmen, deswegen setzen wir sie gleich :

Wir berechnen den Mittelwert mit der üblichen Formel:

= $\frac{1}{3}$ ${\left[2\ln \left(\left|$ x-2$\right|\right)\right]}_4^7$

$=$ $\frac{1}{3}\left(2\ln \left(\left|$ 7-2$\right|\right)-2\ln \left(\left|$ 4-2$\right|\right)\right)$

= $\frac{1}{3}\left(2\ln \left(5\right)-2\ln \left(\left|$ 4-2$\right|\right)\right)$

= $\frac{1}{3}\left(2\ln \left(5\right)-2\ln \left(2\right)\right)$

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

= ${\left[\sqrt{2x-6}\right]}_u^{\mathrm{3,5}}$

$=$ $\sqrt{2\cdot \mathrm{3,5}-6}-\sqrt{2u-6}$

= $\sqrt{7-6}-\sqrt{2u-6}$

= $\sqrt{1}-\sqrt{2u-6}$

= $1-\sqrt{2u-6}$

= $-\sqrt{2u-6}+1$

Für u → 3 (u>3, also von rechts) gilt: A(u) = $-\sqrt{2u-6}+1$ → $0+1$ = $1$ ≈ 1

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