= ${\left[\frac{4}{3}{\left(3x-4\right)}^{\frac{3}{2}}\right]}_{\frac{13}{3}}^{\frac{29}{3}}$

= ${\left[\frac{4}{3}{\left(\sqrt{3x-4}\right)}^3\right]}_{\frac{13}{3}}^{\frac{29}{3}}$

$=$ $\frac{4}{3}{\left(\sqrt{3\cdot \left(\frac{29}{3}\right)-4}\right)}^3-\frac{4}{3}{\left(\sqrt{3\cdot \left(\frac{13}{3}\right)-4}\right)}^3$

= $\frac{4}{3}{\left(\sqrt{29-4}\right)}^3-\frac{4}{3}{\left(\sqrt{13-4}\right)}^3$

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

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

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

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

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

= $\frac{392}{3}$

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

$=$ $e^{2\cdot 3-2}-e^{2\cdot 2-2}$

= $e^{6-2}-e^{4-2}$

= $e^4-e^2$

= ${\left[e^{\mathrm{0,3}x-\mathrm{0,5}}\right]}_0^u$

$=$ $e^{\mathrm{0,3}u-\mathrm{0,5}}-e^{\mathrm{0,3}\cdot 0-\mathrm{0,5}}$

= $e^{\mathrm{0,3}u-\mathrm{0,5}}-e^{0-\mathrm{0,5}}$

= $e^{\mathrm{0,3}u-\mathrm{0,5}}-e^{-\mathrm{0,5}}$

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

Wir berechnen den Mittelwert mit der üblichen Formel:

= $1$ ${\left[\frac{5}{6}{\left(2x-3\right)}^3\right]}_1^2$

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

= $\frac{5}{6}\cdot {\left(4-3\right)}^3-\frac{5}{6}\cdot {\left(2-3\right)}^3$

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

= $\frac{5}{6}\cdot 1-\frac{5}{6}\cdot \left(-1\right)$

= $\frac{5}{6}+\frac{5}{6}$

= $\frac{5}{3}$

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

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

$=$ $-\sqrt{2\cdot 5-6}+\sqrt{2u-6}$

= $-\sqrt{10-6}+\sqrt{2u-6}$

= $-\sqrt{4}+\sqrt{2u-6}$

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

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

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

Für den Flächeninhalt (immer positiv) gilt also I = 2