= ${\left[\frac{4}{3}{\left(3x-4\right)}^{\frac{3}{2}}\right]}_{\frac{8}{3}}^{\frac{13}{3}}$

= ${\left[\frac{4}{3}{\left(\sqrt{3x-4}\right)}^3\right]}_{\frac{8}{3}}^{\frac{13}{3}}$

$=$ $\frac{4}{3}{\left(\sqrt{3\cdot \left(\frac{13}{3}\right)-4}\right)}^3-\frac{4}{3}{\left(\sqrt{3\cdot \left(\frac{8}{3}\right)-4}\right)}^3$

= $\frac{4}{3}{\left(\sqrt{13-4}\right)}^3-\frac{4}{3}{\left(\sqrt{8-4}\right)}^3$

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

= $\frac{4}{3}\cdot 3^3-\frac{4}{3}\cdot 2^3$

= $\frac{4}{3}\cdot 27-\frac{4}{3}\cdot 8$

= $36-\frac{32}{3}$

= $\frac{108}{3}-\frac{32}{3}$

= $\frac{76}{3}$

= ${\left[\frac{4}{9}{\left(3x-4\right)}^{\frac{3}{2}}\right]}_{\frac{8}{3}}^{\frac{13}{3}}$

= ${\left[\frac{4}{9}{\left(\sqrt{3x-4}\right)}^3\right]}_{\frac{8}{3}}^{\frac{13}{3}}$

$=$ $\frac{4}{9}{\left(\sqrt{3\cdot \left(\frac{13}{3}\right)-4}\right)}^3-\frac{4}{9}{\left(\sqrt{3\cdot \left(\frac{8}{3}\right)-4}\right)}^3$

= $\frac{4}{9}{\left(\sqrt{13-4}\right)}^3-\frac{4}{9}{\left(\sqrt{8-4}\right)}^3$

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

= $\frac{4}{9}\cdot 3^3-\frac{4}{9}\cdot 2^3$

= $\frac{4}{9}\cdot 27-\frac{4}{9}\cdot 8$

= $12-\frac{32}{9}$

= $\frac{108}{9}-\frac{32}{9}$

= $\frac{76}{9}$

= ${\left[-6e^{-\mathrm{0,5}x+\mathrm{0,5}}\right]}_0^u$

$=$ $-6e^{-\mathrm{0,5}u+\mathrm{0,5}}+6e^{-\mathrm{0,5}\cdot 0+\mathrm{0,5}}$

= $-6e^{-\mathrm{0,5}u+\mathrm{0,5}}+6e^{0+\mathrm{0,5}}$

= $-6e^{-\mathrm{0,5}u+\mathrm{0,5}}+6e^{\mathrm{0,5}}$

Diese Integralfunktion soll ja den Wert $2$ annehmen, deswegen setzen wir sie gleich :

Wir berechnen den Mittelwert mit der üblichen Formel:

= $\frac{1}{\pi }$ ${\left[6\cdot \sin (x+\pi )\right]}_{\frac{1}{2}\pi }^{\frac{3}{2}\pi }$

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

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

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

= $\frac{1}{\pi }·\left(6+6\right)$

= $\frac{1}{\pi }·12$

= $\frac{12}{\pi }$

= ${\left[3{\left(2x-6\right)}^{\frac{1}{2}}\right]}_u^{11}$

= ${\left[3\sqrt{2x-6}\right]}_u^{11}$

$=$ $3\sqrt{2\cdot 11-6}-3\sqrt{2u-6}$

= $3\sqrt{22-6}-3\sqrt{2u-6}$

= $3\sqrt{16}-3\sqrt{2u-6}$

= $3\cdot 4-3\sqrt{2u-6}$

= $12-3\sqrt{2u-6}$

Für u → 3 (u>3, also von rechts) gilt: A(u) = $-3\sqrt{2u-6}+12$ → $0+12$ = $12$ ≈ 12

Für den Flächeninhalt (immer positiv) gilt also I = 12