= ${\left[\frac{2}{9}{\left(3x-4\right)}^{\frac{3}{2}}\right]}_{\frac{13}{3}}^{\frac{20}{3}}$

= ${\left[\frac{2}{9}{\left(\sqrt{3x-4}\right)}^3\right]}_{\frac{13}{3}}^{\frac{20}{3}}$

$=$ $\frac{2}{9}{\left(\sqrt{3\cdot \left(\frac{20}{3}\right)-4}\right)}^3-\frac{2}{9}{\left(\sqrt{3\cdot \left(\frac{13}{3}\right)-4}\right)}^3$

= $\frac{2}{9}{\left(\sqrt{20-4}\right)}^3-\frac{2}{9}{\left(\sqrt{13-4}\right)}^3$

= $\frac{2}{9}{\left(\sqrt{16}\right)}^3-\frac{2}{9}{\left(\sqrt{9}\right)}^3$

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

= $\frac{2}{9}\cdot 64-\frac{2}{9}\cdot 27$

= $\frac{128}{9}-6$

= $\frac{128}{9}-\frac{54}{9}$

= $\frac{74}{9}$

= ${\left[4e^{x-3}\right]}_2^4$

$=$ $4e^{4-3}-4e^{2-3}$

= $4e^1-4e^{-1}$

= $4e-4e^{-1}$

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

$=$ $-u^2+2^2$

= $-u^2+4$

Diese Integralfunktion soll ja den Wert $-5$ annehmen, deswegen setzen wir sie gleich :

Da u= $-3$ < 2 ist u= $3$ die einzige Lösung.

Wir berechnen den Mittelwert mit der üblichen Formel:

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

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

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

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

= $\frac{2}{\pi }·\left(2+2\right)$

= $\frac{2}{\pi }·4$

= $\frac{8}{\pi }$

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

= ${\left[4\sqrt{x-3}\right]}_u^4$

$=$ $4\sqrt{4-3}-4\sqrt{u-3}$

= $4\sqrt{1}-4\sqrt{u-3}$

= $4\cdot 1-4\sqrt{u-3}$

= $4-4\sqrt{u-3}$

Für u → 3 (u>3, also von rechts) gilt: A(u) = $-4\sqrt{u-3}+4$ → $0+4$ = $4$ ≈ 4

Für den Flächeninhalt (immer positiv) gilt also I = 4