= ${\left[\frac{4}{9}{\left(3x-6\right)}^{\frac{3}{2}}\right]}_5^{\frac{22}{3}}$

= ${\left[\frac{4}{9}{\left(\sqrt{3x-6}\right)}^3\right]}_5^{\frac{22}{3}}$

$=$ $\frac{4}{9}{\left(\sqrt{3\cdot \left(\frac{22}{3}\right)-6}\right)}^3-\frac{4}{9}{\left(\sqrt{3\cdot 5-6}\right)}^3$

= $\frac{4}{9}{\left(\sqrt{22-6}\right)}^3-\frac{4}{9}{\left(\sqrt{15-6}\right)}^3$

= $\frac{4}{9}{\left(\sqrt{16}\right)}^3-\frac{4}{9}{\left(\sqrt{9}\right)}^3$

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

= $\frac{4}{9}\cdot 64-\frac{4}{9}\cdot 27$

= $\frac{256}{9}-12$

= $\frac{256}{9}-\frac{108}{9}$

= $\frac{148}{9}$

= ${\left[e^{x-1}\right]}_1^2$

$=$ $e^{2-1}-e^{1-1}$

= $e^1-e^0$

= $e-1$

= ${\left[-4x^2\right]}_2^u$

$=$ $-4u^2+4\cdot 2^2$

= $-4u^2+4\cdot 4$

= $-4u^2+16$

Diese Integralfunktion soll ja den Wert $-273$ annehmen, deswegen setzen wir sie gleich :

Da u= $-\frac{17}{2}$ < 2 ist u= $\frac{17}{2}$ die einzige Lösung.

Wir berechnen den Mittelwert mit der üblichen Formel:

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

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

= $\frac{1}{\pi }·\left(-\cos \left(\pi \right)+\cos \left(0\right)\right)$

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

= $\frac{1}{\pi }·\left(1+1\right)$

= $\frac{1}{\pi }·2$

= $\frac{2}{\pi }$

= ${\left[{\left(x-3\right)}^{-2}\right]}_4^u$

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

$=$ $\frac{1}{{\left(u-3\right)}^2}-\frac{1}{{\left(4-3\right)}^2}$

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

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

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

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