= ${\left[\frac{2}{9}{\left(3x-3\right)}^{\frac{3}{2}}\right]}_{\frac{7}{3}}^{\frac{19}{3}}$

= ${\left[\frac{2}{9}{\left(\sqrt{3x-3}\right)}^3\right]}_{\frac{7}{3}}^{\frac{19}{3}}$

$=$ $\frac{2}{9}{\left(\sqrt{3\cdot \left(\frac{19}{3}\right)-3}\right)}^3-\frac{2}{9}{\left(\sqrt{3\cdot \left(\frac{7}{3}\right)-3}\right)}^3$

= $\frac{2}{9}{\left(\sqrt{19-3}\right)}^3-\frac{2}{9}{\left(\sqrt{7-3}\right)}^3$

= $\frac{2}{9}{\left(\sqrt{16}\right)}^3-\frac{2}{9}{\left(\sqrt{4}\right)}^3$

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

= $\frac{2}{9}\cdot 64-\frac{2}{9}\cdot 8$

= $\frac{128}{9}-\frac{16}{9}$

= $\frac{112}{9}$

= ${\left[-5{\left(x-2\right)}^{-1}\right]}_3^6$

= ${\left[-\frac{5}{x-2}\right]}_3^6$

$=$ $-\frac{5}{6-2}+\frac{5}{3-2}$

= $-\frac{5}{4}+\frac{5}{1}$

= $-5\cdot \left(\frac{1}{4}\right)+5\cdot 1$

= $-\frac{5}{4}+5$

= $-\frac{5}{4}+\frac{20}{4}$

= $\frac{15}{4}$

= ${\left[-4x^2\right]}_3^u$

$=$ $-4u^2+4\cdot 3^2$

= $-4u^2+4\cdot 9$

= $-4u^2+36$

Diese Integralfunktion soll ja den Wert $-325$ annehmen, deswegen setzen wir sie gleich :

Da u= $-\frac{19}{2}$ < 3 ist u= $\frac{19}{2}$ die einzige Lösung.

Wir berechnen den Mittelwert mit der üblichen Formel:

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

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

= $\frac{2}{\pi }·\left(-3\cdot \cos \left(\frac{5}{2}\pi \right)+3\cdot \cos \left(\frac{3}{2}\pi \right)\right)$

= $\frac{2}{\pi }·\left(-3\cdot 0+3\cdot 0\right)$

= $\frac{2}{\pi }·\left(0+0\right)$

= $\frac{2}{\pi }·0$

= 0

= ${\left[-2{\left(3x-5\right)}^{\frac{1}{2}}\right]}_7^u$

= ${\left[-2\sqrt{3x-5}\right]}_7^u$

$=$ $-2\sqrt{3u-5}+2\sqrt{3\cdot 7-5}$

= $-2\sqrt{3u-5}+2\sqrt{21-5}$

= $-2\sqrt{3u-5}+2\sqrt{16}$

= $-2\sqrt{3u-5}+2\cdot 4$

= $-2\sqrt{3u-5}+8$

Für u → ∞ gilt: A(u) = $-2\sqrt{3u-5}+8$ → $-\infty$