= ${\left[\frac{5}{3}{e}^{3x-4}\right]}_0^2$

$=$ $\frac{5}{3}{e}^{3\cdot 2-4}-\frac{5}{3}{e}^{3\cdot 0-4}$

= $\frac{5}{3}{e}^{6-4}-\frac{5}{3}{e}^{0-4}$

= $\frac{5}{3}{e}^2-\frac{5}{3}{e}^{-4}$

= ${\left[\frac{4}{3}{\left(3x-6\right)}^{\frac{3}{2}}\right]}_{\frac{22}{3}}^{\frac{31}{3}}$

= ${\left[\frac{4}{3}{\left(\sqrt{3x-6}\right)}^3\right]}_{\frac{22}{3}}^{\frac{31}{3}}$

$=$ $\frac{4}{3}{\left(\sqrt{3\cdot \left(\frac{31}{3}\right)-6}\right)}^3-\frac{4}{3}{\left(\sqrt{3\cdot \left(\frac{22}{3}\right)-6}\right)}^3$

= $\frac{4}{3}{\left(\sqrt{31-6}\right)}^3-\frac{4}{3}{\left(\sqrt{22-6}\right)}^3$

= $\frac{4}{3}{\left(\sqrt{25}\right)}^3-\frac{4}{3}{\left(\sqrt{16}\right)}^3$

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

= $\frac{4}{3}\cdot 125-\frac{4}{3}\cdot 64$

= $\frac{500}{3}-\frac{256}{3}$

= $\frac{244}{3}$

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

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

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

= $2u^2-8$

Diese Integralfunktion soll ja den Wert $154$ annehmen, deswegen setzen wir sie gleich :

Da u= $-9$ < 2 ist u= $9$ die einzige Lösung.

Wir berechnen den Mittelwert mit der üblichen Formel:

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

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

= $\frac{1}{\pi }·\left(\frac{1}{2}\cdot \sin \left(2\pi \right)-\frac{1}{2}\cdot \sin \left(0\right)\right)$

= $\frac{1}{\pi }·\left(\frac{1}{2}\cdot 0-\frac{1}{2}\cdot 0\right)$

= $\frac{1}{\pi }·\left(0+0\right)$

= $\frac{1}{\pi }·0$

= 0

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

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

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

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

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

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

Für u → 1 (u>1, also von rechts) gilt: A(u) = $-\frac{1}{{\left(u-1\right)}^2}+1$ → $-\infty$