= ${\left[-\frac{1}{6}{\left(3x-5\right)}^{-2}\right]}_2^4$

= ${\left[-\frac{1}{6{\left(3x-5\right)}^2}\right]}_2^4$

$=$ $-\frac{1}{6{\left(3\cdot 4-5\right)}^2}+\frac{1}{6{\left(3\cdot 2-5\right)}^2}$

= $-\frac{1}{6{\left(12-5\right)}^2}+\frac{1}{6{\left(6-5\right)}^2}$

= $-\frac{1}{6\cdot 7^2}+\frac{1}{6\cdot 1^2}$

= $-\frac{1}{6}\cdot \left(\frac{1}{49}\right)+\frac{1}{6}\cdot 1$

= $-\frac{1}{294}+\frac{1}{6}$

= $-\frac{1}{294}+\frac{49}{294}$

= $\frac{8}{49}$

= ${\left[\frac{2}{3}{\left(x-2\right)}^{\frac{3}{2}}\right]}_{11}^{27}$

= ${\left[\frac{2}{3}{\left(\sqrt{x-2}\right)}^3\right]}_{11}^{27}$

$=$ $\frac{2}{3}{\left(\sqrt{27-2}\right)}^3-\frac{2}{3}{\left(\sqrt{11-2}\right)}^3$

= $\frac{2}{3}{\left(\sqrt{25}\right)}^3-\frac{2}{3}{\left(\sqrt{9}\right)}^3$

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

= $\frac{2}{3}\cdot 125-\frac{2}{3}\cdot 27$

= $\frac{250}{3}-18$

= $\frac{250}{3}-\frac{54}{3}$

= $\frac{196}{3}$

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

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

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

= $2u^2-18$

Diese Integralfunktion soll ja den Wert $144$ annehmen, deswegen setzen wir sie gleich :

Da u= $-9$ < 3 ist u= $9$ die einzige Lösung.

Wir berechnen den Mittelwert mit der üblichen Formel:

= $\frac{2}{\pi }$ ${\left[-\frac{2}{3}\cdot \cos \left(3x+\frac{1}{2}\pi \right)\right]}_0^{\frac{1}{2}\pi }$

$=$ $\frac{2}{\pi }·\left(-\frac{2}{3}\cdot \cos \left(3\cdot \left(\frac{1}{2}\pi \right)+\frac{1}{2}\pi \right)+\frac{2}{3}\cdot \cos \left(3\cdot \left(0\right)+\frac{1}{2}\pi \right)\right)$

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

= $\frac{2}{\pi }·\left(-\frac{2}{3}\cdot 1+\frac{2}{3}\cdot 0\right)$

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

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

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

= $-\frac{4}{3\pi }$

= ${\left[2e^{x-1}\right]}_0^u$

$=$ $2e^{u-1}-2e^{0-1}$

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

Für u → ∞ gilt: A(u) = $2e^{u-1}-2e^{-1}$ → $\infty$