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

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

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

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

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

= $2\cdot 5^3-2\cdot 3^3$

= $2\cdot 125-2\cdot 27$

= $250-54$

= $196$

= ${\left[-6{\left(x-1\right)}^{-1}\right]}_3^4$

= ${\left[-\frac{6}{x-1}\right]}_3^4$

$=$ $-\frac{6}{4-1}+\frac{6}{3-1}$

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

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

= $-2+3$

= $1$

= ${\left[3x^2\right]}_0^u$

$=$ $3u^2-3\cdot 0^2$

= $3u^2-3\cdot 0$

= $3u^2+0$

= $3u^2$

Diese Integralfunktion soll ja den Wert $27$ annehmen, deswegen setzen wir sie gleich :

Da u= $-3$ < 0 ist u= $3$ die einzige Lösung.

Wir berechnen den Mittelwert mit der üblichen Formel:

= $\frac{1}{3}$ ${\left[\frac{5}{6}{\left(2x-5\right)}^3+6x\right]}_0^3$

$=$ $\frac{1}{3}(\frac{5}{6}\cdot {\left(2\cdot 3-5\right)}^3+6\cdot 3-\left(\frac{5}{6}\cdot {\left(2\cdot 0-5\right)}^3+6\cdot 0\right))$

= $\frac{1}{3}(\frac{5}{6}\cdot {\left(6-5\right)}^3+18-\left(\frac{5}{6}\cdot {\left(0-5\right)}^3+0\right))$

= $\frac{1}{3}(\frac{5}{6}\cdot 1^3+18-\left(\frac{5}{6}\cdot {\left(-5\right)}^3+0\right))$

= $\frac{1}{3}(\frac{5}{6}\cdot 1+18-\left(\frac{5}{6}\cdot \left(-125\right)+0\right))$

= $\frac{1}{3}(\frac{5}{6}+18-\left(-\frac{625}{6}+0\right))$

= $\frac{1}{3}(\frac{5}{6}+\frac{108}{6}-\left(-\frac{625}{6}+0\right))$

= $\frac{1}{3}\left(\frac{113}{6}+\frac{625}{6}\right)$

= $\frac{1}{3}·123$

= $41$

= ${\left[\frac{1}{2}{e}^{2x-4}\right]}_0^u$

$=$ $\frac{1}{2}{e}^{2u-4}-\frac{1}{2}{e}^{2\cdot 0-4}$

= $\frac{1}{2}{e}^{2u-4}-\frac{1}{2}{e}^{0-4}$

= $\frac{1}{2}{e}^{2u-4}-\frac{1}{2}{e}^{-4}$

Für u → ∞ gilt: A(u) = $\frac{1}{2}{e}^{2u-4}-\frac{1}{2}{e}^{-4}$ → $\infty$