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

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

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

= ${\left(\sqrt{10-1}\right)}^3-{\left(\sqrt{5-1}\right)}^3$

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

= $3^3-2^3$

= $27-8$

= $19$

= ${\left[4{\left(x-2\right)}^{\frac{3}{2}}\right]}_{18}^{27}$

= ${\left[4{\left(\sqrt{x-2}\right)}^3\right]}_{18}^{27}$

$=$ $4{\left(\sqrt{27-2}\right)}^3-4{\left(\sqrt{18-2}\right)}^3$

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

= $4\cdot 5^3-4\cdot 4^3$

= $4\cdot 125-4\cdot 64$

= $500-256$

= $244$

= ${\left[2x^2\right]}_1^u$

$=$ $2u^2-2\cdot 1^2$

= $2u^2-2\cdot 1$

= $2u^2-2$

Diese Integralfunktion soll ja den Wert $\mathrm{142,5}$ annehmen, deswegen setzen wir sie gleich :

Da u= $-\mathrm{8,5}$ < 1 ist u= $\mathrm{8,5}$ die einzige Lösung.

Wir berechnen den Mittelwert mit der üblichen Formel:

= $\frac{1}{3}$ ${\left[\frac{2}{9}{\left(3x-5\right)}^3\right]}_0^3$

$=$ $\frac{1}{3}\left(\frac{2}{9}\cdot {\left(3\cdot 3-5\right)}^3-\frac{2}{9}\cdot {\left(3\cdot 0-5\right)}^3\right)$

= $\frac{1}{3}\left(\frac{2}{9}\cdot {\left(9-5\right)}^3-\frac{2}{9}\cdot {\left(0-5\right)}^3\right)$

= $\frac{1}{3}\left(\frac{2}{9}\cdot 4^3-\frac{2}{9}\cdot {\left(-5\right)}^3\right)$

= $\frac{1}{3}\left(\frac{2}{9}\cdot 64-\frac{2}{9}\cdot \left(-125\right)\right)$

= $\frac{1}{3}\left(\frac{128}{9}+\frac{250}{9}\right)$

= $\frac{1}{3}·42$

= $14$

= ${\left[\frac{3}{4}{x}^{-1}\right]}_u^3$

= ${\left[\frac{3}{4x}\right]}_u^3$

$=$ $\frac{3}{4\cdot 3}-\frac{3}{4u}$

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

= $\frac{1}{4}-\frac{3}{4u}$

Für u → 0 (u>0, also von rechts) gilt: A(u) = $\frac{1}{4}-\frac{3}{4u}$ → $-\infty$