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

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

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

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

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

= $-\frac{3}{2}\cdot \left(\frac{1}{25}\right)+\frac{3}{2}\cdot \left(\frac{1}{9}\right)$

= $-\frac{3}{50}+\frac{1}{6}$

= $-\frac{9}{150}+\frac{25}{150}$

= $\frac{8}{75}$

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

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

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

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

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

= $4^3-2^3$

= $64-8$

= $56$

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

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

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

= $2u^2-18$

Diese Integralfunktion soll ja den Wert $32$ annehmen, deswegen setzen wir sie gleich :

Da u= $-5$ < 3 ist u= $5$ die einzige Lösung.

Wir berechnen den Mittelwert mit der üblichen Formel:

= $\frac{1}{2}$ ${\left[\frac{5}{8}{\left(2x-2\right)}^4\right]}_0^2$

$=$ $\frac{1}{2}\left(\frac{5}{8}\cdot {\left(2\cdot 2-2\right)}^4-\frac{5}{8}\cdot {\left(2\cdot 0-2\right)}^4\right)$

= $\frac{1}{2}\left(\frac{5}{8}\cdot {\left(4-2\right)}^4-\frac{5}{8}\cdot {\left(0-2\right)}^4\right)$

= $\frac{1}{2}\left(\frac{5}{8}\cdot 2^4-\frac{5}{8}\cdot {\left(-2\right)}^4\right)$

= $\frac{1}{2}\left(\frac{5}{8}\cdot 16-\frac{5}{8}\cdot 16\right)$

= $\frac{1}{2}\left(10-10\right)$

= $\frac{1}{2}·0$

= 0

= ${\left[6{\left(x-2\right)}^{\frac{1}{2}}\right]}_{18}^u$

= ${\left[6\sqrt{x-2}\right]}_{18}^u$

$=$ $6\sqrt{u-2}-6\sqrt{18-2}$

= $6\sqrt{u-2}-6\sqrt{16}$

= $6\sqrt{u-2}-6\cdot 4$

= $6\sqrt{u-2}-24$

Für u → ∞ gilt: A(u) = $6\sqrt{u-2}-24$ → $\infty$