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

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

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

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

= $2{\left(\sqrt{16}\right)}^3-2{\left(\sqrt{4}\right)}^3$

= $2\cdot 4^3-2\cdot 2^3$

= $2\cdot 64-2\cdot 8$

= $128-16$

= $112$

= ${\left[\frac{5}{2}\ln \left(\left|$ 2x-2$\right|\right)\right]}_2^3$

$=$ $\frac{5}{2}\ln \left(\left|$ 2\cdot 3-2$\right|\right)-\frac{5}{2}\ln \left(\left|$ 2\cdot 2-2$\right|\right)$

= $\frac{5}{2}\ln \left(\left|$ 6-2$\right|\right)-\frac{5}{2}\ln \left(\left|$ 4-2$\right|\right)$

= $\frac{5}{2}\ln \left(4\right)-\frac{5}{2}\ln \left(\left|$ 4-2$\right|\right)$

= $\frac{5}{2}\ln \left(4\right)-\frac{5}{2}\ln \left(2\right)$

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

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

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

= $-2u^2+0$

= $-2u^2$

Diese Integralfunktion soll ja den Wert $-\mathrm{12,5}$ annehmen, deswegen setzen wir sie gleich :

Da u= $-\mathrm{2,5}$ < 0 ist u= $\mathrm{2,5}$ die einzige Lösung.

Wir berechnen den Mittelwert mit der üblichen Formel:

= $\frac{1}{2}$ ${\left[\frac{1}{4}{\left(3x-6\right)}^4\right]}_0^2$

$=$ $\frac{1}{2}\left(\frac{1}{4}\cdot {\left(3\cdot 2-6\right)}^4-\frac{1}{4}\cdot {\left(3\cdot 0-6\right)}^4\right)$

= $\frac{1}{2}\left(\frac{1}{4}\cdot {\left(6-6\right)}^4-\frac{1}{4}\cdot {\left(0-6\right)}^4\right)$

= $\frac{1}{2}\left(\frac{1}{4}\cdot 0^4-\frac{1}{4}\cdot {\left(-6\right)}^4\right)$

= $\frac{1}{2}\left(\frac{1}{4}\cdot 0-\frac{1}{4}\cdot 1296\right)$

= $\frac{1}{2}\left(0-324\right)$

= $-162$

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

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

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

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

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

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

Für u → ∞ gilt: A(u) = $\frac{1}{2{\left(u-1\right)}^2}-\frac{1}{2}$ → $0-\frac{1}{2}$ = $-\frac{1}{2}$ ≈ -0.5

Für den Flächeninhalt (immer positiv) gilt also I = 0.5