= ${\left[\frac{5}{3}{\left(2x-4\right)}^{\frac{3}{2}}\right]}_{\frac{13}{2}}^{10}$

= ${\left[\frac{5}{3}{\left(\sqrt{2x-4}\right)}^3\right]}_{\frac{13}{2}}^{10}$

$=$ $\frac{5}{3}{\left(\sqrt{2\cdot 10-4}\right)}^3-\frac{5}{3}{\left(\sqrt{2\cdot \left(\frac{13}{2}\right)-4}\right)}^3$

= $\frac{5}{3}{\left(\sqrt{20-4}\right)}^3-\frac{5}{3}{\left(\sqrt{13-4}\right)}^3$

= $\frac{5}{3}{\left(\sqrt{16}\right)}^3-\frac{5}{3}{\left(\sqrt{9}\right)}^3$

= $\frac{5}{3}\cdot 4^3-\frac{5}{3}\cdot 3^3$

= $\frac{5}{3}\cdot 64-\frac{5}{3}\cdot 27$

= $\frac{320}{3}-45$

= $\frac{320}{3}-\frac{135}{3}$

= $\frac{185}{3}$

= ${\left[e^{2x-3}\right]}_2^4$

$=$ $e^{2\cdot 4-3}-e^{2\cdot 2-3}$

= $e^{8-3}-e^{4-3}$

= $e^5-e^1$

= $e^5-e$

= ${\left[-4x^2\right]}_2^u$

$=$ $-4u^2+4\cdot 2^2$

= $-4u^2+4\cdot 4$

= $-4u^2+16$

Diese Integralfunktion soll ja den Wert $-9$ annehmen, deswegen setzen wir sie gleich :

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

Wir berechnen den Mittelwert mit der üblichen Formel:

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

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

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

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

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

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

= $5+\frac{2}{3}$

= $\frac{15}{3}+\frac{2}{3}$

= $\frac{17}{3}$

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

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

$=$ $\frac{1}{2u-3}-\frac{1}{2\cdot 2-3}$

= $\frac{1}{2u-3}-\frac{1}{4-3}$

= $\frac{1}{2u-3}-\frac{1}{1}$

= $\frac{1}{2u-3}-1$

Für u → ∞ gilt: A(u) = $\frac{1}{2u-3}-1$ → $0-1$ = $-1$ ≈ -1

Für den Flächeninhalt (immer positiv) gilt also I = 1