= ${\left[4{\left(x-1\right)}^{\frac{3}{2}}\right]}_5^{26}$

= ${\left[4{\left(\sqrt{x-1}\right)}^3\right]}_5^{26}$

$=$ $4{\left(\sqrt{26-1}\right)}^3-4{\left(\sqrt{5-1}\right)}^3$

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

= $4\cdot 5^3-4\cdot 2^3$

= $4\cdot 125-4\cdot 8$

= $500-32$

= $468$

= ${\left[\frac{4}{3}{\left(3x-5\right)}^{\frac{3}{2}}\right]}_7^{10}$

= ${\left[\frac{4}{3}{\left(\sqrt{3x-5}\right)}^3\right]}_7^{10}$

$=$ $\frac{4}{3}{\left(\sqrt{3\cdot 10-5}\right)}^3-\frac{4}{3}{\left(\sqrt{3\cdot 7-5}\right)}^3$

= $\frac{4}{3}{\left(\sqrt{30-5}\right)}^3-\frac{4}{3}{\left(\sqrt{21-5}\right)}^3$

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

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

= $\frac{4}{3}\cdot 125-\frac{4}{3}\cdot 64$

= $\frac{500}{3}-\frac{256}{3}$

= $\frac{244}{3}$

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

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

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

= $-3u^2+0$

= $-3u^2$

Diese Integralfunktion soll ja den Wert $-\mathrm{18,75}$ 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:

= $\frac{1}{2}$ ${\left[\frac{5}{6}{\left(2x-4\right)}^3\right]}_1^3$

$=$ $\frac{1}{2}\left(\frac{5}{6}\cdot {\left(2\cdot 3-4\right)}^3-\frac{5}{6}\cdot {\left(2\cdot 1-4\right)}^3\right)$

= $\frac{1}{2}\left(\frac{5}{6}\cdot {\left(6-4\right)}^3-\frac{5}{6}\cdot {\left(2-4\right)}^3\right)$

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

= $\frac{1}{2}\left(\frac{5}{6}\cdot 8-\frac{5}{6}\cdot \left(-8\right)\right)$

= $\frac{1}{2}\left(\frac{20}{3}+\frac{20}{3}\right)$

= $\frac{1}{2}·\frac{40}{3}$

= $\frac{20}{3}$

= ${\left[\frac{2}{3}{e}^{-3x+5}\right]}_1^u$

$=$ $\frac{2}{3}{e}^{-3u+5}-\frac{2}{3}{e}^{-3\cdot 1+5}$

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

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

Für u → ∞ gilt: A(u) = $\frac{2}{3}{e}^{-3u+5}-\frac{2}{3}{e}^2$ → $0-\frac{2}{3}{e}^2$ = $-\frac{2}{3}{e}^2$ ≈ -4.926

Für den Flächeninhalt (immer positiv) gilt also I = 4.926