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

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

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

= $-\frac{3}{2^2}+\frac{3}{1^2}$

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

= $-\frac{3}{4}+3$

= $-\frac{3}{4}+\frac{12}{4}$

= $\frac{9}{4}$

= ${\left[4{\left(x-3\right)}^{\frac{3}{2}}\right]}_{19}^{28}$

= ${\left[4{\left(\sqrt{x-3}\right)}^3\right]}_{19}^{28}$

$=$ $4{\left(\sqrt{28-3}\right)}^3-4{\left(\sqrt{19-3}\right)}^3$

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

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

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

= $500-256$

= $244$

= ${\left[-7e^{-\mathrm{0,1}x+\mathrm{0,2}}\right]}_0^u$

$=$ $-7e^{-\mathrm{0,1}u+\mathrm{0,2}}+7e^{-\mathrm{0,1}\cdot 0+\mathrm{0,2}}$

= $-7e^{-\mathrm{0,1}u+\mathrm{0,2}}+7e^{0+\mathrm{0,2}}$

= $-7e^{-\mathrm{0,1}u+\mathrm{0,2}}+7e^{\mathrm{0,2}}$

Diese Integralfunktion soll ja den Wert $6$ annehmen, deswegen setzen wir sie gleich :

Wir berechnen den Mittelwert mit der üblichen Formel:

= $\frac{2}{\pi }$ ${\left[\frac{5}{3}\cdot \sin (3x+\pi )\right]}_{\frac{1}{2}\pi }^{\pi }$

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

= $\frac{2}{\pi }·\left(\frac{5}{3}\cdot \sin \left(4\pi \right)-\frac{5}{3}\cdot \sin \left(\frac{5}{2}\pi \right)\right)$

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

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

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

= $\frac{2}{\pi }·\left(-\frac{5}{3}\right)$

= $-\frac{10}{3\pi }$

= ${\left[-\frac{1}{2}{e}^{-2x+4}\right]}_0^u$

$=$ $-\frac{1}{2}{e}^{-2u+4}+\frac{1}{2}{e}^{-2\cdot 0+4}$

= $-\frac{1}{2}{e}^{-2u+4}+\frac{1}{2}{e}^{0+4}$

= $-\frac{1}{2}{e}^{-2u+4}+\frac{1}{2}{e}^4$

Für u → ∞ gilt: A(u) = $-\frac{1}{2}{e}^{-2u+4}+\frac{1}{2}{e}^4$ → $0+\frac{1}{2}{e}^4$ = $\frac{1}{2}{e}^4$ ≈ 27.299

Für den Flächeninhalt (immer positiv) gilt also I = 27.299