= ${\left[-\frac{5}{2}{\left(2x-2\right)}^{-1}\right]}_2^4$

= ${\left[-\frac{5}{2\left(2x-2\right)}\right]}_2^4$

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

= $-\frac{5}{2\left(8-2\right)}+\frac{5}{2\left(4-2\right)}$

= $-\frac{5}{2\cdot 6}+\frac{5}{2\cdot 2}$

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

= $-\frac{5}{12}+\frac{5}{4}$

= $-\frac{5}{12}+\frac{15}{12}$

= $\frac{5}{6}$

= ${\left[\frac{3}{2}{e}^{2x-4}\right]}_2^5$

$=$ $\frac{3}{2}{e}^{2\cdot 5-4}-\frac{3}{2}{e}^{2\cdot 2-4}$

= $\frac{3}{2}{e}^{10-4}-\frac{3}{2}{e}^{4-4}$

= $\frac{3}{2}{e}^6-\frac{3}{2}{e}^0$

= $\frac{3}{2}{e}^6-\frac{3}{2}$

= ${\left[-4e^{-\mathrm{0,5}x+\mathrm{0,2}}\right]}_0^u$

$=$ $-4e^{-\mathrm{0,5}u+\mathrm{0,2}}+4e^{-\mathrm{0,5}\cdot 0+\mathrm{0,2}}$

= $-4e^{-\mathrm{0,5}u+\mathrm{0,2}}+4e^{0+\mathrm{0,2}}$

= $-4e^{-\mathrm{0,5}u+\mathrm{0,2}}+4e^{\mathrm{0,2}}$

Diese Integralfunktion soll ja den Wert $4$ annehmen, deswegen setzen wir sie gleich :

Wir berechnen den Mittelwert mit der üblichen Formel:

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

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

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

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

= $\frac{2}{\pi }·\left(0+0\right)$

= $\frac{2}{\pi }·0$

= 0

= ${\left[2{\left(x-3\right)}^{\frac{1}{2}}\right]}_u^{19}$

= ${\left[2\sqrt{x-3}\right]}_u^{19}$

$=$ $2\sqrt{19-3}-2\sqrt{u-3}$

= $2\sqrt{16}-2\sqrt{u-3}$

= $2\cdot 4-2\sqrt{u-3}$

= $8-2\sqrt{u-3}$

Für u → 3 (u>3, also von rechts) gilt: A(u) = $-2\sqrt{u-3}+8$ → $0+8$ = $8$ ≈ 8

Für den Flächeninhalt (immer positiv) gilt also I = 8