= ${\left[2e^{3x-3}\right]}_0^1$

$=$ $2e^{3\cdot 1-3}-2e^{3\cdot 0-3}$

= $2e^{3-3}-2e^{0-3}$

= $2e^0-2e^{-3}$

= $2-2e^{-3}$

= ${\left[-2{\left(2x-5\right)}^{-1}\right]}_3^6$

= ${\left[-\frac{2}{2x-5}\right]}_3^6$

$=$ $-\frac{2}{2\cdot 6-5}+\frac{2}{2\cdot 3-5}$

= $-\frac{2}{12-5}+\frac{2}{6-5}$

= $-\frac{2}{7}+\frac{2}{1}$

= $-2\cdot \left(\frac{1}{7}\right)+2\cdot 1$

= $-\frac{2}{7}+2$

= $-\frac{2}{7}+\frac{14}{7}$

= $\frac{12}{7}$

= ${\left[3e^{\mathrm{0,8}x-\mathrm{0,2}}\right]}_0^u$

$=$ $3e^{\mathrm{0,8}u-\mathrm{0,2}}-3e^{\mathrm{0,8}\cdot 0-\mathrm{0,2}}$

= $3e^{\mathrm{0,8}u-\mathrm{0,2}}-3e^{0-\mathrm{0,2}}$

= $3e^{\mathrm{0,8}u-\mathrm{0,2}}-3e^{-\mathrm{0,2}}$

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

Wir berechnen den Mittelwert mit der üblichen Formel:

= $\frac{2}{\pi }$ ${\left[\sin (x-\pi )\right]}_{\frac{1}{2}\pi }^{\pi }$

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

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

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

= $\frac{2}{\pi }·\left(0+1\right)$

= $\frac{2}{\pi }·1$

= $\frac{2}{\pi }$

= ${\left[-2{\left(x-2\right)}^{-1}\right]}_3^u$

= ${\left[-\frac{2}{x-2}\right]}_3^u$

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

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

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

= $-\frac{2}{u-2}+2$

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

Für den Flächeninhalt (immer positiv) gilt also I = 2