= ${\left[e^{x-3}\right]}_1^4$

$=$ $e^{4-3}-e^{1-3}$

= $e^1-e^{-2}$

= $e-e^{-2}$

= ${\left[-3{\left(2x-1\right)}^{-1}\right]}_2^5$

= ${\left[-\frac{3}{2x-1}\right]}_2^5$

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

= $-\frac{3}{10-1}+\frac{3}{4-1}$

= $-\frac{3}{9}+\frac{3}{3}$

= $-3\cdot \left(\frac{1}{9}\right)+3\cdot \left(\frac{1}{3}\right)$

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

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

= $\frac{2}{3}$

= ${\left[6e^{\mathrm{0,3}x-\mathrm{0,7}}\right]}_0^u$

$=$ $6e^{\mathrm{0,3}u-\mathrm{0,7}}-6e^{\mathrm{0,3}\cdot 0-\mathrm{0,7}}$

= $6e^{\mathrm{0,3}u-\mathrm{0,7}}-6e^{0-\mathrm{0,7}}$

= $6e^{\mathrm{0,3}u-\mathrm{0,7}}-6e^{-\mathrm{0,7}}$

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

Wir berechnen den Mittelwert mit der üblichen Formel:

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

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

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

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

= $\frac{2}{\pi }·\left(0-2\right)$

= $\frac{2}{\pi }·\left(-2\right)$

= $-\frac{4}{\pi }$

= ${\left[-\frac{1}{3}{\left(-3x+3\right)}^{-2}\right]}_3^u$

= ${\left[-\frac{1}{3{\left(-3x+3\right)}^2}\right]}_3^u$

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

= $-\frac{1}{3{\left(-3u+3\right)}^2}+\frac{1}{3{\left(-9+3\right)}^2}$

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

= $-\frac{1}{3{\left(-3u+3\right)}^2}+\frac{1}{3}\cdot \left(\frac{1}{36}\right)$

= $-\frac{1}{3{\left(-3u+3\right)}^2}+\frac{1}{108}$

Für u → ∞ gilt: A(u) = $-\frac{1}{3{\left(-3u+3\right)}^2}+\frac{1}{108}$ → $0+\frac{1}{108}$ = $\frac{1}{108}$ ≈ 0.009

Für den Flächeninhalt (immer positiv) gilt also I = 0.009