= ${\left[-{\left(x-1\right)}^{-3}\right]}_3^6$

= ${\left[-\frac{1}{{\left(x-1\right)}^3}\right]}_3^6$

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

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

= $-\left(\frac{1}{125}\right)+\frac{1}{8}$

= $\frac{117}{1000}$

= ${\left[2{\left(2x-2\right)}^{\frac{3}{2}}\right]}_{\frac{11}{2}}^{\frac{27}{2}}$

= ${\left[2{\left(\sqrt{2x-2}\right)}^3\right]}_{\frac{11}{2}}^{\frac{27}{2}}$

$=$ $2{\left(\sqrt{2\cdot \left(\frac{27}{2}\right)-2}\right)}^3-2{\left(\sqrt{2\cdot \left(\frac{11}{2}\right)-2}\right)}^3$

= $2{\left(\sqrt{27-2}\right)}^3-2{\left(\sqrt{11-2}\right)}^3$

= $2{\left(\sqrt{25}\right)}^3-2{\left(\sqrt{9}\right)}^3$

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

= $2\cdot 125-2\cdot 27$

= $250-54$

= $196$

= ${\left[-8e^{-\mathrm{0,4}x+\mathrm{0,9}}\right]}_0^u$

$=$ $-8e^{-\mathrm{0,4}u+\mathrm{0,9}}+8e^{-\mathrm{0,4}\cdot 0+\mathrm{0,9}}$

= $-8e^{-\mathrm{0,4}u+\mathrm{0,9}}+8e^{0+\mathrm{0,9}}$

= $-8e^{-\mathrm{0,4}u+\mathrm{0,9}}+8e^{\mathrm{0,9}}$

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

Wir berechnen den Mittelwert mit der üblichen Formel:

= $\frac{1}{2}$ ${\left[{\left(x-2\right)}^3+3x\right]}_2^4$

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

= $\frac{1}{2}(2^3+12-\left(0^3+6\right))$

= $\frac{1}{2}(8+12-\left(0+6\right))$

= $\frac{1}{2}\left(8+12-6\right)$

= $\frac{1}{2}·14$

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

= ${\left[\frac{2}{\sqrt{2x-1}}\right]}_5^u$

$=$ $\frac{2}{\sqrt{2u-1}}-\frac{2}{\sqrt{2\cdot 5-1}}$

= $\frac{2}{\sqrt{2u-1}}-\frac{2}{\sqrt{10-1}}$

= $\frac{2}{\sqrt{2u-1}}-\frac{2}{\sqrt{9}}$

= $\frac{2}{\sqrt{2u-1}}-\frac{2}{3}$

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

= $\frac{2}{\sqrt{2u-1}}-\frac{2}{3}$

Für u → ∞ gilt: A(u) = $\frac{2}{\sqrt{2u-1}}-\frac{2}{3}$ → $0-\frac{2}{3}$ = $-\frac{2}{3}$ ≈ -0.667

Für den Flächeninhalt (immer positiv) gilt also I = 0.667