= ${\left[-4{\left(x-2\right)}^{-1}\right]}_4^6$

= ${\left[-\frac{4}{x-2}\right]}_4^6$

$=$ $-\frac{4}{6-2}+\frac{4}{4-2}$

= $-\frac{4}{4}+\frac{4}{2}$

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

= $-1+2$

= $1$

= ${\left[\frac{5}{3}{\left(2x-4\right)}^{\frac{3}{2}}\right]}_{\frac{13}{2}}^{10}$

= ${\left[\frac{5}{3}{\left(\sqrt{2x-4}\right)}^3\right]}_{\frac{13}{2}}^{10}$

$=$ $\frac{5}{3}{\left(\sqrt{2\cdot 10-4}\right)}^3-\frac{5}{3}{\left(\sqrt{2\cdot \left(\frac{13}{2}\right)-4}\right)}^3$

= $\frac{5}{3}{\left(\sqrt{20-4}\right)}^3-\frac{5}{3}{\left(\sqrt{13-4}\right)}^3$

= $\frac{5}{3}{\left(\sqrt{16}\right)}^3-\frac{5}{3}{\left(\sqrt{9}\right)}^3$

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

= $\frac{5}{3}\cdot 64-\frac{5}{3}\cdot 27$

= $\frac{320}{3}-45$

= $\frac{320}{3}-\frac{135}{3}$

= $\frac{185}{3}$

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

$=$ $-4u^2+u-\left(-4\cdot 0^2+0\right)$

= $-4u^2+u-\left(-4\cdot 0+0\right)$

= $-4u^2+u-\left(0+0\right)$

= $-4u^2+u+0$

= $-4u^2+u$

Diese Integralfunktion soll ja den Wert $-\mathrm{217,5}$ annehmen, deswegen setzen wir sie gleich :

$-4u^2+u+\mathrm{217,5}$ = 0

eingesetzt in die Mitternachtsformel (a-b-c-Formel):

u_1,2 = $\frac{-1\pm \sqrt{1^2-4·\left(-4\right)·\mathrm{217,5}}}{2\cdot \left(-4\right)}$

u_1,2 = $\frac{-1\pm \sqrt{1+3480}}{-8}$

u_1,2 = $\frac{-1\pm \sqrt{3481}}{-8}$

u₁ = $\frac{-1+\sqrt{3481}}{-8}$ = $\frac{\mathrm{<math\; xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>-</mo><mn>1</mn></mrow>\; </math>\; <mo>+</mo><math\; xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>59</mn></mrow></math>}}{\mathrm{<math\; xmlns="http://www.w3.org/1998/Math/MathML">\; <mrow><mrow><mo>-</mo><mn>8</mn></mrow>\; </mrow>\; </math>}}$ = $\frac{58}{-8}$ = $-\mathrm{7,25}$

u₂ = $\frac{-1-\sqrt{3481}}{-8}$ = $\frac{\mathrm{<math\; xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>-</mo><mn>1</mn></mrow>\; </math>\; <mo>-</mo><math\; xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>59</mn></mrow></math>}}{\mathrm{<math\; xmlns="http://www.w3.org/1998/Math/MathML">\; <mrow><mrow><mo>-</mo><mn>8</mn></mrow>\; </mrow>\; </math>}}$ = $\frac{-60}{-8}$ = $\mathrm{7,5}$

Da u= $-\mathrm{7,25}$ < 0 ist u= $\mathrm{7,5}$ die einzige Lösung.

Wir berechnen den Mittelwert mit der üblichen Formel:

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

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

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

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

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

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

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

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

= ${\left[-\frac{1}{4}{\left(2x-4\right)}^{-2}\right]}_3^u$

= ${\left[-\frac{1}{4{\left(2x-4\right)}^2}\right]}_3^u$

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

= $-\frac{1}{4{\left(2u-4\right)}^2}+\frac{1}{4{\left(6-4\right)}^2}$

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

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

= $-\frac{1}{4{\left(2u-4\right)}^2}+\frac{1}{16}$

Für u → ∞ gilt: A(u) = $-\frac{1}{4{\left(2u-4\right)}^2}+\frac{1}{16}$ → $0+\frac{1}{16}$ = $\frac{1}{16}$ ≈ 0.063

Für den Flächeninhalt (immer positiv) gilt also I = 0.063