= ${\left[\frac{2}{3}{\left(x-2\right)}^{\frac{3}{2}}\right]}_{18}^{27}$

= ${\left[\frac{2}{3}{\left(\sqrt{x-2}\right)}^3\right]}_{18}^{27}$

$=$ $\frac{2}{3}{\left(\sqrt{27-2}\right)}^3-\frac{2}{3}{\left(\sqrt{18-2}\right)}^3$

= $\frac{2}{3}{\left(\sqrt{25}\right)}^3-\frac{2}{3}{\left(\sqrt{16}\right)}^3$

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

= $\frac{2}{3}\cdot 125-\frac{2}{3}\cdot 64$

= $\frac{250}{3}-\frac{128}{3}$

= $\frac{122}{3}$

= ${\left[\frac{8}{9}{\left(3x-3\right)}^{\frac{3}{2}}\right]}_{\frac{7}{3}}^{\frac{19}{3}}$

= ${\left[\frac{8}{9}{\left(\sqrt{3x-3}\right)}^3\right]}_{\frac{7}{3}}^{\frac{19}{3}}$

$=$ $\frac{8}{9}{\left(\sqrt{3\cdot \left(\frac{19}{3}\right)-3}\right)}^3-\frac{8}{9}{\left(\sqrt{3\cdot \left(\frac{7}{3}\right)-3}\right)}^3$

= $\frac{8}{9}{\left(\sqrt{19-3}\right)}^3-\frac{8}{9}{\left(\sqrt{7-3}\right)}^3$

= $\frac{8}{9}{\left(\sqrt{16}\right)}^3-\frac{8}{9}{\left(\sqrt{4}\right)}^3$

= $\frac{8}{9}\cdot 4^3-\frac{8}{9}\cdot 2^3$

= $\frac{8}{9}\cdot 64-\frac{8}{9}\cdot 8$

= $\frac{512}{9}-\frac{64}{9}$

= $\frac{448}{9}$

= ${\left[-4x^2-2x\right]}_1^u$

$=$ $-4u^2-2u-\left(-4\cdot 1^2-2\cdot 1\right)$

= $-4u^2-2u-\left(-4\cdot 1-2\right)$

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

= $-4u^2-2u-1·\left(-6\right)$

= $-4u^2-2u+6$

Diese Integralfunktion soll ja den Wert $-176$ annehmen, deswegen setzen wir sie gleich :

$-2u^2-u+91$ = 0

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

u_1,2 = $\frac{+1\pm \sqrt{{\left(-1\right)}^2-4·\left(-2\right)·91}}{2\cdot \left(-2\right)}$

u_1,2 = $\frac{+1\pm \sqrt{1+728}}{-4}$

u_1,2 = $\frac{+1\pm \sqrt{729}}{-4}$

u₁ = $\frac{1+\sqrt{729}}{-4}$ = $\frac{\mathrm{<math\; xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>1</mn></mrow>\; </math>\; <mo>+</mo><math\; xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>27</mn></mrow></math>}}{\mathrm{<math\; xmlns="http://www.w3.org/1998/Math/MathML">\; <mrow><mrow><mo>-</mo><mn>4</mn></mrow>\; </mrow>\; </math>}}$ = $\frac{28}{-4}$ = $-7$

u₂ = $\frac{1-\sqrt{729}}{-4}$ = $\frac{\mathrm{<math\; xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>1</mn></mrow>\; </math>\; <mo>-</mo><math\; xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>27</mn></mrow></math>}}{\mathrm{<math\; xmlns="http://www.w3.org/1998/Math/MathML">\; <mrow><mrow><mo>-</mo><mn>4</mn></mrow>\; </mrow>\; </math>}}$ = $\frac{-26}{-4}$ = $\mathrm{6,5}$

Da u= $-7$ < 1 ist u= $\mathrm{6,5}$ die einzige Lösung.

Wir berechnen den Mittelwert mit der üblichen Formel:

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

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

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

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

= $0+3-\left(-\frac{1}{3}+0\right)$

= $3-\left(-\frac{1}{3}+0\right)$

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

= $\frac{9}{3}+\frac{1}{3}$

= $\frac{10}{3}$

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

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

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

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

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

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

Für u → ∞ gilt: A(u) = $-\frac{3}{2{\left(-u+2\right)}^2}+\frac{3}{2}$ → $0+\frac{3}{2}$ = $\frac{3}{2}$ ≈ 1.5

Für den Flächeninhalt (immer positiv) gilt also I = 1.5