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

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

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

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

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

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

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

= $\frac{122}{3}$

= ${\left[3e^{2x-5}\right]}_0^3$

$=$ $3e^{2\cdot 3-5}-3e^{2\cdot 0-5}$

= $3e^{6-5}-3e^{0-5}$

= $3e^1-3e^{-5}$

= $3e-3e^{-5}$

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

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

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

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

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

= $-2u^2+2u$

Diese Integralfunktion soll ja den Wert $-\mathrm{49,5}$ annehmen, deswegen setzen wir sie gleich :

$-2u^2+2u+\mathrm{49,5}$ = 0

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

u_1,2 = $\frac{-2\pm \sqrt{2^2-4·\left(-2\right)·\mathrm{49,5}}}{2\cdot \left(-2\right)}$

u_1,2 = $\frac{-2\pm \sqrt{4+396}}{-4}$

u_1,2 = $\frac{-2\pm \sqrt{400}}{-4}$

u₁ = $\frac{-2+\sqrt{400}}{-4}$ = $\frac{\mathrm{<math\; xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>-</mo><mn>2</mn></mrow>\; </math>\; <mo>+</mo><math\; xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>20</mn></mrow></math>}}{\mathrm{<math\; xmlns="http://www.w3.org/1998/Math/MathML">\; <mrow><mrow><mo>-</mo><mn>4</mn></mrow>\; </mrow>\; </math>}}$ = $\frac{18}{-4}$ = $-\mathrm{4,5}$

u₂ = $\frac{-2-\sqrt{400}}{-4}$ = $\frac{\mathrm{<math\; xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>-</mo><mn>2</mn></mrow>\; </math>\; <mo>-</mo><math\; xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>20</mn></mrow></math>}}{\mathrm{<math\; xmlns="http://www.w3.org/1998/Math/MathML">\; <mrow><mrow><mo>-</mo><mn>4</mn></mrow>\; </mrow>\; </math>}}$ = $\frac{-22}{-4}$ = $\mathrm{5,5}$

Da u= $-\mathrm{4,5}$ < 0 ist u= $\mathrm{5,5}$ die einzige Lösung.

Wir berechnen den Mittelwert mit der üblichen Formel:

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

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

= $\frac{1}{\pi }·\left(-\frac{2}{3}\cdot \cos \left(\frac{7}{2}\pi \right)+\frac{2}{3}\cdot \cos \left(\frac{1}{2}\pi \right)\right)$

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

= $\frac{1}{\pi }·\left(0+0\right)$

= $\frac{1}{\pi }·0$

= 0

= ${\left[\frac{1}{6}{\left(3x-3\right)}^{-2}\right]}_3^u$

= ${\left[\frac{1}{6{\left(3x-3\right)}^2}\right]}_3^u$

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

= $\frac{1}{6{\left(3u-3\right)}^2}-\frac{1}{6{\left(9-3\right)}^2}$

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

= $\frac{1}{6{\left(3u-3\right)}^2}-\frac{1}{6}\cdot \left(\frac{1}{36}\right)$

= $\frac{1}{6{\left(3u-3\right)}^2}-\frac{1}{216}$

Für u → ∞ gilt: A(u) = $\frac{1}{6{\left(3u-3\right)}^2}-\frac{1}{216}$ → $0-\frac{1}{216}$ = $-\frac{1}{216}$ ≈ -0.005

Für den Flächeninhalt (immer positiv) gilt also I = 0.005