= ${\left[4{\left(x-1\right)}^{\frac{3}{2}}\right]}_{10}^{26}$

= ${\left[4{\left(\sqrt{x-1}\right)}^3\right]}_{10}^{26}$

$=$ $4{\left(\sqrt{26-1}\right)}^3-4{\left(\sqrt{10-1}\right)}^3$

= $4{\left(\sqrt{25}\right)}^3-4{\left(\sqrt{9}\right)}^3$

= $4\cdot 5^3-4\cdot 3^3$

= $4\cdot 125-4\cdot 27$

= $500-108$

= $392$

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

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

= $\frac{3}{2}{e}^{4-5}-\frac{3}{2}{e}^{0-5}$

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

= ${\left[-3x^2+3x\right]}_2^u$

$=$ $-3u^2+3u-\left(-3\cdot 2^2+3\cdot 2\right)$

= $-3u^2+3u-\left(-3\cdot 4+6\right)$

= $-3u^2+3u-\left(-12+6\right)$

= $-3u^2+3u-1·\left(-6\right)$

= $-3u^2+3u+6$

Diese Integralfunktion soll ja den Wert $-162$ annehmen, deswegen setzen wir sie gleich :

$-u^2+u+56$ = 0

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

u_1,2 = $\frac{-1\pm \sqrt{1^2-4·\left(-1\right)·56}}{2\cdot \left(-1\right)}$

u_1,2 = $\frac{-1\pm \sqrt{1+224}}{-2}$

u_1,2 = $\frac{-1\pm \sqrt{225}}{-2}$

u₁ = $\frac{-1+\sqrt{225}}{-2}$ = $\frac{\mathrm{<math\; xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>-</mo><mn>1</mn></mrow>\; </math>\; <mo>+</mo><math\; xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>15</mn></mrow></math>}}{\mathrm{<math\; xmlns="http://www.w3.org/1998/Math/MathML">\; <mrow><mrow><mo>-</mo><mn>2</mn></mrow>\; </mrow>\; </math>}}$ = $\frac{14}{-2}$ = $-7$

u₂ = $\frac{-1-\sqrt{225}}{-2}$ = $\frac{\mathrm{<math\; xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>-</mo><mn>1</mn></mrow>\; </math>\; <mo>-</mo><math\; xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>15</mn></mrow></math>}}{\mathrm{<math\; xmlns="http://www.w3.org/1998/Math/MathML">\; <mrow><mrow><mo>-</mo><mn>2</mn></mrow>\; </mrow>\; </math>}}$ = $\frac{-16}{-2}$ = $8$

Da u= $-7$ < 2 ist u= $8$ die einzige Lösung.

Wir berechnen den Mittelwert mit der üblichen Formel:

= $\frac{1}{16}$ ${\left[\frac{4}{3}{\left(x-1\right)}^{\frac{3}{2}}\right]}_{10}^{26}$

= $\frac{1}{16}$ ${\left[\frac{4}{3}{\left(\sqrt{x-1}\right)}^3\right]}_{10}^{26}$

$=$ $\frac{1}{16}\left(\frac{4}{3}{\left(\sqrt{26-1}\right)}^3-\frac{4}{3}{\left(\sqrt{10-1}\right)}^3\right)$

= $\frac{1}{16}\left(\frac{4}{3}{\left(\sqrt{25}\right)}^3-\frac{4}{3}{\left(\sqrt{9}\right)}^3\right)$

= $\frac{1}{16}\left(\frac{4}{3}\cdot 5^3-\frac{4}{3}\cdot 3^3\right)$

= $\frac{1}{16}\left(\frac{4}{3}\cdot 125-\frac{4}{3}\cdot 27\right)$

= $\frac{1}{16}\left(\frac{500}{3}-36\right)$

= $\frac{1}{16}\left(\frac{500}{3}-\frac{108}{3}\right)$

= $\frac{1}{16}·\frac{392}{3}$

= $\frac{49}{6}$

= ${\left[-2{\left(2x-6\right)}^{\frac{1}{2}}\right]}_u^5$

= ${\left[-2\sqrt{2x-6}\right]}_u^5$

$=$ $-2\sqrt{2\cdot 5-6}+2\sqrt{2u-6}$

= $-2\sqrt{10-6}+2\sqrt{2u-6}$

= $-2\sqrt{4}+2\sqrt{2u-6}$

= $-2\cdot 2+2\sqrt{2u-6}$

= $-4+2\sqrt{2u-6}$

Für u → 3 (u>3, also von rechts) gilt: A(u) = $2\sqrt{2u-6}-4$ → $0-4$ = $-4$ ≈ -4

Für den Flächeninhalt (immer positiv) gilt also I = 4