= ${\left[4e^{x-3}\right]}_0^2$

$=$ $4e^{2-3}-4e^{0-3}$

= $4e^{-1}-4e^{-3}$

= ${\left[\frac{4}{3}{\left(x-1\right)}^{\frac{3}{2}}\right]}_5^{10}$

= ${\left[\frac{4}{3}{\left(\sqrt{x-1}\right)}^3\right]}_5^{10}$

$=$ $\frac{4}{3}{\left(\sqrt{10-1}\right)}^3-\frac{4}{3}{\left(\sqrt{5-1}\right)}^3$

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

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

= $\frac{4}{3}\cdot 27-\frac{4}{3}\cdot 8$

= $36-\frac{32}{3}$

= $\frac{108}{3}-\frac{32}{3}$

= $\frac{76}{3}$

= ${\left[x^2-2x\right]}_3^u$

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

= $u^2-2u-\left(9-6\right)$

= $u^2-2u-9+6$

= $u^2-2u-3$

Diese Integralfunktion soll ja den Wert $\mathrm{68,25}$ annehmen, deswegen setzen wir sie gleich :

$u^2-2u-\mathrm{71,25}$ = 0

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

u_1,2 = $\frac{+2\pm \sqrt{{\left(-2\right)}^2-4·1·\left(-\mathrm{71,25}\right)}}{2\cdot 1}$

u_1,2 = $\frac{+2\pm \sqrt{4+285}}{2}$

u_1,2 = $\frac{+2\pm \sqrt{289}}{2}$

u₁ = $\frac{2+\sqrt{289}}{2}$ = $\frac{\mathrm{<math\; xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>2</mn></mrow>\; </math>\; <mo>+</mo><math\; xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>17</mn></mrow></math>}}{\mathrm{<math\; xmlns="http://www.w3.org/1998/Math/MathML">\; <mrow><mrow><mn>2</mn></mrow>\; </mrow>\; </math>}}$ = $\frac{19}{2}$ = $\mathrm{9,5}$

u₂ = $\frac{2-\sqrt{289}}{2}$ = $\frac{\mathrm{<math\; xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>2</mn></mrow>\; </math>\; <mo>-</mo><math\; xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>17</mn></mrow></math>}}{\mathrm{<math\; xmlns="http://www.w3.org/1998/Math/MathML">\; <mrow><mrow><mn>2</mn></mrow>\; </mrow>\; </math>}}$ = $\frac{-15}{2}$ = $-\mathrm{7,5}$

Da u= $-\mathrm{7,5}$ < 3 ist u= $\mathrm{9,5}$ die einzige Lösung.

Wir berechnen den Mittelwert mit der üblichen Formel:

= $\frac{1}{9}$ ${\left[\frac{8}{3}{\left(x-1\right)}^{\frac{3}{2}}\right]}_{17}^{26}$

= $\frac{1}{9}$ ${\left[\frac{8}{3}{\left(\sqrt{x-1}\right)}^3\right]}_{17}^{26}$

$=$ $\frac{1}{9}\left(\frac{8}{3}{\left(\sqrt{26-1}\right)}^3-\frac{8}{3}{\left(\sqrt{17-1}\right)}^3\right)$

= $\frac{1}{9}\left(\frac{8}{3}{\left(\sqrt{25}\right)}^3-\frac{8}{3}{\left(\sqrt{16}\right)}^3\right)$

= $\frac{1}{9}\left(\frac{8}{3}\cdot 5^3-\frac{8}{3}\cdot 4^3\right)$

= $\frac{1}{9}\left(\frac{8}{3}\cdot 125-\frac{8}{3}\cdot 64\right)$

= $\frac{1}{9}\left(\frac{1000}{3}-\frac{512}{3}\right)$

= $\frac{1}{9}·\frac{488}{3}$

= $\frac{488}{27}$

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

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

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

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

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

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

= $\frac{3}{8}-\frac{3}{2\left(2u-2\right)}$

Für u → 1 (u>1, also von rechts) gilt: A(u) = $-\frac{3}{2\left(2u-2\right)}+\frac{3}{8}$ → $-\infty$