= ${\left[{\left(2x-5\right)}^{\frac{3}{2}}\right]}_7^{\frac{21}{2}}$

= ${\left[{\left(\sqrt{2x-5}\right)}^3\right]}_7^{\frac{21}{2}}$

$=$ ${\left(\sqrt{2\cdot \left(\frac{21}{2}\right)-5}\right)}^3-{\left(\sqrt{2\cdot 7-5}\right)}^3$

= ${\left(\sqrt{21-5}\right)}^3-{\left(\sqrt{14-5}\right)}^3$

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

= $4^3-3^3$

= $64-27$

= $37$

= ${\left[\frac{5}{2}{e}^{2x-4}\right]}_1^4$

$=$ $\frac{5}{2}{e}^{2\cdot 4-4}-\frac{5}{2}{e}^{2\cdot 1-4}$

= $\frac{5}{2}{e}^{8-4}-\frac{5}{2}{e}^{2-4}$

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

= ${\left[-4x^2+4x\right]}_2^u$

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

= $-4u^2+4u-\left(-4\cdot 4+8\right)$

= $-4u^2+4u-\left(-16+8\right)$

= $-4u^2+4u-1·\left(-8\right)$

= $-4u^2+4u+8$

Diese Integralfunktion soll ja den Wert $-216$ 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:

= $\frac{1}{\pi }$ ${\left[-\frac{1}{2}\cdot \cos (2x-\pi )\right]}_0^{\pi }$

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

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

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

= $\frac{1}{\pi }·\left(\frac{1}{2}-\frac{1}{2}\right)$

= $\frac{1}{\pi }·\left(\mathrm{0,5}-\mathrm{0,5}\right)$

= $\frac{1}{\pi }·0$

= 0

= ${\left[e^{3x-3}\right]}_1^u$

$=$ $e^{3u-3}-e^{3\cdot 1-3}$

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

= $e^{3u-3}-e^0$

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

Für u → ∞ gilt: A(u) = $e^{3u-3}-1$ → $\infty$