= ${\left[e^{3x-4}\right]}_2^4$

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

= $e^{12-4}-e^{6-4}$

= $e^8-e^2$

= ${\left[\frac{2}{9}{\left(3x-6\right)}^{\frac{3}{2}}\right]}_5^{\frac{22}{3}}$

= ${\left[\frac{2}{9}{\left(\sqrt{3x-6}\right)}^3\right]}_5^{\frac{22}{3}}$

$=$ $\frac{2}{9}{\left(\sqrt{3\cdot \left(\frac{22}{3}\right)-6}\right)}^3-\frac{2}{9}{\left(\sqrt{3\cdot 5-6}\right)}^3$

= $\frac{2}{9}{\left(\sqrt{22-6}\right)}^3-\frac{2}{9}{\left(\sqrt{15-6}\right)}^3$

= $\frac{2}{9}{\left(\sqrt{16}\right)}^3-\frac{2}{9}{\left(\sqrt{9}\right)}^3$

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

= $\frac{2}{9}\cdot 64-\frac{2}{9}\cdot 27$

= $\frac{128}{9}-6$

= $\frac{128}{9}-\frac{54}{9}$

= $\frac{74}{9}$

= ${\left[4x^2+x\right]}_0^u$

$=$ $4u^2+u-\left(4\cdot 0^2+0\right)$

= $4u^2+u-\left(4\cdot 0+0\right)$

= $4u^2+u-\left(0+0\right)$

= $4u^2+u+0$

= $4u^2+u$

Diese Integralfunktion soll ja den Wert $\mathrm{52,5}$ annehmen, deswegen setzen wir sie gleich :

$4u^2+u-\mathrm{52,5}$ = 0

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

u_1,2 = $\frac{-1\pm \sqrt{1^2-4·4·\left(-\mathrm{52,5}\right)}}{2\cdot 4}$

u_1,2 = $\frac{-1\pm \sqrt{1+840}}{8}$

u_1,2 = $\frac{-1\pm \sqrt{841}}{8}$

u₁ = $\frac{-1+\sqrt{841}}{8}$ = $\frac{\mathrm{<math\; xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>-</mo><mn>1</mn></mrow>\; </math>\; <mo>+</mo><math\; xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>29</mn></mrow></math>}}{\mathrm{<math\; xmlns="http://www.w3.org/1998/Math/MathML">\; <mrow><mrow><mn>8</mn></mrow>\; </mrow>\; </math>}}$ = $\frac{28}{8}$ = $\mathrm{3,5}$

u₂ = $\frac{-1-\sqrt{841}}{8}$ = $\frac{\mathrm{<math\; xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>-</mo><mn>1</mn></mrow>\; </math>\; <mo>-</mo><math\; xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>29</mn></mrow></math>}}{\mathrm{<math\; xmlns="http://www.w3.org/1998/Math/MathML">\; <mrow><mrow><mn>8</mn></mrow>\; </mrow>\; </math>}}$ = $\frac{-30}{8}$ = $-\mathrm{3,75}$

Da u= $-\mathrm{3,75}$ < 0 ist u= $\mathrm{3,5}$ die einzige Lösung.

Wir berechnen den Mittelwert mit der üblichen Formel:

= $\frac{2}{\pi }$ ${\left[\sin \left(2x+\frac{1}{2}\pi \right)\right]}_{\frac{1}{2}\pi }^{\pi }$

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

= $\frac{2}{\pi }·\left(\sin \left(\frac{5}{2}\pi \right)-\sin \left(\frac{3}{2}\pi \right)\right)$

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

= $\frac{2}{\pi }·\left(1+1\right)$

= $\frac{2}{\pi }·2$

= $\frac{4}{\pi }$

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

= ${\left[-2\sqrt{x-1}\right]}_u^{10}$

$=$ $-2\sqrt{10-1}+2\sqrt{u-1}$

= $-2\sqrt{9}+2\sqrt{u-1}$

= $-2\cdot 3+2\sqrt{u-1}$

= $-6+2\sqrt{u-1}$

Für u → 1 (u>1, also von rechts) gilt: A(u) = $2\sqrt{u-1}-6$ → $0-6$ = $-6$ ≈ -6

Für den Flächeninhalt (immer positiv) gilt also I = 6