= ${\left[-\frac{1}{2}{\left(2x-3\right)}^{-1}\right]}_2^4$

= ${\left[-\frac{1}{2\left(2x-3\right)}\right]}_2^4$

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

= $-\frac{1}{2\left(8-3\right)}+\frac{1}{2\left(4-3\right)}$

= $-\frac{1}{2\cdot 5}+\frac{1}{2}$

= $-\frac{1}{2}\cdot \left(\frac{1}{5}\right)+\frac{1}{2}\cdot 1$

= $-\frac{1}{10}+\frac{1}{2}$

= $-\frac{1}{10}+\frac{5}{10}$

= $\frac{2}{5}$

= ${\left[\frac{5}{3}{e}^{3x-7}\right]}_0^2$

$=$ $\frac{5}{3}{e}^{3\cdot 2-7}-\frac{5}{3}{e}^{3\cdot 0-7}$

= $\frac{5}{3}{e}^{6-7}-\frac{5}{3}{e}^{0-7}$

= $\frac{5}{3}{e}^{-1}-\frac{5}{3}{e}^{-7}$

= ${\left[-x^2+3x\right]}_0^u$

$=$ $-u^2+3u-\left(-0^2+3\cdot 0\right)$

= $-u^2+3u-\left(-0+0\right)$

= $-u^2+3u$

Diese Integralfunktion soll ja den Wert $-10$ annehmen, deswegen setzen wir sie gleich :

$-u^2+3u+10$ = 0

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

u_1,2 = $\frac{-3\pm \sqrt{3^2-4·\left(-1\right)·10}}{2\cdot \left(-1\right)}$

u_1,2 = $\frac{-3\pm \sqrt{9+40}}{-2}$

u_1,2 = $\frac{-3\pm \sqrt{49}}{-2}$

u₁ = $\frac{-3+\sqrt{49}}{-2}$ = $\frac{\mathrm{<math\; xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>-</mo><mn>3</mn></mrow>\; </math>\; <mo>+</mo><math\; xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>7</mn></mrow></math>}}{\mathrm{<math\; xmlns="http://www.w3.org/1998/Math/MathML">\; <mrow><mrow><mo>-</mo><mn>2</mn></mrow>\; </mrow>\; </math>}}$ = $\frac{4}{-2}$ = $-2$

u₂ = $\frac{-3-\sqrt{49}}{-2}$ = $\frac{\mathrm{<math\; xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>-</mo><mn>3</mn></mrow>\; </math>\; <mo>-</mo><math\; xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>7</mn></mrow></math>}}{\mathrm{<math\; xmlns="http://www.w3.org/1998/Math/MathML">\; <mrow><mrow><mo>-</mo><mn>2</mn></mrow>\; </mrow>\; </math>}}$ = $\frac{-10}{-2}$ = $5$

Da u= $-2$ < 0 ist u= $5$ die einzige Lösung.

Wir berechnen den Mittelwert mit der üblichen Formel:

= $1$ ${\left[\frac{1}{2}{\left(x-2\right)}^4+\frac{5}{2}{x}^2\right]}_1^2$

$=$ $\frac{1}{2}\cdot {\left(2-2\right)}^4+\frac{5}{2}\cdot 2^2-\left(\frac{1}{2}\cdot {\left(1-2\right)}^4+\frac{5}{2}\cdot 1^2\right)$

= $\frac{1}{2}\cdot 0^4+\frac{5}{2}\cdot 4-\left(\frac{1}{2}\cdot {\left(-1\right)}^4+\frac{5}{2}\cdot 1\right)$

= $\frac{1}{2}\cdot 0+10-\left(\frac{1}{2}\cdot 1+\frac{5}{2}\right)$

= $0+10-\left(\frac{1}{2}+\frac{5}{2}\right)$

= $10-\left(\mathrm{0,5}+\mathrm{2,5}\right)$

= $10-1·3$

= $10-3$

= $7$

= ${\left[-\frac{2}{3}{\left(-3x+3\right)}^{-1}\right]}_2^u$

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

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

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

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

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

= $-\frac{2}{3\left(-3u+3\right)}-\frac{2}{9}$

Für u → ∞ gilt: A(u) = $-\frac{2}{3\left(-3u+3\right)}-\frac{2}{9}$ → $0-\frac{2}{9}$ = $-\frac{2}{9}$ ≈ -0.222

Für den Flächeninhalt (immer positiv) gilt also I = 0.222