= ${\left[\frac{3}{2}{e}^{2x-3}\right]}_2^5$

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

= $\frac{3}{2}{e}^{10-3}-\frac{3}{2}{e}^{4-3}$

= $\frac{3}{2}{e}^7-\frac{3}{2}{e}^1$

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

= ${\left[\frac{4}{3}{\left(x-3\right)}^{\frac{3}{2}}\right]}_{19}^{28}$

= ${\left[\frac{4}{3}{\left(\sqrt{x-3}\right)}^3\right]}_{19}^{28}$

$=$ $\frac{4}{3}{\left(\sqrt{28-3}\right)}^3-\frac{4}{3}{\left(\sqrt{19-3}\right)}^3$

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

= $\frac{4}{3}\cdot 5^3-\frac{4}{3}\cdot 4^3$

= $\frac{4}{3}\cdot 125-\frac{4}{3}\cdot 64$

= $\frac{500}{3}-\frac{256}{3}$

= $\frac{244}{3}$

= ${\left[2x^2-2x\right]}_1^u$

$=$ $2u^2-2u-\left(2\cdot 1^2-2\cdot 1\right)$

= $2u^2-2u-\left(2\cdot 1-2\right)$

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

= $2u^2-2u-1·0$

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

= $2u^2-2u$

Diese Integralfunktion soll ja den Wert $60$ annehmen, deswegen setzen wir sie gleich :

$u^2-u-30$ = 0

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

u_1,2 = $\frac{+1\pm \sqrt{{\left(-1\right)}^2-4·1·\left(-30\right)}}{2\cdot 1}$

u_1,2 = $\frac{+1\pm \sqrt{1+120}}{2}$

u_1,2 = $\frac{+1\pm \sqrt{121}}{2}$

u₁ = $\frac{1+\sqrt{121}}{2}$ = $\frac{\mathrm{<math\; xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>1</mn></mrow>\; </math>\; <mo>+</mo><math\; xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>11</mn></mrow></math>}}{\mathrm{<math\; xmlns="http://www.w3.org/1998/Math/MathML">\; <mrow><mrow><mn>2</mn></mrow>\; </mrow>\; </math>}}$ = $\frac{12}{2}$ = $6$

u₂ = $\frac{1-\sqrt{121}}{2}$ = $\frac{\mathrm{<math\; xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>1</mn></mrow>\; </math>\; <mo>-</mo><math\; xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>11</mn></mrow></math>}}{\mathrm{<math\; xmlns="http://www.w3.org/1998/Math/MathML">\; <mrow><mrow><mn>2</mn></mrow>\; </mrow>\; </math>}}$ = $\frac{-10}{2}$ = $-5$

Da u= $-5$ < 1 ist u= $6$ die einzige Lösung.

Wir berechnen den Mittelwert mit der üblichen Formel:

= $\frac{1}{\pi }$ ${\left[-\cos (3x+\pi )\right]}_0^{\pi }$

$=$ $\frac{1}{\pi }·\left(-\cos (3\cdot \pi +\pi )+\cos (3\cdot \left(0\right)+\pi )\right)$

= $\frac{1}{\pi }·\left(-\cos \left(4\pi \right)+\cos \left(\pi \right)\right)$

= $\frac{1}{\pi }·\left(-1-1\right)$

= $\frac{-1-1}{\pi }$

= $\frac{-2}{\pi }$

= $-\frac{2}{\pi }$

= ${\left[\frac{3}{2}{\left(x-3\right)}^{-2}\right]}_4^u$

= ${\left[\frac{3}{2{\left(x-3\right)}^2}\right]}_4^u$

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

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

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

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

Für u → ∞ gilt: A(u) = $\frac{3}{2{\left(u-3\right)}^2}-\frac{3}{2}$ → $0-\frac{3}{2}$ = $-\frac{3}{2}$ ≈ -1.5

Für den Flächeninhalt (immer positiv) gilt also I = 1.5