= ${\left[-\frac{3}{2}{\left(x-2\right)}^{-2}\right]}_3^5$

= ${\left[-\frac{3}{2{\left(x-2\right)}^2}\right]}_3^5$

$=$ $-\frac{3}{2{\left(5-2\right)}^2}+\frac{3}{2{\left(3-2\right)}^2}$

= $-\frac{3}{2\cdot 3^2}+\frac{3}{2\cdot 1^2}$

= $-\frac{3}{2}\cdot \left(\frac{1}{9}\right)+\frac{3}{2}\cdot 1$

= $-\frac{1}{6}+\frac{3}{2}$

= $-\frac{1}{6}+\frac{9}{6}$

= $\frac{4}{3}$

= ${\left[3\ln \left(\left|$ x-2$\right|\right)\right]}_3^4$

$=$ $3\ln \left(\left|$ 4-2$\right|\right)-3\ln \left(\left|$ 3-2$\right|\right)$

= $3\ln \left(2\right)-3\ln \left(\left|$ 3-2$\right|\right)$

= $3\ln \left(2\right)-3\ln \left(1\right)$

= $3\ln \left(2\right)+0$

= $3\ln \left(2\right)$

= ${\left[4x^2+5x\right]}_3^u$

$=$ $4u^2+5u-\left(4\cdot 3^2+5\cdot 3\right)$

= $4u^2+5u-\left(4\cdot 9+15\right)$

= $4u^2+5u-\left(36+15\right)$

= $4u^2+5u-1·51$

= $4u^2+5u-51$

Diese Integralfunktion soll ja den Wert $245$ annehmen, deswegen setzen wir sie gleich :

$4u^2+5u-296$ = 0

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

u_1,2 = $\frac{-5\pm \sqrt{5^2-4·4·\left(-296\right)}}{2\cdot 4}$

u_1,2 = $\frac{-5\pm \sqrt{25+4736}}{8}$

u_1,2 = $\frac{-5\pm \sqrt{4761}}{8}$

u₁ = $\frac{-5+\sqrt{4761}}{8}$ = $\frac{\mathrm{<math\; xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>-</mo><mn>5</mn></mrow>\; </math>\; <mo>+</mo><math\; xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>69</mn></mrow></math>}}{\mathrm{<math\; xmlns="http://www.w3.org/1998/Math/MathML">\; <mrow><mrow><mn>8</mn></mrow>\; </mrow>\; </math>}}$ = $\frac{64}{8}$ = $8$

u₂ = $\frac{-5-\sqrt{4761}}{8}$ = $\frac{\mathrm{<math\; xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>-</mo><mn>5</mn></mrow>\; </math>\; <mo>-</mo><math\; xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>69</mn></mrow></math>}}{\mathrm{<math\; xmlns="http://www.w3.org/1998/Math/MathML">\; <mrow><mrow><mn>8</mn></mrow>\; </mrow>\; </math>}}$ = $\frac{-74}{8}$ = $-\mathrm{9,25}$

Da u= $-\mathrm{9,25}$ < 3 ist u= $8$ die einzige Lösung.

Wir berechnen den Mittelwert mit der üblichen Formel:

= $1$ ${\left[\frac{5}{9}{\left(3x-4\right)}^3+\frac{3}{2}{x}^2\right]}_0^1$

$=$ $\frac{5}{9}\cdot {\left(3\cdot 1-4\right)}^3+\frac{3}{2}\cdot 1^2-\left(\frac{5}{9}\cdot {\left(3\cdot 0-4\right)}^3+\frac{3}{2}\cdot 0^2\right)$

= $\frac{5}{9}\cdot {\left(3-4\right)}^3+\frac{3}{2}\cdot 1-\left(\frac{5}{9}\cdot {\left(0-4\right)}^3+\frac{3}{2}\cdot 0\right)$

= $\frac{5}{9}\cdot {\left(-1\right)}^3+\frac{3}{2}-\left(\frac{5}{9}\cdot {\left(-4\right)}^3+0\right)$

= $\frac{5}{9}\cdot \left(-1\right)+\frac{3}{2}-\left(\frac{5}{9}\cdot \left(-64\right)+0\right)$

= $-\frac{5}{9}+\frac{3}{2}-\left(-\frac{320}{9}+0\right)$

= $-\frac{10}{18}+\frac{27}{18}-\left(-\frac{320}{9}+0\right)$

= $\frac{17}{18}+\frac{320}{9}$

= $\frac{17}{18}+\frac{640}{18}$

= $\frac{17}{18}+\frac{320}{9}$

= $\frac{73}{2}$

= ${\left[{\left(x-1\right)}^{-2}\right]}_3^u$

= ${\left[\frac{1}{{\left(x-1\right)}^2}\right]}_3^u$

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

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

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

Für u → ∞ gilt: A(u) = $\frac{1}{{\left(u-1\right)}^2}-\frac{1}{4}$ → $0-\frac{1}{4}$ = $-\frac{1}{4}$ ≈ -0.25

Für den Flächeninhalt (immer positiv) gilt also I = 0.25