= ${\left[-3{\left(x-1\right)}^{-1}\right]}_3^4$

= ${\left[-\frac{3}{x-1}\right]}_3^4$

$=$ $-\frac{3}{4-1}+\frac{3}{3-1}$

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

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

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

= $-1+\mathrm{1,5}$

= $\mathrm{0,5}$

= ${\left[5e^{x-3}\right]}_2^4$

$=$ $5e^{4-3}-5e^{2-3}$

= $5e^1-5e^{-1}$

= $5e-5e^{-1}$

= ${\left[-e^{-\mathrm{0,1}x+\mathrm{0,3}}\right]}_0^u$

$=$ $-e^{-\mathrm{0,1}u+\mathrm{0,3}}+e^{-\mathrm{0,1}\cdot 0+\mathrm{0,3}}$

= $-e^{-\mathrm{0,1}u+\mathrm{0,3}}+e^{0+\mathrm{0,3}}$

= $-e^{-\mathrm{0,1}u+\mathrm{0,3}}+e^{\mathrm{0,3}}$

Diese Integralfunktion soll ja den Wert $1$ annehmen, deswegen setzen wir sie gleich :

Wir berechnen den Mittelwert mit der üblichen Formel:

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

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

= ${\left(6-1\right)}^3+\frac{1}{2}\cdot 9-\left({\left(4-1\right)}^3+\frac{1}{2}\cdot 4\right)$

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

= $125+\frac{9}{2}-\left(27+2\right)$

= $125+\frac{9}{2}-27-2$

= $\frac{201}{2}$

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

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

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

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

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

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

Für u → 1 (u>1, also von rechts) gilt: A(u) = $\frac{1}{{\left(u-1\right)}^2}-\frac{1}{4}$ → $\infty$