= ${\left[2e^{2x-5}\right]}_0^1$

$=$ $2e^{2\cdot 1-5}-2e^{2\cdot 0-5}$

= $2e^{2-5}-2e^{0-5}$

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

= ${\left[{\left(2x-2\right)}^{\frac{3}{2}}\right]}_9^{\frac{27}{2}}$

= ${\left[{\left(\sqrt{2x-2}\right)}^3\right]}_9^{\frac{27}{2}}$

$=$ ${\left(\sqrt{2\cdot \left(\frac{27}{2}\right)-2}\right)}^3-{\left(\sqrt{2\cdot 9-2}\right)}^3$

= ${\left(\sqrt{27-2}\right)}^3-{\left(\sqrt{18-2}\right)}^3$

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

= $5^3-4^3$

= $125-64$

= $61$

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

$=$ $4u^2-4\cdot 0^2$

= $4u^2-4\cdot 0$

= $4u^2+0$

= $4u^2$

Diese Integralfunktion soll ja den Wert $196$ annehmen, deswegen setzen wir sie gleich :

Da u= $-7$ < 0 ist u= $7$ die einzige Lösung.

Wir berechnen den Mittelwert mit der üblichen Formel:

= $1$ ${\left[\frac{5}{12}{\left(3x-4\right)}^4+\frac{1}{2}{x}^2\right]}_1^2$

$=$ $\frac{5}{12}\cdot {\left(3\cdot 2-4\right)}^4+\frac{1}{2}\cdot 2^2-\left(\frac{5}{12}\cdot {\left(3\cdot 1-4\right)}^4+\frac{1}{2}\cdot 1^2\right)$

= $\frac{5}{12}\cdot {\left(6-4\right)}^4+\frac{1}{2}\cdot 4-\left(\frac{5}{12}\cdot {\left(3-4\right)}^4+\frac{1}{2}\cdot 1\right)$

= $\frac{5}{12}\cdot 2^4+2-\left(\frac{5}{12}\cdot {\left(-1\right)}^4+\frac{1}{2}\right)$

= $\frac{5}{12}\cdot 16+2-\left(\frac{5}{12}\cdot 1+\frac{1}{2}\right)$

= $\frac{20}{3}+2-\left(\frac{5}{12}+\frac{1}{2}\right)$

= $\frac{20}{3}+\frac{6}{3}-\left(\frac{5}{12}+\frac{6}{12}\right)$

= $\frac{26}{3}-1·\frac{11}{12}$

= $\frac{26}{3}-\frac{11}{12}$

= $\frac{31}{4}$

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

$=$ $-e^{2u-3}+e^{2\cdot 0-3}$

= $-e^{2u-3}+e^{0-3}$

= $-e^{2u-3}+e^{-3}$

Für u → ∞ gilt: A(u) = $-e^{2u-3}+e^{-3}$ → $-\infty$