= ${\left[e^{2x-2}\right]}_2^4$

$=$ $e^{2\cdot 4-2}-e^{2\cdot 2-2}$

= $e^{8-2}-e^{4-2}$

= $e^6-e^2$

= ${\left[\frac{2}{3}{\left(3x-3\right)}^{\frac{3}{2}}\right]}_{\frac{7}{3}}^{\frac{19}{3}}$

= ${\left[\frac{2}{3}{\left(\sqrt{3x-3}\right)}^3\right]}_{\frac{7}{3}}^{\frac{19}{3}}$

$=$ $\frac{2}{3}{\left(\sqrt{3\cdot \left(\frac{19}{3}\right)-3}\right)}^3-\frac{2}{3}{\left(\sqrt{3\cdot \left(\frac{7}{3}\right)-3}\right)}^3$

= $\frac{2}{3}{\left(\sqrt{19-3}\right)}^3-\frac{2}{3}{\left(\sqrt{7-3}\right)}^3$

= $\frac{2}{3}{\left(\sqrt{16}\right)}^3-\frac{2}{3}{\left(\sqrt{4}\right)}^3$

= $\frac{2}{3}\cdot 4^3-\frac{2}{3}\cdot 2^3$

= $\frac{2}{3}\cdot 64-\frac{2}{3}\cdot 8$

= $\frac{128}{3}-\frac{16}{3}$

= $\frac{112}{3}$

= ${\left[3e^{\mathrm{0,8}x-\mathrm{0,8}}\right]}_0^u$

$=$ $3e^{\mathrm{0,8}u-\mathrm{0,8}}-3e^{\mathrm{0,8}\cdot 0-\mathrm{0,8}}$

= $3e^{\mathrm{0,8}u-\mathrm{0,8}}-3e^{0-\mathrm{0,8}}$

= $3e^{\mathrm{0,8}u-\mathrm{0,8}}-3e^{-\mathrm{0,8}}$

Diese Integralfunktion soll ja den Wert $7$ annehmen, deswegen setzen wir sie gleich :

Wir berechnen den Mittelwert mit der üblichen Formel:

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

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

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

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

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

= $\frac{1}{3}·9$

= $3$

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

= ${\left[-\sqrt{2x-2}\right]}_u^3$

$=$ $-\sqrt{2\cdot 3-2}+\sqrt{2u-2}$

= $-\sqrt{6-2}+\sqrt{2u-2}$

= $-\sqrt{4}+\sqrt{2u-2}$

= $-2+\sqrt{2u-2}$

= $\sqrt{2u-2}-2$

Für u → 1 (u>1, also von rechts) gilt: A(u) = $\sqrt{2u-2}-2$ → $0-2$ = $-2$ ≈ -2

Für den Flächeninhalt (immer positiv) gilt also I = 2