= ${\left[-\frac{5}{4}{\left(2x-1\right)}^{-2}\right]}_2^4$

= ${\left[-\frac{5}{4{\left(2x-1\right)}^2}\right]}_2^4$

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

= $-\frac{5}{4{\left(8-1\right)}^2}+\frac{5}{4{\left(4-1\right)}^2}$

= $-\frac{5}{4\cdot 7^2}+\frac{5}{4\cdot 3^2}$

= $-\frac{5}{4}\cdot \left(\frac{1}{49}\right)+\frac{5}{4}\cdot \left(\frac{1}{9}\right)$

= $-\frac{5}{196}+\frac{5}{36}$

= $-\frac{45}{1764}+\frac{245}{1764}$

= $\frac{50}{441}$

= ${\left[e^{3x-7}\right]}_0^1$

$=$ $e^{3\cdot 1-7}-e^{3\cdot 0-7}$

= $e^{3-7}-e^{0-7}$

= $e^{-4}-e^{-7}$

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

$=$ $-3u^2+3\cdot 0^2$

= $-3u^2+3\cdot 0$

= $-3u^2+0$

= $-3u^2$

Diese Integralfunktion soll ja den Wert $-48$ annehmen, deswegen setzen wir sie gleich :

Da u= $-4$ < 0 ist u= $4$ die einzige Lösung.

Wir berechnen den Mittelwert mit der üblichen Formel:

= $\frac{1}{21}$ ${\left[2{\left(x-1\right)}^{\frac{3}{2}}\right]}_5^{26}$

= $\frac{1}{21}$ ${\left[2{\left(\sqrt{x-1}\right)}^3\right]}_5^{26}$

$=$ $\frac{1}{21}\left(2{\left(\sqrt{26-1}\right)}^3-2{\left(\sqrt{5-1}\right)}^3\right)$

= $\frac{1}{21}\left(2{\left(\sqrt{25}\right)}^3-2{\left(\sqrt{4}\right)}^3\right)$

= $\frac{1}{21}\left(2\cdot 5^3-2\cdot 2^3\right)$

= $\frac{1}{21}\left(2\cdot 125-2\cdot 8\right)$

= $\frac{1}{21}\left(250-16\right)$

= $\frac{1}{21}·234$

= $\frac{78}{7}$

= ${\left[2{\left(2x-6\right)}^{\frac{1}{2}}\right]}_u^{11}$

= ${\left[2\sqrt{2x-6}\right]}_u^{11}$

$=$ $2\sqrt{2\cdot 11-6}-2\sqrt{2u-6}$

= $2\sqrt{22-6}-2\sqrt{2u-6}$

= $2\sqrt{16}-2\sqrt{2u-6}$

= $2\cdot 4-2\sqrt{2u-6}$

= $8-2\sqrt{2u-6}$

Für u → 3 (u>3, also von rechts) gilt: A(u) = $-2\sqrt{2u-6}+8$ → $0+8$ = $8$ ≈ 8

Für den Flächeninhalt (immer positiv) gilt also I = 8