Bestimme die Nullstellen um die Grenzen zu bestimmen:
$$-\frac12x^2+2cx = x\left(-\frac12x+2c\right) = 0$$
$$x_1 = 0$$
$$x_2 = 4c$$
$$ \int_0^{4c} -\frac12x^2+2cx \;dx = 144 \\ \left[-\frac16x^3+cx^2\right]_0^{4c} = 144 \\ -\frac16(4c)^3 + c*(4c)^2 = 144 \\ -\frac{64}{6}c^3 + 16c^3 = 144 \\ -\frac{64}{6}c^3 + \frac{16·6}{6}c^3 = 144 \\ \frac{32}{6}c^3 = 144 \\ \frac{16}{3}c^3 = 144 \\ c^3 = 27 \\ c = 3 $$
Alles klar?
Grüße