Question:
The area of the region
$S=\left\{(x, y): 3 x^{2} \leq 4 y \leq 6 x+24\right\}$ is
Solution:
For A \& B
$3 x^{2}=6 x+24 \Rightarrow x^{2}-2 x-8=0$
$\Rightarrow x=-2,4$
Area $=\int_{-2}^{4}\left(\frac{3}{2} x+6-\frac{3}{4} x^{2}\right) d x$
$=\left[\frac{3 x^{2}}{4}+6 x-\frac{x^{3}}{4}\right]_{-2}^{4}=27$