The area of the region
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 $=\int_{-2}^{4}\left(\frac{3}{2} x+6-\frac{3}{4} x^{2}\right) d x$

$=\left[\frac{3 x^{2}}{4}+6 x-\frac{x^{3}}{4}\right]_{-2}^{4}=27$

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