Q. The area (in sq. units) of the region $\left\{\left(x, y\right)\epsilon\,R^{2}|4x^{2} \le y \le 8x+12\right\}$ is :
Solution:
$4x^{2}-y \le 0$ and $8x-y+12 \ge 0$
On solving $y = 4x^{2}$
and $y = 8x + 12$
We get $A \left(-1, 4\right) \& B\left(3, 36\right)$
Required area = area of the shaded region
$=\int\limits^{3}_{{-1}}(8x + 12 - 4x^2 )dx=\frac{128}{3}$
