The area (in sq. units) of the region $\{(x, y) \in R^2 |4x^2 \le y \le 8x + 12\}$ is :

Shaded are is the area between curve

$Y = 4x^2$ __(I)

and

$Y = 8x + 12$ __(II)

on solving equation (I) & (II) (to find intersection point of the curves).

$4x^2 - 8x - 12 = 0$

$4(x +1 )(x - 3) = 0$

$x = -1$ or $x = 3$

$\therefore$ shaded Area $= \displaystyle \int_{-1}^3 [(8x + 12) - (4x^2)] dx$

$= \displaystyle \int_{-1}^3 8x dx + \int_{-1}^3 12 dx - 4 \int_{-1}^3 x^2 dx$

$= \left[4x^2 + 12 x - \dfrac{4}{3} x^3 \right]_{-1}^3$

$=4\cdot 9+12\cdot 3-\dfrac 43\cdot 27-4+12-\dfrac 43$

$= \dfrac{128}{3}$