The area (in sq. units) of the region

Solution:

Given curves are

$4 x^{2}=y$

$y=8 x+12$

From eqns. (i) and (ii),

$4 x^{2}=8 x+12$

$\Rightarrow x^{2}-x-3=0$

$\Rightarrow x^{2}-2 x-3=0$

$\Rightarrow x^{2}-3 x+x-3=0$

$\Rightarrow(x+1)(x-3)=0$

$\Rightarrow x=-1,3$

Required area bounded by curves is given by

$A=\int_{-1}^{3}\left(8 x+12-4 x^{2}\right) d x$

$A=\frac{8 x^{2}}{2}+12 x-\left.\frac{4 x^{3}}{3}\right|_{-1} ^{3}$

$=(4(9)+36-36)-\left(4-12+\frac{4}{3}\right)$

$=36+8-\frac{4}{3}=44-\frac{4}{3}=\frac{132-4}{3}=\frac{128}{3}$