Let $A=\left\{\left(x,y\right):y^2\le 4x, y-2x\ge -4\right\}$. The area of the region $A$ in square units is

Given, $y^2=4x ...\left(1\right)$

$y=2x-4\Rightarrow 2x=y+4 ...\left(2\right)$

Solving equations $\left(1\right) \& \left(2\right)$, we get

$\Rightarrow y^2=2y+8$

$\Rightarrow y^2-2y-8=0$

$\Rightarrow \left(y-4\right)\left(y+2\right)=0$

$\Rightarrow y=4, -2$

From equation $\left(2\right)$,

If $y=4$ then $x=4$ and if $y=-2$ then $x=1$.

$A=\left[\underset{0}{\overset{4}{\int }}\left(\sqrt{4x}\right) dx-\frac{1}{2}\times 2\times 4\right]+\left[\underset{0}{\overset{1}{\int }}\sqrt{4x}dx\right]+\frac{1}{2}\times 1\times 2$

$=2{\left[\frac{x^{3/2}}{\frac{3}{2}}\right]}_0^4-4+2{\left[\frac{x^{3/2}}{\frac{3}{2}}\right]}_0^1+1$

$=\frac{4}{3}\left[8\right]-4+\frac{4}{3}\left[1\right]+1$

$=\frac{36}{3}-3$

$=9$