SOLUTION :

Shaded portion is the area bounded. For the point of intersection A and B solving

$ y^2 = x $ & $ |y| = x - 2$

$(x - 2 )^2 = x \Rightarrow x^2 - 5x + 4 = 0 \Rightarrow x = 1, 4 $

$\Rightarrow x = 4 (as \space x \neq1) $

$|y| = x-2= 2\Rightarrow y = ±2$

Area bounded = $2\int\limits_0^{2} = [(y+2) -y^2] dy =2\bigg[\dfrac {y^2}2 +2y-\dfrac {y^3}3\bigg]_0^2 =2\Big[2+4-\dfrac 83\Big]=\dfrac{20}3$ sq.units