JEE Main 2014 · 9 Apr · Shift 1, Mathematics question 77

The parabola $y^2=4x$ meets the line $x=\frac{y+4}2$ where $y^2=2y+8$, i.e. $y=-2$ and $y=4$. The region lies between them:
$$\int_{-2}^{4}\left(\frac{y+4}2-\frac{y^2}4\right)dy=\left[\frac{y^2}4+2y-\frac{y^3}{12}\right]_{-2}^{4}=\frac{20}3+\frac73=9.$$