Find the area of the following region using integration: {( x, y ): y^2 ≤ 2 x and y ≥ x -4}

Solving $\displaystyle \mathrm{y}^{2}=2 \mathrm{x}$ and $\displaystyle \mathrm{y}=\mathrm{x}-4$, we get $\displaystyle \mathrm{y}=4$ or -$\displaystyle 2$ Required area $\displaystyle =\int_{-2}^{4}\left[(\mathrm{y}+4)-\frac{\mathrm{y}^{2}}{2}\right] d \mathrm{y}$ $$=\left|\frac{\mathrm{y}^{2}}{2}+4 \mathrm{y}-\frac{1}{6} \mathrm{y}^{3}\right|_{-2}^{4} $$ = $\displaystyle 18$