The area (in sq. units) of the region

Given region, $A=\left\{(x, y): \frac{y^{2}}{2} \leq x \leq y+4\right\}$

Hence, area $=\int_{-2}^{4} x d y=\int_{-2}^{4}\left(y+4-\frac{y^{2}}{2}\right) d y$

$=\left[\frac{y^{2}}{2}+4 y-\frac{y^{3}}{6}\right]_{-2}^{4}=\left(8+16-\frac{64}{6}\right)-\left(2-8+\frac{8}{6}\right)$

$=\left(24-\frac{32}{3}\right)-\left(-6+\frac{4}{3}\right)=\frac{40}{3}+\frac{14}{3}=\frac{54}{3}=18$