The area of the region enclosed  by $y=x+2$ and $y=x^2+x-2$ is :

$\frac{32}{3}units^2$

The correct answer is Option (2) → $\frac{32}{3}units^2$

$y=x^2+x-2⇒y=(x+2)(x-1)$   ...(1)

$y=x+2$  ...(2)

finding intersection points

$(x+2)=(x+2)(x-1)$

$(x+2)(x-2)=0$

$x=2,-2$

$y=4,0$

area = $-\int\limits_{-2}^1(x^2+x-2)dx+\int\limits_{-2}^2x+2dx-\int\limits_{1}^2x^2+x-2dx$

$=-\left[\frac{x^3}{3}+\frac{x^2}{2}-2x\right]_{-2}^1+\left[\frac{x^2}{2}+2x\right]_{-2}^2-\left[\frac{x^3}{3}+\frac{x^2}{2}-2x\right]_1^2$

$=\frac{32}{3}units$