The area enclosed between the parabola $y=x^2-x+2$ and the line $y=x+2$ in (sq. units) equals 

Given curves are $y=x^2-x+2,y=x+2$, intersect each other at $(0,2),(2,4)$

Now, required area $={\int }_0^2 \left(x+2-x^2+x-2\right)dx$

$={\int }_0^2 \left(2x-x^2\right)dx={\left[x^2-\frac{x^3}{3}\right]}_0^2=4-\frac{8}{3}=\frac{4}{3}$ sq. units