The area (in sq. units) of the region

Point of intersection of $y=x^{2}$ and $y=-2 x+3$ is obtained by $x^{2}+2 x-3=0$

$\Rightarrow \quad x=-3,1$

So, required area $=\int_{-3}^{1}($ line $-$ parabola $) d z$

$=\int_{-2}^{1}\left(3-2 x-x^{2}\right) d x$

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

$=(3) 4-2\left(\frac{1^{2}-3^{2}}{2}\right)-\left(\frac{1^{3}+3^{3}}{3}\right)$

$=12+8-\frac{28}{3}=\frac{32}{3}$