Let P be the set of points (x, y) satisfying $x^2 ≤y≤-2x+3$. Then the area of region bounded by points in set P, is 

$\frac{32}{3}$ sq. units

We have,

$P=\{(x, y): x^2 <y≤-2x+3\}$

Now, $x^2 ≤y≤-2x+3$

$⇒x^2≤y$ and $y ≤-2x+3$

$⇒x^2≤y$ and $2x+y-3≤0$

Thus, the set P is the set of points bounded by the parabola $x^2=y$ and the line $2x+y-3=0$ as shown by shaded region in Fig.

Area A of the shaded region is given by

$A=\int\limits_{-3}^{1}(y_2-y_1)dx$

$⇒A=\int\limits_{-3}^{1}(3-2x-x^2)dx=\left[3x-x^2-\frac{x^3}{3}\right]_{-3}^{1}$

$⇒A=\left(3-1-\frac{1}{3}\right)-9(-9-9+9)=\frac{32}{3}$ sq. units