Question

The area (in sq.units) of the region $\left\{\right(x,y)\in R^2:x^2\le y\le 3-2x\}$, is:

• A. $\frac{31}{3}$
• B. $\frac{32}{3}$
• C. $\frac{29}{3}$
• D. $\frac{34}{3}$

Answer 1

The correct option is B $\frac{32}{3}$

We have $x^2\le y\le -2x+3$


For point of intersection of two curves $y=x^2$ and $y=-2x+3$
$\Rightarrow x^2+2x-3=0$
$\Rightarrow x=-3,1$
$\Rightarrow \text{Area}=\underset{-3}{\overset{1}{\int }}\left(\right(-2x+3)-x^2)dx$
$={[-x^2+3x-\frac{x^3}{3}]}_{-3}^1=\frac{32}{3}$sq.units.