Question: The area (in sq. units) of the region $A=\left\{(x, y): x^{2} \leq y \leq x+2\right\}$ is
$\frac{10}{3}$
$\frac{9}{2}$
$\frac{31}{6}$
$\frac{13}{6}$
Correct Option: , 2
Solution:
$x^{2} \leq y \leq x+2$
$x^{2}=y ; y=x+2$
$x^{2}=x+2$
$x^{2}-x-2=0$
$(x-2)(x-1)=0$
$x=2,-1$
Area $=\int_{-1}^{2}(x+2)-x^{2} d x=\frac{9}{2}$
