Solve this following


The area (in sq. units) of the region $A=\left\{(x, y): x^{2} \leq y \leq x+2\right\}$ is

  1. $\frac{10}{3}$

  2. $\frac{9}{2}$

  3. $\frac{31}{6}$

  4. $\frac{13}{6}$

Correct Option: , 2


$x^{2} \leq y \leq x+2$

$x^{2}=y ; y=x+2$





Area $=\int_{-1}^{2}(x+2)-x^{2} d x=\frac{9}{2}$


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