The area (in sq. units) of the region $\{x \in R: x \geq 0, y \geq 0, y \geq x-2$ and $y \leq \sqrt{x}\}$, is
The area (in sq. units) of the region $\{x \in R: x \geq 0, y \geq 0, y \geq x-2$ and $y \leq \sqrt{x}\}$, is
$\frac{13}{3}$
$\frac{10}{3}$
$\frac{5}{3}$
$\frac{8}{3}$
Solution
The intersection point of $y=x-2$ and $y=\sqrt{x}$ is $(4,2)$.
The required area
$
=\int_0^4 \sqrt{x} d x-\frac{1}{2} \times 2 \times 2=\frac{16}{3}-2=\frac{10}{3}
$
