Question

The area (in sq. units) of the region $\{x\in R:x⩾0,y⩾0,y⩾x-2$ and $y⩽\sqrt{x}\}$ is :

Answer 1

$\begin{array}{c}y\ge x-2\\ y\le \sqrt{x}\\ x^2-4x+4=x\\ x^2-5x+4=0\\ \left(x-4\right)\left(x-1\right)=0\\ x=1,4\\ Area={\int }_0^4\sqrt{x}dx-\frac{1}{2}\times 2\times 2\\ ={\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]}_0^4-2\\ =\frac{2}{3}{\left[x\sqrt{x}\right]}_0^4-2\\ =\frac{2}{3}\times 4\times 2-2\\ =\frac{16}{3}-2\\ =\frac{10}{3}squnit.\end{array}$

Hence, this is the answer.