Question

The area of the region(s) enclosed by the curves $y=x^2$ and $y=\sqrt{\left|x\right|}$ is:

• A. 1/3
• B. 2/3
• C. 1/6
• D. 1

Answer 1

Answer: (A)

1/3

The curve $y=x^2$ and $y=\sqrt{x}$ intersect at $O(0,0)$ and $P(1,1)$
Therefore required area
$={\int }_0^1(\sqrt{x}-x^2)dx={[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}-\frac{x^3}{3}]}_0^1=\frac{1}{3}$