Question

The area enclosed by the curves $y^2=x$ and $y=|x|$ is

• A. $\frac{2}{3}$
• B. $1$
• C. $\frac{1}{6}$
• D. $\frac{1}{3}$

Answer 1

Answer: (C)

$\frac{1}{6}$

Required area $=A={\int }_0^1(\sqrt{x}-x)dx$
$={[\frac{2}{3}{x}^{\frac{3}{2}}-\frac{x^2}{2}]}_0^1=\frac{2}{3}-\frac{1}{2}=\frac{1}{6}$