Question

The area enclosed between the curves $y^2=x$ and $\left|y\right|=\left|x\right|$ is

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

Answer 1

The correct option is D $\frac{1}{3}$



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