Question

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

• A. $2/3$
• B. $1$
• C. $1/6$
• D. $1/3$

Answer 1

Answer: (C)

$A=\underset{0}{\overset{1}{\int }}$$\left(\sqrt{x}-x\right)dx$
$={\left[\frac{2}{3}{x}^{3/2}-\frac{x^2}{2}\right]}_{{}_{{}_0}}^{{}^1}$
$=\frac{2}{3}-\frac{1}{2}=\frac{1}{6}.$