Question

The area enclosed between 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

The correct option is C $\frac{1}{6}$

POint of intersection P

$y^2=x{x}^2=xx=1and0$

Area enclosed $={\int }_0^1\sqrt{x}-x\Rightarrow \frac{2}{3}{x}^{2/3}-\frac{x^2}{2}{]}_0^1\Rightarrow \frac{2}{3}-\frac{1}{2}\Rightarrow \frac{4-3}{6}=\frac{1}{6}$