Question

The area enclosed between the curve $y^2=xandy=|x|$ is

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

Answer 1

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

$y=\mid x\mid$$\{x;x\ge 0-x;x<0\}$

to find $P$

$y^2=x$ and $y=x$

$x^2=xx\left(x_1\right)=0$

$x=0,1P=1$

$A={\int }_0^1\sqrt{x}-xdxA={[\frac{2x^{3/2}}{3}-\frac{x^2}{2}]}_0^{1}A=[\frac{2}{3}-\frac{1}{2}]A=\frac{1}{6}$