The area (in sq. units) of the region $\{(x,y): 3x^2≤ y ≤ |x|\}$ is equal to

$\frac{1}{27}$

The correct answer is Option (3) → $\frac{1}{27}$

Region is defined by

$3x^2\le y\le |x|$

Points of intersection are obtained from

$3x^2=|x|$

$|x|=3x^2$

For $x\ge0$

$x=3x^2$

$3x^2-x=0$

$x(3x-1)=0$

$x=0,\;\frac{1}{3}$

For $x\le0$

$-x=3x^2$

$3x^2+x=0$

$x(3x+1)=0$

$x=0,\;-\frac{1}{3}$

Area is symmetric about $y$-axis

$\text{Area}=2\int_{0}^{1/3}\left(|x|-3x^2\right)dx$

For $x\ge0,\;|x|=x$

$=2\int_{0}^{1/3}(x-3x^2)\,dx$

$=2\left[\frac{x^2}{2}-x^3\right]_{0}^{1/3}$

$=2\left(\frac{1}{18}-\frac{1}{27}\right)$

$=2\cdot\frac{1}{54}$

$=\frac{1}{27}$

The required area is $\frac{1}{27}$ square units.