The area enclosed between the curves $y^2 =x$ and $y =|x|$, is
Answer & explanation
Correct answer: option 2
The correct answer is Option 2: $\frac{1}{3}$
Let A be the area enclosed between the given curves, then
$A=\int\limits_{0}^{1}(\sqrt{x}-x)dx=\left[\frac{2}{3}x^{3/2}-\frac{x^2}{2}\right]_{0}^{1}=\frac{1}{3}$
