Q. The area enclosed between the curves $y^2 = x$ and $y = |x|$ is
AIEEEAIEEE 2007Application of Integrals
Solution:
$A=\int\limits^{1}_{{0}}$$\left(\sqrt{x}-x\right)dx$
$=\left[\frac{2}{3}x^{3 / 2}-\frac{x^{2}}{2}\right]^{^1}_{_{_0}}$
$=\frac{2}{3}-\frac{1}{2}=\frac{1}{6}.$
