The area enclosed between the curves $y^2=x$ and $y=|x|$ is
The area enclosed between the curves $y^2=x$ and $y=|x|$ is
-
$2 / 3$
-
$1$
-
$1 / 6$
-
$1 / 3$
Solution
$A=\int_0^1(\sqrt{x}-x) d x$
$=\left[\frac{2}{3} x^{3 / 2}-\frac{x^2}{2}\right]_0^1$
$=\frac{2}{3}-\frac{1}{2}=\frac{1}{6}$.
Asked in: JEE Main 2007
Practice more Area Under Curves questions on Aicharya