AIEEE 2007 (Offline)Question 1
The area enclosed between the curves $y^{2}=x$ and $y=|x|$ is :
Application of Integrals
Solution
$y=|x|\ge0$ meets $y^2=x$ only for $x\ge0$, at $x=0,1$, enclosing $\int_0^1(\sqrt x-x)dx=\frac23-\frac12=\frac16$.Application of Integrals - JEE Mathematics question (AIEEE 2007 (Offline))
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