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Area of the region enclosed by the curves $y=\sqrt{x}$ and $y=x^2$ is : |
$\frac{1}{2}$ units2 $\frac{1}{3}$ units2 1 units2 $\frac{1}{4}$ units2 |
$\frac{1}{3}$ units2 |
The correct answer is Option (2) - $\frac{1}{3}$ units2
at $\sqrt{x}=x^2$ $x=0,1$ point of intersection area = $\int\limits_0^1\sqrt{x}-x^2dx$ $=\left[\frac{2}{3}x\sqrt{x}-\frac{x^3}{3}\right]_0^1=\frac{1}{3}unit^2$ |
