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Q.
The area enclosed between the curves $y = x^3$ and $y = \sqrt{x}$ is, (in sq unit) :
Application of Integrals
Solution:
Given curves are $y = x^3$ and $y = \sqrt{x}$
From the above equation one can found that:
$x^6 = x \, \Rightarrow \, x = 0, 1$
Required area $ = \int^1_0 ( \sqrt{x} - x^3 )dx$
$= \left[\frac{x^{3/2}}{3}. 2 - \frac{x^{4}}{4}\right]^1_0 = \frac{2}{3} - \frac{1}{4}$
$ = \frac{5}{12} $ Sq. unit.