The area enclosed between the curves $y = x^{3}$ and $y = \sqrt{x}$ is (in square unit)

Question

Dr. Abubacker · Accepted Answer

(c)
On solving $y = \sqrt{x}$ or $y^{2} = x,(y \geq 0)$ and $y = x^{3}$
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We get points of intersection which are (0, 0) and (1, 1)
$	herefore$ Required area $= \int_{0}^{1}{\left( \sqrt{x} - x^{3} ight)dx}$
$= \left\lbrack \frac{x^{3/2}}{3/2} - \frac{x^{4}}{4} ightbrack_{0}^{1} = \frac{5}{12}\ sq\ unit$

Q. The area enclosed between the curves $y = x^{3}$ and $y = \sqrt{x}$ is (in square unit)

Explanation

Reviewed by Dr. Abubacker

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