The area enclosed between the curves y = x^3 | vedclass

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(C) The given curves are $y = x^3$ and $y = \sqrt{x}$.To find the points of intersection,set $x^3 = \sqrt{x}$.Squaring both sides,$x^6 = x$,which impl

Vedclass · Accepted Answer

$\frac{5}{12}$

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(C) The given curves are $y = x^3$ and $y = \sqrt{x}$.To find the points of intersection,set $x^3 = \sqrt{x}$.Squaring both sides,$x^6 = x$,which implies $x(x^5 - 1) = 0$.Thus,the intersection points are $x = 0$ and $x = 1$.In the interval $[0, 1]$,$\sqrt{x} \geq x^3$.Therefore,the required area $A = \int_{0}^{1} (\sqrt{x} - x^3) \, dx$.$A = \left[ \frac{x^{3/2}}{3/2} - \frac{x^4}{4} \right]_{0}^{1}$.$A = \left[ \frac{2}{3}x^{3/2} - \frac{x^4}{4} \right]_{0}^{1}$.$A = \left( \frac{2}{3} - \frac{1}{4} \right) - (0 - 0)$.$A = \frac{8 - 3}{12} = \frac{5}{12}$ square units.

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

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