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

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

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$\frac{5}{12}$

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(C) The given curves are $y = \sqrt{x}$ and $y = x^3$.To find the points of intersection,set $x^3 = \sqrt{x}$.Squaring both sides,we get $x^6 = x$,which implies $x(x^5 - 1) = 0$.Thus,$x = 0$ or $x = 1$.The corresponding $y$-values are $y = 0$ and $y = 1$.So,the points of intersection are $(0, 0)$ and $(1, 1)$.In the interval $[0, 1]$,$\sqrt{x} \geq x^3$.The required area is given by the integral:$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{1}{4}x^4 \right]_{0}^{1}$$A = \left( \frac{2}{3}(1) - \frac{1}{4}(1) \right) - (0 - 0)$$A = \frac{2}{3} - \frac{1}{4} = \frac{8 - 3}{12} = \frac{5}{12}$.

The area between the curves $y = x^3$ and $y = \sqrt{x}$ is

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