The area enclosed by the curves y = x^2,y = x | vedclass

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(D) The given curves are $y = x^2$ and $y = x^3$.The intersection point of $y = x^2$ and $y = x^3$ is found by setting $x^2 = x^3$,which gives $x^2(x

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$4/3$

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(D) The given curves are $y = x^2$ and $y = x^3$.The intersection point of $y = x^2$ and $y = x^3$ is found by setting $x^2 = x^3$,which gives $x^2(x - 1) = 0$,so $x = 0$ or $x = 1$.For $0  x^3$,and for $x > 1$,$x^3 > x^2$.Given $p > 1$,the area is the sum of the areas in the intervals $[0, 1]$ and $[1, p]$:Area $= \int_{0}^{1} (x^2 - x^3) \, dx + \int_{1}^{p} (x^3 - x^2) \, dx = \frac{1}{6}$.Calculating the first integral:$\int_{0}^{1} (x^2 - x^3) \, dx = \left[ \frac{x^3}{3} - \frac{x^4}{4} \right]_{0}^{1} = \frac{1}{3} - \frac{1}{4} = \frac{1}{12}$.Calculating the second integral:$\int_{1}^{p} (x^3 - x^2) \, dx = \left[ \frac{x^4}{4} - \frac{x^3}{3} \right]_{1}^{p} = \left( \frac{p^4}{4} - \frac{p^3}{3} \right) - \left( \frac{1}{4} - \frac{1}{3} \right) = \frac{p^4}{4} - \frac{p^3}{3} + \frac{1}{12}$.Summing the areas:$\frac{1}{12} + \frac{p^4}{4} - \frac{p^3}{3} + \frac{1}{12} = \frac{1}{6}$.$\frac{p^4}{4} - \frac{p^3}{3} + \frac{2}{12} = \frac{1}{6}$.$\frac{p^4}{4} - \frac{p^3}{3} + \frac{1}{6} = \frac{1}{6}$.$\frac{p^4}{4} - \frac{p^3}{3} = 0$.Multiplying by $12$:$3p^4 - 4p^3 = 0$.$p^3(3p - 4) = 0$.Since $p > 1$,we have $p = 4/3$.

The area enclosed by the curves $y = x^2$,$y = x^3$,$x = 0$,and $x = p$,where $p > 1$,is $1/6$. The value of $p$ is:

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