Area of the region bounded by two parabolas y | vedclass

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(A) The given parabolas are $y=x^{2}$ and $x=y^{2}$.First,we find the points of intersection by substituting $y=x^{2}$ into $x=y^{2}$:$x=(x^{2})^{2} \

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$1/3$

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(A) The given parabolas are $y=x^{2}$ and $x=y^{2}$.First,we find the points of intersection by substituting $y=x^{2}$ into $x=y^{2}$:$x=(x^{2})^{2} \implies x=x^{4} \implies x^{4}-x=0 \implies x(x^{3}-1)=0$.This gives $x=0$ and $x=1$.For $x=0$,$y=0$,and for $x=1$,$y=1$. The intersection points are $(0,0)$ and $(1,1)$.The area $A$ is given by the integral of the upper curve minus the lower curve from $x=0$ to $x=1$:$A = \int_{0}^{1} (\sqrt{x} - x^{2}) dx$.Evaluating the integral:$A = [\frac{x^{3/2}}{3/2} - \frac{x^{3}}{3}]_{0}^{1} = [\frac{2}{3}x^{3/2} - \frac{1}{3}x^{3}]_{0}^{1}$.$A = (\frac{2}{3} - \frac{1}{3}) - (0 - 0) = \frac{1}{3}$ square units.

Area of the region bounded by two parabolas $y=x^{2}$ and $x=y^{2}$ is

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