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(A) The points of intersection of the two parabolas $y=x^{2}$ and $y^{2}=x$ are found by substituting $y=x^{2}$ into $y^{2}=x$,which gives $(x^{2})^{2

Vedclass · Accepted Answer

$1/3$

Vedclass · Answer

(A) The points of intersection of the two parabolas $y=x^{2}$ and $y^{2}=x$ are found by substituting $y=x^{2}$ into $y^{2}=x$,which gives $(x^{2})^{2}=x$,or $x^{4}-x=0$. This factors as $x(x^{3}-1)=0$,giving $x=0$ and $x=1$. Thus,the intersection points are $O(0, 0)$ and $A(1, 1)$.In the interval $[0, 1]$,the curve $y=\sqrt{x}$ lies above the curve $y=x^{2}$.The required area is given by the integral of the upper curve minus the lower curve:Area $= \int_{0}^{1} (\sqrt{x} - x^{2}) dx$$= \left[ \frac{x^{3/2}}{3/2} - \frac{x^{3}}{3} \right]_{0}^{1}$$= \left[ \frac{2}{3}x^{3/2} - \frac{x^{3}}{3} \right]_{0}^{1}$$= (\frac{2}{3}(1) - \frac{1}{3}(1)) - (0 - 0)$$= \frac{2}{3} - \frac{1}{3} = \frac{1}{3}$ square units.

Find the area of the region bounded by the two parabolas $y=x^{2}$ and $y^{2}=x$.

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