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

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(A) The curves are $y^2 = x$ and $y = |x|$.Since $y = |x|$,the curves are $y^2 = x$ and $y = x$ (for $x \ge 0$) and $y = -x$ (for $x < 0$).However,$y^

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$\frac{1}{6}$

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(A) The curves are $y^2 = x$ and $y = |x|$.Since $y = |x|$,the curves are $y^2 = x$ and $y = x$ (for $x \ge 0$) and $y = -x$ (for $x However,$y^2 = x$ implies $x \ge 0$,so we only consider the region where $x \ge 0$ and $y = x$.The intersection points are found by $x^2 = x$,which gives $x(x-1) = 0$,so $x = 0$ and $x = 1$.In the interval $[0, 1]$,$\sqrt{x} \ge x$.The required area is given by $\int_{0}^{1} (\sqrt{x} - x) dx$.$= \left[ \frac{x^{3/2}}{3/2} - \frac{x^2}{2} \right]_{0}^{1}$.$= \left( \frac{2}{3} - \frac{1}{2} \right) - (0 - 0) = \frac{4-3}{6} = \frac{1}{6}$.

The area enclosed between the curves $y^2 = x$ and $y = |x|$ is

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