The area of the region bounded by y^2=x and y | vedclass

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(B) The given curves are $y^2 = x$ (a parabola opening to the right) and $y = |x|$ (a $V$-shaped graph). To find the points of intersection,we set $y^

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$\frac{1}{6} 	ext{ sq.unit}$

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(B) The given curves are $y^2 = x$ (a parabola opening to the right) and $y = |x|$ (a $V$-shaped graph). To find the points of intersection,we set $y^2 = x$ and $y = |x|$. Since $y = |x|$,we have $y^2 = x^2$. Substituting $y^2 = x$ into $x^2 = y^2$,we get $x^2 = x$,which implies $x(x - 1) = 0$. Thus,the intersection points are at $x = 0$ and $x = 1$. In the interval $[0, 1]$,the curve $y = \sqrt{x}$ lies above the line $y = x$. The area $A$ is given by the integral: $A = \int_{0}^{1} (\sqrt{x} - x) \, dx$ $A = \left[ \frac{x^{3/2}}{3/2} - \frac{x^2}{2} ight]_{0}^{1}$ $A = \left( \frac{2}{3}(1)^{3/2} - \frac{1^2}{2} ight) - (0 - 0)$ $A = \frac{2}{3} - \frac{1}{2} = \frac{4 - 3}{6} = \frac{1}{6} 	ext{ sq. unit}$.

The area of the region bounded by $y^2=x$ and $y=|x|$ is

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