The area lying between the curves y^{2}=4x an | vedclass

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(B) To find the area between the curves $y^{2}=4x$ and $y=2x$,we first find their points of intersection.Substituting $y=2x$ into $y^{2}=4x$,we get $(

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$\frac{1}{3}$

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(B) To find the area between the curves $y^{2}=4x$ and $y=2x$,we first find their points of intersection.Substituting $y=2x$ into $y^{2}=4x$,we get $(2x)^{2}=4x$,which simplifies to $4x^{2}-4x=0$,or $4x(x-1)=0$.Thus,the points of intersection are $x=0$ and $x=1$.For $x=0$,$y=0$,and for $x=1$,$y=2$. So the intersection points are $(0,0)$ and $(1,2)$.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} (	ext{upper curve} - 	ext{lower curve}) dx$$A = \int_{0}^{1} (\sqrt{4x} - 2x) dx$$A = \int_{0}^{1} (2\sqrt{x} - 2x) dx$$A = 2 \left[ \frac{x^{3/2}}{3/2} ight]_{0}^{1} - 2 \left[ \frac{x^{2}}{2} ight]_{0}^{1}$$A = 2 \left( \frac{2}{3} ight) - 1 = \frac{4}{3} - 1 = \frac{1}{3}$ square units.Thus,the correct answer is $B$.

The area lying between the curves $y^{2}=4x$ and $y=2x$ is

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