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(A) To find the area enclosed between the parabolas $y^{2}=2x-1$ and $y^{2}=4x-3$,we first find their points of intersection.Equating the two expressi

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$\frac{1}{3}$

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(A) To find the area enclosed between the parabolas $y^{2}=2x-1$ and $y^{2}=4x-3$,we first find their points of intersection.Equating the two expressions for $x$:From $y^{2}=2x-1$,we get $x = \frac{y^{2}+1}{2}$.From $y^{2}=4x-3$,we get $x = \frac{y^{2}+3}{4}$.Setting them equal: $\frac{y^{2}+1}{2} = \frac{y^{2}+3}{4} \implies 2y^{2}+2 = y^{2}+3 \implies y^{2}=1 \implies y = \pm 1$.The area is symmetric about the $x$-axis,so we calculate the area for $y \in [0, 1]$ and multiply by $2$.Required area $= 2 \int_{0}^{1} \left( \frac{y^{2}+3}{4} - \frac{y^{2}+1}{2} ight) dy$$= 2 \int_{0}^{1} \left( \frac{y^{2}+3 - 2y^{2}-2}{4} ight) dy$$= 2 \int_{0}^{1} \frac{1-y^{2}}{4} dy$$= \frac{1}{2} \left[ y - \frac{y^{3}}{3} ight]_{0}^{1}$$= \frac{1}{2} \left( 1 - \frac{1}{3} ight) = \frac{1}{2} 	imes \frac{2}{3} = \frac{1}{3}$.

The area of the region enclosed between the parabolas $y^{2}=2x-1$ and $y^{2}=4x-3$ is

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