The area enclosed between the parabola y^2=4x | vedclass

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(D) To find the area enclosed between the parabola $y^2=4x$ and the line $y=2x-4$,we first find the points of intersection. Substituting $x = \frac{y^

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$9 	ext{ sq. units}$

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(D) To find the area enclosed between the parabola $y^2=4x$ and the line $y=2x-4$,we first find the points of intersection. Substituting $x = \frac{y^2}{4}$ into the equation of the line $y = 2x - 4$: $y = 2\left(\frac{y^2}{4}ight) - 4$ $y = \frac{y^2}{2} - 4$ $y^2 - 2y - 8 = 0$ $(y - 4)(y + 2) = 0$ Thus,$y = 4$ and $y = -2$. The required area is given by the integral of the difference between the line and the parabola with respect to $y$: $	ext{Area} = \int_{-2}^{4} \left( \frac{y+4}{2} - \frac{y^2}{4} ight) dy$ $= \left[ \frac{y^2}{4} + 2y - \frac{y^3}{12} ight]_{-2}^{4}$ $= \left( \frac{16}{4} + 8 - \frac{64}{12} ight) - \left( \frac{4}{4} - 4 - \frac{-8}{12} ight)$ $= \left( 4 + 8 - \frac{16}{3} ight) - \left( 1 - 4 + \frac{2}{3} ight)$ $= \left( 12 - \frac{16}{3} ight) - \left( -3 + \frac{2}{3} ight)$ $= \frac{20}{3} - \left( -\frac{7}{3} ight) = \frac{27}{3} = 9 	ext{ sq. units}$.

The area enclosed between the parabola $y^2=4x$ and the line $y=2x-4$ is

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