The area bounded between the parabola y^{2}=4 | vedclass

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(C) To find the area bounded between the parabola $y^{2}=4x$ and the line $y=2x-4$,we first find their points of intersection by substituting $x = \fr

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$9$ sq unit

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(C) To find the area bounded between the parabola $y^{2}=4x$ and the line $y=2x-4$,we first find their points of intersection by substituting $x = \frac{y+4}{2}$ into the parabola equation:$y^{2} = 4\left(\frac{y+4}{2}ight)$$y^{2} = 2(y+4)$$y^{2} - 2y - 8 = 0$$(y-4)(y+2) = 0$Thus,the intersection points occur at $y = 4$ and $y = -2$.The required area is given by the integral of the difference between the line and the parabola with respect to $y$:$	ext{Area} = \int_{-2}^{4} \left( \frac{y+4}{2} - \frac{y^{2}}{4} ight) dy$$= \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 unit}$.

$x$	$0$	$1$	$2$	$3$	$4$	$5$
$f(x)$	$2$	$3$	$6$	$11$	$18$	$27$

The area bounded between the parabola $y^{2}=4x$ and the line $y=2x-4$ is equal to

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