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

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(C) To find the area bounded by the parabola $y^2 = 4x$ and the line $2x + y - 4 = 0$,we first find the points of intersection.Substituting $x = \frac

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$9$

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(C) To find the area bounded by the parabola $y^2 = 4x$ and the line $2x + y - 4 = 0$,we first find the points of intersection.Substituting $x = \frac{y^2}{4}$ into the line equation $2x + y - 4 = 0$,we get $2(\frac{y^2}{4}) + y - 4 = 0$,which simplifies to $\frac{y^2}{2} + y - 4 = 0$,or $y^2 + 2y - 8 = 0$.Factoring the quadratic equation,we get $(y + 4)(y - 2) = 0$,so $y = -4$ and $y = 2$.The required area is given by the integral of the difference between the line and the parabola with respect to $y$ from $y = -4$ to $y = 2$.Area $= \int_{-4}^{2} (x_{line} - x_{parabola}) dy = \int_{-4}^{2} (\frac{4 - y}{2} - \frac{y^2}{4}) dy$.Area $= \int_{-4}^{2} (2 - \frac{y}{2} - \frac{y^2}{4}) dy$.Evaluating the integral: $[2y - \frac{y^2}{4} - \frac{y^3}{12}]_{-4}^{2}$.At $y = 2$: $2(2) - \frac{4}{4} - \frac{8}{12} = 4 - 1 - \frac{2}{3} = 3 - \frac{2}{3} = \frac{7}{3}$.At $y = -4$: $2(-4) - \frac{16}{4} - \frac{-64}{12} = -8 - 4 + \frac{16}{3} = -12 + \frac{16}{3} = \frac{-36 + 16}{3} = -\frac{20}{3}$.Area $= \frac{7}{3} - (-\frac{20}{3}) = \frac{7 + 20}{3} = \frac{27}{3} = 9$ sq. units.

The area bounded between the parabola $y^2 = 4x$ and the line $2x + y - 4 = 0$ is

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