The area enclosed by the curves y^2+4x=4 and | vedclass

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(C) Given curves are $y^2 = -4(x-1)$ and $x = \frac{y-2}{2}$.To find the points of intersection,substitute $x = \frac{y-2}{2}$ into the parabola equat

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$9$

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(C) Given curves are $y^2 = -4(x-1)$ and $x = \frac{y-2}{2}$.To find the points of intersection,substitute $x = \frac{y-2}{2}$ into the parabola equation:$y^2 = -4(\frac{y-2}{2} - 1) = -2(y-2-2) = -2(y-4) = -2y + 8$.$y^2 + 2y - 8 = 0 \implies (y+4)(y-2) = 0$.So,the intersection points are at $y = -4$ and $y = 2$.The area $A$ is given by the integral of the right curve minus the left curve with respect to $y$:$A = \int_{-4}^{2} [x_{right} - x_{left}] dy = \int_{-4}^{2} [\frac{4-y^2}{4} - \frac{y-2}{2}] dy$.$A = \int_{-4}^{2} [1 - \frac{y^2}{4} - \frac{y}{2} + 1] dy = \int_{-4}^{2} [2 - \frac{y}{2} - \frac{y^2}{4}] dy$.$A = [2y - \frac{y^2}{4} - \frac{y^3}{12}]_{-4}^{2}$.$A = (2(2) - \frac{4}{4} - \frac{8}{12}) - (2(-4) - \frac{16}{4} - \frac{-64}{12}) = (4 - 1 - \frac{2}{3}) - (-8 - 4 + \frac{16}{3}) = (3 - \frac{2}{3}) - (-12 + \frac{16}{3}) = \frac{7}{3} - (\frac{-36+16}{3}) = \frac{7}{3} - (\frac{-20}{3}) = \frac{27}{3} = 9$.

The area enclosed by the curves $y^2+4x=4$ and $y-2x=2$ is:

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