Let A = {(x, y) : y^2 le 4x, y - 2x ge -4}. T | vedclass

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(B) The region $A$ is bounded by the parabola $y^2 = 4x$ and the line $y = 2x - 4$.To find the points of intersection,substitute $x = \frac{y^2}{4}$ i

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$9$

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(B) The region $A$ is bounded by the parabola $y^2 = 4x$ and the line $y = 2x - 4$.To find the points of intersection,substitute $x = \frac{y^2}{4}$ into the line equation $x = \frac{y+4}{2}$:$\frac{y^2}{4} = \frac{y+4}{2} \implies y^2 = 2y + 8 \implies y^2 - 2y - 8 = 0$.$(y - 4)(y + 2) = 0$,so $y = 4$ and $y = -2$.For $y = 4$,$x = 4$. For $y = -2$,$x = 1$.The area is given by $\int_{-2}^{4} (x_{line} - x_{parabola}) dy = \int_{-2}^{4} (\frac{y+4}{2} - \frac{y^2}{4}) dy$.$= \left[ \frac{y^2}{4} + 2y - \frac{y^3}{12} \right]_{-2}^{4}$.$= (\frac{16}{4} + 8 - \frac{64}{12}) - (\frac{4}{4} - 4 - \frac{-8}{12})$.$= (4 + 8 - \frac{16}{3}) - (1 - 4 + \frac{2}{3}) = (12 - \frac{16}{3}) - (-3 + \frac{2}{3}) = \frac{20}{3} - (-\frac{7}{3}) = \frac{27}{3} = 9$ sq units.

Let $A = \{(x, y) : y^2 \le 4x, y - 2x \ge -4\}$. The area of the region $A$ is

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