The area of the region given by A = {(x, y) : | vedclass

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(B) To find the area of the region $A = \{(x, y) : x^{2} \leq y \leq \min \{x+2, 4-3x\}\}$,we first find the intersection points of the curves.$1$. In

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$\frac{17}{6}$

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(B) To find the area of the region $A = \{(x, y) : x^{2} \leq y \leq \min \{x+2, 4-3x\}\}$,we first find the intersection points of the curves.$1$. Intersection of $y = x^2$ and $y = x+2$:$x^2 = x+2 \implies x^2 - x - 2 = 0 \implies (x-2)(x+1) = 0$. So,$x = -1$ and $x = 2$.$2$. Intersection of $y = x^2$ and $y = 4-3x$:$x^2 = 4-3x \implies x^2 + 3x - 4 = 0 \implies (x+4)(x-1) = 0$. So,$x = -4$ and $x = 1$.$3$. Intersection of $y = x+2$ and $y = 4-3x$:$x+2 = 4-3x \implies 4x = 2 \implies x = \frac{1}{2}$.Based on the region definition,the area is bounded by $y = x^2$ from below and the minimum of the two lines from above. The transition occurs at $x = \frac{1}{2}$.The area is given by:$Area = \int_{-1}^{\frac{1}{2}} (x+2 - x^2) dx + \int_{\frac{1}{2}}^{1} (4-3x - x^2) dx$Evaluating the first integral:$\int_{-1}^{\frac{1}{2}} (x+2 - x^2) dx = [\frac{x^2}{2} + 2x - \frac{x^3}{3}]_{-1}^{\frac{1}{2}} = (\frac{1}{8} + 1 - \frac{1}{24}) - (\frac{1}{2} - 2 + \frac{1}{3}) = (\frac{3+24-1}{24}) - (\frac{3-12+2}{6}) = \frac{26}{24} - (-\frac{7}{6}) = \frac{13}{12} + \frac{14}{12} = \frac{27}{12} = \frac{9}{4}$.Evaluating the second integral:$\int_{\frac{1}{2}}^{1} (4-3x - x^2) dx = [4x - \frac{3x^2}{2} - \frac{x^3}{3}]_{\frac{1}{2}}^{1} = (4 - \frac{3}{2} - \frac{1}{3}) - (2 - \frac{3}{8} - \frac{1}{24}) = (\frac{24-9-2}{6}) - (\frac{48-9-1}{24}) = \frac{13}{6} - \frac{38}{24} = \frac{52-38}{24} = \frac{14}{24} = \frac{7}{12}$.Total Area = $\frac{9}{4} + \frac{7}{12} = \frac{27+7}{12} = \frac{34}{12} = \frac{17}{6}$.

The area of the region given by $A = \{(x, y) : x^{2} \leq y \leq \min \{x+2, 4-3x\}\}$ is:

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