The area (in sq. units) of the region A = {(x | vedclass

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(C) The given region is bounded by the parabola $y = x^2$ and the line $y = x + 2$.To find the points of intersection,we set $x^2 = x + 2$.This gives

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$\frac{9}{2}$

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(C) The given region is bounded by the parabola $y = x^2$ and the line $y = x + 2$.To find the points of intersection,we set $x^2 = x + 2$.This gives $x^2 - x - 2 = 0$.Factoring the quadratic equation,we get $(x - 2)(x + 1) = 0$,which implies $x = 2$ and $x = -1$.The area is given by the integral of the upper curve minus the lower curve from $x = -1$ to $x = 2$:$	ext{Area} = \int_{-1}^{2} ((x + 2) - x^2) dx$$= [\frac{x^2}{2} + 2x - \frac{x^3}{3}]_{-1}^{2}$$= (\frac{4}{2} + 4 - \frac{8}{3}) - (\frac{1}{2} - 2 - \frac{-1}{3})$$= (2 + 4 - \frac{8}{3}) - (\frac{1}{2} - 2 + \frac{1}{3})$$= (6 - \frac{8}{3}) - (\frac{3 - 12 + 2}{6})$$= \frac{10}{3} - (-\frac{7}{6})$$= \frac{20 + 7}{6} = \frac{27}{6} = \frac{9}{2}$ sq. units.

The area (in sq. units) of the region $A = \{(x, y) : x^2 \le y \le x + 2\}$ is

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