The area (in sq. units) of the region bounded | vedclass

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(C) To find the area of the region bounded by the line $y = x + 2$ and the parabola $y = x^2$,we first find their points of intersection by setting $x

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$\frac{9}{2}$

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(C) To find the area of the region bounded by the line $y = x + 2$ and the parabola $y = x^2$,we first find their points of intersection by setting $x^2 = x + 2$.$x^2 - x - 2 = 0$$(x - 2)(x + 1) = 0$Thus,$x = 2$ or $x = -1$.The corresponding $y$-values are $y = 4$ and $y = 1$.The points of intersection are $(-1, 1)$ and $(2, 4)$.The required area is given by the integral of the upper curve minus the lower curve from $x = -1$ to $x = 2$:$	ext{Area} = \int_{-1}^{2} ((x + 2) - x^2) dx$$= \left[ \frac{x^2}{2} + 2x - \frac{x^3}{3} ight]_{-1}^{2}$$= \left( \frac{4}{2} + 2(2) - \frac{8}{3} ight) - \left( \frac{1}{2} + 2(-1) - \frac{-1}{3} ight)$$= \left( 2 + 4 - \frac{8}{3} ight) - \left( \frac{1}{2} - 2 + \frac{1}{3} ight)$$= \left( 6 - \frac{8}{3} ight) - \left( \frac{3 - 12 + 2}{6} ight)$$= \frac{10}{3} - \left( -\frac{7}{6} ight)$$= \frac{20}{6} + \frac{7}{6} = \frac{27}{6} = \frac{9}{2} 	ext{ sq. units.}$

The area (in sq. units) of the region bounded by $y-x=2$ and $x^2=y$ is equal to

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