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

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(D) The given curves are $y=x^2+1$ and $y=2x-2$.The region is bounded by the vertical lines $x=-1$ and $x=2$.For the interval $[-1, 2]$, we check if t

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$9$

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(D) The given curves are $y=x^2+1$ and $y=2x-2$.The region is bounded by the vertical lines $x=-1$ and $x=2$.For the interval $[-1, 2]$, we check if the curves intersect. Setting $x^2+1 = 2x-2$, we get $x^2-2x+3=0$. The discriminant $D = (-2)^2 - 4(1)(3) = 4-12 = -8  2x-2$ for all $x$.The required area is given by:$	ext{Area} = \int_{-1}^{2} [(x^2+1) - (2x-2)] \, dx$$	ext{Area} = \int_{-1}^{2} (x^2 - 2x + 3) \, dx$$	ext{Area} = \left[ \frac{x^3}{3} - x^2 + 3x ight]_{-1}^{2}$$	ext{Area} = \left( \frac{8}{3} - 4 + 6 ight) - \left( \frac{-1}{3} - 1 - 3 ight)$$	ext{Area} = \left( \frac{8}{3} + 2 ight) - \left( -\frac{1}{3} - 4 ight)$$	ext{Area} = \frac{14}{3} - \left( -\frac{13}{3} ight)$$	ext{Area} = \frac{14}{3} + \frac{13}{3} = \frac{27}{3} = 9$ sq units.

The area (in sq units) of the region bounded by $x=-1$, $x=2$, $y=x^2+1$ and $y=2x-2$ is

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