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

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(D) The given curves are $y=x^2+1$ and the line $y=2x-2$.We need to find the area bounded by these curves between $x=-1$ and $x=2$.The required area $

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$9$

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(D) The given curves are $y=x^2+1$ and the line $y=2x-2$.We need to find the area bounded by these curves between $x=-1$ and $x=2$.The required area $A$ is given by the integral of the upper curve minus the lower curve:$A = \int_{-1}^{2} [(x^2+1) - (2x-2)] dx$$A = \int_{-1}^{2} (x^2 - 2x + 3) dx$Now,evaluate the integral:$A = \left[ \frac{x^3}{3} - x^2 + 3x \right]_{-1}^{2}$Substitute the upper limit $x=2$:$\left( \frac{2^3}{3} - 2^2 + 3(2) \right) = \left( \frac{8}{3} - 4 + 6 \right) = \frac{8}{3} + 2 = \frac{14}{3}$Substitute the lower limit $x=-1$:$\left( \frac{(-1)^3}{3} - (-1)^2 + 3(-1) \right) = \left( -\frac{1}{3} - 1 - 3 \right) = -\frac{1}{3} - 4 = -\frac{13}{3}$Subtract the lower limit value from the upper limit value:$A = \frac{14}{3} - \left( -\frac{13}{3} \right) = \frac{14}{3} + \frac{13}{3} = \frac{27}{3} = 9$Thus,the area is $9$ 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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