The area of the region bounded by y-x=2 and x | vedclass

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(B) Given curves are $y = x+2$ and $y = x^2$.To find the points of intersection,set $x^2 = x+2$.$x^2 - x - 2 = 0$$(x-2)(x+1) = 0$So,$x = 2$ and $x = -

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$\frac{9}{2}$

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(B) Given curves are $y = x+2$ and $y = x^2$.To find the points of intersection,set $x^2 = x+2$.$x^2 - x - 2 = 0$$(x-2)(x+1) = 0$So,$x = 2$ and $x = -1$.The area $A$ is given by the integral of the upper curve minus the lower curve from $x = -1$ to $x = 2$:$A = \int_{-1}^{2} (x+2 - x^2) \, dx$$A = \left[ \frac{x^2}{2} + 2x - \frac{x^3}{3} \right]_{-1}^{2}$$A = \left( \frac{4}{2} + 4 - \frac{8}{3} \right) - \left( \frac{1}{2} - 2 - \frac{-1}{3} \right)$$A = \left( 2 + 4 - \frac{8}{3} \right) - \left( \frac{1}{2} - 2 + \frac{1}{3} \right)$$A = \left( 6 - \frac{8}{3} \right) - \left( \frac{3 - 12 + 2}{6} \right)$$A = \frac{10}{3} - \left( -\frac{7}{6} \right) = \frac{20}{6} + \frac{7}{6} = \frac{27}{6} = \frac{9}{2}$.

The area of the region bounded by $y-x=2$ and $x^{2}=y$ is equal to:

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