The area enclosed by the parabolas y = x^2 - | vedclass

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(D) The given parabolas are $y = x^2 - 1$ and $y = 1 - x^2$.To find the points of intersection,set $x^2 - 1 = 1 - x^2$,which gives $2x^2 = 2$,so $x^2

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$8$

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(D) The given parabolas are $y = x^2 - 1$ and $y = 1 - x^2$.To find the points of intersection,set $x^2 - 1 = 1 - x^2$,which gives $2x^2 = 2$,so $x^2 = 1$,implying $x = \pm 1$.The area enclosed by the two curves is given by the integral of the upper curve minus the lower curve from $x = -1$ to $x = 1$.Area $= \int_{-1}^{1} ((1 - x^2) - (x^2 - 1)) \, dx$$= \int_{-1}^{1} (2 - 2x^2) \, dx$$= 2 \int_{-1}^{1} (1 - x^2) \, dx$Since the function is even,Area $= 2 	imes 2 \int_{0}^{1} (1 - x^2) \, dx$$= 4 \left[ x - \frac{x^3}{3} ight]_0^1$$= 4 \left( 1 - \frac{1}{3} ight) = 4 \left( \frac{2}{3} ight) = \frac{8}{3}$ square units.

The area enclosed by the parabolas $y = x^2 - 1$ and $y = 1 - x^2$ is (in $/3$)

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