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

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(B) To find the area enclosed by the curves $y=x^{2}-1$ and $y=1-x^{2}$,we first find their points of intersection by setting $x^{2}-1 = 1-x^{2}$.This

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$\frac{8}{3}$

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(B) To find the area enclosed by the curves $y=x^{2}-1$ and $y=1-x^{2}$,we first find their points of intersection by setting $x^{2}-1 = 1-x^{2}$.This gives $2x^{2} = 2$,so $x^{2} = 1$,which means $x = -1$ and $x = 1$.The area $A$ is given by the integral of the upper curve minus the lower curve from $x = -1$ to $x = 1$.$A = \int_{-1}^{1} ((1-x^{2}) - (x^{2}-1)) dx$$A = \int_{-1}^{1} (2 - 2x^{2}) dx$Since the function $f(x) = 2 - 2x^{2}$ is an even function,we can write:$A = 2 \int_{0}^{1} (2 - 2x^{2}) dx = 4 \int_{0}^{1} (1 - x^{2}) dx$$A = 4 [x - \frac{x^{3}}{3}]_{0}^{1}$$A = 4 (1 - \frac{1}{3}) = 4 (\frac{2}{3}) = \frac{8}{3}$ sq. units.

The area (in sq. units) of the region enclosed by the curves $y=x^{2}-1$ and $y=1-x^{2}$ is equal to

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