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The area enclosed by the parabolas $y=x^2-1$ and $y=1-x^2$ is
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The correct answer is:
$8 / 3$
Given parabolas are $x^2=1+y, x^2=1-y$
Required area = $=4 \int_0^1\left(1-x^2\right) d x$ = $4\left[x-\frac{x^3}{3}\right]_0^1=\frac{8}{3}$
Required area = $=4 \int_0^1\left(1-x^2\right) d x$ = $4\left[x-\frac{x^3}{3}\right]_0^1=\frac{8}{3}$
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