Find the area bounded by the curves y = x^2 a | vedclass

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(A) Given curves are $y = x^2$ $(i)$ and $y = 2 - x^2$ $(ii)$.To find the points of intersection,set $x^2 = 2 - x^2$,which gives $2x^2 = 2$,so $x^2 =

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$8/3$

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(A) Given curves are $y = x^2$ $(i)$ and $y = 2 - x^2$ $(ii)$.To find the points of intersection,set $x^2 = 2 - x^2$,which gives $2x^2 = 2$,so $x^2 = 1$,implying $x = \pm 1$.The curves intersect at $x = -1$ and $x = 1$.The required area is the integral of the upper curve minus the lower curve between the intersection points:$	ext{Area} = \int_{-1}^{1} ((2 - x^2) - x^2) \, dx$$= \int_{-1}^{1} (2 - 2x^2) \, dx$Since the function is even,we can write:$= 2 \int_{0}^{1} (2 - 2x^2) \, dx$$= 2 [2x - \frac{2x^3}{3}]_{0}^{1}$$= 2 [2(1) - \frac{2(1)^3}{3}]$$= 2 [2 - \frac{2}{3}]$$= 2 [\frac{4}{3}] = \frac{8}{3}$ square units.

Find the area bounded by the curves $y = x^2$ and $y = 2 - x^2$.

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