A) 8/3
B) 3/8
C) 3/2
D) None of these
Correct Answer: A
Solution :
\[y={{x}^{2}}\] .....(i) \[y=2-{{x}^{2}}\] .....(ii) \[\therefore \] By equation (i) and (ii) , we get, \[x=\pm 1\] \[\therefore \] \[y=\pm 1\] \[\therefore \] Required area \[=2\left[ \int_{0}^{1}{(2-{{x}^{2}})dx-\int_{0}^{1}{{{x}^{2}}dx}} \right]\] \[=2\,\left[ 2x-\frac{2{{x}^{3}}}{3} \right]_{0}^{1}=4\left[ x-\frac{{{x}^{3}}}{3} \right]_{0}^{1}=4\left( \frac{2}{3} \right)=\frac{8}{3}\].You need to login to perform this action.
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