A) \[\frac{1}{3}\]
B) \[\frac{3}{4}\]
C) \[\frac{3}{5}\]
D) \[\frac{4}{3}\]
Correct Answer: D
Solution :
Solving \[y+2{{x}^{2}}=0\] \[y+3{{x}^{2}}=1\] Points of intersection (1 , -2) and (-1 , -2) Area \[=2\int\limits_{0}^{1}{\left( \left( 1-3{{x}^{2}} \right)-\left( -2{{x}^{2}} \right) \right)dx}\] \[=2\int\limits_{0}^{1}{\left( 1-3{{x}^{2}} \right)dx=2\left( x-\frac{{{x}^{3}}}{3} \right)_{0}^{1}}\]\[=\frac{4}{3}\]You need to login to perform this action.
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