question_answer

The area (in square units) of the region bounded by the curves \[y+2{{x}^{2}}=0\]and \[y+3{{x}^{2}}=1,\]is equal to : JEE Main Online Paper (Held On 10 April 2015)

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 \[=2\int\limits_{0}^{1}{\left( \left( 1-3{{x}^{2}} \right)-\left( -2{{x}^{2}} \right) \right)dx}\] \[=2\int\limits_{0}^{1}{\left( 1-3{{x}^{2}} \right)dx=2\left( x-\frac{{{x}^{3}}}{3} \right)_{0}^{1}}\]\[=\frac{4}{3}\]