question_answer

The area bounded by the curve \[y={{x}^{2}}\] and \[y=\frac{2}{1+{{x}^{2}}}\] is \[\lambda \] sq. units, then the value of \[[\lambda ]\] is [Note: \[[k]\] denotes greatest integer less than or equal to k.]

A)  2         

B)  3

C)  4                                            

D)  5

Correct Answer: A

Solution :

Required area \[=\int\limits_{-1}^{1}{\left( \frac{2}{1+{{x}^{2}}}-{{x}^{2}} \right)}\,dx=2\int\limits_{0}^{1}{\left( \frac{2}{1+{{x}^{2}}}-{{x}^{2}} \right)}\,dx\] \[=2\left( 2{{\tan }^{-1}}x-\frac{{{x}^{3}}}{3} \right)_{0}^{1}=2\left( \frac{\pi }{2}-\frac{1}{3} \right)=\pi -\frac{2}{3}\] \[\therefore \left[ \pi -\frac{2}{3} \right]=2\]