Question

Area bounded by $y=2x^2$ and $y=\frac{4}{(1+x^2)}$ will be (in sq units)

• A. $(2\pi +4/3)$
• B. $(2\pi -4/3)$
• C. $4/3-2{\tan }^{-1}2+\pi /2$
• D. $4/3-8{\tan }^{-1}2+2\pi$

Answer 1

The correct option is B $(2\pi -4/3)$

$\begin{array}{c}Pointofintersection:\\ 2x^2=\frac{4}{1+x^2}\\ \Rightarrow x^4+x^2-2=0\\ \Rightarrow x^4+2x^2-x^2-2=0\\ \Rightarrow \left(x^2-1\right)\left(x^2+2\right)=0\\ Then\\ Areaboundedbyy=2x^{2}andy=\frac{4}{1+x^2}\\ 2{\int }_0^1\left(\frac{4}{1+x^2}-2x^2\right)dx=2{\left[4{\tan }^{-1}x-\frac{2x^3}{3}\right]}_0^1\\ =2\left[4\times \frac{\pi }{4}-\frac{2}{3}\right]\\ =2\left[\pi -\frac{2}{3}\right]=2\pi -\frac{4}{3}\\ Hence,optionBisthecorrectanswer.\end{array}$