Question

The area of the region $\left\{\right(x,y):\frac{2}{1+x^2}\ge y\ge x^2\}$ is (in sq. units):

• A. $\pi -\frac{1}{3}$
• B. $\pi -\frac{2}{3}$
• C. $\frac{\pi }{2}-\frac{1}{3}$
• D. $2\pi -\frac{2}{3}$

Answer 1

The correct option is B $\pi -\frac{2}{3}$

$y=2^2,y=\frac{2}{1+x^2}\Rightarrow x^2=\frac{2}{1+x^2}\Rightarrow x=\pm 1$
$\therefore$ area of region $={\int }_{-1}^1|x^2-\frac{2}{x^2+1}|dx=2{\int }_0^1|x^2-\frac{2}{x^2+1}|dx=2{[2ta{n}^{-1}-\frac{x^3}{3}]}_0^1=2[(2.\frac{\pi }{4}-\frac{1}{3})-0]=\pi -\frac{2}{3}$