Question

The area between the curves y=$x^2$ and $y=\frac{2}{1+x^2}$ is-

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

Answer 1

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

we have,

$y=x^2......\left(1\right)$

$y=\frac{2}{x^2+1}.......\left(2\right)$

Taking limit then,

By equation (1) and (2) to, and get,

$x^2=\frac{2}{x^2+1}$

Put x=1 and we get,

${\left(1\right)}^2=\frac{2}{{\left(1\right)}^2+1}$

$1=\frac{2}{2}$

$1=1$

Now, taking’

$x=-1$ and we get,

${(-1)}^2=\frac{2}{{(-1)}^2+1}$

$1=1$

Then, the limit of this function of $$1to-1$$.

The required area$$={\int }_{-1}^1(\frac{2}{x^2+1}-x^2)dx$$

$={\int }_{-1}^1(\frac{2}{x^2+1}-x^2)dx=2{\int }_0^1(\frac{2}{x^2+1}-x^2)dx$

$=2{{[2{\tan }^{-1}x-\frac{x^3}{3}]}_0}^1$

$=2[2{\tan }^{-1}\left(1\right)-2{\tan }^{-1}0-\frac{1-0}{3}]$

$=2[2\frac{\pi }{4}-2\times 0-\frac{1}{3}]$

$=\pi -\frac{2}{3}$

Hence, this is the answer.