Question

Sketch the region bounded by the curves $y=x^2$ & $y=2/(1+x^2)$. Find the area:

• A. $\pi -\frac{2}{3}$
• B. $\pi -\frac{1}{3}$
• C. $\pi -\frac{5}{3}$
• D. $\pi -\frac{7}{3}$

Answer 1

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

$y=x^2$ and $y=\frac{2}{1+x^2}$
Therefore
$x^2=\frac{2}{1+x^2}$
$x^4+x^2-2=0$
$(x^2+2)(x^2-1)=0$
Now x is real.
Therefore $x^2+2=0$ is not possible.
Hence
$x=\pm 1$.
Hence the required area is
$={\int }_{-1}^1{x}^2-\frac{2}{1+x^2}$
$=2{\int }_0^1{x}^2-\frac{2}{1+x^2}$
$=2[\frac{x^3}{3}-2ta{n}^{-1}(x){]}_0^1$
$=2[\frac{1}{3}-2.\frac{\pi }{4}]$
$=|\frac{2}{3}-\pi |$
$=\pi -\frac{2}{3}$ sq.units.