Question

Sketch the region bounded by the curves $y=x^2$ and $y=\frac{2}{1+x^2}$. Find the area.

Answer 1

The given curves are $y=x^2$
and $y=\frac{2}{1+x^2}$
Solving (1) and (2), we have
$x^2=\frac{2}{1+x^2}$
$\Rightarrow x^2+x^2-2=0$
$\Rightarrow (x^2-1)(x^2+2)=0$
$\Rightarrow x=\pm 1$
Also, $y=\frac{2}{1+x^2}$ is an even function.
Hence, its graph is symmetrical about y-axis.
At $x=0,y=2$, by increasing the values of $x,y$ decreases and when $x\to \infty ,y\to 0$.
$\therefore y=0$ is an asymptote of the given curve.
Thus, the graph of the two curves is as follows
Ref. image
$\therefore$ The required area
$=2{\int }_0^1(\frac{2}{1+x^2}-x^2)dx$
$={(4{\tan }^{-1}x-\frac{2x^3}{3})}_0^1$
$=\pi -\frac{2}{3}sq.units$