Question

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

Answer 1

The curve $y=x^2$ is a a parabola. It is symmetric about
$7$-axis and has its vertex at $(0,0)$ and the curve
$y=\frac{2}{1+x^2}$ is a bell shaped curve. X-axis is its asymptote
and it is symmetric about 7-axis and its vertex is $(0,2)$.
Since, $y=x^2$ ....(i)
and $y=\frac{2}{1+x^2}$ ....(ii)
$\Rightarrow y=\frac{2}{1+y}$
$\Rightarrow y^2+y-2=0$
$\Rightarrow (y-1)(y+2)=0$
$\Rightarrow y=-2,1$
But $y\ge 0$ , so $y=1$ $\Rightarrow x=\pm ;1$
Therefore, coordinates of $C$ are $(-1,1)$ and coordinates of $B$ are (1,1).
$\therefore$ Required area $OBACO=2x$ Area of curve $OBAO$
$=2[\underset{0}{\overset{1}{\int }}\frac{2}{1+x^2}dx-\underset{0}{\overset{1}{\int }}{x}^{2}dx]$
$=2[[2{\tan }^{-1}x{]}_0^1-\left[\frac{x^3}{3}{]}_0^1\right]$
$=2[\frac{2\pi }{4}-\frac{1}{3}]=(\pi -\frac{2}{3})$
sq unit