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 $y \ge0$ , 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 = 2 x $ Area of curve $OBAO$
$=2 \Big[ \int\limits_0^1 \frac {2}{1+x^2} dx - \int\limits_0^1 x^2 dx \Big]$
$=2 \Big[ [2 \tan^{-1} x ]_0^1 - \Big[ \frac {x^3}{3}\Big]_0^1\Big]$
$=2 \Big[ \frac {2\pi}{4}-\frac {1}{3} \Big]= \Big( \pi-\frac {2}{3} \Big)$
sq unit