Area between $y = \frac{2}{1+x^2}$ and $y = |x|$

Question

The exercise is from Anton 8th, page 443, question 15. The problem asks for the area between:

$f(x) = \frac{2}{1+x^2}$ and $g(x) = |x|$

It is not said what interval the area should be calculated, so assume that it is when both equations 'touch' each other, and here is my question.

Usually I would do:

$$ f(x) = g(x) $$ $$ \frac{2}{1+x^2} = |x| $$

But then I can't proceed. I've tried to put $1+x^2$ multiplying $|x|$, as something like:

$$ 2 = |x|.(1+x^2) $$ then: $$ 2 = |x|+x^3$$ $$ x^3+|x|-2 = 0$$ But I can't have three roots as the graph shows (and that operation does not look good to me, so it's probably not right) (please don't mind my awesome drawing skill to identify the area):

I did find the area based on the graph, but I have to justify where (points) both equations are touching each other to define the interval of the integral.

How could I proof my points (-1,1) and (1,1) by doing $f(x) = g(x)$?

Answer 1

A good point to start when dealing with equations with absolute values is to "fork" the equation in two, dependig on the sign of what is inside of the absolute value.

The equation

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

forks to $$\frac{2}{x^2+1}=x$$ when $x\geq 0$. The roots of the polynomial $x^3+x-2$ are $1$ and two complex numbers that are not real. The other fork is $$\frac{2}{x^2+1}=-x$$ when $x<0$. Now the polynomial is $x^3+x+2$, whose roots are $-1$ and tow nonreal complex numbers.

So the solutions are $-1$ and $1$.

Answer 2

You can just say by inspection you found $x=-1,1$ to be the points of intersection.

$$ Area = \int_{-1}^0 f -g + \int_0^1 f - g = 2 \int_0^1 (f -g)dx \;\;\; \text{by symmetry}$$