Find $P\left[\cfrac 13<X^2< 1\right]$

Question

For $c > 0$, let the function $f : R→R$ be defined as $f(x) =\frac{c} {x^2 + 1}$

Find $ P\left[\cfrac 13 < X^2 < 1\right]$ when X has a density of f.

First off, How can you find c without intervals? I understand that I need c to be able to find P but how can I start without an interval for the integration?

Answer 1

Note: $$\int_{-\infty}^{\infty} \frac{c}{x^2+1}dx=c\arctan{x}\bigg{|}_{-\infty}^{\infty}=c\left(\frac{\pi}{2}-\left(-\frac{\pi}{2}\right)\right)=c\pi=1 \Rightarrow c=\frac{1}{\pi}.$$ Also note: $$\frac13<x^2<1 \Rightarrow -1<x<-\frac{1}{\sqrt{3}} \ \ \text{or} \ \ \frac{1}{\sqrt{3}}<x<1;$$ $$P\left(\frac13<x<1\right)=\int_{-1}^{-\frac{1}{\sqrt{3}}} \frac{\frac{1}{\pi}}{x^2+1}dx+\int_{\frac{1}{\sqrt{3}}}^{1} \frac{\frac{1}{\pi}}{x^2+1}dx=\\ \frac{1}{\pi}\left(-\frac{\pi}{6}-\left(-\frac{\pi}{4}\right)\right)+\frac{1}{\pi}\left(\frac{\pi}{4}-\frac{\pi}{6}\right)=\frac16.$$