The problem given was:
Let $X$ be a continuous random variable with probability density function $$f(x) = \dfrac 1 4 \min \left( 1, \dfrac 1 {x^2} \right)$$ Find $P(−2 \le X \le 4)$.
The $\min$ part has totally thrown me off, any suggestions or steps how to find $P$?
Hint: $$ f(x)=\begin{cases} \frac14 \quad &\text{if }\, |x|\leq 1,\\ \frac{1}{4}x^{-2}\quad&\text{if }\, |x|>1. \end{cases} $$