# Symmetry between integrals including absolute value

So I came across below symmetry in my probability course that I can't understand.

I understand how the lower bound changes when removing the absolute value operator, but how does the 2 disappear?

• Draw the graphs of the integrands. Sketch in the areas represented by the integrals. The result will be obvious. – whuber Nov 20 '19 at 16:17

In the left hand you have the integral of a function of $$|v|$$ (considering the quadratic term too), hence that function is symmetric on $$v$$'s domain. This comes simply from the fact that $$f(|-v|) = f(|v|)$$.
Integrating that function on positive values of $$v$$ yields the same result as integrating it on negative values of $$v$$, so, to compute the whole integral, it is sufficent to only perform the integral on positive $$v$$ and to double it (if you don't you will not account for negative $$v$$).