Splitting up integral when there is an absolute value

Question

For evaluating an integral with an integrand that has an absolute value, will it work if I split up the integral into sections where the integrand equals zero and then take the absolute value of each part? (a specific easy example below)

$\int_0^4|(x-1)(x-3)|dt=\left|\int_0^1(x-1)(x-3)dt \right|+\left|\int_1^3(x-1)(x-3)dt \right|+\left|\int_3^4(x-1)(x-3)dt \right|$

I know that it's normally done where one just takes the integral of the function when it's positive and then takes the negative of the function when it's negative (so that it turns out positive in the end). But I'm wondering if this way above is equivalent to that all the time?

I haven't been able to find an equality of it, apart from maybe the 6th one here: https://en.wikipedia.org/wiki/Integral#Inequalities

Answer 1

Yes, you can do that. The equality you wrote is fairly easy to prove:

1. Since $$\int_{0}^4f(x)dx = \int_0^1 f(x)dx+\int_1^3 f(x)dx+\int_3^4 f(x)dx$$ is true for any function that is integrable on $[0,4]$, this is also true for $f(x)=|(x-1)(x-3)|$, which means that $$\int_{0}^4|(x-1)(x-3)|dx =\\= \int_0^1 |(x-1)(x-3)|dx+\int_1^3 |(x-1)(x-3)|dx+\int_3^4 |(x-1)(x-3)|dx$$
2. For any function $f$ that is either always non-negative or always non-positive on $[a,b]$, you have $$\int_a^b |f(x)|dx=\left|\int_a^bf(x)dx\right|$$ and if you apply this to all three summands from point 1 (and you can apply it, since the function is entirely non-negative in the first and last summands and entirely negative in the second one), and get the result you need.