prove the absolute value of the integral from $a$ to $b$ of $f$ is less or equal than integral from $a$ to $b$ of the absolute value of $f$ if $f$ is integrable on $[a,b]$ , then
$$\bigg\lvert\,\int_a^b{f(x) dx}\,\bigg\rvert \leq \int_a^b{\big\lvert\,f(x)\,\big\rvert\,dx}$$
I know how this works because of the area of the integrals, but I can't express it as a proof.
There's a hint: $-\lvert\,f(x)\,\rvert\leq\,f(x)\leq\lvert\,f(x)\,\rvert$

Hint: if $\forall x \in [a,b], f(x) \leq g(x)$, then $\int_a^b f(x)dx \leq \int_a^b g(x)dx$
How to actually do it:
We apply that statement in the following two cases


*

*$f$ and $\lvert\,f\,\rvert$: So $\int_a^b f(x)dx \leq \int_a^b \lvert\,f(x)\,\rvert\,dx$

*$-f$ and $\lvert\,f\,\rvert$ So $\int_a^b -f(x)dx = -\int_a^b f(x)dx \leq \int_a^b \lvert\,f(x)\,\rvert\,dx$
Now we apply that $a<b$ and $-a<b$ together imply $\lvert\,a\,\rvert<b$
