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: $-\lvert\,f(x)\,\rvert\leq\,f(x)\leq\lvert\,f(x)\,\rvert$
 A: Some other methods :
Method 1
Set $f^+(x)=\max\{f(x),0\}$ and $f^{-}(x)=-\min\{f(x),0\}$. Then $$\lvert\,f\,\rvert=f^++f^-\quad \text{and}\quad f=f^+-f^-.$$
Both $f^+$ and $f^-$ are integrable. Then $$\left|\int_a^b f\right|=\left|\int_a^b f^+-\int_a^b f^-\right|\leq \int_a^b f^++\int_a^bf^-=\int_a^b(f^++f^-)=\int_a^b |f|.$$
Method 2
Let $(x_i^n)_i$ a subdivision of $[a,b]$ s.t. $h^n\to 0$ when $n\to \infty $ where $h_i^n=x_{i+1}^n-x_i$. Then, $$\left|\int_a^b f\right|=\left|\lim_{n\to \infty }\sum_{i}f(x_i^n)h_i^n\right|\leq \lim_{n\to \infty }\sum_{i}|f(x_i^n)|h_i^n=\int_a^b|f|.$$
A: 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 $\lvert\,a\,\rvert<b$
