Can someone help me with this proof?
Let $f:[a,b]\rightarrow\mathbb{R}$ be a continuous function defined on a non-empty closed interval.
Prove: $|\int_{a}^{b}f(x)dx|=\int_{a}^{b}|f(x)|dx$ $\iff$ $f$ is either non-negative or non-positive.
(Hint: See $f$ as the difference between its positive part and its negative part)
So far I've got:
$\impliedby:|\int_{a}^{b}f(x)dx| = |\int_{a}^{b}f_+(x)dx -\int_a^bf_-(x)dx|\leq\int_{a}^{b}f_+(x)dx+\int_a^bf_-(x)dx=\int_{a}^{b}|f(x)|dx$
We have equality if either $\int_{a}^{b}f_+(x)dx=0$ or $\int_{a}^{b}f_-(x)dx=0$.
If $\left|\int_a^bf(x)\,\mathrm dx\right|=\int_a^b|f(x)|\,\mathrm dx$, then $\int_a^bf(x)\,\mathrm dx=\pm\int_a^b|f(x)|\,\mathrm dx$. Suppose that$$\int_a^bf(x)\,\mathrm dx=\int_a^b|f(x)|\,\mathrm dx.$$Then $\int_a^b|f(x)|-f(x)\,\mathrm dx=0$. But $|f|-f$ is a continuous non-negative function, and therefore$$\int_a^b|f(x)|-f(x)\,\mathrm dx=0\iff|f|-f=0\iff f=|f|.$$The case in which $\int_a^bf(x)\,\mathrm dx=-\int_a^b|f(x)|\,\mathrm dx$ is similar: you can deduce that $f=-|f|$.