Proof or counterexample to $ \int_a^t |f(x)|dx \leq \sup_t|\int f(x)dx| $

Question

I'm trying to finish off an exercise and the following identity would come in handy, if it were true

$$\int_0^t |f(x)|dx \leq \sup_{0 \leq t \leq T} \left| \int_0^t f(x)dx \right|$$

Let's assume $f:[0,T] \rightarrow \mathbb{R}$ is at least continuous. Does it hold? I was trying to imagine when it could fail, perhaps when there's lots of cancellation in the integral from large negative and large positive values of $f$, which would add up to become big in the $|f(x)|$ integral but would end up being zero for the $f(x)$ integral, but then this would be prevented by the supremum, so I have a hunch it should work.

Answer 1

I'm supposing the integral with the absolute value inside should have an upper limit of $T$. If so, then the inequality is reversed, and equality holds if and only if $f$ is either non-negative [almost everywhere] on $[0,T]$ or non-positive [almost everywhere] on $[0,T]$.

The reversed inequality is essentially the triangle inequality.