$f:[0,2]\to\mathbb{R}$ continuous, then if $\int_0^x |f(t)| \ dt = \int_x^2 |f(t)| \ dt$ then $f(x) = 0$ I have that
$f:[0,2]\to\mathbb{R}$ continuous, then if $\int_0^x |f(t)| \ dt = \int_x^2 |f(t)| \ dt$, $f(x) = 0$ for $x\in [0,2]$
I tried to substitute $x=0$ in the equality to get:
$$0 = \int_0^0 |f(t)| \ dt = \int_0^2 |f(t)| \ dt$$
So we have that $$\int_0^2 |f(t)| \ dt = 0$$
then $f(x) = 0$ because we're integrating something that's always positive or $0$.
Am I right?

Right so far.  To conclude, reason as follows.  Since $f(x)$ is continuous, so is $|f(x)|$. Therefore, if there existed a $x_* \in [0, 2]$ at which $|f(x_*)| > 0$, then (because of the said continuity) $x_*$ would have a neighborhood in which $|f| > $ some positive constant.  This, in turn, would imply
$$
\int_{0}^{2} |f(t)| \; dt > \int_{\mbox{that neighborhood}} |f(t)| \; dt > 0,
$$
a contradiction.
