$f:[0,2]\to\mathbb{R}$ 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?

For an alternative approach, using the Fundamental theorem of calculus and the given identity:
$$|f(x)| = \left(\int_0^x |f(t)|\;dt\right)' = \left(\int_x^2 |f(t)|\;dt\right)' = -|f(x)|$$
Therefore $\;\;|f(x)| = -|f(x)|$ $\;\;\implies\;\; |f(x)| = 0$ $\;\;\implies\;\; f(x) = 0\;\;$ for $\;\; x \in [0,2]$.
