Does $\int_0^1 f(x)=\int_0^1 xf(x)$ imply $\int_0^x f(t)$ has a root My question is whether or not the following is true:
If $f:[0,1]\to \mathbb{R}$ is a continous function such that
$$\int_0^1 f(x)dx=\int_0^1 xf(x)dx$$
then there exist $c\in(0,1)$ such that
$$\int_0^c f(x)dx=0$$
It is quite clear that in the interval $(0,1)$ there must be some points $a$ such that $f(1)f(a)<0$ but I can't say from this if the statement above is true or not.

Hint: Note that
$$
\int_0^1 F(x)dx =\int_0^1\left(\int_0^x f(t)dt\right)dx=\int_0^1\left(\int_t^1dx\right)f(t)dt= \int_0^1 (1-t)f(t)dt
$$ where $F(x) = \int_0^x f(t)dt$.
