Integrals conditioning relations Let be $f:[0,1]\to\mathbb R$ a continuous function such that:
$$\int_0^1 f(x) dx = 0 $$
Prove that there exists $c\in(0,1)$ such that:
$$\int_0^c xf(x) dx = 0 $$                     
I tried to go the integration by parts way but got stuck. I'm also curious if the converse is true.
 A: Suppose not.
Then for all $c \in (0,1)$, we can say without loss of generality that $\displaystyle \int_0^c xf(x) > 0$. Let $\displaystyle F(t) := \int_0^t f(x)dx$. Then integrating by parts, as you suggested, yields 
$$ 0 < \int_0^t xf(x) dx = tF(t) - \int_0^t F(x)dx \quad \forall t \in (0,1) \tag{1}$$
One can justify that the limit as $t \to 1$ exists, and going to that limit in $(1)$ we get that 
$$\int_0^1 F(x)dx \leq 0 \tag{2}$$
Now, we get a little witty. Define $G: [0,1] \to \mathbb{R}$ by $\displaystyle G: = \frac{\int_0^tF(x)dx}{t}$ if $t \neq 0$, and $G:= 0$ if $t = 0$. Then you can check that $G$ is differentiable, and $\displaystyle G' = \frac{tF(t) - \int_0^t F(x)dx}{t^2} > 0$.
Thus $G$ is increasing on $(0,1)$, and thus nondecreasing on $[0,1]$. And so, as $G(0) = 0$, we have that $G(t) > 0$ if $t > 0$. But then $\displaystyle \int_0^1 F(x)dx > 0$, contradicting $(2)$.
Thus there exists such a $c$.
With respect to the converse - it's very much false. If you let $f$ be a smooth function supported (and positive) on $[1/2, 1]$, then choosing any $c$ in $(0, 1/2)$ will force $\displaystyle \int_0^c xf(x) dx = 0$, although $\int_0^1 f(x)dx = \int_{1/2}^1 f(x)dx > 0$ as I chose $f$ postive. 
