If $\int_0^{x/3} f(t)dt =\int_0^xf(t)dt$, prove $f$ is identically $0$ $f:[0,1] \to \mathbf R$ is continuous. If $$\int_0^{x/3} f(t)dt =\int_0^xf(t)dt$$ for all $x$ in $[0,1]$, prove that $f$ is identically $0$. 
My thought is to prove that the maximum and minimum of $f$ are equal then $f$ is constant and this constant can only be zero. But I can't think of a way to do that. Can somebody help and give me some hints. Thanks.
 A: Hint: 
FTC $\implies \frac13 f(x/3) = f(x)$ for all $x \in [0,1]$
Added:
Which in turn gives:
$$\frac1{3^n} f\left(\frac1{3^n}x\right) = f(x), \ \forall \ x \in [0,1], \ n \in \mathbb N$$
A: Let $F(x) = \int_0^x f(t) \mathrm{dt}, x \in [0,1]$ be a primitive of $f$.
Taking  $x \in [0, 1]$, from $F(x)=F(x/3)$ we get (by recursively changing $x \rightarrow x/3$) $F(x)=F(x/3^{n}), \forall n \in \mathbf N$. Because F continous and $(x/3^n) \rightarrow 0$ when $n \rightarrow \infty$ we have  $F(x)=F(0)$. Therefore F is constant and $f = F'= 0$
A: Expand @BolzWeir's answer.  
First I propose to show that $f(0)=0$:
Due to the equality we have
$$\int_{x/3}^x f =0$$
for all $x\in[0,1]$. Note that $f$ is uniformly continuous on $[0,1]$, which is to say, for an $\epsilon>0$ we can find a uniform $\delta>0$ that suits all $x\in [0,1]$. Suppose $f(0)=a> 0$, then let $\epsilon=a/2$, and take an arbitrary $x\in(0,\delta)$, we would have
$$\int_{x/3}^x f >a/2\cdot 2x/3>0$$
a contradiction. The case where $a<0$ is handled similarly. 
By continuity, it means $f(x)\to 0$ as $x\to 0^+$. And for all $x\in[0,1]$, based on @BolzWeir's answer 
$$3^nf(x)=f(\frac x{3^n})\to 0\quad \text{as}\, n\to\infty$$
which indicates $f(x)=0$. 
A: Here is another way:
It is straightforward to see from the hypotheses that
$\int_0^x f = \int_0^{x \over 3^n} f$ for any $n$ and, taking limits, that
$\int_0^x f = 0$ for any $x$. In particular, it follows that
$\int_a^b f = \int_0^b f -\int_0^a f = 0$ and hence (reasoning by contradiction on a suitably small interval) that $f = 0$.
If $f$ is just integrable, the above reasoning can be extended to show that $f(x) = 0$ for ae. $x$.
