Suppose $f:[0,1] \Rightarrow \mathbb{R}$ is continuous and $\int_0^x f(x)dx = \int_x^1 f(x)dx$. Prove that $f(x) = 0$ for all $x$ Suppose $f:[0,1] \Rightarrow \mathbb{R}$ is continuous and $\int_0^x f(x)dx = \int_x^1 f(x)dx$. Prove that $f(x) = 0$ for all $x$.
So, I can intuitively see that this is true. My proof mostly makes sense, I think, but I'm not sure if it covers the case where there are negative and positive values in each segment, resulting in a mean value of 0, but still having nonzero values. Can someone tell me how to cover that, or how this does?
Suppose $f(x) \neq 0$ for all x $\in$ [0,1]. By the mean value theorem, there exist some c and d for which $f(c)(x) = f(d)(1-x) = \int_0^x f(x)dx$. Since x $\neq$ (1-x) for all x $\in$ [0,1], it follows that $f(c) = f(d) = 0$. Therefore, $f(x) = 0$ for all $x \in [0,1]$. 
 A: Here is another line of attack:
Since $\int_0^x f(t)dt = \int_x^1 f(t)dt$, we also have $\int_0^y f(t)dt = \int_y^1 f(t)dt$, so subtracting gives
$\int_0^y f(t)dt-\int_0^x f(t)dt = \int_y^1 f(t)dt-\int_x^1 f(t)dt$, or
$\int_x^y f(t)dt = \int_y^x f(t)dt = -\int_x^y f(t)dt$.
In other words, $\int_x^y f(t)dt = 0$ for all $x,y$.
Since $f$ is continuous, if $f(x_0) \neq 0 $ for some $x_0$, then there is some interval $[a,b]$ containing $x_0$ for which $|f(x)| \ge \frac{1}{2} |f(x_0)|$ for all $x \in [a,b]$. If follows that $\int_a^b f(t)dt \neq 0$, a contradiction.
A: The Fundamental Theorem of Calculus proof suggested in a comment by Peter Tamaroff is one short line, and one cannot do better.
Here is a more awkward proof that does not use the FTC. Suppose that $f(x)\ne 0$ for some $x$. Say for example that $f(a)=c\gt 0$ for some $a$. By continuity we can assume that $a$ is not $0$ or $1$. Then there is an interval $(a-\epsilon,a+\epsilon)$ contained in $(0,1)$ such that $f(x)\gt c/2$ in this interval. 
Note that $\int_{a-\epsilon}^{a+\epsilon}\gt c\epsilon\gt 0$. 
Let $x_1=a-\epsilon$, and $x_2=a+\epsilon$. Then if $\int_0^{x_1} f(t)\,dt=\int_{x_1}^1 f(t)\,dt$, we must have $\int_{0}^{x_2} f(t)\,dt \gt \int_{x_2}^1 f(t)\,dt$. 
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A: I think this works:
$\int_0^0 f(x)dx =0= \int_1^0 f(x)dx$ ; so that:
$(i) \int_0^x-\int_x^1=0$
Now, we also have:
$(ii)$ $\int_0^x+\int_x^1=\int_0^1=0$ (since $\int_1^0 f(x)dx=0)$
Taking these two:
$\int_0^x-\int_1^x=\int_x^1+\int_0^x$, we get $\int_0^x f(t)dt=0$ for all $x$
So that the function $F(x)=\int_0^xf(t)dt$ is constant in $x$.
Then $\frac{d}{dx}\int_0^xf(t)dt=f(x)=0$ for all $x$
