$\int_{0}^{x} f(t) dt = \int_{x}^{1} f(t) dt \implies f(x) = 0 \forall x \in [0,1]$ $f:[0,1] \to \mathbb{R}$ is continuous.
$\int_{0}^{x} f(t) dt = \int_{x}^{1} f(t) dt \implies f(x) =  0 \forall x \in [0,1]$
I don't understand why I can say $f(1)=f(0)=0$.
My attempt is using FTC:
$\int_{0}^{x} f(t) dt = \int_{x}^{1} f(t) dt  \implies \int_{0}^{x} f(t) dt= \int_{1}^{x} -f(t) dt \implies f(x) = -f(x) \forall x \in (0,1) \implies f(x) =  0 \forall x \in (0,1)$. However, the statement is about $[0,1]$. How can I conclude for $f(0)$ and $f(1)$?
Thanks
 A: Assume $f$ is continuous on $[0,1]$. If $\int_{0}^{x}f(t)dt=\int_{x}^{1}f(t)dt$ for all $0 \le x \le 1$, then the Fundamental Theorem of Calculus gives
$$
       f(x)=\frac{d}{dx}\int_0^x f(t)dt=\frac{d}{dx}\int_x^1f(t)dt = -f(x)
$$
So $f(x)=0$ for $0 \le x \le 1$. (You take a right-hand derivative at $x=0$ and a left-hand derivative at $x=1$.)
A: I agree that the Fundamental Theorem of Calculus is the weapon to use here. However, if you want to get your hands a little bit dirtier, notice that 
$$\int_0^x f(t)dt = \int_x^1f(t)dt = \int^1_0f(t)dt - \int_0^xf(t)dt$$
implies that
$$\int_0^x f(t)dt = \frac 12 \int_0^1 f(t)dt.$$
In particular, this also means (write the previous identity for $x=a$ and $x=b$, and substract them) that
$$\int_a^bf(t)dt = 0,\quad\forall a,b\in(0,1).$$
Finally, assume there exists $x_0\in(0,1)$ such that $f(x_0)>0$. Since $f$ is continuous, there exists $\delta>0$ such that $f(x)>\frac 12 f(x_0) > 0$ for each $|x-x_0|<\delta$ (definition of continuity with $\epsilon = \frac 12f(x_0)$). This implies that the integral of $f$ over $(x_0-\delta,x_0+\delta)$ must be strictly positive, contradicting the previous equation. If you assume $f(x_0)<0$, an analogous reasoning follows.
