Let $f$ be continuous on $[0,1]$, and suppose that for all $x$, $0<x<1$,
$\int_0^x f = \int_x^1 f$
Can you determine $f$?
I've argued the following:
Since $f$ is continuous on $[0,1]$, it follows that $f$ has an antiderivative $F$ on $I=[0,1]$.
Evaluating both sides of the equation we get
$F(x)-F(0) = F(1)-F(x)$=
$2F(x) = F(1) + F(0)$
$F(x) = \frac{F(1) + F(0)}{2}$
Taking the derivative of both sides we get
$F'(x) = f(x) = \frac{F'(1) + F'(0)}{2} = 0$ since $F(1) + F(0)$ is a constant.
I'm not sure if I'm going about this problem correctly.
Best Answer
Your method is generally correct, but I think this problem is easier than you think. We have:
$$\int_0^x f(t)dt=\int_x^1 f(t)dt=-\int_1^x f(t)dt$$
Use the fundamental theorem of calculus to take the derivative of both sides:
$$f(x)=-f(x)\rightarrow 2f(x)=0\rightarrow f(x)=0$$
Thus, $f(x)=0$ for all $x\in [0, 1]$.