Find the continuous function such that the Riemann integrable is the same

Question

Find all functions $f$ such that $f$ is continuous on $[0,1]$ and

$\int_0^x f(t) dt = \int_x^1 f(t) dt$

for every $x \in (0,1)$

I can't think of any function that would satisfy this property! Please help!

Answer 1

Hint: If $g(x) := \displaystyle\int_{0}^{x}f(t)\,dt - \int_{x}^{1}f(t)\,dt = 0$ for all $x \in (0,1)$, then $g'(x) = 0$ for all $x \in (0,1)$. Now apply the fundamental theorem of calculus.

Answer 2

You have $\int_0^1 f = \int_0^x f+ \int_x^1 = 2 \int_0^x f$, and so $x \mapsto \int_0^x f$ is a constant (and so we have $\int_a^b f = 0$ for any $0<a <b <1$). Since $f$ is continuous, it must be zero.