An interesting real analysis problem involving integrals

There's a little trick you can use here. Let $g(x) = e^{-x}\int_0^x f(y)\,dy$. Note that $g(0) = g(1) = 0$. Now apply Rolle's Theorem.

Lemma 1: Let $g$ be a differentiable function $[0,1]\rightarrow\mathbb{R}$ such that $g'-g>0$ (resp. $<0$) on $(0,1)$ and $g(0)=0$, then $g>0$ (resp. $<0$) on $(0,1]$.

Proof: (Zarrax's trick) Let $h=\exp(-x)g$. Then $h'$ is strictly positive on $(0,1)$ and $h(0)=0$, whence $h>0$ on $(0,1]$ and therefore $g>0$ on $(0,1]$.

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Now let $g(x)=\int_0^xf(t)dt$. Let $\lambda(x)=g'-g$. Either $\exists c:\lambda(c)=0$ (in which case we are done) or $\lambda$ doesn't change sign on $[0,1]$. By lemma 1, $g(1)\neq 0$, which is a contradiction.