When are definite integrals of continuous functions pairwise distinct? Let $f_1, \ldots, f_n : [0,1]\rightarrow\mathbb{R}_{>0}$ be a finite family of (positive) continuous functions.
I was wondering about the weakest condition to impose on the $f_i$ so as to  guarantee that
$$\tag{1}\exists\, \text 0\leq s < t \leq 1 \quad \text{ s.t. } \ \text{ the numbers} \quad q_i(s,t):=\int_s^t\!f_i(u)\,\mathrm{d}u, \ \ i=1,\ldots, n, \quad \text{ are pairwise distinct}? $$
(In case someone's in doubt: This is not a homework question.)
 A: I would think of a statement like this one:

Assume there exists a $x\in[0,1]$ such that the $f_i(x)$, $i=1,\ldots,n$ are pairwise distinct. Then there is an $\varepsilon>0$, such that the numbers
$$ \int_{\max(0,x-\varepsilon)}^{\min(1,x+\varepsilon)}f_i(t)dt \qquad\mbox{for}\ i=1,\ldots,n $$
are pairwise distinct.

The proof goes like this: W.l.o.g. assume $f_1(x)< f_2(x)<\cdots \le f_n(x)$. Then, by continuity of the functions $f_i$, there is an $\varepsilon>0$, such that
$$f_1(t)<f_2(t)<\cdots<f_n(t) \qquad\mbox{holds for all}\ t\in(x-\varepsilon,x+\varepsilon)\cap[0,1].$$
This immediately yields
$$ \int_{\max(0,x-\varepsilon)}^{\min(1,x+\varepsilon)}f_1(t)dt\le \int_{\max(0,x-\varepsilon)}^{\min(1,x+\varepsilon)}f_2(t)dt\le\cdots\le\int_{\max(0,x-\varepsilon)}^{\min(1,x+\varepsilon)}f_n(t)dt,  $$
by monotonicity of the integral. To prove these inequalities are strict, assume that $$\int_{\max(0,x-\varepsilon)}^{\min(1,x+\varepsilon)}f_k(t)dt = \int_{\max(0,x-\varepsilon)}^{\min(1,x+\varepsilon)}f_{k+1}(t)dt \quad\Longrightarrow \int_{\max(0,x-\varepsilon)}^{\min(1,x+\varepsilon)}(f_{k+1}(t)-f_k(t))dt=0. $$
Therefore, $f_{k+1}(t)=f_k(t)$ for almost every $t\in(x-\varepsilon,x+\varepsilon)\cap[0,1]$, which is a contradiction to the above strict inequality.
