Show that there is an $x$ such that $\sum_{k = 1}^{n}a_{k}\int_{x}^{x_{k}}f(t)\,dt = 0$ 
Let $f(x)$ be continuous in $[a, b]$ and let $a_{1}, a_{2}, \ldots, a_{n}$ be positive constants. Also let $x_{1}, x_{2}, \ldots, x_{n}$ be any $n$ points in $[a, b]$. Show that there is a point $x \in [a, b]$ for which $$a_{1}\int_{x}^{x_{1}}f(t)\,dt + a_{2}\int_{x}^{x_{2}}f(t)\,dt + \cdots + a_{n}\int_{x}^{x_{n}}f(t)\,dt = 0$$

The case $n = 2$ is not that difficult. For $n = 2$ let's remove subscripts (to save labor in typing) and suppose $p, q$ play role of positive constants and $u, v$ play the role of two points in $[a, b]$. We need to find an $x \in [a, b]$ for which $$g(x) = p\int_{x}^{u}f(t)\,dt + q\int_{x}^{v}f(t)\,dt = 0$$ Clearly if $u = v$ then $x = u = v$ is a solution of the above equation. Hence let $u < v$. Now $g(u) = q\int_{u}^{v}f(t)\,dt$. If $g(u) = 0$ then $x = u$ and we are done. If $g(u) \neq 0$ then we can see that $g(v) = p\int_{v}^{u}f(t)\,dt$ and then $g(u)g(v) = -pq\left(\int_{u}^{v}f(t)\,dt\right)^{2} < 0$. By the intermediate value property of continuous function $g(x)$ there is a point $x \in (u, v)$ where $g(x) = 0$. This solves the problem when $n = 2$.
But extending the same logic seems difficult and clumsy. Considering the value of function $g$ at all the points $x_{i}$ and then finding out those two points where $g$ has different sign seems tricky. I think there is some alternative approach which can be applied directly for the general case rather than dealing with $n = 2$. Any hints would be greatly appreciated.
 A: Let $g(x)=\sum_{k=1}^na_k\int_x^{x_k}f(t)\,dt$. We want to prove that $g(x)=0$ for some $x\in[a,b]$. If we define $x_0=x$, then we have
$$\begin{align*}
g(x)=&\,\sum_{k=1}^na_k\sum_{j=0}^{k-1}\int_{x_j}^{x_{j+1}}f(t)\,dt\\
=&\,\sum_{j=0}^{n-1}\sum_{k=j+1}^{n}\biggl[a_k\int_{x_j}^{x_{j+1}}f(t)\,dt\biggr]\\
=&\,\sum_{j=0}^{n-1}\,b_j\int_{x_j}^{x_{j+1}}f(t)\,dt\,,
\end{align*}$$
where $b_j=\sum_{k=j+1}^{n}a_k$, for $j=0,\dots,n-1$. Thus, we have
$$g(x)=b_0\int_x^{x_1}f(t)\,dt+\sum_{j=1}^{n-1}\,b_j\int_{x_j}^{x_{j+1}}f(t)\,dt\,,$$
and now I will show that for some $x$ between $x_1$ and $x_n$ we have
$$\sum_{j=1}^{n-1}\,b_j\int_{x_j}^{x_{j+1}}f(t)\,dt=b_0\int_{x_1}^xf(t)\,dt\,.$$
Redefine $x_0$ as $x_0=x_1$ (forget the previous definition $x_0=x$), and for $j=0,\dots,n-1$ let $c_j=\int_{x_j}^{x_{j+1}}f(t)\,dt$. Then the desired result is:
$$\sum_{j=0}^{n-1}b_jc_j=b_0\int_{x_1}^xf(t)\,dt,\ \style{font-family:inherit;}{\text{for some}}\ x\ \style{font-family:inherit;}{\text{between}}\ x_1\ \style{font-family:inherit;}{\text{and}}\ x_n\,.$$
Recall the summation by parts formula (prove it!):
$$\sum_{j=0}^{n-1}b_jc_j=s_{n-1}b_{n-1}+\sum_{j=0}^{n-2}s_j(b_j-b_{j+1}),\ \style{font-family:inherit;}{\text{where}}\ s_j=c_0+c_1+\cdots+c_j\,.$$
Note that $s_j=\int_{x_1}^{x_{j+1}}f(t)\,dt$ for $j=0,\dots,n-1$, so all the values $s_j$ are assumed by the continuous function $h(x)=\int_{x_1}^xf(t)\,dt$. On the other hand, let $m$ and $M$ be, respectively, the smallest and the largest of the values $s_0,s_1,\dots,s_{n-1}$. Since the $a_i$ are positive, it follows that the $b_j$ form a strictly decreasing sequence of positive  numbers, so from the inequalities $m\leq s_j\leq M$ we obtain
$$mb_{n-1}+\sum_{j=0}^{n-2}m(b_j-b_{j+1})\leq s_{n-1}b_{n-1}+\sum_{j=0}^{n-2}s_j(b_j-b_{j+1})\leq Mb_{n-1}+\sum_{j=0}^{n-2}M(b_j-b_{j+1})\,,$$
that is
$$mb_0\leq\sum_{j=0}^{n-1}b_jc_j\leq Mb_0\,.$$
Finally, since both $m$ and $M$ are in the range of $h$, it follows from the intermediate value theorem that $h(x)=\frac1{b_0}\,\sum_{j=0}^{n-1}b_jc_j$ for some $x$ between $x_1$ and $x_n$, which is the desired result.
A: Well this one has been dormant for a while, but a recent edit popped it up again. It seems rather too simple for an elaborate answer, and so I will supply a trivial one.
Note, first of all, that we may assume that the positive constants satisfy
$$\sum_{k=1}^n a_k = 1$$
since we can divide by that sum without changing the problem.
Let $F(x)=\int_a^x f(t)\,dt$   on the interval $[a,b]$.  Suppose that $M$ is the
maximum value of $F$ and that $m$ is the minimum.
Write the required condition 
$$a_{1}\int_{x}^{x_{1}}f(t)\,dt + a_{2}\int_{x}^{x_{2}}f(t)\,dt + \cdots + a_{n}\int_{x}^{x_{n}}f(t)\,dt = 0$$
in the form
$$a_{1}[F(x_1)-F(x)] + a_{2}[F(x_2)-F(x)]+ \cdots + a_{n}[F(x_n)-F(x)] = 0$$
and finally as
$$F(x) = \sum_{k=1}^n a_k F(x_k).\ \ \ \ \ (\ast)$$
Since $$m \leq  \sum_{k=1}^n a_k F(x_k) \leq M$$
there is a point $x$ in the interval where the continuous function  $F$ assumes
exactly the value in $(\ast)$.
