# Prove that there exists $x_0\in [a,b]$ such that $\sum_{i=1}^{n} k_i \int_{x_0}^{x_i} fdt=0$

Let $f$ is a continuous function on $[a,b]$, $x_1,x_2,\ldots,x_n\in [a,b]$, $k_1,k_2,\ldots,k_n>0$. Prove that there exists $x_0\in [a,b]$ such that $$k_1\displaystyle \int_{x_0}^{x_1} fdt+k_2\displaystyle \int_{x_0}^{x_2} fdt+\cdots+k_n\displaystyle \int_{x_0}^{x_n} fdt=0$$

The solution says that assume $g(x)=\displaystyle \sum_{i=1}^{n} k_i\displaystyle \int_{x}^{x_i} fdt$ and then consider the equality $$k_1g(x_1)+k_2g(x_2)+\cdots+k_ng(x_n)=0$$

I just can not understand why they can come up with such an equality? Is there any special property for the function $g$?

• The second term should be....$k_2g(x_2)$... – DeepSea Jan 29 '15 at 3:30
• – Paramanand Singh Jan 29 '15 at 7:34

Assume you can prove their shown equality, then it says like $k_ig(x_i)$ and $k_jg(x_j)$ must have opposite signs, and since $k_i > 0$ for all $i$, this means $g(x_i)$ and $g(x_j)$ have opposite signs, and you can apply IVT to conclude.

• Thanks for your answer, but my question is how they can find that equality. Is that just a trick or based on something more general? – Tien Kha Pham Jan 29 '15 at 3:32
• you just simplify it and you should get it. It looks simple. – DeepSea Jan 29 '15 at 3:35

Note that

$$\sum_{j=1}^nk_jg(x_j) = \sum_{j=1}^nk_j\sum_{i=1}^nk_i\int_{x_j}^{x_i}f(t)\,dt \\= \sum_{j=1}^n\sum_{i=1}^nk_ik_j\int_{x_j}^{x_i}f(t)\,dt \\=-\sum_{j=1}^n\sum_{i=1}^nk_ik_j\int_{x_i}^{x_j}f(t)\,dt.$$

Interchanging indexes $i$ and $j$ it follows that

$$\sum_{j=1}^nk_jg(x_j) = -\sum_{j=1}^n\sum_{i=1}^nk_ik_j\int_{x_j}^{x_i}f(t)\,dt= -\sum_{j=1}^nk_jg(x_j),$$

and

$$\sum_{j=1}^nk_jg(x_j) = 0.$$

Then apply the argument of @Back2Basic.