Prove that the equation: $c_0+c_1x+\ldots+c_nx^n=0$ has a real solution between 0 and 1.

Let $c_0,c_1,c_2,\ldots ,c_n$ be constants such that :

$$c_0+\frac{c_1}{2}+\ldots+\frac{c_{n-1}}{n}+\frac{c_n}{n+1}=0$$

I have to prove that the equation: $$c_0+c_1x+\ldots+c_nx^n=0$$

Has a real solution between 0 and 1.

Didn't know how to start...I thought that maybe I could use something about derivative...But I'm lost... Any help,much appreciated!!!

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HINT 1: $$c_0 + \dfrac{c_1}{2} + \cdots + \dfrac{c_n}{n+1} = \int_0^1 \left( c_0 + c_1x + \cdots + c_nx^n\right) dx$$

HINT 2:

If $f(x)$ is continuous and $\displaystyle \int_a^b f(t) dt = 0$, then can you conclude that there is a root of $f(x)$ between $a$ and $b$?

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Thanks so much! –  Charlie Sep 21 '12 at 15:19

This is an exercise from the differentiation chapter in Baby Rudin, and knowledge of integration is not expected. You can solve it without integrals using the following function:

$$f(x) = c_0x+\frac{c_1x^2}{2}+\ldots+\frac{c_{n-1}x^n}{n}+\frac{c_nx^{n+1}}{n+1}$$

What value does it take at $0$ and $1$? What value must the derivative take in $(0, 1)$ then?

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Thanks a lot!!!Very helpful! –  Charlie Sep 21 '12 at 15:16
This exercise also appears in Spivak's Calculus, if you should know. –  Pedro Tamaroff Sep 29 '12 at 21:39
This is very much the same approach, you just wrote the integral. –  Sawarnik Mar 5 at 20:16