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Given $a_0, a_1, .., a_n$ are the real numbers satisfying $$\dfrac {a_0}{n+1} + \dfrac {a_1}{n} +......+\dfrac {a_{n-1}}{2}+a_n=0$$ then prove that there exists at least one real root of the equation $a_0 x^n + a_1 x^{n-1} +....+a_n=0$ such that $x\in (0,1)$.

My teacher told me to integrate the left hand side of the given equation and call it as $g(x)$, then after verifying the requirements of Rolle's Theorem again differentaite $g(x)$ to get the required equation and this completes the proof. However, I don't understand that first integrating and then differentiating the same thing works here.

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  • $\begingroup$ Rolle's Theorem allows you to conclude something about a value of $f'(x)$ given some information about some values of $f(x)$. You want to conclude something about $a_0x^n+a_1x^{n-1}+\cdots +a_n$, so you need to view this as $f'(x)$ for some $f(x)$. Hence you need to integrate to figure out what $f(x)$ is. $\endgroup$ – kccu Mar 31 at 17:39
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Let $$p(x) = a_0x^n + a_1x^{n-1}+\ldots + a_n$$

and let $g$ be an antiderivative of $p$ which satisfies $g(0)=0$, that is

$$g(x) = \frac{a_0x^{n+1}}{n+1}+\frac{a_1x^n}{n}+\ldots +a_nx$$

Check that $g(1)=g(0)=0$, now apply Rolle's theorem to conclude about root of $p$.

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