Let p(x) be a polynomial so that int(a-->a+1){p(x)dx} = 0 for all a in R.
Prove p(x)=0 for all x in R.
Suppose $\displaystyle p(x)\neq 0$ and let $\displaystyle a\in\mathbb{R}$ be the largest real root of $\displaystyle p(x)$ (if the pol. has no real roots then the contradiction is immediate since then the pol's values are all positive or all negative), and get a contradiction evaluating the integral from a to a+1.
Tonio
"...and get a contradiction evaluating the integral from a to a+1."
How I get this kind of contradiction(for the maximality of a as root of p(x))?
My try:
let deg(p(x))=n
so int(a-->a+1){p(x)dx} = q(a+1) - q(a) = 0 , when deg(q(x))=n+1
so, deg{q(x+1) - q(x)}=n
q(a+1) - q(a)= m(a+1)=0
contradiction? (m(x)!=p(x))
Thanks!
So, $\displaystyle I_n=\sum_{j=1}^{n}\int_{j}^{j+1}p(x)dx=\int_{1}^{n +1}p(x)=0$. Thus, $\displaystyle \lim_{n\to\infty}I_n=0$. Thus, if $\displaystyle p(x)=a_0+\cdots+a_nx^n$ then $\displaystyle \int p(x)=a_0x+\cdots+ a_nx^{n+1}$ and so $\displaystyle \lim_{n\to\infty}\left(a_0x+\cdots+a_nx^{n+1}\righ t)=0$. Do your stuff.