If $a_0+\frac{a_1}{2}+\dots + \frac{a_n}{n+1}=0$ then $a_nx^n+\dots + a_0$ has a root in $[0,1]$ [duplicate]

Let $P(x)=a_nx^n+a_{n-1}x^{n-1}+\dots +a_0$ be a real polynomial such that $a_0+\frac{a_1}{2}+\frac{a_2}{3}+\dots + \frac{a_n}{n+1}=0$.

Show that $P$ has a root in $[0,1]$. I would like proof verification and alternative proofs.

My solution:

notice that the antiderivative of $P(x)$ is $\frac{a_nx^{n+1}}{n+1}+\frac{a_{n-1}x^n}{n}+\dots + \frac{a_1x^2}{2}+a_0x+C$. From here $\int_{0}^1P(x)dx=a_0+\frac{a_1}{2}+\frac{a_2}{3}+\dots + \frac{a_n}{n+1}=0$. If $P$ had no roots then $P$ would be a strictly positive(or negative) continuous function, so its integral would be strictly positive ( or negative), a contradiction.

This problem was made famous by the Putnam competition many years ago. The customary answer is to apply Rolle's theorem to the polynomial $R(x) = \int_0^x P(t) dt$ on the interval $[0,1]$ where $R(0)=R(1)=0$, and is equivalent to your solution.