TITLE: How can I answer this Putnam question more rigorously?
QUESTION [40 upvotes]: Given real numbers $a_0, a_1, ..., a_n$ such that $\dfrac {a_0}{1} + \dfrac {a_1}{2} + \cdots + \dfrac {a_n}{n+1}=0,$ prove that $a_0 + a_1 x + a_2 x^2 + \cdots + a_n x^n=0$ has at least one real solution.

My solution:

Let $$f(x) = a_0 + a_1 x + a_2 x^2 + \cdots + a_n x^n$$
$$\int f(x) = \dfrac {a_0}{1} x + \dfrac {a_1}{2}x^2 + \cdots + \dfrac {a_n}{n+1} x^{n+1} + C$$
$$\int_0^1 f(x) = \left[ \dfrac {a_0}{1} + \dfrac {a_1}{2} + \cdots + \dfrac {a_n}{n+1} \right]-0$$
$$\int_0^1 f(x) = 0$$
Since $f$ is continuous, by the area interpretation of integration, it must have at least one zero.

My question is, is this rigorous enough? Do I need to prove the last statement, perhaps by contradiction using Riemann sums? Is this a theorem I can/should quote?

REPLY [13 votes]: You can also prove it using the mean value theorem. You showed that $$\int_{0}^{1}f\left(x\right)dx=0
 $$ now since $f$ is continuous by the mean value theorem for integrals we have that exists some $c\in\left(0,1\right)
 $ such that $$f\left(c\right)=\int_{0}^{1}f\left(x\right)dx
 $$ so $$f\left(c\right)=0$$ as wanted.