Minimum number of zeroes of a function $f$ that makes $\int^b_a t^k f(t)dt = 0$ for all $k=0,1,...,n$ 
Question: Let $a < b, n \in N,$ and $f : [a, b] \to \mathbb R$ be a continuous function such that for all
$k = 0,1,...,n$ we have $\int^b_a t^k f(t)dt = 0$.
Determine the minimum number of elements of the set $\{ x \in (a, b):  f(x) = 0\}$.

I don't have any idea how to start. I only know that $f(t)$ must take both positive and negative values for the integral to be $0$, then there is at least one $x$ that makes $f(x)=0$ as the function is continuous.
I also tried to use the MVT for integrals to see that
$$\forall k, \exists c_k, f(c_k) \int^b_a t^kdt = 0$$
but I still couldn't get any extra information after evaluating the definite integral.
 A: Suppose the claim holds up to some $ n $ for a particular continuous function $ f $. We may assume $ f $ only has finitely many zeroes in $ [a, b] $, say $ N $ zeroes, and we may also assume without loss of generality that $ f(a) \geq 0 $.
In this case, the set of zeroes of $ f $ in the interior form a discrete subset of $ (a, b) $, and we may separate the zeroes into two categories: zeroes where $ f $ has a different sign on two sides of the zero, and zeroes where $ f $ has the same sign. Enumerate the zeroes in the former category as $ \alpha_1, \alpha_2, \ldots, \alpha_i $ (of course $ i \leq N $), and consider the integral (multiplied by a factor $ -1 $ if $ f(a) = 0 $ and $ f $ is negative on a neighborhood of $ a $, but this doesn't affect the basic argument):
$$ \int_{a}^{b} f(x) \cdot \prod_{h=1}^i (\alpha_h - x) \, dx $$
It's easy to see that the integrand is $ \geq 0 $, is continuous, and only vanishes on finitely many points. These imply that the integral is $ > 0 $, and since we're integrating $ f $ against a polynomial of degree $ \leq N $, this proves that $ n < N $. In other words, if $ N $ is the number of zeroes of $ f $, then $ N $ must be at least $ n+1 $.
To see that this lower bound on the number of zeroes is tight, it's enough to notice that the orthogonal complement of $ \langle 1, t, t^2, \ldots, t^n \rangle $ in $ \langle 1, t, t^2, \ldots, t^{n+1} \rangle $ is spanned by some polynomial $ P $ of degree $ n+1 $. It clearly has at most $ n+1 $ zeroes in $ [a, b] $, and by the above argument it must have at least $ n+1 $, so we conclude that it has exactly $ n+1 $.
