I've come across the following remark without many details:
For $\alpha \in \{0, 1, \dots\}$, the sequence $\{x^{\alpha}\}$ is total in $L^2[a,b]$.
By total, I mean $X = \{x^{\alpha}\}$, then $\overline{\text{span}}(X) = L^2[a,b]$.
I generally recall we can use some theorems/strategies relating to Hilbert spaces here, but am rusty. Does the following make sense more or less:
i) Since $L^2[a,b]$ forms an infinite-dimensional Hilbert space, we know there exists some orthonormal (is total) basis $\{e_n\}_{n \in \mathbb{N}}$.
ii) Let $f \in L^2[a,b]$. If $\langle f, x^{\alpha} \rangle = 0$ for all $\alpha$, then $f = 0$, and it follows that $x^{\alpha}$ is total in $L^2[a,b]$. We aim to show this.
iii) By Cauchy-Schwarz, $|\langle f, e_n - x^{\alpha} \rangle| \leq ||f||_2||e_n - x^{\alpha}||_2$. Thus, the general Pythagorean Theorem (I forget actual name) implies
$$||f||^2_2 = \sum_{n=1}^{\infty}|\langle f, e_n - x^{\alpha}\rangle|^2 \leq ||f||_2^2 \sum_{n=1}^{\infty}||e_n - x^{\alpha}||_2^2.$$
iv) By Weierstrass, we know for any $\epsilon > 0$ there exists a polynomial $p$ such that $||e_n - p||_{\infty} < \frac{\epsilon}{2^n}$. Therefore, $$||\epsilon_n - p||_2^2 \leq (b-a)||\epsilon_n - p||_{\infty} \quad\Rightarrow\quad ||f||_2^2 \leq ||f||_2^2\sum_{n=1}^{\infty} \frac{\epsilon}{2^n} = ||f||_2^2(b-a)\epsilon,$$
and therefore we must have $f = 0$ for equality.