Is there a case where $\int_{0}^{1} p^4(x) dx>0$ but $\int_{0}^{1} x p^2(x)dx=0$? Is there a case where $$\int_{0}^{1} p^4(x) dx>0$$ but $$\int_{0}^{1} x  p^2(x)dx=0$$, where $p(x)$ is a polynomial? Can anyone prove it? Thanks.
Additionally, here are some of my thoughts,   $\int_{0}^{1} p^4(x) dx>0$ would imply that  $ p^2(x) dx>0$, then $xp^2(x)>0 ,\forall x \in (0,1]$. Can I do something like:$$\int_{0+}^{1} xp^2(x) dx>0$$
Hence, there are no such $p(x)$?

If $\int_{0}^{1} p^4(x) dx>0$, then $p(x)$ is not identically zero on $[0,1]$.
For $x \in [0,1]$, $x\geq 0$ and $p^2(x) \geq 0$ so $xp^2(x) \geq 0$. As a result, $\int_{0}^{1} xp^2(x) dx>0$ holds. If not, then $\int_{0}^{1} xp^2(x) dx=0$ (it can't be negative, so this is the only option), which forces $p(x)=0$ on $[0,1]$ which is a contradiction to $\int_{0}^{1} p^4(x) dx>0$ as explained earlier.
