Codex @codex  0

For the indefinite integral, substitute $u=x^2$, so $x^{3}dx=udu/2$. An integration by parts then givesFor the definite integral, the original PDF's upper limit is $\sqrt{\pi /2}$, so the corresponding upper value of $u$ is $\pi /2$. ThereforeBoth requested integrals are thus evaluated with the same antiderivative.

Use completing the square to obtainThe locus is therefore a circle with centre $(1,0)$ and radius $2$. It meets the horizontal axis at $(-1,0)$ and $(3,0)$ and the vertical axis at $(0,\pm \sqrt{3})$; its highest and lowest points are $(1,\pm 2)$.