A universally closed and quasi-separated morphism $f:X\to S$ admits a factorisation, as $\pi\circ f'$, where $\pi:\underline{\mathbf{Spec}}_S(f_\ast \mathcal{O}_X)\to S$ is an integral morphism, and $f':X\to \underline{\mathbf{Spec}}_S(f_\ast \mathcal{O}_X)$ is a universally closed, quasi-compact, quasi-separated, and surjective morphism.
Item (5) in Tag 03GY at the Stacks Project asserts that $\pi$ is the normalisation of $S$ in $X$. However, the normalisation of $S$ in $X$ is defined to be the relative spectrum of the integral closure of $\mathcal{O}_S$in $f_\ast \mathcal{O}_X$ over $S$.
Is $f^\#:\mathcal{O}_S \to f_\ast \mathcal{O}_X$ integeral, giving that $f$ ia universally closed and quasi-separated, or is item (5) wrong?
Universally closed morphisms are quasi-compact, by Tag 04XU. Since $f$ is universally closed and quasi-separated, $f^\#:\mathcal{O}_S\to f_\ast \mathcal{O}_X$ is integral, by Tag 03GQ.