Is the ring of global functions on an integral scheme integral? Is the ring of global functions on an integral scheme an integral domain (if the scheme is affine then it is try by definition so we are interested in non-affine schemes)? It is necessarily a reduced ring (https://stacks.math.columbia.edu/tag/01OL) so the question is whether it is irreducible.
If $X$ is any reduced scheme and $U$ is a dense open subset, then the restriction map $\mathcal{O}_X(X)\to\mathcal{O}_X(U)$ is injective.  Indeed, an element of the kernel vanishes in the fiber at every point of $U$, but the set where a section of $\mathcal{O}_X$ vanishes in the fiber is closed, so it must vanish in the fiber everywhere.  Since $X$ is reduced, a function which vanishes in the fiber everywhere must be $0$.

In particular, if $X$ is integral, we can consider $\mathcal{O}_X(X)$ as a subring of the domain $\mathcal{O}_X(U)$ for any nonempty affine open $U$, and so $\mathcal{O}_X(X)$ is also a domain.
This is an alternative answer to your question.
Question:
&quot;Q1. Is the ring of global functions on an integral scheme an integral domain (if the scheme is affine then it is try by definition so we are interested in non-affine schemes)?
Q2. It is necessarily a reduced ring?&quot;
Answer: If you use the definition in Hartshorne (page 82) you define a scheme $(X,\mathcal{O}_X)$ to be &quot;integral&quot; iff for any open set $U\subseteq X$ it follows
$\mathcal{O}_X(U)$ is an integral domain. In particular, since $X$ is an open set, it follows $\Gamma(X,\mathcal{O}_X):=\mathcal{O}_X(X)$ is an integral domain. Hence it seems to me the answer to Q1 is &quot;yes, this is true by definition&quot;. Since an integral domain is a reduced ring it follows $\mathcal{O}_X(X)$ is a reduced ring. Hence the answer to Q2 is also &quot;yes&quot;.