Show a scheme is integral if and only if it is irreducible and reduced.

Hint for the forward direction: If $X$ is not irreducible, that means there are two disjoint nonempty open subsets $U,V\subset X$. Consider $\mathscr{O}_X(U\cup V)$.

Hint for the reverse direction: If $X$ is irreducible and reduced, so is any open subscheme of $X$, so every nonempty affine open subscheme of $X$ is Spec of a domain. To show $\mathscr{O}_X(U)$ is a domain for arbitrary $U$, you can show that the restriction $\mathscr{O}_X(U)\to\mathscr{O}_X(V)$ is injective for any nonempty affine open subset $V\subseteq U$.

A stronger hint for the reverse direction is hidden below:

To show the restriction $\mathscr{O}_X(U)\to\mathscr{O}_X(V)$ is injective, suppose $f$ is in its kernel. Then $f$ vanishes on $V$. To show $f$ vanishes on $U$, it suffices to show $f$ vanishes on $W$ for every other affine open $W\subseteq U$. Now use the fact that $W$ is Spec of a domain and $f$ vanishes on $V\cap W$.

If there were any disjoint open subsets $U,V$ then by the sheaf axioms $$\mathcal{O}_X(U \cup V) \cong \mathcal{O}_X(U) \times \mathcal{O}_X(V)$$ via restrictions $s \mapsto (s|_U, s|_V)$. This product has zero divisors of the form $(0,t)$ and $(t,0)$ unless one of $U$ and $V$ is empty. Contradiction since $\mathcal{O}_X(U \cup V)$ is integral.

This appears as Proposition 3.27 in Görtz, Wedhorn's Algebraic Geometry (which is a great reference for the basics of schemes).