 Solution

If $x,y$ are integral over $A$, then $A[x,y]$ is finite over $A$: it is generated by finitely many monomials $x^i{y}^j$. Multiplication by $x+y$, $x-y$, or $xy$ is an endomorphism of this finite module, so the determinant trick gives a monic annihilating polynomial. Hence the integral elements form a subring $C$ containing $A$.

The integral closure of $A$ in $B$ is this subring $C$. The ring $A$ is integrally closed in $B$ when $C=A$, and $B$ is integral over $A$ when $C=B$.