Integral closure of the subring $\mathbb Z(A)$ in $A$.

Question

Let $A$ be a commutative ring with identity. Define the image of the ring homomorphism $h:\mathbb Z\to A$ by $\mathbb{Z}(A)$. It is the smallest subring in $A$ with respect to inclusion. I am trying to compute the integral closure of $\mathbb Z(A)$ in $A$, and I guess it should be $\mathbb Z(A)$ itself. Let me recall a familiar example: $\mathbb Z(\mathbb Q)=\mathbb Z$ and the integral closure of $\mathbb Z$ in $\mathbb Q$ is $\mathbb Z$ itself. Since in general $A$ is not the field of fractions of $\mathbb Z(A)$ anymore, I haven’t figure out how to prove it.