As Slade points out in the comments, this is largely a matter of definition. The basic assumptions that most people make is that your group $G$ is connected and smooth (the latter actually being guaranteed in characteristic $0$ for finite type groups). This automatically implies that $G$ is geometrically integral. To see that it's geometrically connected you can see the discussion of such matters here and irreducibility follows from the whole classical "smooth implies local rings are domains" and then "connected plus local rings are domains implies irreducible".
That said, there's actually a fairly convincing argument that one should consider non-connected groups in the sense that really simple groups can quickly give rise to disconnected groups via simple operations.
For example, consider the following well-known theorem:
Theorem(Steinberg): Let $G$ be a (connected) reductive algebraic group over $k=\overline{k}$. Then, the property that for every semisimple $s\in G(k)$ satisfies that $C_G(s)$ (its centralizer) is connected is equivalent to the claim that its derived subgroup $G^\mathrm{der}$ is simply connected.
So, for example, if you are dealing with adjoint groups like $\mathrm{PGL}_n$, then taking centralizers of semisimple elements needn't give you connected groups! As a simple example, convince yourself that if $s:=\begin{pmatrix}-1 & 0\\ 0 & 1\end{pmatrix}$ then $\pi_0(C_{\mathrm{PGL}_2}(s))=2$.
EDIT: Here's a reference for the claim that in characteristic 0 group schemes locally of finite type are smooth. The point is that it suffices to show that $G$ is reduced since then generic smoothness gives you a dense open locus of smooth points and then you can use the homogeneity of $G$ to show that all closed points of $G$ are regular, so that $G$ is smooth. The reducedness then comes from a trick/theorem of Cartier.