Let $R$ be a ring and $x \in R$. The Koszul complex $K_\bullet(x)$ is then $0 \rightarrow R \stackrel{x}{\rightarrow} R \rightarrow 0$. Given $x_1,\dots,x_n \in R$ the Koszul complex $K_\bullet(x_1,\dots,x_n)$ is defined to be $K_\bullet(x_1) \otimes \cdots \otimes K_\bullet(x_n)$, where $\otimes$ is the tensor product of complexes. Then the Koszul homology of an $R$-module $M$ is the homology of the complex $K_\bullet(x_1,\dots,x_n) \otimes M$.
In his notes, Craig Huneke is defining a different Koszul complex $K^\bullet(x)$ by $0 \rightarrow R \rightarrow R_x \rightarrow 0$, where the middle arrow is the canonical map of localization. As above, he defines $K^\bullet(x_1,\dots,x_n) = K^\bullet(x_1) \otimes \cdots \otimes K^\bullet(x_n)$ and the Koszul cohomology of an $R$-module $M$ is the cohomology of the complex $K^\bullet(x_1,\dots,x_n) \otimes M$. In these notes it is then shown that the local cohomology $H^i_I(M)$ of $M$ with respect to an ideal $I=(x_1,\dots,x_n)$ is in fact the Koszul cohomology of $M$.
Question: I have seen before the Koszul homology in Matsumura and Bruns and Herzog, but Koszul cohomology is new to me. The dual denominations Koszul homology-cohomology suggest that the two Koszul complexes $K_\bullet(x_1,\dots,x_n)$ and
$K^\bullet(x_1,\dots,x_n)$ described above are in some sense dual. Is this the case and in what sense?