In Differential Geometry class I have been given the following definition for the integral of an $n$-form:

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> Let $(M,\nu)$ be an oriented manifold of dimension $n$. Let $\mathfrak{U} = \{(U\_\alpha, \phi\_\alpha)\}\_{\alpha \in A}$ be an atlas of $M$ whose transition functions have positive jacobians and such that $\nu|\_{U\_\alpha} = f\_\alpha dx^1 \wedge \dots \wedge dx^n$ with $f\_\alpha > 0$ in $C^\infty(U\_\alpha)$ for each local chart. Let $\{h\_\alpha:\alpha \in A\}$ be a smooth, locally finite partition of unity such that the support of $h\_\alpha$ is contained in $U\_\alpha$ for each $\alpha \in A$. If $\omega \in \mathcal{A}^n(M)$, then $\omega = \sum\_\alpha h\_\alpha \omega$. It is not hard to see that the sum $\sum\_\alpha \int\_{U\_\alpha}(h\_\alpha \omega)$ does not depend on the partition of unity that is used. Therefore, we define the integral of the $n$-form $\omega$ over $M$ (relative to the orientation given) as follows:
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> $$\int\_M \omega := \sum\_\alpha \int\_{U\_\alpha} (h\_\alpha \omega).$$
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The issue I have with this definition is that I don't understand what role $\nu$ is playing in here. This is supposed to be a volume form in $M$, i.e., a nowhere-vanishing $n$-form. But it is not clear to me how things change if I use a different volume form. For example, if I change $\nu$ for $-\nu$, I know this is supposed to change the sign of the integral, but I don't understand how, where is $\nu$ being used in this definition?