Let $M$ be a compact Kähler manifold with Kähler form $\omega$. For any line bundle $\mathcal{L}$ on $M$, we define the $\omega$-degree of $\mathcal{L}$ to be $deg(\mathcal{L}) :=\int_{M}c_{1}(\mathcal{L})\wedge \omega ^{n-1}$.
How does one show that if $D$ is an effective cartier divisor on $M$ and $\mathcal{O}(D)$ the corresponding line bundle, then $\deg \mathcal{O}(D) \geq 0$? The book I'm reading (Kobayashi's Differential geometry on complex vector bundles) states that the integral above actually becomes $\int_{D} \omega^{n-1}$ and thus concludes from there, but I can't figure out why it's true. Thanks in advance!
Note that $c_1(\mathcal{O}(D)) \in H^2(M; \mathbb{Z})$ is Poincaré dual to $[D] \in H_{2n-2}(M; \mathbb{Z})$; that is, $[D] = [M]\cap c_1(\mathcal{O}(D))$. Therefore,
\begin{align*} \int_M c_1(\mathcal{O}(D))\cup\omega^{n-1} &= \langle c_1(\mathcal{O}(D))\cup\omega^{n-1}, [M]\rangle\\ &= \langle \omega^{n-1}, [M]\cap c_1(\mathcal{O}(D))\rangle\\ &= \langle\omega^{n-1}, [D]\rangle\\ &= \int_D\omega^{n-1} \end{align*}
where the angled brackets indicate the pairing between cohomology and homology of the appropriate degree.