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Use the standard convention that these compact smooth manifolds have no boundary. The interior product identity for a one-form gives $df\wedge {\iota }_X{\omega }_g=df(X){\omega }_g=X(f){\omega }_g$. For example, in the coframe from part (b), only the matching index survives in this wedge product, with the same pair of cancelling signs as in part (c). Thus the graded Leibniz rule and part (c) giveIntegrate and apply the Generalized Stokes theorem. Since $\partial M=\varnothing$, the integral of $d(f\nu )$ is zero. Rearranging proves integration by parts for Riemannian divergence:If a boundary were allowed, the exact general formula would instead have the additional term ${\int }_{\partial M}f\nu$ on the right; compactness alone does not remove that boundary contribution.