= Solution
Let $M$ be an oriented smooth $m$-dimensional <manifold with boundary> and $\alpha$ a smooth compactly supported $(m-1)$-<differential form>. Give $\partial M$ the outward-normal-first <orientation>. A <partition of unity> subordinate to oriented interior and boundary charts reduces the proof to a compactly supported form on $\mathbb R^m$ or on a half-space. Indeed $\alpha=\sum_j\chi_j\alpha$, so $d\alpha=\sum_jd(\chi_j\alpha)$; the sum of terms involving $d\chi_j$ is zero. <Exterior derivative> commutes with chart <pullbacks>, so each summand can be calculated in coordinates.
Write the local form as
$$
\alpha=\sum_{j=1}^m(-1)^{j-1}f_j\,dx^1\wedge\cdots\wedge\widehat{dx^j}\wedge\cdots\wedge dx^m,
\qquad d\alpha=\left(\sum_j\partial_jf_j\right)dx^1\wedge\cdots\wedge dx^m.
$$
For an interior chart, integrate each derivative using the <fundamental theorem of calculus>: its integral is zero because $f_j$ has compact support. For a boundary chart modeled on $x^m\geq0$, all tangential derivatives again integrate to zero, while $\int_0^\infty\partial_mf_m\,dx^m=-f_m(x',0)$. The outward normal is $-\partial_m$; its contraction with the volume form gives exactly this sign in the induced boundary integral of $\alpha$. Summing the chart identities proves the <Generalized Stokes theorem>:
$$
\boxed{\int_Md\alpha=\int_{\partial M}\alpha.}
$$
Compact support can be replaced by compactness of $M$; on a noncompact <manifold>, sufficient support or convergence conditions are necessary.
Back to article page
