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> Let $\omega$ be a $k$-form on $\mathbb{R}^n$ and suppose that $\omega=d\alpha$ for some $(k-1)$-form $\alpha$. Show that, for any singular $k$-cube $c$ on $\mathbb{R}^n$ with $\partial c=0$,
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> $$\displaystyle \int\_{c}\omega=0$$
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I get that $\displaystyle \int\_{c}\omega = \int\_{c}d\alpha= \int\_{\partial c}\alpha=0?$

I just cannot explain why the last equality should equal $0$, ie $\int\_{\partial c}\alpha=0$.

Note that : $\displaystyle \partial c = \sum\_{j=1}^{k} \sum\_{\alpha=0,1}(-1)^{j+\alpha}c\_{(j,\alpha)} $ and

$\displaystyle \int\_{\partial c} \omega = \sum\_{j=1}^k \sum\_{\alpha=0,1} (-1)^{j+\alpha}\int\_{c(j,\alpha)}\omega$

Please comment if I have left unexplained terminology you might need.