Let $M$ be a compact $m$ dimensional manifold with boundary $\partial M$. 


Assume that $I\_{1}, I\_{2}$ are two linear functionals on $\Omega^{m}(M), \Omega^{m-1}(\partial M)$, respectively.
Assume that we have $I\_{1}(d\alpha )=I\_{2} ( \alpha )$ for every $m-1$ differential form $\alpha$ on $M$.



> 
> Are $I\_{1},I\_{2}$ necessarily equal to a constant multiple of the usual integral?
> 
> 
> 



An Stokes type theorem for some operations other than integral
If you assume that $M$ is oriented, then up to a multiple $I\_1$ and $I\_2$ are the usual integral. In this case $\partial M$ is oriented, and since this is a manifold without boundary, the integral induces a linear isomorphism $\Omega^{m-1}(\partial M)/d(\Omega^{m-2}(\partial M))\to\mathbb R$. Now for any $\beta\in\Omega^{m-2}(\partial M)$ there exists an extension $\tilde\beta\in\Omega^{m-2}(M)$ and by naturality of $d$, we get $d\tilde\beta|\_{\partial M}=d\beta$. Since $0=d(d\tilde\beta)$ the defining equation tells you that $0=I\_1(d(d\tilde\beta))=I\_2(d\beta)$. Thus you see that $I\_2$ factorizes to the quotient $\Omega^{m-1}(\partial M)/d(\Omega^{m-2}(\partial M)$ and hence that is a number $a\in\mathbb R$ such that $I\_2(\alpha)=a\int\_{\partial M}\alpha$ for all $\alpha\in \Omega^{m-1}(\partial M)$. 


Now consider $\alpha\in\Omega^{m-1}(M)$ and form $I\_1(d\alpha)-a\int\_M(d\alpha)$. By your defining equation the first term gives $I\_2(\alpha)=a\int\_{\partial M}\alpha$, so by Stokes $I\_1-a\int\_M$ vanishes on any exact form. But on a manifold with boundary any top degree form is exact, so $I\_1=a\int\_M$. 


