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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 $\Omega^{m-1}(\partial M)/d(\Omega^{m-2}(\partial M))\to\mathbb R$. By definition, you must have $I_2(d\beta)=0$ for any $\beta$, so up to a multiple, $I_2$ is just the usual integral. By Stokes, $I_1$ has to coincide with the same multiple of the integral on exact $m$-form. For a manifold with boundary $H^m(M)=0$, so any $m$-form is exact.