Does $f^{\ast}$ homotopic to $g^{\ast}$ imply $\int f^{\ast} w = \int g^{\ast} w$? Let $f,g: M^{k} \to N$ ($M$ and $N$ with out boundary ) such that they are homotopic then for $\omega$ a $k$-form on $N$ do we have that 
$$ \int_M f^{\ast} \omega = \int_M g^{\ast} \omega$$ 
as conclusion? I  can't figure out a proof so I am starting to think that it is not true. I can't use Homology or Cohomology and all that fancy stuff . I think is just a matter of rearranging singular cubes but I can't see how. 
The way I was trying to approach this is the following: Consider first
the case where support of $\omega$ is in the interior of the image of a singular cube $f \circ c$, where $c$ is a singular cube into $M$. 
Then we look at the new singular cube $f \circ c : I \to N$. But I don't know what to do from here and how to approach the general case. 

Attempt

Using @TedShifrin's answer I've managed the following 
$$\int_{M \times [0,1]}d(H^{*}w)=\int_{\partial(M \times [0,1])}H^{*}w = \int_{\partial M}g^{\ast}w-f^{\ast}w=0$$
but I am not sure.
Thanks a lot in advance.

You had better assume $M$ is a compact manifold without boundary and $\omega$ is a closed $k$-form on $N$. (The latter is automatic if $\dim N = k$ as well.) Then you can consider the homotopy mapping $H\colon M\times [0,1]\to N$ and apply Stokes's Theorem to $H^*\omega$. 
