Let $M$ be a manifold, $\omega$ is a differential $k$-form compactly supported in a chart $(U,\phi)$. Then the integral $\int_{M} \omega$ is defined to be:
$\int_{M} \omega := \int_{\phi(U)} (\phi^{-1})^{*}(\omega)$
$(\phi^{-1})^{*}(\omega)$ is the pullback of $\omega$ by $\phi^{-1}$ and it is a differential form on $\mathbb{R}^k$ defined by:
$(\phi^{-1})^{*}(\omega)(x)(v_1,...,v_k) := \omega(\phi^{-1}(x))(D\phi^{-1} v_1,...,D\phi^{-1} v_k)$.
The only way for me to make sense of an integral in some nice subset of $\mathbb{R}^k$ would be something like $\int_{\phi(U)} f(x) dx_1...dx_k$, so somehow $(\phi^{-1})^{*}(\omega)(x) = f(x) dx_1 \wedge...\wedge dx_k$ for some continuous function $f$. My thinking is that the reason we can obtain such an $f$ because the vector space of $k$-tensors in $\mathbb{R}^k$ has dimension 1 with the basis $dx_1 \wedge ... \wedge dx^k$, so when we fix a point $x$ in $\phi(U)$, $(\phi^{-1})^{*}(\omega)(x)$ is a $k$-tensor and must be equal to $\lambda dx_1 \wedge ... dx_k$ for some $\lambda$ a constant dependent on $x$, and we set $f(x) = \lambda$ to obtain a function on $\phi(U)$. Is my understanding correct?