Derivative of the integral of a pull-back form Let $\omega$ be a $n$-form on the smooth compact manifold $M$ without boundary.
Let $X$ be a smooth vector field on $M$ and $\phi_t$ the associated flow.
Let $A(t)=\int_M \phi_t^* \omega$. How can we compute $dA/dt$?
 A: Since $M$ is compact, every smooth top-dimensional form is integrable and furthermore the mapping $t\mapsto\omega(t):=\phi_t^*\omega$ is also smooth, so we can just differentiate the integrand. This is in principle justified by measure theory but you could also verify this by choosing an atlas and a partition of unity and then just use the according result for $\mathbb R^n$. 
As a side remark, note that $\Omega^n(M)$ is a vector space (actually a $C^\infty(M)$-module), so the derivative of the curve $t\mapsto\omega(t)$ can be seen as an element of $\Omega^n(M)$ and thus can be integrated.
It is sufficient, however, to compute the derivative at $t=0$, due to the group action property of the flow. Then this is actually the definition of the Lie derivative:
$$
\frac{d}{dt}\Big|_{t=0}\phi_t^*\omega =: \mathcal{L}_X\omega
$$
By Cartan's formula, we have
$$
\mathcal{L}_X\omega = i_Xd\omega + di_X\omega = di_X\omega,
$$
since $\omega$ is top-dimensional. Here the interiour product is $i_X\omega=\omega(X,\cdot,\dots,\cdot)$. Hence we have
$$
A'(0) = \int_M \mathcal{L}_X\omega = \int_M di_X\omega = \int_{\partial M} i_X\omega,
$$
which is zero if $M$ has no boundary.
