Is $\int_0^1 d \lambda_t dt = d(\int_0^1 \lambda_t dt) ? $ Let $M$ be a smooth manifold. Let $\lambda_t$ be a  smooth family of differential forms (say of degree 1), where $t$ is a real parameter.
What can we say about the integral $\int_0^1 d \lambda_t dt $, where $d$ denotes the exterior differential ? Is it $d(\int_0^1 \lambda_t dt) $ ?
 A: Yes, if one interprets everything appropriately. Saying $\lambda_t$ is a smooth family of differential forms wrt $t$ means that in every coordinate chart, we can write $\lambda_t=\sum_If_I(x,t)\,dx^I$, where $I=(i_1,\dots, i_k)$ is an increasing tuple, so explicitly, I mean $\lambda_t=\sum_If_{i_1\dots, i_k}(x,t)\,dx^{i_1}\wedge \cdots \wedge dx^{i_k}$, where we assume each $f_I$ is a smooth function on its domain (an open subset of $\Bbb{R}^{\dim M}\times \Bbb{R}$).
Because I'm a nitpicky individual, here are some remarks. Note that a-priori, just writing down the symbol $\int_0^1\lambda_t\,dt$ is meaningless, because for each $t\in [0,1]$, $\lambda_t$ is a $k$-form on $M$, so $\lambda$ is a mapping $[0,1]\to \Omega^k(M)$, and the latter is not a simple space (i.e a finite-dimensional real vector space, or a Banach space) to work with. I'm sure one can come up with theories of integration which deal with integrating maps with very general target spaces (eg topological vector spaces with perhaps some additional/fewer hypotheses blablabla). I'm not super well-versed in those matters, but at the end of the day, even if one does formulate such a theory properly (again, I'm not saying it can't be done, just that I don't have all the details), it has to boil down to the following:

By definition, $\omega:=\int_0^1\lambda_t\,dt$ shall mean the (a-priori not necessarily smooth) $k$-form on $M$, whose value at a point $p\in M$ and when evaluated on $k$-tangent vectors $v_1,\dots, v_k\in T_pM$ is $\omega_p(v_1,\dots, v_k):=\int_0^1(\lambda_t)_p(v_1,\dots, v_k)\,dt$.

i.e we are forcing in our definition that evaluation commutes with integrals.
Note that once we fix $p\in M$ and tangent vectors $v_1,\dots, v_k\in T_pM$, the integral $\int_0^1(\lambda_t)_p(v_1,\dots, v_k)\,dt$ on the RHS is the integral of a smooth function $[0,1]\to\Bbb{R}$, which is a basic thing we learn about in elementary calculus. Now, because we defined things pointwise, everything we'd like to hold true, does indeed work out.
For instance, fix a coordinate chart on $M$, and label the coordinates as $x$. Write $\lambda_t=f_I(x,t)\,dx^I$ (where the sum is over increasing $k$-tuples $I$). Then,
\begin{align}
\omega:=\int_0^1\lambda_t\,dt&=\int_0^1(f_I(x,t)dx^I)\,dt=\left(\int_0^1f_I(x,t)\,dt\right)dx^I\tag{$*$}
\end{align}
(the last equal sign we pulled out the $dx^I$ from the integral because we defined things pointwise, and we essentially forced the evaluation to commute with integrals)
In words, $\omega$ is a $k$-form on $M$ whose local expression is $g_I(x)\,dx^I$, where $g_I(x)=\int_0^1f_I(x,t)\,dt$ is a smooth function on (an open subset of) $M$ (or rather the corresponding portion of $\Bbb{R}^n$ when we identify using coordinate charts) obtained by  integrating a smooth function of $(x,t)$ with respect to $t$. In particular, this shows $\omega$ is a smooth $k$-form on $M$.
So, putting it all together,
\begin{align}
d\omega&=\left[\frac{\partial}{\partial x^j}\left(\int_0^1f_I(x,t)\,dt\right)\right]\,\,dx^j\wedge dx^I\tag{from $*$}\\
&=\left(\int_0^1\frac{\partial f_I}{\partial x^j}(x,t)\,dt\right)\,dx^j\wedge dx^I\tag{Leibniz Integral Rule}\\
&=\int_0^1\left(\frac{\partial f_I}{\partial x^j}(x,t)\,dx^j\wedge dx^I\right)\,dt\\
&=\int_0^1(d\lambda_t)\,dt
\end{align}
So, to summarize, once we clarify the definition, the only thing left to do is write things out in local coordinates, and use Leibniz's integral rule in its most basic form (because here we're integrating smooth functions on the compact interval $[0,1]$ so we have all the regularity we need to exchange the partial derivative with the integral). Thus, the identity you're asking about holds.
