Proving a property of a real differential form and integrating it I tried to solve the following exercise, but I'm unsure if my solution is correct and if possible I'd like to get some background information on the exercise.
Exercise: Let $$\omega = \sum_{i = 1}^{n} (-1)^{i-1} x_i \cdot {\rm d}x_1 \wedge \dots \wedge \widehat{{\rm d}x_i} \wedge \dots \wedge {\rm d}x_n$$ be a differential $(n-1)$-form over $\mathbb{R}^n$. The hat notation is supposed to mean that the form ${\rm d}x_i$ is dropped from the wedge product in the $i$-th summand.
a) Show that ${\rm d}\omega = n \cdot {\rm d}x_1 \wedge \dots \wedge {\rm d}x_n$.
b) Let $n = 3$. Calculate
$${\rm d}\omega\left(\begin{pmatrix}1 \\ 1 \\ 0\end{pmatrix}, \begin{pmatrix}0 \\ 2 \\ 1\end{pmatrix},\begin{pmatrix}1 \\ 1 \\ 0\end{pmatrix} \right)$$
c) Calculate $\int_{[0,1]^n} {\rm d}\omega$.
My solution: a) I tried to prove the claim by induction. For $n = 2$ we have $\omega = x_1{\rm d}x_2 - x_2{\rm d}x_1$ and thus $${\rm d}\omega = {\rm d}(x_1)\wedge {\rm d}x_2 - {\rm d}(x_2)\wedge {\rm d}x_1 = {\rm d}x_1 \wedge {\rm d}x_2 + {\rm d}x_1 \wedge {\rm d}x_2 = 2 {\rm d}x_1 \wedge {\rm d}x_2.$$
where the second equality follows from the anticommutativity of $\wedge$. Now for the induction step we have
\begin{align*}
{\rm d} \omega &= {\rm d}\left( \sum_{i = 1}^{n+1} (-1)^{i-1} x_i \cdot {\rm d}x_1 \wedge \dots \wedge \widehat{{\rm d}x_i} \wedge \dots \wedge {\rm d}x_{n+1}\right)\\
&= {\rm d}\left(\sum_{i = 1}^{n} (-1)^{i-1} x_i \cdot {\rm d}x_1 \wedge \dots \wedge \widehat{{\rm d}x_i} \wedge \dots \wedge {\rm d}x_{n+1} + (-1)^n x_{n+1} \cdot {\rm d}x_1 \wedge \dots \wedge {\rm d}x_{n}\right)\\
&= {\rm d}\left(\left[\sum_{i = 1}^{n} (-1)^{i-1} x_i \cdot {\rm d}x_1 \wedge \dots \wedge \widehat{{\rm d}x_i} \wedge \dots \wedge {\rm d}x_{n}\right]\wedge{\rm d}x_{n+1} + (-1)^n x_{n+1} \cdot {\rm d}x_1 \wedge \dots \wedge {\rm d}x_{n}\right)\\
\end{align*}
where in the last line I factored out ${\rm d}x_{n+1}$ as it is present in each of the terms of the sum. Now, to tidy up the notation a bit let the sum be denoted by $\omega_n$. Then by linearity and the product rule of ${\rm d}$ we have
\begin{align*}
{\rm d} \omega = {\rm d}(\omega_n)\wedge{\rm d}x_{n+1} + (-1)^{n-1}\omega_n{\rm d}^2x_{n+1} + (-1)^{n}{\rm d}x_{n+1} \wedge {\rm d}x_1 \wedge \dots \wedge {\rm d}x_{n}
\end{align*}
Now we can use the induction hypothesis on the first term, the second term is equal to zero, because ${\rm d}^2x_i = 0$. So
\begin{align*}
{\rm d} \omega &= n \cdot {\rm d}x_1 \wedge \dots \wedge {\rm d}x_n\wedge{\rm d}x_{n+1} + (-1)^{n}{\rm d}x_{n+1} \wedge {\rm d}x_1 \wedge \dots \wedge {\rm d}x_{n}\\
 &= n \cdot {\rm d}x_1 \wedge \dots \wedge{\rm d}x_{n+1} + (-1)^{2n}{\rm d}x_1 \wedge \dots \wedge{\rm d}x_{n+1}\\
&= (n+1)\cdot{\rm d}x_1 \wedge \dots \wedge{\rm d}x_{n+1}.
\end{align*}
where I have used the $\wedge$-anticommutativity $n$ times to get ${\rm d}x_{n+1}$ to the correct position.
b) In this part the notation confuses me a bit. Strictly speaking ${\rm d}\omega$ is a $3$-differential form and thus i would expect something like ${\rm d}\omega(x)(v_1,v_2,v_3)$ where $x, v_1, v_2, v_3 \in \mathbb{R}^3$. I suppose the first argument was dropped since we showed that ${\rm d}\omega$ yields a constant alternating $3$-form for fixed $n$. Since two inputs are equal and ${\rm d}\omega$ is alternating we should then have ${\rm d}\omega(v_1, v_2, v_1) = 0$.
c) I'm still a bit confused when it comes to integrating differential forms, but i think this should work:
$$\int_{[0,1]^n} {\rm d}\omega = \int_{[0,1]^n} n \cdot {\rm d}x_1 \wedge \dots \wedge{\rm d}x_{n} = n \cdot \int_{[0,1]^n} {\rm d}\lambda^n(x) = n \cdot \lambda^n([0,1]^n) = n.$$
Here $\lambda^n$ is supposed to denote the $n$-dim Lebesgue measure on $\mathbb{R}^n$.
Additional questions: Does the given differential form $\omega$ have any specific use or meaning? Is there a shorter solution to part b) that I missed? Thanks!
 A: Part (a) has a much quicker solution, induction is not needed at all. One of the possible definitions of $d$ is to first write $\omega = \sum_I a_I dx^I$, where $I$ is an injective tuple of numbers between $1$ and $n$, $a_I = a_{i_1 \dots i_k}$ and $dx_I:= dx_{i_1}\wedge \cdots \wedge dx_{i_k}$, then we define $d\omega := \sum_I (da_I)\wedge dx_I$. So, in your case,
\begin{align}
d\omega &:= \sum_{i=1}^nd((-1)^{i-1}x_i) \wedge dx_1 \wedge\cdots \wedge \widehat{dx_i}\wedge \cdots dx_n \\
&= \sum_{i=1}^n(-1)^{i-1}dx_i \wedge dx_1 \wedge\cdots \wedge \widehat{dx_i}\wedge \cdots dx_n \\
&= \sum_{i=1}^n dx_1 \wedge \cdots\wedge dx_n \\
&= n \cdot dx_1 \cdots \wedge \wedge dx_n
\end{align}
(with some practice, this calculation becomes as "obvious" as $(a+b)^3 = a^3+3a^2b + 3ab^2 + b^3$)
For part (b), yes, what's written is technically an abuse of notation, because $d\omega$ being a differential $n$-form on a manifold $M$ means you have to first plug in a point $p\in M$, to get $d\omega(p)$, and then given tangent vectors $\xi_1, \dots, \xi_n \in T_pM$, you can plug these in to get a number $d\omega(p)[\xi_1, \dots, \xi_n] \in \Bbb{R}$. But your solution is correct (which I think is as short as it can get) because of the alternating nature of differential forms.
Part (c) is right.

As for uses of $\omega$, one thing I can think of is that if you let $\iota:S^{n-1}\to \Bbb{R}^n$ be the inclusion mapping, then the pull-back $\iota^*\omega$ is the volume form on the unit sphere $S^{n-1}$. For example, if $n=2$, this is $\omega = x dy - y dx$, while for $n=3$ this becomes
\begin{align}
\omega &= x\, dy \wedge dz - y\, dx \wedge dz + z\, dx\wedge dy \\
&= x\, dy \wedge dz + y\, dz \wedge dx + z\, dx\wedge dy
\end{align}
More generally if you take an $m$-dimensional oriented manifold $M$ with volume form $\mu$, and an $m-1$-dimensional embedded submanifold $N\subset M$ (i.e a hypersurface), with unit outward normal vector field $\nu$, then by taking (the pullback to $N$ of) the interior product $\iota_{\nu}\mu$, you get the volume form on $N$.
In more common notation (and by suppressing the pullback from notation), we write this as $d^{n-1}V = \iota_{\nu}(d^nV) \equiv \nu \lrcorner d^nV$, or in the case of $n=3$, we write this as $dA = \nu \lrcorner dV$.
A: I don't know of a specific use of $\omega$. It seems to be constructed just for part (a) to hold. I think your solutions for part (b) and (c) are correct and fine. You could probably do part (a) for induction as you did, but I think if you just used the formula
$${\rm d} \left(\alpha_I {\rm d}x^I\right) = \sum_{i=1}^{n} \frac{\partial \alpha_I}{\partial x^i} {\rm d}x^i\wedge {\rm d}x^I$$
it follows directly.
