What to follow is a summary of part of the sixth chapter of the text Analysis on Manifolds by James Munkres.
Definition
Given $x\in\Bbb R^n$ we define a tangent vector to $\Bbb R^n$ at $x$ to be a pair $(x;\vec v)$ where $\vec v\in\Bbb R^n$. The set of tangent vector to $\Bbb R^n$ at $x$ forms a vector space if we define
$$
(x;\vec v)+(x;\vec w):=(x;\vec v+\vec w)\\
c(x;v\vec):=(x;c\vec v)
$$
It is called the tangent space to $\Bbb R^n$ at $x$ and is denoted by the symbol $\mathcal T_x(\Bbb R^n)$.Definition
Let $A$ be open set in $\Bbb R^n$. A $k$-tensor field in $A$ is a function $\omega$ assigning to each $x\in A$ a $k$-tensor defined on the vector space $\mathcal T_x(\Bbb R^n)$, that is
$$
\omega(x)\in\mathcal L^k\big(\mathcal T_x(\Bbb R^n)\big)
$$
for each $x\in A$. In particular if it happens that $\omega(x)$ is an alternanting $k$-tensor for each $x\in A$ then $\omega$ is called a differential form of order $k$ on $A$.Now if the vectors $\vec v_1,…,\vec v_n$ form a basis of $\Bbb R^n$ then the vectors $(x;\vec v_1),…,(x;\vec v_n)$ form a bsis of $\mathcal T_x(\Bbb R^n)$ and thus we define a $1$-form $dx^i$ on $\Bbb R^n$ by letting that
$$
[dx^i(x)](x;\vec v_j)=\delta^i_j
$$
for any $x\in\Bbb R^n$ and for any $j=1,…,n$. Well it is possible to prove that if $\omega$ is a $k$-form defined in any open set $A$ then
$$
\sum_{i_1<…<i_k}f_{i_1,..,i_k}dx^{i_1}\curlywedge…\curlywedge dx^{i_k}
$$
where $f_{i_1,…,i_k}$ is for any ascending $k$-tuple of indices $i_1,…,i_k$ is a scalar function defined on $A$.Definition
Let $A$ be open in $\Bbb R^k$; let be $\alpha:A\rightarrow\Bbb R^n$ be of class $C^\infty$; let be $B$ and open set of $\Bbb R^n$ containing $\alpha[A]$. We define a dual transformation of forms
$$
\alpha^*:\Omega^l(B)\rightarrow\Omega^l(A)
$$
as follows: given a $0$-form $f:B\rightarrow\Bbb R$ on $B$ we define a $0$-form $\alpha^* f$ on $A$ by setting $\alpha^* f:=f\circ\alpha$ and given an $l$-form on $B$ with $l>0$ we define an $l$-form $\alpha^*\omega$ on $A$ by the equation
$$
[\alpha^*\omega(x)]\big((x;v\vec v_1),…,(x;\vec v_l)\big):=[\omega(\alpha(x)]\big((\alpha(x);D\alpha(x)\cdot\vec v_1),…,(\alpha(x);D\alpha(x)\cdot\vec v_l)\big)
$$
for any $x\in A$ and for any $\vec v_1,…,\vec v_k\in\Bbb R^n$.Definition
Let $A$ be an open set in $\Bbb R^k$ and let be $\eta$ a $k$-form defined in $A$ so that it can be written uniquely in the form
$$
\eta=fdx^1\curlywedge…\curlywedge dx^k
$$
for any scalar function $f$ defined on $A$. So we define the integral of $\eta$ over $A$ by the equation
$$
\int_A\eta:=\int_A f
$$
provided the latter integral exists.This definition seems to be coordinate-dependent: indeed the function $f$ dependes to the choice of a basis in $\Bbb R^n$. One can, however, formulate the definition in a coordinate-free fashion: specifically if $\vec v_1,…,\vec v_k$ is any right-handed orthonormal basis for $\Bbb R^k$ tehn it is an lemenetary exercise to show that
$$
\int_A\eta=\int_{x\in A}[\eta(x)]\big((x;\vec v_1),…,(x;\vec v_k)\big)
$$
Thus the integral of $\eta$ does not depend on the choice of any orthonormal basis in $\Bbb R^k$ although it does depend on the orientation of $\Bbb R^k$.Definition
Let $A$ be open in $\Bbb R^k$; let $\alpha A\rightarrow\Bbb R^n$ be a $k$-parametrized manifold. If $\omega$ is a $k$-form defined in an open set of $\Bbb R^n$ containing $Y_\alpha:=\alpha[A]$ then we define the integral of $\omega$ on $Y_\alpha$ by letting
$$
\int_{Y_\alpha}\omega:=\int_A\alpha^*\omega
$$
So now I would like to know if it is possible to define the integral of a differential form on an open set independently of any basis and not only for orthonormal bases.
So if $dx^{1}\curlywedge…\curlywedge dx^{k}$ and $dy^{1}\curlywedge…\curlywedge dy^k$ are the $k$-elementary form with respect two any different bases $\mathcal V:=\{\vec v_1,…,\vec v_k\}$ and $\mathcal W:=\{\vec w_1,…,\vec w_k\}$ then there exist two scalar functions $f$ and $g$ defined on $A$ such that
$$
f dx^1\curlywedge…\curlywedge dx^k=\eta=g dy^1\curlywedge…\curlywedge dy^k
$$
so that they are related by the equation
$$
g(x)=[\eta(x)]\big((x;\vec w_1),…,(x;\vec w_k)\big)=\\ [(f\cdot dx^1\curlywedge…\curlywedge dx^k)(x)]\big((x;\vec w_1),…,(x;\vec w_k)\big)=…=f(x)[dx^1(x)\wedge…\wedge dx^k(x)]\big((x;\vec w_1),…,(x;\vec w_k)\big)=\\
f(x)\det\begin{pmatrix}[dx^1(x)](x;\vec w_1)&&\cdots&&[dx^1(x)](x;\vec w_k)\\
\vdots&&\ddots&&\vdots\\
[dx^k(x)](x;\vec w_1)&&\cdots&&[dx^k(x)](x;\vec w_k)\end{pmatrix}=f(x)\det\begin{pmatrix}w^1_1&&\cdots&&w^1_k\\
\vdots&&\ddots&&\vdots\\
w^k_1&&\cdots&&w^k_k\end{pmatrix}
$$
where $w^i_j$ are the components of the basis $\mathcal W$ with respect the basis $\mathcal V$. So by the previous equation clearly we conclude that generally if the two bases are not orthonormal then
$$
\int_A g=\det\begin{pmatrix}w^1_1&&\cdots&&w^1_k\\
\vdots&&\ddots&&\vdots\\
w^k_1&&\cdots&&w^k_k\end{pmatrix}\cdot\int_A f\neq\int _A f
$$
and thus the integral of $\eta$ it is not uniquely determined. Anyway I tried to expalin this using the following theorem
Theorem
Let $g:A\rightarrow B$ be a diffeomorphism of open sets in $\Bbb R^k$. Assume $\det Dg$ does not change sign on $A$. Let $\beta: B\rightarrow\Bbb R^n$ be a map of class $C^\infty$; let $Y_\beta:=\beta[B]$. Let $\alpha:=\beta\circ g$ and thus $Y_\alpha:=\alpha[A]$. If $\omega$ is a $k$-form defined in an open set of $\Bbb R^n$ containing $Y$ then $\omega$ is integrable over $Y_\beta$ if and only if it is integrable over $Y_\alpha$ and in this case
$$
\int_{Y_\alpha}\omega=\pm\int_{Y_\beta}\omega
$$
but unfortunately I did not conclude anything. However I am sure that using the change of variables theorem it is possible to prove that the above two quantities are equal unless the sign because the above matrix is the matrix of the identity map with respect the basis $\mathcal V$ and $\mathcal W$. Indeed we suppose that $\mathcal W$ is the standard basis of $\Bbb R^k$ and thus we define the diffeormorphism $\phi:\Bbb R^k\rightarrow\Bbb R^k$ through the position
$$
\phi(x):=x^iw^j_i\vec v_j=x^i\vec w_i=x
$$
for any $x\in\Bbb R^n$ so that by the change of variable theorem it follows that
$$
\int_{x\in A} f=\int_{y\in\phi^{-1}[A]}(f\circ \phi)|\det D\phi|=\int_{y\in A} f|\det[w^i_j]|=\pm\int_{y\in A} g
$$
as we desired.
However, I fear that there is some error in the above identity. So could anyone give some clarification about this, please?
Best Answer
Consider the function $f=1$ and the region $A\subset\Bbb R^2$ to be the rectangle spanned by $(1,0)$ and $(0,2)$. I consider the linear mapping $T(x^1,x^2)=(x^1,2x^2)=(y^1,y^2)$ mapping the unit square $Q$ to $A$. Then we have $\omega = dy^1\wedge dy^2 = 2\,dx^1\wedge dx^2$. What the second equals sign means, more carefully, is that $T^*\omega = 2\,dx^1\wedge dx^2$, as officially these $2$-forms are defined on different versions of $\Bbb R^2$. Now, \begin{multline*} 2 = \int_A dy^1\wedge dy^2 = \int_{T(Q)} dy^1\wedge dy^2 = \int_Q T^*(dy^1\wedge dy^2) \\= \int_Q 2\,dx^1\wedge dx^2 = 2\cdot 1. \end{multline*} This generalizes to any invertible linear map, of course.