Change of Variable in $\mathbb{R}^k$. Assume that $f: V \to U$ is a diffeomorphism of open sets in $\mathbb{R^k}$ and $a$ is an integrable function on $U$. Then $$\int_U a dx_1 \cdots dx_k = \int_V (a \circ f) \left|\det(df)\right|dy_1 \cdots dy_k.$$
Is $dx_1 \cdots dx_k$ is a short hand for $dx_1 \wedge \cdots \wedge dx_k$? I just don't see a reason why the author suddenly decide to abbreviate it.
Thank you.
No, $dx_1\dots dx_k$ here is not a short hand for $dx_1\wedge \dots \wedge dx_k$. It is a longhand for the $k$-dimensional Lebesgue measure.
There are two similar kinds of integrals in $\mathbb R^k$:
The first one uses the measure space structure. The second uses the manifold structure.
Since a $k$-form $\omega$ is visibly associated with a function via $\omega=f\,dx_1\wedge\dots\wedge dx_k$, integrals 1 and 2 can look alike. The difference between them transpires in the change of variables formula.
For integrals of type 1 it is $$\int_U a\, dx_1 \cdots dx_k = \int_V (a \circ f)\, \left|\det(df)\right|\,dy_1 \cdots dy_k.\tag1$$ Here $\left|\det(df)\right|$ arises as the Radon-Nikodym derivative of the pushforward of the measure $ dx_1 \cdots dx_k$ under $f^{-1}$.
For integrals of type 2 it is $$\int_U a\, dx_1 \wedge\cdots\wedge dx_k = \int_V (a \circ f)\, \det(df) \,dy_1\wedge \cdots\wedge dy_k.\tag2$$ Here $\det(df)$ arises from the computation of $f^*(a\, dx_1 \wedge\cdots\wedge dx_k)$, the pullback of $k$-form $dx_1 \wedge\cdots\wedge dx_k$ under $f$.
It is rare to see an honest explanation of the above difference in a calculus textbook. Usually, the authors find it convenient to use (2) in $\mathbb R^1$ and (1) in higher dimensions, with little to no explanation of the reason. Wikipedia follows suit.