I wonder how to reconcile the transformation rules for integration by substitution in a single variable vs several variables.
In the single-variable case
the Jacobian factor $\varphi'(x)$ is just a derivative and may take on negative as well as positive values depending on the particular function and integration range.
However, in the multi-variable case
the Jacobian determinant factor Det$(D\varphi)(u)$ is taken in absolute value, which allows this transformation Jacobian to take on positive values only.
If we take the multi-variable expression and consider the single-variable special case, then the partial derivative determinant again reduces to simply $\varphi'(x)$. However, the additional absolute value of the determinant seems to introduce a discrepancy between the purely single-variable treatment and this special case.
Moreover, it is easy to find examples showing that the absolute value indeed should not be there in the single-variable case. How to reconcile the two? Is the absolute value truly needed and correct in the multi-variable case?
EDIT:
Since some doubt was expressed in the comments on whether $\varphi'(x)$ may or may not be negative in the single-variable case, consider the following example:
$$\int_0^1 du\,u^2=\frac{1}{3}$$
Now take $u\equiv\varphi(x)=-x$. If we consider the absolute value of the Jacobian to be true, we get:
$$\int_{0}^{-1}\left|\frac{d\varphi(x)}{dx}\right|dx\,(-x)^2=\int_0^{-1}dx\,x^2=-\frac{1}{3}$$
which gives a wrong result; while if we do not take the absolute value, we recover:
$$\int_{0}^{-1} \frac{d\varphi(x)}{dx}dx\,(-x)^2=-\int_0^{-1}dx\,x^2=\frac{1}{3}$$
which is correct.
In the case $U=(a,b)$ and $n=1$ the first formula is actually the more general version, because the second statement requires $\varphi$ to be injective, whereas the first does not. If we restrict to injective functions, both statements agree. This is clear for $\varphi > 0$. So assume that $\varphi <0$. Then $\varphi((a,b)) = (\varphi(b),\varphi(a))$ and: $$\begin{align}\int_{\varphi(U)} f(v) dv &= \int_{\varphi(b)}^{\varphi(a)} f(u) d(u) \\ &= -\int_{\varphi(a)}^{\varphi(b)} f(u) d(u) \\ &= -\int_{a}^{b} f(\varphi(x))\varphi'(x) dx \\ &= \int_{a}^{b} f(\varphi(x)) |\varphi'(x)| dx \end{align}$$