If this question is too elementary, I apologize to ask here.
Suppose that we have a multiple integral under $n$-dimensional simplex as follows: \begin{align*} \underbrace{\int_0^1 dx_n \int_0^{x_n} dx_{n-1} \cdots \int_0^{x_2} dx_1}_{n\text{ times}}\, f(x_i) \underbrace{g(x_1,x_2,\ldots,x_{i-1},x_{i+1},\ldots,x_n)}_{\text{not depend on }x_i}, \end{align*} where $f$ only depends on $x_i$ for some index $i$, and $g$ is depends on every index except for $x_i$.
Here I try to simplify the above integral by integrating $f$ first, so that the question is
Question What would be the domain of interchanged integral if we integrate $dx_i$ first, that is, \begin{align*} \int_0^1 dx_n \int_0^{x_n} dx_{n-1} \cdots \int_0^{x_{i+2}}dx_{i+1} \int_?^? dx_{i-1} \cdots \int_?^? dx_1 \underbrace{\int_?^? f_{x_i}dx_i}_{\text{interchanged integral}}\, g(x_1,\ldots,x_{i-1},x_{i+1},\ldots,x_n). \end{align*} (Here, of course you can see that $x_n$ to $x_{i+1}$ are not affected by the interchanging.)
For a simple $3$-dimensional case, we set $x_i = x_3$ (otherwise it would be trivial), \begin{align*} &\int_0^1 dx_3 \int_0^{x_3}dx_2 \int_0^{x_2} dx_1 f(x_3)g(x_1,x_3) \\ &=\int_0^1 dx_2 \int_{0}^{x_2}dx_1 \int_{x_2}^{1}f(x_3)dx_3\, g(x_1,x_3). \end{align*} So I suspect that the domain will be shifted one by one for $x_{i-1}$ to $x_1$, and the domain of $x_i$'s integral would be $[x_{i-1},x_{i+1}]$, i.e., \begin{align*} \int_0^1 dx_n \int_0^{x_n} dx_{n-1} \cdots \int_0^{x_{i+2}}dx_{i+1} \int_0^{x_{i+1}} dx_{i-1} \cdots \int_0^{x_2} dx_1 \underbrace{\int_{x_{i-1}}^{x_{i+1}} f_{x_i}dx_i}_{\text{interchanged integral}}\, g(x_1,\ldots,x_{i-1},x_{i+1},\ldots,x_n), \end{align*} but I have no idea of how to prove this over any $n$th dimension.
Any help would be appreciated.
Thanks,
