A geometric interpretation of this integral could be submitted by the Stieltjes integral. But this could be a Riemann approach: Let us suppose a line segment (~an idealized rod) from $0$ to $L$ of x-axis. Define a "mass function" $m:[0,L]\to\Bbb R^+$, which for every $x\in [0,L]$ it gives us the mass $m(x)$ contained in the segment from $0$ to $x$. The mass function is increasing, for we suppose that we have mass everywhere in the rod. This means that the mass function in invertible, and its inverse is $x=x(m):[0,M]\to[0,L]$, where $m(0)=0$ (we have no mass from $0$ to $0$) and $m(L)=M$, the whole mass of the rod. Given a partition of the "mass segment" $[0,M]$, $P=\{0=m_0,m_1, ..., m_n=M\}$, and a set of intermediate points $E_P=\{m_1^*,...,m_n^* \}$ we acquire the sums $$ \sum_{i=1}^nx(m_i^*)(m_{i}-m_{i-1})\approx\int_0^Mxdm.$$
If we desire the more formal approach by Stieltjes integral, we can consider like this: Assume the mass function as above and the identity function on the rod $id:[0,L]\to\Bbb R$. Given a partition of the line segment $[0,L]$, $P=\{0=x_0,x_1, ..., x_n=L\}$, and a set of intermediate points $E_P=\{x_1^*,...,x_n^* \}$, by Stieltjes definition we get $$ \sum_{i=1}^nid(x_i^*)(m(x_{i})-m(x_{i-1}))=\sum_{i=1}^nx_i^*(m(x_{i})-m(x_{i-1}))\approx\int_0^Lid(x)dm(x)=\int_0^Lxdm.$$