In Rivasseau's and Wang's "How to Resum Feynman Graphs", the weights of a spanning tree corresponding to a connected graph are defined as
$$w(G,T) = \frac{N(G,T)}{|E|!},$$
where $N(G,T)$ is the number of Hepp sectors $\sigma = \{\sigma(1), \dots, \sigma(|E|)\}$ such that $T(\sigma) = T$, where $T(\sigma)$ denotes the leading tree (i.e. the minimal spanning tree found with the Kruskal algorithm).
A theorem in the paper gives the integral form
$$w(G,T) = \int_0^1 \prod_{\ell \in T} \, \mathrm dw_\ell \, \prod_{l \notin T} x_\ell^T (\{w\}).$$
I tried understanding the formula by looking at examples given. In Rivasseau's lecture notes about QFT found online, there is the following example:
I assume $T_{12}$ is the spanning tree of the graph consisting of the edges $\ell_1, \ell_2$. Therefore, the integral form gives us the integral shown in the picture. However, I don't see how to get the value $\frac{1}{6}$ from this. As the integral is symmetric, let, w.l.o.g. $w_1 \leq w_2$. Then the integral $\int_0^1 \, \mathrm dw_2$ just gives the constant $1$. We have remaining
$$\int_0^1 \, \mathrm dw_1 \, w_1^2 = \left. \frac{w_1^3}{3} \right|_0^1 = \frac{1}{3}.$$
What am I doing wrong?