I'm computing integrals $u_i$, where $$u_i = \int_a^b s^i G(s) ds$$ $G$ and $s$ are given numerically; in other words, I have a number of ($s$, $G(s)$) pairs. The integral is evaluated approximately using quadrature (in this case, trapezoidal rule).
Now I want to estimate the error in $u_i$; I have estimates for the error $\sigma_G$ of each $G$, and there is no error in $s$.
If I apply the propagation-of-errors rule for a weighted sum, considering $s^i$ as weights, I get something like $$\sigma_{u_i}^2= \sum_j (s^i_j)^2\sigma_{G_j}^2$$
Is this analysis correct?
I'd also like to express the error in $u_i$ based on the original integral, and then derive the approximation using quadrature from that, but I have no idea how to do this.
Just a thought on how to connect your integral to the error. In your original integral, you have an error $\delta G(s)$ for each $s$. Therefore, the absolute error in the integral follows from replacing $G(s)$ by $G(s)+\delta G(s)$ and taking only the resididual term: $$\delta u_i = \int_a^b s^i \delta G(s) ds.$$ If we approximate this with the trapezoidal rule, we have (denoting $\delta G(s_n) = \sigma_n$) $$ \delta u_i \approx \sum_{n=1}^N \frac{s_n^i \sigma_n + s_{n-1}^n\sigma_{n-1}}{2}\Delta s_n.$$ Provided the mesh size $\Delta s_n$ is constant for all $n$, we have $$ \delta u_i \approx \frac{\Delta s}{2}\big(s_0^i\sigma_0 + 2 s_1^i\sigma_1 + 2 s_2^i\sigma_2 + \dots + 2 s_{N-1}^i \sigma_{N-1}+ s_N^i \sigma_N\big).$$ Now, the problem with this formula is that the $\sigma_n$ can be positive or negative, so they will tend to cancel one another out. It is possible that certain samples of the $\sigma_n$ will therefore given an unrealistically small error.
If we want an upper bound for the error, we can replace these quantities by their absolute values: $$ \delta u_i \approx \frac{\Delta s}{2}\big( |s_0^i\sigma_0| + 2 |s_1^i \sigma_1|+ \dots + 2 |s_{N-1}^i \sigma_{N-1}| + |s_N^i \sigma_N| \big).$$ This should be a fine upper bound for the error in $\delta u_i$. Notably it does not rely on the assumption that $\delta G(s)/G(s)\ll 1 $ like the formula for the error in the weighted mean you shared does. The relative error then follows from $\delta u_i/u_i$.