Typically, the function that one wants to integrate numerically, $f$, is given, i.e. its values for various points $\{x_i\}$ are known precisely. The resulting error is due to the fact that we chose a finite number of points. Let's call this error $numerical$.
I have a different situation. I know the values of $f$, $\{f(x_i)\}_{i=1}^N$, but with corresponding variances, $\{var(f(x_i))\}$. If I use, for example, the trapezoidal rule: $$F(b)-F(a)=\int_a^b f(x) dx \approx \frac{\Delta x}{2} [f(x_1) + 2 f(x_2) + \ldots 2f(x_{N-1}) + f(x_N) ] $$ (assuming that $x_i$ are equally separated by $\Delta x$), I calculate the variance of this estimate as: $$\left(\frac{\Delta x}{2}\right)^2 \left[ var(f(x_1)) + 4\cdot var(f(x_2)) + \ldots + 4\cdot var(f(x_{N-1})) + var(f(x_N)) \right] $$ but what about the $numerical$ error?
How can I combine these two errors? What is the "final" error of my estimate?
EDIT:
I've calculated the variance the same way I would do it for a sum of $N$ independent random variables, $X_i$, with coefficients $a_i$. That is, if: $$ S :=\sum_{i=1}^{N}a_i X_i,\qquad \forall_{i\neq j}Cov(X_i,X_j)=0 $$ then $$ Var(S)=\sum_{i=1}^{N} a_i^2 Var(X_i). $$ If this error analysis is not correct, please let me know.
Also, I noticed a problem with the variance of partial sums, $S_k:=\sum_{i=1}^k a_i X_i$. It seems that the error propagates, and therefore $S_N$ has the largest variance, while $S_2$ -- the lowest. But if I were to reverse the order in the sum (which would correspond to estimating a reversed integral, $\int_b^a f dx$), the error would again propagate, but in the opposite direction. I would get two different estimates of "partial integrals", depending on the direction of integration.
Is there a way of combining information from numerical integration in both ways?