 Solution

With independent samples and finite variance, Monte Carlo integration has a root-mean-square stochastic error proportional to $N^{-1/2}$. The estimated standard error measures that random sampling uncertainty; it is not a deterministic bound.

For a sufficiently smooth one-dimensional integrand, the composite trapezoidal rule has errorfor some $\xi \in (a,b)$, hence $O(n^{-2})$. It is normally much more efficient in one dimension. The stated continuity assumption alone guarantees convergence, but does not guarantee this second-order rate; bounded second derivatives are an additional assumption.

The dimensional advantage is different: a tensor grid with $N$ total points in $d$ dimensions has only about $N^{1/d}$ points per direction and typically error $O(N^{-2/d})$. The Monte Carlo exponent remains $-1/2$, although its variance prefactor can still deteriorate with dimension. This explains why the stochastic method becomes attractive for high-dimensional integrals.