= Solution
For a generic integrand in 20 dimensions, <Monte Carlo integration> gives <root mean square> error $O(N^{-1/2})$, whereas <tensor-product quadrature> with <composite Simpson's rule> gives only $O(N^{-4/20})=O(N^{-1/5})$. One can sample uniformly in the <Euclidean unit ball>, for example by an independent uniform direction and radius $U^{1/20}$ with $U$ uniform on $[0,1]$.
For this particular integrand, the most efficient elementary choice uses <radial integration over a ball> first. The area of the unit 19-sphere is $2\pi^{10}/9!$, as follows from the <volume of an n-ball>. Hence
$$
I=\frac{2\pi^{10}}{9!}\int_0^1 e^{r^2}r^{19}\,dr=\frac{\pi^{10}}{9!}\int_0^1 e^t t^9\,dt.
$$
The reduced integrand is smooth. Apply <composite Simpson's rule> to obtain \b[$O(N^{-4})$] deterministic error in <big O notation>, with one original radial integrand evaluation per node. If the selection is restricted to direct multidimensional methods, \b[<Monte Carlo integration> with $O(N^{-1/2})$ <root mean square> error] is the appropriate choice.
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