First Semester in Numerical Analysis with Julia, 2020a

CHAPTER 4. NUMERICAL QUADRATURE AND DIFFERENTIATION 162
Apply Simpson’s rule again to this integral with n =2to obtain the final approximation:
[
1
6
1
6
(
f(0, 0) + 4f(0, 0.5) + f(0, 1) + 4(f(0.5, 0) + 4f(0.5, 0.5) + f(0.5, 1))
+ f(1, 0) + 4f(1, 0.5) + f(1, 1)) ] . (4.7)
Figure (4.2) displays the nodes used in the above calculation.
1
0.5
0 0.5
1
Figure 4.2: Nodes of double Simpson’s rule
For a specific example, consider the integral
∫ 1 ∫ 1 ( π
)( π
)
2 sin πx 2 sin πy dydx.
0
0
This integral can be evaluated exactly, and its value is 1. It is used as a test integral
for numerical quadrature rules. Evaluating equations (4.6) and(4.7) with f(x, y) =
( π sin πx)( π
sin πy) , we obtain the approximations given in the table below:
2 2
Simpson’s rule (9 nodes) Gauss-Legendre (4 nodes) Exact integral
1.0966 0.93685 1
The Gauss-Legendre rule gives a slightly better estimate than Simpson’s rule, but using less
than half the number of nodes.
The approach we have discussed can be extended to regions that are not rectangular,
and to higher dimensions. More details, including algorithms for double and triple integrals
using Simpson’s and Gauss-Legendre rule, can be found in Burden, Faires, Burden [4].
There is however, an obvious disadvantage of the way we have generalized onedimensional
quadrature rules to higher dimensions. Imagine a numerical integration problem
where the dimension is 360; such high dimensions appear in some problems from financial
engineering. Even if we used two nodes for each dimension, the total number of nodes would