First Semester in Numerical Analysis with Julia, 2020a

CHAPTER 4. NUMERICAL QUADRATURE AND DIFFERENTIATION 161
to get
1
4
∫ 1
0
(
f
(
) (
))
s +1
2 , −1/√ 3+1 s +1
+ f
2
2 , 1/√ 3+1
ds.
2
Apply the Gauss-Legendre rule again to get
[ (
1
f
4
+ f
−1/ √ 3+1
2
(
) (
, −1/√ 3+1
+ f
2
)
1/ √ 3+1
, −1/√ 3+1
2 2
Figure (4.1) displays the nodes used in this calculation.
−1/ √ )
3+1
, 1/√ 3+1
2 2
(
1/ √ )]
3+1
+ f
, 1/√ 3+1
. (4.6)
2 2
( − 1 ,
3
1
3 ) 1 (
1
,
3
1
3 )
-1 1
-1
( − 1 ,− 1
3 3 ) (
1
,− 1
3 3 )
Figure 4.1: Nodes of double Gauss-Legendre rule
Next we derive the two-dimensional Simpson’s rule for the same integral,
∫ 1 ∫ 1
f(x, y)dydx, using n = 2, which corresponds to three nodes in the Simpson’s rule
0 0
(recall that n is the number of nodes in Gauss-Legendre rule, but n +1is the number of
nodes in Newton-Cotes formulas).
The inner integral is approximated as
∫ 1
0
f(x, y)dy ≈ 1 (f(x, 0) + 4f(x, 0.5) + f(x, 1)) .
6
Substitute this approximation for the inner integral in ∫ (
1 ∫ )
1
f(x, y)dy dx to get
0 0
1
6
∫ 1
0
(f(x, 0) + 4f(x, 0.5) + f(x, 1)) dx.