Civil Engineering Reference
In-Depth Information
If, for example, we have a linear polynomial
n
=
1
, the first two of Equa-
tion 6.102 are applicable and lead to the conclusions that we need only one sam-
pling point and that the appropriate values of the weighting factor and sample
point to satisfy the two equations (in this case) are
W
1
=
2
and
r
1
=
0
. Next, con-
sider the case of a cubic polynomial,
n
=
3
. In this case, we have
m
W
i
=
2
i
=
1
m
W
i
r
i
=
0
i
=
1
m
2
3
W
i
r
i
=
i
=
1
representing three equations in
2
m
+
1
unknowns. If we let
m
=
1
, the first two
equations lead to
W
1
=
2,
r
1
=
0
, but the third equation cannot be satisfied. On
the other hand, if
m
=
2
, we have
W
1
+
W
2
=
2
W
1
r
1
+
W
2
r
2
=
0
2
3
W
1
r
1
+
W
2
r
2
=
a system of three equations in four unknowns. We cannot directly solve these
equations, but if we examine the case
W
1
=
W
2
=
1
and
r
1
=−
r
2
, the first two
equations are satisfied and the third equation becomes
1
3
=
√
3
3
2
3
⇒
r
1
+
r
2
2
r
1
=
=
r
1
=
=
0
.
57735
...
corresponding exactly to the second entry in Table 6.1. These weighting factors
and Gauss points also integrate a quadratic polynomial exactly. The reader is
urged to note that, because of the zero result from integrating the odd powers in
the polynomial, exact results are obtained for
two
polynomial orders for each set
of sampling points and weighting factors.
This discussion is by no means intended to be mathematically rigorous in
terms of the theory underlying numerical integration. The intent is to give some
insight as to the rationale behind the numerical values presented in Table 6.1.
EXAMPLE 6.7
Evaluate the integral
1
(
r
2
f
(
r
)
=
−
3
r
+
7) d
r
−
1
using Gaussian quadrature so that the result is exact.
