Civil Engineering Reference
In-Depth Information
Lest the reader be lulled into the false impression that numerical integration
can always be made exact, we present the following example to illustrate that
(1) numerical integration is not always exact but (2) numerical integration con-
verges to exactness as the number of integration points is increased.
EXAMPLE 6.10
Evaluate the integral
1
r
2
−
1
(
r
+
3)
2
I
=
d
r
−
1
using Gaussian integration with one, two, and three integration points.
■
Solution
The integration procedure requires that we evaluate the integrand at discrete points and
sum the results as follows (we do not present all the calculation details: the reader is urged
to check our calculations)
One Integration Point
W
1
=
2
r
1
=
0
≈
2
−
1
3
I
=−
0
.
666667
Two Integration Points
√
3
3
W
1
=
W
2
=
1
r
1
=
r
2
=−
r
1
I
≈−
0
.
16568
Three Terms
W
1
=
0
.
8888888
W
2
=
0
.
5555555
...
W
3
=−
0
.
5555555
...
r
1
=
0
r
2,3
=±
0
.
7745966
I
≈−
0
.
15923
Continuing with the four-point integration results in
I
≈−
0
.
15891
Hence, we see a convergence to a value. If the exact result is obtained by formal integra-
tion, the result is
I
=−
0
.
15888
.
As illustrated by this example, the Gaussian integration
procedure is not always exact but does, indeed, converge to exact solutions as the num-
ber of sampling or Gauss points is increased.
