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FIGURE 6.37
Transformation from
a
≤
x
≤
b
to
−
1
≤
ξ
≤+
1, Example 6.12.
Hence,
1
√
3
1
√
3
ξ
1
=−
,
ξ
2
=
(3)
The weighting coefficients are obtained from Eq. (6.124)
1
ξ
−
ξ
−
2
ξ
W
(
2
)
1
2
2
=
ξ
=
2
=
.
d
1
0
ξ
−
ξ
ξ
−
ξ
−
1
1
2
1
(4)
1
ξ
−
ξ
1
ξ
2
−
ξ
1
ξ
1
ξ
2
−
ξ
1
=
−
2
W
(
2
)
2
=
d
ξ
=
1
.
0
−
1
Note that the integration points and weighting factors are all symmetric. Thus, the Gaussian
quadrature for
n
=
2 has the form
1
F
F
1
1
√
3
F
(ξ )
d
ξ
≈
F
(ξ
)
+
F
(ξ
)
=
−
+
√
3
(5)
1
2
−
1
EXAMPLE 6.12 Application of Gaussian Quadrature
Evaluate the integral
2
.
8
e
x
2
dx
1
.
2
using
n
2.
The integration interval
a
=
≤
≤
−
≤
ξ
≤
x
b
is transformed to
1
1 using (Fig. 6.37)
=
(
b
−
a
)ξ
+
(
b
+
a
)
1
.
6
ξ
+
4
x
=
(1)
2
2
Then
e
1
.
6
ξ
+
4
e
x
=
(2)
2
and
=
(
b
−
a
)
1
6
2
.
dx
d
ξ
=
d
ξ
(3)
2
since
a
=
1
.
2 and
b
=
2
.
8. Thus, the integral is reduced to
2
.
8
1
1
2
e
x
dx
1
6
4
.
e
1
.
6
ξ
+
4
=
d
ξ
(4)
2
1
.
2
−
1
2, from Eq. (4) of the previous
e
xample,
W
(
2
)
1
0 and
W
(
2
)
2
When
n
=
=
1
.
=
1
.
0, and from Eq.
/
√
3
/
√
3
(3), the integration points are
ξ
=−
1
=−
0
.
57735 and
ξ
=
1
=
0
.
57735. Then
1
2
1
4
e
1
.
6
ξ
1
+
4
e
1
.
6
ξ
2
+
2
1
6
4
.
1
.
6
e
1
.
6
ξ
+
4
d
ξ
≈
+
2
2
−
1
.
1
6
4
e
2
.
4619
e
1
.
5381
=
(
+
)
=
6
.
5531
(5)
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