It's nice fit to Gauss quadrature formula

1. 4/19/2024 http://numericalmethods.eng.usf.edu 1
Gauss Quadrature Rule of
Integration
Major: All Engineering Majors
Authors: Autar Kaw, Charlie Barker
http://numericalmethods.eng.usf.edu
Transforming Numerical Methods Education for STEM
Undergraduates

2. Gauss Quadrature Rule of
Integration
http://numericalmethods.eng.usf.edu

3. http://numericalmethods.eng.usf.edu
3
What is Integration?
Integration


b
a
dx
)
x
(
f
I
The process of measuring
the area under a curve.
Where:
f(x) is the integrand
a= lower limit of integration
b= upper limit of integration
f(x)
a b
y
x

b
a
dx
)
x
(
f

4. http://numericalmethods.eng.usf.edu
4
Two-Point Gaussian
Quadrature Rule

5. http://numericalmethods.eng.usf.edu
5
Basis of the Gaussian
Quadrature Rule
Previously, the Trapezoidal Rule was developed by the method
of undetermined coefficients. The result of that development is
summarized below.
)
(
2
)
(
2
)
(
)
(
)
( 2
1
b
f
a
b
a
f
a
b
b
f
c
a
f
c
dx
x
f
b
a








6. http://numericalmethods.eng.usf.edu
6
Basis of the Gaussian
Quadrature Rule
The two-point Gauss Quadrature Rule is an extension of the
Trapezoidal Rule approximation where the arguments of the
function are not predetermined as a and b but as unknowns
x1 and x2. In the two-point Gauss Quadrature Rule, the
integral is approximated as


b
a
dx
)
x
(
f
I )
x
(
f
c
)
x
(
f
c 2
2
1
1 


7. http://numericalmethods.eng.usf.edu
7
Basis of the Gaussian
Quadrature Rule
The four unknowns x1, x2, c1 and c2 are found by assuming that
the formula gives exact results for integrating a general third
order polynomial, .
x
a
x
a
x
a
a
)
x
(
f 3
3
2
2
1
0 



Hence
 

 



b
a
b
a
dx
x
a
x
a
x
a
a
dx
)
x
(
f 3
3
2
2
1
0
b
a
x
a
x
a
x
a
x
a 









4
3
2
4
3
3
2
2
1
0
  






 








 








 



4
3
2
4
4
3
3
3
2
2
2
1
0
a
b
a
a
b
a
a
b
a
a
b
a

8. http://numericalmethods.eng.usf.edu
8
Basis of the Gaussian
Quadrature Rule
It follows that
   
3
2
3
2
2
2
2
1
0
2
3
1
3
2
1
2
1
1
0
1 x
a
x
a
x
a
a
c
x
a
x
a
x
a
a
c
dx
)
x
(
f
b
a









Equating Equations the two previous two expressions yield
  






 








 








 


4
3
2
4
4
3
3
3
2
2
2
1
0
a
b
a
a
b
a
a
b
a
a
b
a
   
3
2
3
2
2
2
2
1
0
2
3
1
3
2
1
2
1
1
0
1 x
a
x
a
x
a
a
c
x
a
x
a
x
a
a
c 







       
3
2
2
3
1
1
3
2
2
2
2
1
1
2
2
2
1
1
1
2
1
0 x
c
x
c
a
x
c
x
c
a
x
c
x
c
a
c
c
a 








9. http://numericalmethods.eng.usf.edu
9
Basis of the Gaussian
Quadrature Rule
Since the constants a0, a1, a2, a3 are arbitrary
2
1 c
c
a
b 

 2
2
1
1
2
2
2
x
c
x
c
a
b



2
2
2
2
1
1
3
3
3
x
c
x
c
a
b


 3
2
2
3
1
1
4
4
4
x
c
x
c
a
b




10. http://numericalmethods.eng.usf.edu
10
Basis of Gauss Quadrature
The previous four simultaneous nonlinear Equations have
only one acceptable solution,
2
1
a
b
c


2
2
a
b
c


2
3
1
2
1
a
b
a
b
x














 

2
3
1
2
2
a
b
a
b
x













 


11. http://numericalmethods.eng.usf.edu
11
Basis of Gauss Quadrature
Hence Two-Point Gaussian Quadrature Rule
   







 

















 














2
3
1
2
2
2
3
1
2
2
)
( 2
2
1
1
a
b
a
b
f
a
b
a
b
a
b
f
a
b
x
f
c
x
f
c
dx
x
f
b
a

12. http://numericalmethods.eng.usf.edu
12
Higher Point Gaussian
Quadrature Formulas

13. http://numericalmethods.eng.usf.edu
13
Higher Point Gaussian
Quadrature Formulas
)
(
)
(
)
(
)
( 3
3
2
2
1
1 x
f
c
x
f
c
x
f
c
dx
x
f
b
a




is called the three-point Gauss Quadrature Rule.
The coefficients c1, c2, and c3, and the functional arguments x1, x2, and x3
are calculated by assuming the formula gives exact expressions for
 
 




b
a
dx
x
a
x
a
x
a
x
a
x
a
a 5
5
4
4
3
3
2
2
1
0
General n-point rules would approximate the integral
)
x
(
f
c
.
.
.
.
.
.
.
)
x
(
f
c
)
x
(
f
c
dx
)
x
(
f n
n
b
a




 2
2
1
1
integrating a fifth order polynomial

14. http://numericalmethods.eng.usf.edu
14
Arguments and Weighing Factors
for n-point Gauss Quadrature
Formulas
In handbooks, coefficients and
Gauss Quadrature Rule are
 
 

1
1 1
n
i
i
i )
x
(
g
c
dx
)
x
(
g
as shown in Table 1.
Points Weighting
Factors
Function
Arguments
2 c1 = 1.000000000
c2 = 1.000000000
x1 = -0.577350269
x2 = 0.577350269
3 c1 = 0.555555556
c2 = 0.888888889
c3 = 0.555555556
x1 = -0.774596669
x2 = 0.000000000
x3 = 0.774596669
4 c1 = 0.347854845
c2 = 0.652145155
c3 = 0.652145155
c4 = 0.347854845
x1 = -0.861136312
x2 = -0.339981044
x3 = 0.339981044
x4 = 0.861136312
arguments given for n-point
given for integrals
Table 1: Weighting factors c and function
arguments x used in Gauss Quadrature
Formulas.

15. http://numericalmethods.eng.usf.edu
15
Arguments and Weighing Factors
for n-point Gauss Quadrature
Formulas
Points Weighting
Factors
Function
Arguments
5 c1 = 0.236926885
c2 = 0.478628670
c3 = 0.568888889
c4 = 0.478628670
c5 = 0.236926885
x1 = -0.906179846
x2 = -0.538469310
x3 = 0.000000000
x4 = 0.538469310
x5 = 0.906179846
6 c1 = 0.171324492
c2 = 0.360761573
c3 = 0.467913935
c4 = 0.467913935
c5 = 0.360761573
c6 = 0.171324492
x1 = -0.932469514
x2 = -0.661209386
x3 = -0.2386191860
x4 = 0.2386191860
x5 = 0.661209386
x6 = 0.932469514
Table 1 (cont.) : Weighting factors c and function arguments x used in
Gauss Quadrature Formulas.

16. http://numericalmethods.eng.usf.edu
16
Arguments and Weighing Factors
for n-point Gauss Quadrature
Formulas
So if the table is given for 

1
1
dx
)
x
(
g integrals, how does one solve

b
a
dx
)
x
(
f ? The answer lies in that any integral with limits of  
b
,
a
can be converted into an integral with limits  
1
1,
 Let
c
mt
x 

If then
,
a
x  1


t
,
b
x 
If then 1

t
Such that:
2
a
b
m



17. http://numericalmethods.eng.usf.edu
17
Arguments and Weighing Factors
for n-point Gauss Quadrature
Formulas
2
a
b
c


Then Hence
2
2
a
b
t
a
b
x



 dt
a
b
dx
2


Substituting our values of x, and dx into the integral gives us

 






 



1
1
2
2
2
)
( dt
a
b
a
b
t
a
b
f
dx
x
f
b
a

18. http://numericalmethods.eng.usf.edu
18
Example 1
For an integral derive the one-point Gaussian Quadrature
Rule.
,
dx
)
x
(
f
b
a

Solution
The one-point Gaussian Quadrature Rule is
 
1
1 x
f
c
dx
)
x
(
f
b
a



19. http://numericalmethods.eng.usf.edu
19
Solution
The two unknowns x1, and c1 are found by assuming that the
formula gives exact results for integrating a general first order
polynomial,
.
)
( 1
0 x
a
a
x
f 

 

 

b
a
b
a
dx
x
a
a
dx
x
f 1
0
)
(
b
a
x
a
x
a 







2
2
1
0
  






 



2
2
2
1
0
a
b
a
a
b
a

20. http://numericalmethods.eng.usf.edu
20
Solution
It follows that
 
1
1
0
1
)
( x
a
a
c
dx
x
f
b
a



Equating Equations, the two previous two expressions yield
  






 


2
2
2
1
0
a
b
a
a
b
a  
1
1
0
1 x
a
a
c 
 )
(
)
( 1
1
1
1
0 x
c
a
c
a 


21. http://numericalmethods.eng.usf.edu
21
Basis of the Gaussian
Quadrature Rule
Since the constants a0, and a1 are arbitrary
1
c
a
b 

1
1
2
2
2
x
c
a
b


a
b
c 

1
2
1
a
b
x


giving

22. http://numericalmethods.eng.usf.edu
22
Solution
Hence One-Point Gaussian Quadrature Rule
  




 



 2
)
(
)
( 1
1
a
b
f
a
b
x
f
c
dx
x
f
b
a

23. http://numericalmethods.eng.usf.edu
23
Example 2
Use two-point Gauss Quadrature Rule to approximate the distance
covered by a rocket from t=8 to t=30 as given by
 














30
8
8
9
2100
140000
140000
2000 dt
t
.
t
ln
x
Find the true error, for part (a).
Also, find the absolute relative true error, for part (a).
a

a)
b)
c)
t
E

24. http://numericalmethods.eng.usf.edu
24
Solution
First, change the limits of integration from [8,30] to [-1,1]
by previous relations as follows








 




1
1
30
8 2
8
30
2
8
30
2
8
30
dx
x
f
dt
)
t
(
f
 




1
1
19
11
11 dx
x
f

25. http://numericalmethods.eng.usf.edu
25
Solution (cont)
Next, get weighting factors and function argument values from Table 1
for the two point rule,
000000000
1
1 .
c 
577350269
0
1 .
x 

000000000
1
2 .
c 
577350269
0
2 .
x 

26. http://numericalmethods.eng.usf.edu
26
Solution (cont.)
Now we can use the Gauss Quadrature formula
     
19
11
11
19
11
11
19
11
11 2
2
1
1
1
1







x
f
c
x
f
c
dx
x
f
   
19
5773503
0
11
11
19
5773503
0
11
11 



 )
.
(
f
)
.
(
f
)
.
(
f
)
.
(
f 35085
25
11
64915
12
11 

)
.
(
)
.
( 4811
708
11
8317
296
11 

m
.44
11058


27. http://numericalmethods.eng.usf.edu
27
Solution (cont)
since
)
.
(
.
)
.
(
ln
)
.
(
f 64915
12
8
9
64915
12
2100
140000
140000
2000
64915
12 








8317
296.

)
.
(
.
)
.
(
ln
)
.
(
f 35085
25
8
9
35085
25
2100
140000
140000
2000
35085
25 








4811
708.


28. http://numericalmethods.eng.usf.edu
28
Solution (cont)
The absolute relative true error, t
 , is (Exact value = 11061.34m)
%
.
.
.
t 100
34
11061
44
11058
34
11061




%
.0262
0

c)
The true error, , is
b) t
E
Value
e
Approximat
Value
True
Et 

44
.
11058
34
.
11061 

m
9000
.
2


29. Additional Resources
For all resources on this topic such as digital audiovisual
lectures, primers, textbook chapters, multiple-choice
tests, worksheets in MATLAB, MATHEMATICA, MathCad
and MAPLE, blogs, related physical problems, please
visit
http://numericalmethods.eng.usf.edu/topics/gauss_qua
drature.html

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