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Lecture Note of Gaussian Quadrature Method

الكلية كلية الهندسة     القسم  الهندسة البيئية     المرحلة 3
أستاذ المادة وليد علي حسن       13/03/2017 07:22:15
University of Babylon
College of Engineering
Department of Environmental Engineering
Engineering Analysis I (ENAN 103)








Gaussian Quadrature Method
Undergraduate Level, 3th Stage



Mr. Waleed Ali Tameemi
Engineer/ College of Engineering/ Babylon University
M.Sc. Civil Engineering/ the University of Kansas/ USA



2016-2017
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4.0 – Gaussian Quadrature Method
The integration of a function can be approximately calculated using Gaussian Quadrature Method as follows:
x=1/2 (b+a)+1/2 (b-a)t
dx=1/2 (b-a)dt
I=?_a^b??f(x)dx=? ?_(-1)^1?f(t)dt=C_1 f(t_1 )+C_2 f(t_2 )+ C_3 f(t_3 )
Then this can be solved as follows:
t_1=(-1)/?3
t_2=0
t_3=1/?3
For two points quadrature,
C_1=1
C_2=0
C_3=1
For three points quadrature,
C_1=0.555
C_2=0.888
C_3=0.555

Ex1: Calculate the value of the given integration.
?_0^10??300x/(1+e^x ) dx?
Use two points and three points quadrature.
Compare your solution with the exact solution (I=246.59).

Solution:
a=0 b=10
x=1/2 (b+a)+1/2 (b-a)t=1/2 (10+0)+1/2 (10-0)t=5+5t
dx=1/2 (b-a)dt=1/2 (10-0)dt=5dt
Sub. x and dx into f(x)
I=?_0^10??300x/(1+e^x ) dx?=?_(-1)^1??(300×(5+5t))/(1+e^((5+5t)) ) 5dt?
Then this can be solved as follows:
I=?_(-1)^1??(300×(5+5t))/(1+e^((5+5t)) ) 5dt?=C_1 f(t_1 )+C_2 f(t_2 )+ C_3 f(t_3 )
t_1=(-1)/?3?f(t_1 )=(300×(5+5×((-1)/?3)))/(1+e^((5+5×((-1)/?3))) )×5=341.8
t_2=0?f(t_2 )=(300×(5+5×(0)))/(1+e^((5+5×(0))) )×5=50.2
t_3=1/?3?f(t_3 )=(300×(5+5×(1/?3)))/(1+e^((5+5×(1/?3))) )×5=4.4
For two points quadrature,
C_1=1
C_2=0
C_3=1
I=?_(-1)^1??(300×(5+5t))/(1+e^((5+5t)) ) 5dt?=1×(341.8)+0×50.2+ 1×4.4=346.2
Compare with the true value:
?=|(True Value-Approximate Value)/(True Value)|×100%=|(246.59-346.2)/246.59|×100%=40.4%

For three points quadrature,
C_1=0.555
C_2=0.888
C_3=0.555
I=?_(-1)^1??(300×(5+5t))/(1+e^((5+5t)) ) 5dt?=0.555×(341.8)+0.888×50.2+ 0.555×4.4=236.71
Compare with the true value:
?=|(True Value-Approximate Value)/(True Value)|×100%=|(246.59-236.71)/246.59|×100%=4%


Ex2: Calculate the value of the given integration.
?_0^1??x/sin?x dx?
Use two points and three points quadrature.

Solution:
a=0 b=1
x=1/2 (b+a)+1/2 (b-a)t=1/2 (1+0)+1/2 (1-0)t=1/2+t/2
dx=1/2 (b-a)dt=1/2 (1-0)dt=dt/2
Sub. x and dx into f(x)
I=?_0^1??x/sin?x dx?=?_(-1)^1??(1/2+t/2)/sin??(1/2+t/2)? dt/2?
Then this can be solved as follows:
I=?_(-1)^1??(1/2+t/2)/sin??(1/2+t/2)? dt/2?=C_1 f(t_1 )+C_2 f(t_2 )+ C_3 f(t_3 )
t_1=(-1)/?3?f(t_1 )=(1/2+(((-1)/?3))/2)/sin??(1/2+(((-1)/?3))/2)? 1/2=0.50
t_2=0?f(t_2 )=(1/2+((0))/2)/sin??(1/2+((0))/2)? 1/2=0.52
t_3=1/?3=?f(t_3 )=(1/2+((1/?3))/2)/sin??(1/2+((1/?3))/2)? 1/2 0.56
For two points quadrature,
C_1=1
C_2=0
C_3=1
I=?_(-1)^1??(1/2+t/2)/sin??(1/2+t/2)? dt/2?=1×0.50+0×0.52+ 1×0.56=1.06

For three points quadrature,
C_1=0.555
C_2=0.888
C_3=0.555
I=?_(-1)^1??(1/2+t/2)/sin??(1/2+t/2)? dt/2?=0.555×0.50+0.888×0.52+ 0.555×0.56=1.05

Ex3: Calculate the value of the given integration.
?_1^5???(1+x^2 ) dx?
Use two points and three points quadrature.

Solution:
a=1 b=5
x=1/2 (b+a)+1/2 (b-a)t=1/2 (1+5)+1/2 (5-1)t=3+2t
dx=1/2 (b-a)dt=1/2 (5-1)dt=2dt
Sub. x and dx into f(x)
I=?_1^5???(1+x^2 ) dx?=?_(-1)^1???(1+?(3+2t)?^2 )×2dt?
Then this can be solved as follows:
I=?_(-1)^1???(1+?(3+2t)?^2 )×2dt?=C_1 f(t_1 )+C_2 f(t_2 )+ C_3 f(t_3 )
t_1=(-1)/?3?f(t_1 )=?(1+?(3+2((-1)/?3)?^2 )×2=4.2
t_2=0?f(t_2 )=?(1+?(3+2(0))?^2 )×2=6.3
t_3=1/?3?f(t_3 )=?(1+?(3+2(1/?3))?^2 )×2=8.5
For two points quadrature,
C_1=1
C_2=0
C_3=1
I=?_(-1)^1???(1+?(3+2t)?^2 )×2?=1×4.2+0×6.3+ 1×8.5=12.7

For three points quadrature,
C_1=0.555
C_2=0.888
C_3=0.555
I=?_(-1)^1??(300×(5+5t))/(1+e^((5+5t)) ) 5t?=0.555×4.2+0.888×6.3+ 0.555×8.5=12.6
Homework 15
Use Gaussian Quadrature method to estimate the value of the following integration.
?_0^4???xe?^2x dx?
Use two points and three points quadrature.
Compare your solution with the exact solution (I=5216.92).
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5.0 – Summary

References
http://numericalmethods.eng.usf.edu/topics/trapezoidal_rule.html


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