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The Islamic University of Gaza Faculty of Engineering Civil Engineering Department Numerical Analysis ECIV 3306 Chapter 22. Integration of Equations . Gauss Quadrature. - PowerPoint PPT Presentation
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The Islamic University of GazaFaculty of Engineering
Civil Engineering Department
Numerical Analysis ECIV 3306
Chapter 22
Integration of Equations
Gauss Quadrature
• Gauss quadrature implements a strategy of positioning any two points on a curve to define a straight line that would balance the positive and negative errors.
• Hence, the area evaluated under this straight line provides an improved estimate of the integral.
Two points Gauss-Legendre Formula• Assume that the two Integration points are xo and x1 such that:
• The object of Gauss quadrature is to determine the equations of the form:
• c0 and c1 are constants, the function arguments x0 and x1 are unknowns…….(4 unknowns)
)()( 1100 xfcxfcI
Two points Gauss-Legendre Formula
• Thus, four unknowns to be evaluated require four conditions.
• If this integration is exact for a constant, 1st order, 2nd order, and 3rd order functions:
1
1
31100
1
1
21100
1
11100
1
11100
0)()(
32)()(
0)()(
21)()(
dxxxfcxfc
dxxxfcxfc
dxxxfcxfc
dxxfcxfc
Two points Gauss-Legendre Formula
• Solving these 4 equations, we can determine c1, c2, x1 and x2.
31
31 ffI
5773503.03
1
5773503.03
1:are pointsn Integratio The
1:are factors weightingThe
1
0
10
x
x
cc
Two points Gauss-Legendre Formula• Since we used limits for the previous integration from –1 to 1
and the actual limits are usually from a to b, then we need first to transform both the function and the integration from the x-system to the xd-system
1ax1bx
d2
abdx
2ab
2abx
f(x)
x
a b
f(xo)
f(x1)
xo x1 -1 1
f()
Higher-Points Gauss-Legendre Formula
)(.....)()()(
)(
nn2211
n
1iii
1
1i
fcfcfcfcI
:points Gauss n usingdfI
Multiple Points Gauss-LegendrePoints Weighting factor Function argument Exact
for 2 1.0 -0.577350269 up to
3rd 1.0 0.577350269 degree
3 0.5555556 -0.774596669 up to 5th
0.8888889 0.0 degree0.5555556 0.774596669
4 0.3478548 -0.861136312 up to 7th
0.6521452 -0.339981044 degree0.6521452 0.3399810440.3478548 0.861136312
6 0.1713245 -0.932469514 up to 11th
0.3607616 -0.661209386 degree0.4679139 -0.2386191860.4679139 0.2386191860.3607616 0.6612093860.1713245 0.932469514
Gauss Quadrature - Example
Find the integral of: f(x) = 0.2 + 25 x – 200 x2 + 675 x3 – 900 x4 + 400 x5
Between the limits 0 to 0.8 using:
– 2 points integration points (ans. 1.822578)
– 3 points integration points (ans. 1.640533)
Improper Integral• Improper integrals can be evaluated by making a change
of variable that transforms the infinite range to one that is finite,
b
A
b A
b
a
a
b
dxxfdxxfdxxf
abdtt
ft
dxxf
)()()(
011)(/1
/12
A
A
dtt
ft
dxxf0
/12
11)(Can be evaluated by Newton-Cotes closed formula
Improper Integral - Examples
• .
• .
• .
dtt
dtt
ttxx
dx
5.0
0
5.0
0 2
2 211
2/11)(1
)2(
dyyedyyedyye yyy
2
22
0
2
0
2 sin sin sin
dttet
dyye ty 2/1
0
2/12
2
2 )/1(sin1 sin
dyyedyyedyye yyy
2
2
2
2
dtet
dyye ty 2/1
0
/13
2 1
Multiple Integration
dydxyxfId
cy
b
ax
),(
• Double integral:
Multiple Integration using Gauss Quadrature Technique
• .
• .
• . ddf
2ab
2cddydxyxfI
1
1
1
1
d
cy
b
ax
),(),(
1ax1bx &
d2
abdx2
ab2
abx
&
1cx1dy &
d2
cddy2
cd2
cdy
&
Multiple Integration using Gauss Quadrature Technique
Now we can use the Gauss Quadrature technique:
If we use two points Gauss Formula:
n
1jjiij
n
1i
fcc2
ab2
cdI ),(
)}],(),({
)},(),({[
31
31fc
31
31fcc
31
31fc
31
31fcc
2ab
2cdI
212
211
2
1jjijj
2
1i
fcc2
ab2
cdI ),(
2
1jj2j1j 3
1fc3
1fcc2
ab2
cdI )],(),([
Double integral - Example
72222),( 22 yxxxyyxT
• Compute the average temperature of a rectangular heated plate which is 8m long in the x direction and 6 m wide in the y direction. The temperature is given as:
• (Use 2 segment applications of the trapezoidal rule in each dimension)
Double integral - Example
6667.58)86/(2816,28163/1
56)86/(2688,2688)2(
)72222( 6
0
8
0
22
avg
avg
TIruleSimpson
TInrulelTrapezoidaMultiple
dxdyyxxxyI
HW: Use two points Gauss formula to solve the problem