gauss quadrature rule of integration
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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. Gauss Quadrature Rule of Integration http://numericalmethods.eng.usf.edu. f(x). y. a. - PowerPoint PPT PresentationTRANSCRIPT
04/21/23http://
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Gauss Quadrature Rule of Integration
Major: All Engineering Majors
Authors: Autar Kaw, Charlie Barker
http://numericalmethods.eng.usf.eduTransforming Numerical Methods Education for STEM
Undergraduates
Gauss Quadrature Rule of Integration
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What is Integration?Integration
b
a
dx)x(fI
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
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Two-Point Gaussian Quadrature Rule
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Basis of the Gaussian Quadrature Rule
Previously, the Trapezoidal Rule was developed by the methodof undetermined coefficients. The result of that development issummarized below.
)(2
)(2
)()()( 21
bfab
afab
bfcafcdxxfb
a
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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 unknownsx1 and x2. In the two-point Gauss Quadrature Rule, the integral is approximated as
b
a
dx)x(fI )x(fc)x(fc 2211
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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, .xaxaxaa)x(f 3
32
210 Hence
b
a
b
a
dxxaxaxaadx)x(f 33
2210
b
a
xa
xa
xaxa
432
4
3
3
2
2
10
432
44
3
33
2
22
10
aba
aba
abaaba
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Basis of the Gaussian Quadrature Rule
It follows that
3232222102
313
2121101 xaxaxaacxaxaxaacdx)x(f
b
a
Equating Equations the two previous two expressions yield
432
44
3
33
2
22
10
aba
aba
abaaba
3232222102
313
2121101 xaxaxaacxaxaxaac
3223113
222
211222111210 xcxcaxcxcaxcxcacca
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Basis of the Gaussian Quadrature Rule
Since the constants a0, a1, a2, a3 are arbitrary
21 ccab 2211
22
2xcxc
ab
222
211
33
3xcxc
ab
322
311
44
4xcxc
ab
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Basis of Gauss Quadrature
The previous four simultaneous nonlinear Equations have only one acceptable solution,
21
abc
22
abc
23
1
21
ababx
23
1
22
ababx
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Basis of Gauss Quadrature
Hence Two-Point Gaussian Quadrature Rule
23
1
2223
1
22
)( 2211
ababf
abababf
ab
xfcxfcdxxfb
a
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Higher Point Gaussian Quadrature Formulas
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Higher Point Gaussian Quadrature Formulas
)()()()( 332211 xfcxfcxfcdxxfb
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
dxxaxaxaxaxaa 55
44
33
2210
General n-point rules would approximate the integral
)x(fc.......)x(fc)x(fcdx)x(f nn
b
a
2211
integrating a fifth order polynomial
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Arguments and Weighing Factors for n-point Gauss
Quadrature Formulas
In handbooks, coefficients and
Gauss Quadrature Rule are
1
1 1
n
iii )x(gcdx)x(g
as shown in Table 1.
Points
WeightingFactors
FunctionArguments
2 c1 = 1.000000000
c2 = 1.000000000
x1 = -0.577350269x2 = 0.577350269
3 c1 = 0.555555556c2 = 0.888888889c3 = 0.555555556
x1 = -0.774596669x2 = 0.000000000x3 = 0.774596669
4 c1 = 0.347854845c2 = 0.652145155c3 = 0.652145155c4 = 0.347854845
x1 = -0.861136312x2 = -0.339981044x3 = 0.339981044x4 = 0.861136312
arguments given for n-point
given for integrals
Table 1: Weighting factors c and function arguments x used in Gauss Quadrature Formulas.
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Arguments and Weighing Factors for n-point Gauss
Quadrature Formulas
Points WeightingFactors
FunctionArguments
5 c1 = 0.236926885c2 = 0.478628670c3 = 0.568888889c4 = 0.478628670c5 = 0.236926885
x1 = -0.906179846x2 = -0.538469310x3 = 0.000000000x4 = 0.538469310x5 = 0.906179846
6 c1 = 0.171324492c2 = 0.360761573c3 = 0.467913935c4 = 0.467913935c5 = 0.360761573c6 = 0.171324492
x1 = -0.932469514x2 = -0.661209386
x3 = -0.2386191860
x4 = 0.2386191860x5 = 0.661209386x6 = 0.932469514
Table 1 (cont.) : Weighting factors c and function arguments x used in Gauss Quadrature Formulas.
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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 11, Let
cmtx
If then,ax 1t
,bx If then 1tSuch that:
2
abm
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Arguments and Weighing Factors for n-point Gauss
Quadrature Formulas
2
abc
Then Hence
22
abtab
x
dtab
dx2
Substituting our values of x, and dx into the integral gives us
1
1 222)( dt
ababtab
fdxxfb
a
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Example 1
For an integral
derive the one-point Gaussian Quadrature
Rule.
,dx)x(fb
a
Solution
The one-point Gaussian Quadrature Rule is
11 xfcdx)x(fb
a
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SolutionThe two unknowns x1, and c1 are found by assuming that the formula gives exact results for integrating a general first order polynomial,
.)( 10 xaaxf
b
a
b
a
dxxaadxxf 10)(
b
a
xaxa
2
2
10
2
22
10
abaaba
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Solution
It follows that
1101)( xaacdxxfb
a
Equating Equations, the two previous two expressions yield
2
22
10
abaaba 1101 xaac )()( 11110 xcaca
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Basis of the Gaussian Quadrature Rule
Since the constants a0, and a1 are arbitrary
1cab
11
22
2xc
ab
abc 1
21
abx
giving
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Solution
Hence One-Point Gaussian Quadrature Rule
2)()( 11
abfabxfcdxxf
b
a
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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
892100140000
1400002000 dtt.
tlnx
Find the true error, for part (a).
Also, find the absolute relative true error, for part (a).a
a)
b)
c)
tE
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Solution
First, change the limits of integration from [8,30] to [-1,1]
by previous relations as follows
1
1
30
8 2
830
2
830
2
830dxxfdt)t(f
1
1
191111 dxxf
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Solution (cont)
Next, get weighting factors and function argument values from Table 1
for the two point rule,
00000000011 .c
57735026901 .x
00000000012 .c
57735026902 .x
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Solution (cont.)
Now we can use the Gauss Quadrature formula
191111191111191111 2211
1
1
xfcxfcdxxf
1957735030111119577350301111 ).(f).(f
).(f).(f 350852511649151211
).().( 481170811831729611
m.4411058
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Solution (cont)
since
).(.).(
ln).(f 64915128964915122100140000
14000020006491512
8317296.
).(.).(
ln).(f 35085258935085252100140000
14000020003508525
4811708.
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Solution (cont)
The absolute relative true error, t, is (Exact value = 11061.34m)
%.
..t 100
3411061
44110583411061
%.02620
c)
The true error, , isb) tEValueeApproximatValueTrueEt
44.1105834.11061
m9000.2
Additional ResourcesFor 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_quadrature.html