mws gen int ppt gaussquadrature

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4/5/2012 http://numericalmethods.eng.usf.edu 1

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
http://numericalmethods.eng.usf.edu/http://numericalmethods.eng.usf.edu/


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Gauss Quadrature Rule ofIntegration

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http://numericalmethods.eng.usf.edu3

What is Integration?Integration

b

a

dx)x(fI

The process of measuringthe 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 GaussianQuadrature Rule


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Basis of the Gaussian

Quadrature RulePreviously, the Trapezoidal Rule was developed by the method

of undetermined coefficients. The result of that development is

summarized below.

)(2

)(2

)()()( 21

bfab

afab

bfcafcdxxf

b

a


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Quadrature Rule

The two-point Gauss Quadrature Rule is an extension of theTrapezoidal 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, theintegral is approximated as

b

adx)x(fI )x(fc)x(fc 2211


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Quadrature RuleThe four unknowns x1, x2, c1 and c2 are found by assuming thatthe formula gives exact results for integrating a general thirdorder polynomial, .xaxaxaa)x(f 33

2

210

Hence

b

a

b

a

dxxaxaxaadx)x(f 332

210

b

a

xaxaxaxa

432

4

3

3

2

2

10

432

44

3

33

2

22

10

aba

aba

abaaba


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Quadrature RuleIt follows that

32322221023132121101 xaxaxaacxaxaxaacdx)x(fb

a

Equating Equations the two previous two expressions yield

432

44

3

33

2

22

10

aba

aba

abaaba

3

23

2

222102

3

13

2

121101 xaxaxaacxaxaxaac

322

3

113

2

22

2

11222111210

xcxcaxcxcaxcxcacca


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Quadrature RuleSince the constants a0, a1, a2, a3 are arbitrary

21 ccab 221122

2xcxc

ab

2

22

2

11

33

3 xcxcab

322

311

44

4xcxcab


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Basis of Gauss QuadratureThe previous four simultaneous nonlinear Equations haveonly 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

abab

f

ababab

f

ab

xfcxfcdxxf

b

a


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Higher Point GaussianQuadrature Formulas


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Higher Point Gaussian

Quadrature Formulas)()()()( 332211 xfcxfcxfcdxxf

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 dxxaxaxaxaxaa

5

5

4

4

3

3

2

210

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 Factorsfor n-point Gauss Quadrature

Formulas

In handbooks, coefficients and

Gauss Quadrature Rule are

1

1 1

n

i

ii )x(gcdx)x(g

as shown in Table 1.

Points WeightingFactors

FunctionArguments

2 c1 = 1.000000000c2 = 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 functionarguments x used in Gauss QuadratureFormulas.

d i hi


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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 = -0.661209386x3 = -0.2386191860x4 = 0.2386191860x5 = 0.661209386x6 = 0.932469514

Table 1 (cont.) : Weighting factors c and function arguments x used inGauss Quadrature Formulas.

A d W i hi F


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FormulasSo 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

A t d W i hi F t


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Formulas

2

abc

Then Hence

22

abtabx dtabdx2

Substituting our values of x, and dx into the integral gives us

1

1222

)( dtabab

tab

fdxxf

b

a


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Example 1For 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 theformula gives exact results for integrating a general first orderpolynomial,

.)( 10 xaaxf

b

a

b

a

dxxaadxxf 10)(

b

a

xaxa

2

2

10

2

22

10

abaaba


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SolutionIt follows that

1101

)( xaacdxxf

b

a

Equating Equations, the two previous two expressions yield

2

22

10

abaaba 1101 xaac )()( 11110 xcaca


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Quadrature RuleSince 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 2Use 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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SolutionFirst, 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.)Now we can use the Gauss Quadrature formula

191111191111191111 22111

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)tE

ValueeApproximatValueTrueEt

44.1105834.11061 m9000.2


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Additional ResourcesFor all resources on this topic such as digital audiovisuallectures, primers, textbook chapters, multiple-choicetests, worksheets in MATLAB, MATHEMATICA, MathCad

and MAPLE, blogs, related physical problems, pleasevisit

http://numericalmethods.eng.usf.edu/topics/gauss_qua

drature.html
http://numericalmethods.eng.usf.edu/topics/gauss_quadrature.htmlhttp://numericalmethods.eng.usf.edu/topics/gauss_quadrature.htmlhttp://numericalmethods.eng.usf.edu/topics/gauss_quadrature.htmlhttp://numericalmethods.eng.usf.edu/topics/gauss_quadrature.html


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