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Gauss Quadrature Rule of Integration

http://numericalmethods.eng.usf.

edu

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

TWO-POINT GAUSSIAN

QUADRATURE RULE

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 (

) ( )

( )

(

_{1 2}

b a f

a b a f

b

b f c a

f c dx

x f

b

a

+ −

= −

+

 

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  c

₁

f ( x

₁

) + c

₂

f ( x

₂

)

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,

f ( x ) = a

₀

+ a

₁

x + a

₂

x

²

+ a

₃

x

³

.

Hence

( )



 ⁼

^b

^{+ + +}

a b

a

dx x

a x

a x

a a

dx ) x (

f

_{0 1 2} ² ₃ ³

b

a

a x a x

a x x

a 



 



 + + +

= 2 3 4

4 3 3

2 2

1 0

( ) _



 



 −

 +



 



 −

 +



 



 −

+

−

= 2 3 4

4 4

3 3

3 2

2 2

1 0

a a b

a a b

a a b

a b

a

Basis of the Gaussian Quadrature Rule

It follows that

( ) (

3 2³

)

2 2 2 2

1 0

2 3

1 3 2

1 2 1

1 0

1

a a x a x a x c a a x a x a x

c dx

) x ( f

b a

+ +

+ +

+ +

+

 =

Equating Equations the two previous two expressions yield

( ) _



 



 −

 +



 



 −

 +



 



 −

+

− 2 3 4

4 4

3 3

3 2

2 2

1 0

a a b

a a b

a a b

a b

a

( ) (

3 2³

)

2 2 2 2

1 0

2 3

1 3 2

1 2 1

1 0

1

a a x a x a x c a a x a x a x

c + + + + + + +

=

( ) ( ) ( ) (

2 2³

)

3 1 1 3 2

2 2 2

1 1 2 2

2 1

1 1 2

1

0

c c a c x c x a c x c x a c x c x

a + + + + + + +

=

Basis of the Gaussian Quadrature Rule

Since the constants a₀, a₁, a₂, a₃ are arbitrary

2

1

c

c a

b − = +

^{1 1 2 2}

2 2

2 a c x c x

b − = +

2 2 2 2

1 1 3

3

3 a c x c x

b − = +

₃

2 2 3

1 1 4

4

4 a c x c x

b − = +

Basis of Gauss Quadrature

The previous four simultaneous nonlinear Equations have only one acceptable solution,

1

2

a c b −

=

²

2

a c b −

= 2

3 1

1

2

a b

a

x b  + +



 

 −

 

 



=  −

2 3

1

2

2

a b

a

x b  + +



 



 



 



=  −

Basis of Gauss Quadrature

Hence Two-Point Gaussian Quadrature Rule

( ) ( )

 

 



  + +



 



−  + −

 

 



  + +



 

 −

−

= −

+

 

3 2 1 2

2 3 2

1 2

2 )

(

_{1 1 2 2}

a b a

f b a b a

b a

f b a b

x f c x

f c dx

x f

b

a

HIGHER POINT GAUSSIAN

QUADRATURE FORMULAS

Higher Point Gaussian Quadrature Formulas

) ( )

( )

( )

( x dx c

₁

f x

₁

c

₂

f x

₂

c

₃

f x

₃

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

_{0 1 2} ² ₃ ³ ₄ ⁴ ₅ ⁵

General n-point rules would approximate the integral

) x ( f c .

. . . . . . ) x ( f c ) x ( f c dx

) x (

f

_{n n}

b

 

1 1

+

2 2

+ +

integrating a fifth order polynomial

Arguments and Weighing Factors for n-point Gauss Quadrature

Formulas

In handbooks, coefficients and

Gauss Quadrature Rule are

 

−¹



=

1 1

n

i

c

i

g ( x

i

) 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.

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.

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

  ^{a , b}

can be converted into an integral with limits

 ^{− 1 , 1} 

^Let

c mt

x = +

If then

x = a , t = − 1 ,

b x =

If then

t = 1

^{Such that:}

2 a m b −

=

Arguments and Weighing Factors for n-point Gauss Quadrature

Formulas

2 a c b +

Then

=

Hence

2 2

a t b

a

x b +

− +

= b a dt

dx 2

= −

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





−

 −



 



 − + +

=

1

1

2 2 2

)

( b a b a dt

a t f b

dx x f

b

a

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

f x c

dx ) x ( f

b a

 

Solution

The two unknowns x₁, and c₁ are found by assuming that the formula gives exact results for integrating a general first order polynomial,

. )

(x a₀ a₁x

f = +

( )





^{= b +}

a b

a

dx x a a

dx x

f ( ) _{0 1}

b

a

a x x

a 



 



 +

= 2

2 1 0

( ) _



 



 −

+

−

= 2

2 2

1 0

a a b

a

b

a

Solution

It follows that

(

_{0 1 1}

)

)

1

( x dx c a a x f

b

a

+

 =

Equating Equations, the two previous two expressions yield

( ) _



 



 −

+

− 2

2 2

1 0

a a b

a b

a ^{= c}

₁

( ^a

₀

^{+ a}

₁

^x

₁

) = a

₀

( c

₁

) + a

₁

( c

₁

x

₁

)

Basis of the Gaussian Quadrature Rule

Since the constants a₀, and a₁ are arbitrary

c

1

a b − =

1 1 2

2

2 a c x

b − =

a b

c

₁

= −

1

2

a x = b +

giving

Solution

Hence One-Point Gaussian Quadrature Rule

( ) _



 



−  +

=



^b

^{f ( x ) dx}  ^c

¹

^{f x}

¹

^{( b a ) f b} ₂ ^a

a

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 2100 9

140000

140000

2000 . t dt

ln t x

Find the true error, for part (a).

Also, find the absolute relative true error, for part (a).a

a)

b) c)

Et

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 f x dx

dt ) t ( f

( )

−



+

=

¹

1

19 11

11 f x dx

Solution (cont)

Next, get weighting factors and function argument values from Table 1 for the two point rule,

000000000

1

1 .

c =

577350269

1

0 .

x = −

000000000

2

1 .

c =

577350269

2

0 .

x =

Solution (cont.)

Now we can use the Gauss Quadrature formula

( ^{11 19} ) ¹¹ ( ^{11 19} ) ¹¹ ( ^{11 19} )

11

_{1 1 2 2}

1

1

+ +

+



 +

−

x f

c x

f c dx

x f

( ^{11 0 5773503 19} ) ¹¹ ( ^{11 0 5773503 19} )

11 − + + +

= f ( . ) f ( . )

) .

( f )

. (

f 12 64915 11 25 35085

11 +

=

) .

( )

.

( 296 8317 11 708 4811

11 +

=

m .44 11058

=

Solution (cont)

since

) .

( ) .

. ln (

) .

(

f 9 8 12 64915

64915 12

2100 140000

140000 2000

64915

12 −

 

 

= −

8317

= 296.

) .

( ) .

. ln (

) .

(

f 9 8 25 35085

35085 25

2100 140000

140000 2000

35085

25 −

 

 

= −

4811

= 708.

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)

E

t

Value e

Approximat Value

True

E

_t

= −

44 . 11058 34

.

11061 −

=

m 9000 .

= 2