Mws Gen Int Ppt Simpson3by8

Numerical Methods Part: Simpson Rule For Integration. http://numericalmethods.eng.usf.edu

83

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Chapter 07.08: Simpson Rule For Integration.

Major: All Engineering Majors

Authors: Duc Nguyen

http://numericalmethods.eng.usf.edu Numerical Methods for STEM undergraduates

http://numericalmethods.eng.usf.edu 5 11/14/2011

Lecture # 1

83


Most (if not all) of the developed formulas for integration is based on a simple concept of replacing a given (oftently complicated) function by a simpler function (usually a polynomial function) where represents the order of the polynomial function. 6

Introduction The main objective in this chapter is to develop appropriated formulas for obtaining the integral expressed in the following form:

∫=b

adxxfI )(

where is a given function. )(xf

)(xf)(xfi i

(1)

http://numericalmethods.eng.usf.edu 7

In the previous chapter, it has been explained and illustrated that Simpsons 1/3 rule for integration can be derived by replacing the given function )(xfwith the 2nd –order (or quadratic) polynomial function

)()( 2 xfxfi = , defined as:

22102 )( xaxaaxf ++= (2)

which can also be symbolically represented in Figure 1. 8

In a similar fashion, Simpson rule for integration can be derived by replacing the given function )(xfwith the 3rd-order (or cubic) polynomial (passing through 4 known data points) function )()( 3 xfxfi =defined as

{ }

×=

+++=

3

2

1

0

32

33

22103

,,,1

)(

aaaa

xxx

xaxaxaaxf

(3)

83


Method 1 The unknown coefficients (in Eq. (3)) can be obtained by substituting 4 known coordinate data points

3210 ,, aandaaa

( ) ( ){ } ( ){ } ( ){ }{ }33221100 ,,,,,, xfxandxfxxfxxfxinto Eq. (3), as following

+++=

+++=

+++=

+++=

233

2323103

223

2222102

213

2121101

203

2020100

)(

)(

)(

)(

xaxaxaaxf

xaxaxaaxf

xaxaxaaxf

xaxaxaaxf

(4)


Eq. (4) can be expressed in matrix notation as

( )( )( )( )

=

3

2

1

0

3

2

1

0

33

233

32

222

31

211

30

200

1111

xfxfxfxf

aaaa

xxxxxxxxxxxx

The above Eq. (5) can be symbolically represented as

[ ] 141444 ××× = faA

(5)

(6)


Thus,

[ ] fA

aaaa

a

×=

= −1

3

2

1

0

Substituting Eq. (7) into Eq. (3), one gets

( ) { } [ ] fAxxxxf

××= −1323 ,,,1

(7)

(8)


Remarks As indicated in Figure 1, one has

=−

+=+=

+=

−+=+=

+=

−+=+=

=

babahax

baabahax

baabahax

ax

3333

32

3222

32

3

3

2

1

0

With the help from MATLAB [2], the unknown vector (shown in Eq. 7) can be solved. a

(9)


Method 2 Using Lagrange interpolation, the cubic polynomial function that passes through 4 data points ( )xfi 3=

(see Figure 1) can be explicitly given as

( ) ( )( )( )( )( )( ) ( ) ( )( )( )

( )( )( ) ( )

( )( )( )( )( )( ) ( ) ( )( )( )

( )( )( ) ( )3231303

2103

321202

310

1312101

3200

302010

3213

xfxxxxxx

xxxxxxxfxxxxxx

xxxxxx

xfxxxxxx

xxxxxxxfxxxxxx

xxxxxxxf

×−−−

−−−+×

−−−−−−

+

×−−−−−−

+×−−−

−−−=

(10)


Simpsons Rule For Integration

Thus, Eq. (1) can be calculated as (See Eqs. 8, 10 for Method 1 and Method 2, respectively):

( ) ( )∫∫ ≈=b

a

b

adxxfdxxfI 3

Integrating the right-hand-side of the above equations, one obtains

( ) ( ) ( ) ( ) ( ){ }8

33 3210 xfxfxfxfabI +++×−= (11)

83


Since hence , and the above 3abh −

=

equation becomes:

( ) ( ) ( ) ( ){ }3210 338

3 xfxfxfxfhI +++×≈

The error introduced by the Simpson 3/8 rule can be derived as [Ref. 1]:

( )ζfabEt ′′′′×−

−=6480

)( 5, where ba ≤≤ζ

hab 3=−

(13)

(12)


Example 1 (Single Simpson rule)

Compute

∫=

=

−

−

=30

8,8.9

2100000,140000,140ln2000

b

adxx

xI

by using a single segment Simpson rule Solution In this example:

3333.73

8303

=−

=−

=abh

83

83


( ) 2667.17788.982100140000

140000ln20008 00 =×−

×−=⇒= xfx

( )

=×−

×−=

=+=+=

4629.3723333.158.93333.152100140000

140000ln2000

3333.153333.78

1

01

xf

hxx


( )

=×−

×−=

=+=+=

8976.6086666.228.96666.222100140000

140000ln2000

6666.22)3333.7(282

2

02

xf

hxx

( )

=×−

×−=

=+=+=

6740.901308.9302100140000

140000ln2000

30)3333.7(383

3

03

xf

hxx


Applying Eq. (12), one has:

{ }

3104.11063

6740.9018976.60834629.37232667.1773333.783

=

+×+×+××=

I

I

The “exact” answer can be computed as

34.11061=exactI


3. Multiple Segments for Simpson Rule

Using = number of equal (small) segments, the width can be defined as

3abh −

=

Notes: = multiple of 3 = number of small segments

""n""h

(14)

n ""h

83


( ) ( )∫∫ ≈=b

a

b

adxxfdxxfI 3

( ) ( ) ( )∫∫∫=

= −

+++≈bx

x

x

x

x

ax

n

n

dxxfdxxfdxxfI3

6

3

3

0

333 ........

The integral, shown in Eq. (1), can be expressed as

(15)


Substituting Simpson rule (See Eq. 12) into Eq. (15), one gets

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( )

++++++++++++

=−−− nnnn xfxfxfxf

xfxfxfxfxfxfxfxfhI123

65433210

33.....3333

83

( ) ( ) ( ) ( ) ( )

++++= ∑∑∑−

=

−

=

−

=n

n

ii

n

ii

n

ii xfxfxfxfxfhI

3

,..9,6,3

1

,..8,5,2

2

,..7,4,10 233

83

(16)

(17)

83


Example 2 (Multiple segments Simpson rule)

Compute ∫=

=

−

−

=30

8,8.9

2100000,140000,140ln2000

b

adxx

xI

using Simple multiple segments rule, with number (of ) segments = = 6 (which corresponds to 2 “big” segments).

""h n

83

83


Solution In this example, one has (see Eq. 14):

6666.36

830=

−=h

( ){ } { }2667.177,8, 00 =xfx

( ){ } { }6666.11

6666.38;4104.270,6666.11, 0111

=+=+== hxxwherexfx

( ){ } { } 3333.152;4629.372,3333.15, 0222 =+== hxxwherexfx

( ){ } { } 193;7455.484,19, 0333 =+== hxxwherexfx


( ){ } { } 6666.224;8976.608,6666.22, 0444 =+== hxxwherexfx

( ){ } { } 3333.265;9870.746,3333.26, 0555 =+== hxxwherexfx

( ){ } { } 306;6740.901,30, 0666 =+== hxxwherexfx


Applying Eq. (17), one obtains:

( ) ( ) ( ) ( )

++++= ∑∑∑=−

=

=−

=

=−

=6740.9012332667.1776666.3

83 33

,..6,3

51

,..5,2

42

,..4,1

n

ii

n

ii

n

ii xfxfxfI

( ) ( ) ( )( )

++++++

=6740.9017455.4842

9870.7464629.37238976.6084104.27032667.1773750.1I

4696.601,11=I


Example 3 (Mixed, multiple segments Simpson and rules)

Compute ∫=

=

−

−

=30

8,8.9

2100000,140000,140ln2000

b

adxx

xI

using Simpson 1/3 rule (with 4 small segments), and Simpson 3/8 rule (with 3 small segments).

Solution: In this example, one has:

( ) 1429.334830

21=

+−

=+−

=−

=nnab

nabh

83

31

=1n=2n


( )( )( )

ruleSimpson

hxxhxxhxx

hxxax

31

5714.201429.34844286.171429.33832857.141429.3282

1429.111429.3818

04

03

02

01

0

=+=+==+=+==+=+=

=+=+===

( )( )( ) 301429.3787

8571.261429.36867143.231429.3585

07

06

05

=+=+==+=+==+=+=

hxxhxxhxx


( ) 2667.17788.982100000,140

000,140ln200080 =×−

×−

==xf

Similarly: ( )( )( )( )( )( )( ) 6740.901

9978.7678260.6463909.5362749.4353241.342

5863.2561429.11

7

6

5

4

3

2

1

======

==

xfxfxfxfxfxfxf


For multiple segments ( )segmentsfirstn 41 = using Simpson rule, one obtains (See Eq. 19):

( ) ( ) ( ) ( )

( ) ( ){ }

1197.4364

3909.5363241.34222749.4355863.25642667.1773

1429.3

243

1

1

22

,...2

31

,...3,101 1

11

=

++++

=

+++

= ∑∑

=−

=

=−

=

I

I

xfxfxfxfhI n

n

ii

n

ii

31


For multiple segments using Simpson 3/8 rule, one obtains (See Eq. 17):

( )segmentslastn 32 =

( ) ( ) ( ) ( ) ( )

( ) ( ) ( ){ }

2748.6697

2749.435!3241.34235863.25632667.1771429.383

23383

2

2

03

,...6,3

21

,...2

12

,...3,102 1

222

=

++++

×=

++++

= ∑∑∑

=−

//=

=−

=

=−

=

I

skipI

xfxfxfxfxfhI n

n

ii

n

ii

n

ii

The mixed (combined) Simpson 1/3 and 3/8 rules give:

3946.061,112748.66971197.436421

=+=+=

IIII


Remarks: (a) Comparing the truncated error of Simpson 1/3 rule

( ) ( )ζfabEt ′′′′×−

−=2880

5

With Simple 3/8 rule (See Eq. 13), the latter seems to offer slightly more accurate answer than the former. However, the cost associated with Simpson 3/8 rule (using 3rd order polynomial function) is significant higher than the one associated with Simpson 1/3 rule (using 2nd order polynomial function).

(18)


(b) The number of multiple segments that can be used in the conjunction with Simpson 1/3 rule is 2,4,6,8,.. (any even numbers).

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ){

( ) ( ) ( ) ( )

+++

=

+++++++++

=

∑∑−

=

−

=

−−

n

n

ii

n

ii

nnn

xfxfxfxfhI

xfxfxfxfxfxfxfxfxfhI

2

...6,4,2

1

,...3,102

124322101

243

4.....443

(19)


However, Simpson 3/8 rule can be used with the number of segments equal to 3,6,9,12,.. (can be either certain odd or even numbers).

(c) If the user wishes to use, say 7 segments, then the mixed Simpson 1/3 rule (for the first 4 segments), and Simpson 3/8 rule (for the last 3 segments).


Based on the earlier discussions on (Single and Multiple segments) Simpson 1/3 and 3/8 rules, the following “pseudo” step-by-step mixed Simpson rules can be given as

Step 1 User’s input information, such as

Given function integral limits

= number of small, “h” segments, in conjunction with Simpson 1/3 rule.

1n

4. Computer Algorithm For Mixed Simpson 1/3 and 3/8

rule For Integration

),(xf ,"," ba


= number of small, “h” segments, in conjunction with Simpson 3/8 rule.

2n

Notes:

= a multiple of 2 (any even numbers) 1n

= a multiple of 3 (can be certain odd, or even numbers)

2n


Step 2 Compute 21 nnn +=

nabh −

=

bnhax

ihax

haxhax

ax

n

i

=+=

+=

+=+=

=

.

.

.

.2

1

2

1

0


Step 3 Compute “multiple segments” Simpson 1/3 rule (See Eq. 19)

( ) ( ) ( ) ( )

+++

= ∑∑

−

=

−

=n

n

ii

n

ii xfxfxfxfhI

2

...6,4,2

1

,...3,101

1124

3

(19, repeated)


Step 4 Compute “multiple segments” Simpson 3/8 rule (See Eq. 17)

( ) ( ) ( ) ( ) ( )

++++

= ∑∑∑

−

=

−

=

−

=2

222 3

,...9,6,3

1

...8,5,2

2

...7,4,102 233

83

n

n

ii

n

ii

n

ii xfxfxfxfxfhI

Step 5

21 III +=and print out the final approximated answer for I.

(17, repeated)

(20)

THE END

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Acknowledgement


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