Simpson Rule For Integration.

 

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.2

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

b

a

dxxfI )(

where is a given function.)(xf

)(xf)(xfi i

(1)

3

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 1.4

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

)(

a

a

a

a

xxx

xaxaxaaxf

(3)

83

5

Method 1The 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)

6

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

1

1

1

1

xf

xf

xf

xf

a

a

a

a

xxx

xxx

xxx

xxx

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

141444 faA

(5)

(6)

7

Thus,

fA

a

a

a

a

a

1

3

2

1

0

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

fAxxxxf

1323 ,,,1

(7)

(8)

8

Remarks As indicated in Figure 1, one has

bab

ahax

baabahax

baabahax

ax

3

333

3

2

3

222

3

2

3

3

2

1

0

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

(9)

9

Method 2Using 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

xxxxxxxf

xxxxxx

xxxxxx

xfxxxxxx

xxxxxxxf

xxxxxx

xxxxxxxf

(10)

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

a

dxxfdxxfI 3

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

8

33 3210 xfxfxfxfabI

(11)

83

11

Since hence , and the above3

abh

equation becomes:

3210 338

3xfxfxfxf

hI

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

fab

Et

6480

)( 5

, where ba

hab 3

(13)

(12)

12

Example 1 (Single Simpson rule)

Compute

30

8

,8.92100000,140

000,140ln2000

b

a

dxxx

I

by using a single segment Simpson ruleSolutionIn this example:

3333.73

830

3

abh

83

83

13

2667.17788.982100140000

140000ln20008 00

xfx

4629.3723333.158.93333.152100140000

140000ln2000

3333.153333.78

1

01

xf

hxx

14

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

15

Applying Eq. (12), one has:

3104.11063

6740.9018976.60834629.37232667.1773333.78

3

I

I

The “exact” answer can be computed as

34.11061exactI

16

3. Multiple Segments for Simpson Rule

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

3

abh

Notes: = multiple of 3 = number of small segments

""n""h

(14)

n ""h

83

17

b

a

b

a

dxxfdxxfI 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)

18

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

nnnn xfxfxfxf

xfxfxfxfxfxfxfxfhI

123

65433210

33.....

3333

8

3

n

n

ii

n

ii

n

ii xfxfxfxfxf

hI

3

,..9,6,3

1

,..8,5,2

2

,..7,4,10 233

8

3

(16)

(17)

83

19

Example 2 (Multiple segments Simpson rule)

Compute

30

8

,8.92100000,140

000,140ln2000

b

a

dxxx

I

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

""h n

83

83

20

SolutionIn 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

21

6666.224;8976.608,6666.22, 0444 hxxwherexfx

3333.265;9870.746,3333.26, 0555 hxxwherexfx

306;6740.901,30, 0666 hxxwherexfx

22

Applying Eq. (17), one obtains:

6740.9012332667.1776666.3

8

3 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

4696.601,11I

23

Example 3 (Mixed, multiple segments Simpson and rules)

Compute

30

8

,8.92100000,140

000,140ln2000

b

a

dxxx

I

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

Solution:In this example, one has:

1429.334

830

21

nn

ab

n

abh

83

31

1n2n

24

ruleSimpson

hxx

hxx

hxx

hxx

ax

31

5714.201429.3484

4286.171429.3383

2857.141429.3282

1429.111429.381

8

04

03

02

01

0

301429.3787

8571.261429.3686

7143.231429.3585

07

06

05

hxx

hxx

hxx

25

2667.17788.982100000,140

000,140ln200080

xf

Similarly: 6740.901

9978.767

8260.646

3909.536

2749.435

3241.342

5863.2561429.11

7

6

5

4

3

2

1

xf

xf

xf

xf

xf

xf

xf

26

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

xfxfxfxfh

I n

n

ii

n

ii

31

27

For multiple segments

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

segmentslastn 32

2748.6697

2749.435!3241.34235863.25632667.1771429.38

3

2338

3

2

2

03

,...6,3

21

,...2

12

,...3,102 1

222

I

skipI

xfxfxfxfxfh

I n

n

ii

n

ii

n

ii

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

3946.061,11

2748.66971197.436421

I

III

28

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

fab

Et

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)

29

(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

xfxfxfxfh

I

xfxfxfxfxfxfxfxfxfh

I

2

...6,4,2

1

,...3,102

124322101

243

4.....443

(19)

30

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).

31

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

32

= 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

33

Step 2

Compute 21 nnn

n

abh

bnhax

ihax

hax

hax

ax

n

i

.

.

.

.

2

1

2

1

0

34

Step 3

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

n

n

ii

n

ii xfxfxfxf

hI

2

...6,4,2

1

,...3,101

11

243

(19, repeated)

35

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

8

3n

n

ii

n

ii

n

ii xfxfxfxfxf

hI

Step 5

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

(17, repeated)

(20)