Numerical Methods
Part: Simpson
Integration.
3
8
Rule For
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Lecture # 1
Chapter 07.08: Simpson
For Integration.
3
8
Rule
Major: All Engineering Majors
Authors: Duc Nguyen
http://numericalmethods.eng.usf.edu
Numerical Methods for STEM undergraduates
4/6/2015
http://numericalmethods.eng.usf.edu
5
Introduction
The main objective in this chapter is to develop
appropriated formulas for obtaining
the integral
expressed in the following form:
b
I 

a
f ( x ) dx
(1)
where f ( x ) is a given function.
Most (if not all) of the developed formulas for integration
is based on a simple concept of replacing a given
(oftently complicated) function f ( x ) by a simpler
function (usually a polynomial function) f i ( x ) where i
6 represents the order of the polynomial function.
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 f ( x )
with the 2nd –order (or quadratic) polynomial function
f i ( x )  f 2 ( x ) , defined as:
f 2 ( x )  a 0  a1 x  a 2 x
7
2
(2)
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In a similar fashion, Simpson 3 8 rule for integration
can be derived by replacing the given function f ( x )
with the 3rd-order (or cubic) polynomial (passing
through 4 known data points) function f i ( x )  f 3 ( x )
defined as f ( x )  a  a x  a x 2  a x 3 
3
0

1
 1, x , x , x
8
2
2
3

a0 
 
 a1 
 
a2 
a3 
 
3







(3)
which can also be symbolically represented in Figure 1.
Method 1
The unknown coefficients a 0 , a1 , a 2 and a 3 (in Eq. (3))
can be obtained by substituting 4 known coordinate
data points x 0 , f  x 0 , x1 , f  x1 , x 2 , f  x 2  and x 3 , f  x 3 
into Eq. (3), as following
9
2
2
f ( x 0 )  a 0  a1 x 0  a 2 x 0  a 3 x 0 

2
2
f ( x1 )  a 0  a1 x1  a 2 x1  a 3 x1 
(4)

2
2
f ( x 2 )  a 0  a1 x 2  a 2 x 2  a 3 x 2 
2
2 
f ( x 3 )  a 0  a1 x 3  a 2 x 3 http://numericalmethods.eng.usf.edu
a 3 x3 
Eq. (4) can be expressed in matrix notation as
1

1
1

1
2
x0
x0
x1
x1
x2
2
x2
2
x3
x3
2
3
x 0   a 0   f  x 0 
 

3


a
f
x
x1   1  
1 


3
x 2  a 2   f  x 2 

3 
x 3   a 3   f  x 3  
(5)
The above Eq. (5) can be symbolically represented as


(6)
 A 4  4 a 4 1  f 4 1
10
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Thus,
 a0 
 

  a1 
1
a     A  f
a2 
 a 3 
(7)
Substituting Eq. (7) into Eq. (3), one gets

f 3  x   1, x , x , x
11
2
3
  A 
1

 f
(8)
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Remarks
As indicated in Figure 1, one has
x0  a


b  a 2a  b

x1  a  h  a 

3
3


2b  2 a a  2b 
x2  a  2 h  a 


3
3

3b  3 a

x3  a  3 h  a 
b

3
(9)
With
 the help from MATLAB [2], the unknown vector
a (shown in Eq. 7) can be solved.
12
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Method 2
Using Lagrange interpolation, the cubic polynomial
function f
 x  that passes through 4 data points
i3
(see Figure 1) can be explicitly given as
f3  x  
x 
x1  x  x 2  x  x 3 
 f  x0  
x 
x 0  x  x 2  x  x 3 
 f  x1 
 x 0  x1  x 0  x 2  x 0  x 3 
 x1  x 0  x1  x 2  x1  x 3 
 x  x 0  x  x1  x  x 3 
 x  x 0  x  x1  x  x 2 

 f  x3  
 f  x3 
 x 2  x 0  x 2  x1  x 2  x 3 
 x 3  x 0  x 3  x1  x 3  x 2 
(10)
13
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Simpsons 8 Rule For
Integration
3
Thus, Eq. (1) can be calculated as (See Eqs. 8, 10 for
Method 1 and Method 2, respectively):
b
I 
 f  x dx
a
b


f 3  x dx
a
Integrating the right-hand-side
equations, one obtains
I  b  a  
14
of
the
above
 f  x 0   3 f  x1   3 f  x 2   f  x3 
(11)
8
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Since h 
ba
3
hence b  a  3 h , and the above
equation becomes:
3h
I 
  f  x 0   3 f  x1   3 f  x 2   f  x 3 
(12)
8
The error introduced by the Simpson 3/8 rule can be
derived as [Ref. 1]:
Et  
15
(b  a )
6480
5
 f 
 , where a    b
(13)
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Example 1 (Single Simpson 3 8 rule)
Compute
b  30
I 

a 8


140 , 000


  9 . 8 x  dx ,
 2000 ln 
 140 , 000  2100 x 


by using a single segment Simpson 3 8 rule
Solution
In this example:
h
16
ba
3

30  8
3
 7 . 3333
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140000


x 0  8  f  x 0   2000 ln 
  9 . 8  8  177 . 2667
 140000  2100  8 
 x1  x 0  h  8  7 . 3333  15 . 3333

140000



 f  x1   2000 ln  140000  2100  15 . 3333   9 . 8  15 . 3333  372 . 4629



17
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 x 2  x 0  2 h  8  2 ( 7 . 3333 )  22 . 6666

140000





f
x

2000
ln

  9 . 8  22 . 6666  608 . 8976
2

 140000  2100  22 . 6666 

 x 3  x 0  3 h  8  3 ( 7 . 3333 )  30

140000





f
x

2000
ln

  9 . 8  30  901 . 6740
3

 140000  2100  30 

18
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Applying Eq. (12), one has:
I 
3
8
 7 . 3333  177 . 2667  3  372 . 4629  3  608 . 8976  901 . 6740
I  11063 . 3104
The “exact” answer can be computed as
I exact  11061 . 34
19
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
3. Multiple Segments for
Simpson 3 8 Rule
Using " n " = number of equal (small) segments, the
width " h " can be defined as
h
ba
3
(14)
Notes:
n = multiple of 3 = number of small " h " segments
20
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The integral, shown in Eq. (1), can be expressed as
b
I 
 f  x dx
a
x3
I 

f 3  x dx 
x0  a
21
b


f 3  x dx
a
x6

x3
f 3  x dx  ........ 
xn b

f 3  x dx
(15)
xn3
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Substituting Simpson 3 8 rule (See Eq. 12) into
Eq. (15), one gets
3 h  f  x 0   3 f  x1   3 f  x 2   f  x 3   f  x 3   3 f  x 4   3 f  x 5   f  x 6 
I 


8   .....  f  x n  3   3 f  x n  2   3 f  x n  1   f  x n 

(16)
n2
n 1
n3

3h 
I 
 f  x 0   3  f  x i   3  f  x i   2  f  x i   f  x n 
8 
i  1, 4 , 7 ,..
i  2 , 5 , 8 ,..
i  3 , 6 , 9 ,..

(17)
22
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Example 2 (Multiple segments Simpson 3 8 rule)
b  30
Compute I  
a 8


140 , 000


  9 . 8 x  dx ,
 2000 ln 
 140 , 000  2100 x 


3
using Simple 8 multiple segments rule, with number
(of " h " ) segments = n = 6 (which corresponds to 2
“big” segments).
23
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Solution
In this example, one has (see Eq. 14):
h
30  8
 3 . 6666
6
x 0 , f  x 0   8 ,177 . 2667 
x1 , f  x1   11 . 6666
, 270 . 4104 ; where x1  x 0  h  8  3 . 6666
 11 . 6666
x 2 , f  x 2   15 . 3333 ,372 . 4629 ; where
x 3 , f  x 3   19 , 484 . 7455 ; where
24
x 2  x 0  2 h  15 . 3333
x 3  x 0  3 h  19
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x 4 , f  x 4   22 . 6666
, 608 . 8976 ; where x 4  x 0  4 h  22 . 6666
x 5 , f  x 5   26 . 3333 , 746 . 9870 ; where
x 6 , f  x 6   30 ,901 . 6740 ; where
25
x 5  x 0  5 h  26 . 3333
x 6  x 0  6 h  30
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Applying Eq. (17), one obtains:
n24
n 1  5
n 33


I  3 . 6666 177 . 2667  3  f  x i   3  f  x i   2  f  x i   901 . 6740 
8
i  1, 4 ,..
i  2 , 5 ,..
i  3 , 6 ,..


3
177 . 2667  3  270 . 4104  608 . 8976   3 372 . 4629  746 . 9870 
I  1 . 3750 



  2 484 . 7455  901 . 6740

I  11 , 601 . 4696
26
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3
1
Example 3 (Mixed, multiple segments Simpson 3 and 8
rules)
b  30
Compute I  
a 8


140 , 000


  9 . 8 x  dx ,
 2000 ln 
 140 , 000  2100 x 


using Simpson 1/3 rule (with n1  4 small segments), and
Simpson 3/8 rule (with n 2  3 small segments).
Solution:
In this example, one has:
h
27
ba
n

ba
n1  n 2

30  8
4  3 
 3 . 1429
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x0  a  8


x1  x 0  1h  8  3 . 1429  11 . 1429


x 2  x 0  2 h  8  2 3 . 1429   14 . 2857  Simpson
x 3  x 0  3 h  8  3 3 . 1429   17 . 4286 

x 4  x 0  4 h  8  4 3 . 1429   20 . 5714 
x 5  x 0  5 h  8  5 3 . 1429   23 . 7143
x 6  x 0  6 h  8  6 3 . 1429
x 7  x 0  7 h  8  7 3 . 1429
28
1
3
rule
  26 . 8571
  30
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140 , 000


f  x 0  8   2000 ln 
  9 . 8  8  177 . 2667
 140 , 000  2100  8 
Similarly:
f  x1  11 . 1429
  256 . 5863
f  x 2   342 . 3241
f  x 3   435 . 2749
f  x 4   536 . 3909
f  x 5   646 . 8260
f  x 6   767 . 9978
29
f  x 7   901 . 6740
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For multiple segments  n 1  first 4 segments 
using Simpson 1 rule, one obtains (See Eq. 19):
3
n1  1  3
n1  2  2

 h 
I1     f  x 0   4  f  xi   2  f  xi   f x n 
1
 3 
i  1, 3 ,...
i  2 ,...

 
 3 . 1429 
I1  
 177 . 2667  4  256 . 5863  435 . 2749
3


  2 342 . 3241   536 . 3909 
I 1  4364 . 1197
30
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For multiple segments  n 2  last 3 segments 
using Simpson 3/8 rule, one obtains (See Eq. 17):
n 2  2 1
n 2 1 2
n2 3 0

 3h  
I2  
  f  x 0   3  f  x i   3  f  x i   2  f  x i   f x n1 
 8 
i  1, 3 ,...
i  2 ,...
i  3 , 6 ,...

 
3

I 2    3 . 1429  177 . 2667  3  256 . 5863   3 342 . 3241    skip !  435 . 2749
8


I 2  6697 . 2748
The mixed (combined) Simpson 1/3 and 3/8 rules give:
I  I 1  I 2  4364 . 1197  6697 . 2748
I  11 , 061 . 3946
31
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Remarks:
(a) Comparing the truncated error of Simpson 1/3 rule
5
b  a 
(18)
Et  
 f  
2880
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).
32
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(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).
h
I 1     f  x 0   4 f  x1   f  x 2   f  x 2   4 f  x 3   f  x 4   .....  f  x n  2   4 f  x n  1   f  x n 
3
n 1
n2

 h 
I 2     f  x 0   4  f  x i   2  f  x i   f  x n 
 3 
i  1, 3 ,...
i  2 , 4 , 6 ...

33
(19)
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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).
34
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4. Computer Algorithm For Mixed Simpson
1/3 and 3/8
rule For Integration
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 f ( x ), integral limits " a , b " ,
n1= number of small, “h” segments, in conjunction with
Simpson 1/3 rule.
35
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n 2 = number of small, “h” segments, in conjunction with
Simpson 3/8 rule.
Notes:
n1 = a multiple of 2 (any even numbers)
n 2 = a multiple of 3 (can be certain odd, or even
numbers)
36
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Step 2
Compute n  n1  n 2
h
ba
x0  a
x1  a  1h
x2  a  2 h
n
.
.
x i  a  ih
.
.
37
x n  a  nh  b
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Step 3
Compute “multiple segments” Simpson 1/3 rule (See
Eq. 19)
n1  1
n1  2

 h 
I 1     f  x 0   4  f  x i   2  f  x i   f  x n 
 3 
i  1, 3 ,...
i  2 , 4 , 6 ...

(19, repeated)
38
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Step 4
Compute “multiple segments” Simpson 3/8 rule (See
Eq. 17)
n2  2
n 2 1
n2  3

 3h  
I2  
  f  x 0   3  f  xi   3  f  xi   2  f  xi   f x n 2 
 8 
i  1, 4 , 7 ...
i  2 , 5 , 8 ...
i  3 , 6 , 9 ,...

 
(17, repeated)
Step 5
(20)
I  I1  I 2
and print out the final approximated answer for I.
39
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THE END
http://numericalmethods.eng.usf.edu
Acknowledgement
This instructional power point brought to you by
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http://numericalmethods.eng.usf.edu
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undergraduate
For instructional videos on other topics, go to
http://numericalmethods.eng.usf.edu/videos/
This material is based upon work supported by the National
Science Foundation under Grant # 0717624. Any opinions,
findings, and conclusions or recommendations expressed in
this material are those of the author(s) and do not necessarily
reflect the views of the National Science Foundation.
The End - Really