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Numerical Methods
Part: Simpson
Integration.
3
8
Rule For
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Lecture # 1
Chapter 07.08: Simpson 3 8 Rule
For Integration.
Major: All Engineering Majors
Authors: Duc Nguyen
http://numericalmethods.eng.usf.edu
Numerical Methods for STEM undergraduates
11/14/2011
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 = ∫ f ( x)dx
a
(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) fi (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
fi ( x) = f 2 ( x), defined as:
f 2 ( x) = a0 + a1x + a2 x
7
2
(2)
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8
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 fi ( x ) = f3 ( x )
defined as f ( x) = a + a x + a x 2 + a x 3 
3
0
1
2
3

 a0 


a 
(3)

 1
2
3
= {1, x, x , x }×  

a
 2

a3 

which can also be symbolically represented in Figure 1.
Method 1
The unknown coefficients a0 , a1 , a2 and a3 (in Eq. (3))
can be obtained by substituting 4 known coordinate
data points {x0 , f ( x0 ), {x1 , f ( x1 )}, {x2 , f (x2 )} and {x3 , f (x3 )}}
into Eq. (3), as following
9
f ( x0 ) = a0 + a1x0 + a2 x02 + a3 x02 

f ( x1 ) = a0 + a1x1 + a2 x12 + a3 x12 
(4)

2
2
f ( x2 ) = a0 + a1x2 + a2 x2 + a3 x2 
2
2
f ( x3 ) = a0 + a1x3 + a2 x3 +http://numericalmethods.eng.usf.edu
a3 x3 
Eq. (4) can be expressed in matrix notation as
1

1
1

1
x0
x02
x1
x2
x3
x12
x22
x32
x03  a0   f ( x0 )
3   a   f ( x )
x1   1   1 
=
3  a   f ( x )
x2  2  
2 
3   a   f ( x )
x3   3  
3 
(5)
The above Eq. (5) can be symbolically represented as


[A]4×4 a4×1 = f 4×1
10
(6)
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Thus,
a0 
a 

  1
−1
a =   = [ A] × f
a2 
a3 
(7)
Substituting Eq. (7) into Eq. (3), one gets
{
}
f3 ( x ) = 1, x, x , x × [ A]
11
2
3
−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 − 2a a + 2b 
x2 = a + 2h = a +
=
3
3 

3b − 3a

x3 = a + 3h = 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
i = 3 x that passes through 4 data points
(see Figure 1) can be explicitly given as
( )
( x − x1 )( x − x2 )( x − x3 ) × f ( x ) + ( x − x0 )( x − x2 )( x − x3 ) × f ( x )
0
1
( x0 − x1 )( x0 − x2 )( x0 − x3 )
( x1 − x0 )( x1 − x2 )( x1 − x3 )
(
(
x − x0 )( x − x1 )( x − x3 )
x − x0 )( x − x1 )( x − x2 )
+
× f ( x3 ) +
× f ( x3 )
(x2 − x0 )(x2 − x1 )(x2 − x3 )
(x3 − x0 )(x3 − x1 )(x3 − x2 )
f3 ( x ) =
(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
b
a
a
I = ∫ f ( x )dx ≈ ∫ f3 ( x )dx
Integrating the right-hand-side
equations, one obtains
of
the
above
{
f ( x0 ) + 3 f ( x1 ) + 3 f ( x2 ) + f ( x3 )}
I = (b − a ) ×
(11)
8
14
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b−a
Since h =
hence b − a = 3h , and the above
3
equation becomes:
3h
I ≈ × { f ( x0 ) + 3 f ( x1 ) + 3 f ( x2 ) + f ( x3 )}
8
(12)
The error introduced by the Simpson 3/8 rule can be
derived as [Ref. 1]:
(b − a )
× f ′′′′(ζ ) , where
Et = −
6480
5
15
a ≤ζ ≤b
(13)
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Example 1 (Single Simpson 3 8 rule)
Compute
b = 30


140,000


I = ∫ 2000 ln
 − 9.8 x dx,
 140,000 − 2100 x 

a =8 
16
by using a single segment Simpson 3 8 rule
Solution
In this example:
b − a 30 − 8
h=
=
= 7.3333
3
3
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140000


x0 = 8 ⇒ f ( x0 ) = 2000 ln
 − 9.8 × 8 = 177.2667
 140000 − 2100 × 8 
 x1 = x0 + 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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 x2 = x0 + 2h = 8 + 2(7.3333) = 22.6666

140000



(
)
=
2000
ln
f
x

 − 9.8 × 22.6666 = 608.8976
2

 140000 − 2100 × 22.6666 

 x3 = x0 + 3h = 8 + 3(7.3333) = 30

140000



(
)
=
2000
ln
f
x
 − 9.8 × 30 = 901.6740

 3
 140000 − 2100 × 30 

18
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Applying Eq. (12), one has:
3
I = × 7.3333 × {177.2667 + 3 × 372.4629 + 3 × 608.8976 + 901.6740}
8
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
b−a
h=
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
b
a
a
I = ∫ f ( x )dx ≈ ∫ f3 ( x )dx
I≈
21
x3
x6
xn = b
x0 = a
x3
x n −3
∫ f3 (x )dx + ∫ f3 (x )dx + ........ + ∫ f3 (x )dx
(15)
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Substituting Simpson 3 8 rule (See Eq. 12) into
Eq. (15), one gets
3h  f ( x0 ) + 3 f ( x1 ) + 3 f ( x2 ) + f ( x3 ) + f ( x3 ) + 3 f ( x4 ) + 3 f ( x5 ) + f ( x6 )
I= 

8 + ..... + f ( xn −3 ) + 3 f ( xn − 2 ) + 3 f ( xn −1 ) + f ( xn )

(16)
n−2
n −1
n −3

3h 
I =  f ( x0 ) + 3 ∑ f ( xi ) + 3 ∑ f ( xi ) + 2 ∑ f ( xi ) + f ( xn )
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


140,000


Compute I = ∫ 2000 ln
 − 9.8 x dx,
 140,000 − 2100 x 

a =8 
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):
30 − 8
h=
= 3.6666
6
{x0 , f (x0 )} = {8,177.2667}
{x1, f (x1 )} = {11.6666,270.4104}; where x1 = x0 + h = 8 + 3.6666
= 11.6666
{x2 , f (x2 )} = {15.3333,372.4629}; where x2 = x0 + 2h = 15.3333
{x3 , f (x3 )} = {19,484.7455}; where x3 = x0 + 3h = 19
24
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{x4 , f (x4 )} = {22.6666,608.8976}; where x4 = x0 + 4h = 22.6666
{x5 , f (x5 )} = {26.3333,746.9870}; where x5 = x0 + 5h = 26.3333
{x6 , f (x6 )} = {30,901.6740}; where x6 = x0 + 6h = 30
25
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Applying Eq. (17), one obtains:
n −2=4
n −1= 5
n − 3= 3


3
I = (3.6666)177.2667 + 3 ∑ f ( xi ) + 3 ∑ f ( xi ) + 2 ∑ f ( xi ) + 901.6740
8
i =1, 4,..
i = 2,5,..
i = 3, 6,..


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


140,000


Compute I = ∫ 2000 ln
 − 9.8 x dx,
 140,000 − 2100 x 

a =8 
using Simpson 1/3 rule (with n1 = 4 small segments), and
Simpson 3/8 rule (with n2 = 3 small segments).
Solution:
In this example, one has:
27
b−a b−a
30 − 8
h=
=
=
= 3.1429
n
n1 + n2 (4 + 3)
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x0 = a = 8


x1 = x0 + 1h = 8 + 3.1429 = 11.1429

x2 = x0 + 2h = 8 + 2(3.1429) = 14.2857  Simpson 1 rule
3
x3 = x0 + 3h = 8 + 3(3.1429) = 17.4286 

x4 = x0 + 4h = 8 + 4(3.1429) = 20.5714
x5 = x0 + 5h = 8 + 5(3.1429 ) = 23.7143
x6 = x0 + 6h = 8 + 6(3.1429 ) = 26.8571
x7 = x0 + 7 h = 8 + 7(3.1429 ) = 30
28
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140,000


(
)
8
2000
ln
f x0 = =
 − 9.8 × 8 = 177.2667

 140,000 − 2100 × 8 
Similarly:
f ( x1 = 11.1429 ) = 256.5863
f ( x2 ) = 342.3241
f ( x3 ) = 435.2749
f ( x4 ) = 536.3909
f ( x5 ) = 646.8260
f ( x6 ) = 767.9978
29
f ( x7 ) = 901.6740
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For multiple segments (n1 = first 4 segments )
using Simpson 1 rule, one obtains (See Eq. 19):
3
n1 −1= 3
n1 − 2 = 2

 h 
I1 =   f ( x0 ) + 4 ∑ f ( xi ) + 2 ∑ f ( xi ) + f xn1 
 3 
i =1,3,...
i = 2,...

 3.1429 
I1 = 
{177.2667 + 4(256.5863 + 435.2749) + 2(342.3241) + 536.3909}
 3 
I1 = 4364.1197
( )
30
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For multiple segments (n2 = last 3 segments )
using Simpson 3/8 rule, one obtains (See Eq. 17):
n2 − 2 =1
n2 −1= 2
n2 − 3 = 0

 3h 
I 2 =   f ( x0 ) + 3 ∑ f ( xi ) + 3 ∑ f ( xi ) + 2 ∑ f ( xi ) + f xn1 
 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 = I1 + 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
(
b − a)
=−
5
Et
2880
× f ′′′′(ζ )
(18)
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
I1 =  { f ( x0 ) + 4 f ( x1 ) + f ( x2 ) + f ( x2 ) + 4 f ( x3 ) + f ( x4 ) + ..... + f ( xn − 2 ) + 4 f ( xn −1 ) + f ( xn )
3
n −1
n−2

 h 
I 2 =   f ( x0 ) + 4 ∑ f ( xi ) + 2 ∑ f ( xi ) + f ( xn )
 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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n2 = number of small, “h” segments, in conjunction with
Simpson 3/8 rule.
Notes:
n1 = a multiple of 2 (any even numbers)
n2 = a multiple of 3 (can be certain odd, or even
numbers)
36
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Step 2
Compute n = n1 + n2
b−a
h=
n
x0 = a
x1 = a + 1h
x2 = a + 2h
.
.
xi = a + ih
.
.
37
xn = a + nh = b
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Step 3
Compute “multiple segments” Simpson 1/3 rule (See
Eq. 19)
n1 −1
n1 − 2

 h 
I1 =   f ( x0 ) + 4 ∑ f ( xi ) + 2 ∑ f ( xi ) + f ( xn )
 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
n2 −1
n2 − 3

 3h 
I 2 =   f ( x0 ) + 3 ∑ f ( xi ) + 3 ∑ f ( xi ) + 2 ∑ f ( xi ) + f xn2 
 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
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Acknowledgement
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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
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The End - Really