07.08
Simpson Chapter 3/8 Rule for Integration
After reading this chapter, you should be able to
1. derive the formula for Simpsons 3/8 rule of integration,
2. use Simpsons 3/8 rule it to solve integrals,
3. develop the formula for multiple-segment Simpsons 3/8 rule of integration,
4. use multiple-segment Simpsons 3/8 rule of integration to solve integrals,
5. compare true error formulas for multiple-segment Simpsons 1/3 rule and multiplesegment Simpsons 3/8 rule, and
6. use a combination of Simpsons 1/3 rule and Simpsons 3/8 rule to approximate
integrals.
Introduction
The main objective of this chapter is to develop appropriate formulas for approximating the
integral of the form
b
I=
f ( x)dx
(1)
a
Most (if not all) of the developed formulas for integration are based on a simple concept of
approximating a given function f ( x) by a simpler function (usually a polynomial function)
f i ( x) , where i represents the order of the polynomial function. In Chapter 07.03, Simpsons
1/3 rule for integration was derived by approximating the integrand f ( x ) with a 2nd order
(quadratic) polynomial function. f 2 ( x )
f 2 ( x ) = a 0 + a1 x + a 2 x 2
(2)
07.08.1
07.08.2
Chapter 07.08
~
Figure 1 f ( x ) Cubic function.
In a similar fashion, Simpson 3/8 rule for integration can be derived by approximating the
given function f ( x) with the 3rd order (cubic) polynomial f 3 ( x )
f 3 ( x ) = a0 + a1 x + a 2 x 2 + a 3 x 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 a 0 , a1 , a 2 and a3 in Equation (3) can be obtained by substituting 4
known coordinate data points {x0 , f ( x0 )}, {x1 , f ( x1 )}, {x 2 , f ( x 2 )} and {x3 , f ( x3 )} into
Equation (3) as follows.
2
2
f ( x 0 ) = a 0 + a1 x0 + a 2 x0 + a3 x 0
f ( x1 ) = a 0 + a1 x1 + a 2 x12 + a3 x12
(4)
2
2
f ( x 2 ) = a 0 + a1 x 2 + a 2 x 2 + a3 x 2
2
2
f ( x3 ) = a 0 + a1 x3 + a 2 x3 + a3 x3
Equation (4) can be expressed in matrix notation as
Simpson 3/8 Rule for Integration
2
3
1 x 0 x0 x 0 a 0 f ( x 0 )
2
3
1 x1 x1 x1 a1 = f ( x1 )
2
3
1 x 2 x 2 x 2 a 2 f ( x 2 )
2
3
1 x3 x3 x3 a3 f ( x3 )
The above Equation can symbolically be represented as
(5)
[ A] 44 a 41 = f 41
Thus,
a1
a
1
a = 2 = [ A] f
a3
a 4
07.08.3
(5)
(6)
(7)
Substituting Equation (7) into Equation (3), one gets
1
f 3 ( x ) = 1, x, x 2 , x 3 [ A] f
(8)
As indicated in Figure 1, one has
x0 = a
x1 = a + h
ba
=a+
3
2a + b
=
3
x 2 = a + 2h
(9)
2b 2a
=a+
3
a + 2b
=
3
x3 = a + 3h
3b 3a
=a+
3
=b
With the help from MATLAB [Ref. 2], the unknown vector a (shown in Equation 7) can be
solved for symbolically.
{
}
Method 2
Using Lagrange interpolation, the cubic polynomial function f 3 ( x ) that passes through 4
data points (see Figure 1) can be explicitly given as
07.08.4
Chapter 07.08
( x x1 ) ( x x 2 ) ( x x3 )
( x x0 ) ( x x 2 ) ( x x3 )
f ( x0 ) +
f ( x1 )
( x0 x1 ) ( x0 x 2 ) ( x0 x3 )
( x1 x0 ) ( x1 x 2 ) ( x1 x3 )
( x x0 ) ( x x1 ) ( x x3 )
( x x0 ) ( x x1 ) ( x x 2 )
+
f ( x3 ) +
f ( x3 )
( x 2 x0 ) ( x 2 x1 ) ( x2 x3 )
( x3 x0 ) ( x3 x1 ) ( x3 x 2 )
f3 ( x) =
(10)
Simpsons 3/8 Rule for Integration
Substituting the form of f 3 ( x ) from Method (1) or Method (2),
b
I = f ( x ) dx
a
b
f 3 ( x ) dx
a
= (b a)
{ f ( x0 ) + 3 f ( x1 ) + 3 f ( x2 ) + f ( x3 )}
8
(11)
Since
ba
3
b a = 3h
and Equation (11) becomes
3h
I
{ f ( x 0 ) + 3 f ( x1 ) + 3 f ( x 2 ) + f ( x3 )}
8
Note the 3/8 in the formula, and hence the name of method as the Simpsons 3/8 rule.
The true error in Simpson 3/8 rule can be derived as [Ref. 1]
(b a) 5
Et =
f ( ) , where a b
6480
Example 1
The vertical distance covered by a rocket from x = 8 to x = 30 seconds is given by
30
140000
s = 2000 ln
9.8 x dx
140000 2100t
8
Use Simpson 3/8 rule to find the approximate value of the integral.
h=
(12)
(13)
Simpson 3/8 Rule for Integration
Solution
ba
n
ba
=
3
30 8
=
3
= 7.3333
3h
I
{ f ( x 0 ) + 3 f ( x1 ) + 3 f ( x 2 ) + f ( x3 )}
8
x0 = 8
h=
140000
f ( x0 ) = 2000 ln
9.8 8
140000 2100 8
= 177.2667
x = x + h
0
1
= 8 + 7.3333
= 15.3333
140000
f ( x1 ) = 2000 ln
9.8 15.3333
140000 2100 15.3333
= 372.4629
x = x + 2h
0
2
= 8 + 2(7.3333)
= 22.6666
140000
f ( x 2 ) = 2000 ln
9.8 22.6666
140000 2100 22.6666
= 608.8976
07.08.5
07.08.6
Chapter 07.08
x = x + 3h
0
3
= 8 + 3(7.3333)
= 30
140000
f ( x3 ) = 2000 ln
9.8 30
140000 2100 30
= 901.6740
Applying Equation (12), one has
3
I = 7.3333 {177.2667 + 3 372.4629 + 3 608.8976 + 901.6740}
8
= 11063.3104
The exact answer can be computed as
I exact = 11061.34
Multiple Segments for Simpson 3/8 Rule
Using n = number of equal segments, the width h can be defined as
ba
h=
(14)
n
The number of segments need to be an integer multiple of 3 as a single application of
Simpson 3/8 rule requires 3 segments.
The integral shown in Equation (1) can be expressed as
b
I = f ( x ) dx
a
b
f 3 ( x ) dx
a
x3
xn = b
x6
f ( x ) dx + f ( x ) dx + ........ + f ( x ) dx
3
x0 = a
3
x3
3
Using Simpson 3/8 rule (See Equation 12) into Equation (15), one gets
3h f ( x0 ) + 3 f ( x1 ) + 3 f ( x 2 ) + f ( x3 ) + f ( x3 ) + 3 f ( x 4 ) + 3 f ( x5 ) + f ( x 6 )
I=
8 + ..... + ( f x n 3 ) + 3 f ( x n 2 ) + 3 f ( x n 1 ) + f ( x n )
=
(15)
xn 3
n 2
n 1
n 3
3h
f ( x 0 ) + 3 f ( xi ) + 3 f ( xi ) + 2 f ( xi ) + f ( x n )
8
i =1, 4 , 7 ,..
i = 2 , 5,8,..
i =3, 6, 9 ,..
Example 2
The vertical distance covered by a rocket from x = 8 to x = 30 seconds is given by
(16)
(17)
Simpson 3/8 Rule for Integration
07.08.7
30
140000
s = 2000 ln
9.8 x dx
140000 2100t
8
Use Simpson 3/8 multiple segments rule with six segments to estimate the vertical distance.
Solution
In this example, one has (see Equation 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
{ 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
Applying Equation (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 ,..
= (1.3750 ){177.2667 + 3( 270.4104 + 608.8976) + 3( 372.4629 + 746.9870 ) + 2( 484.7455) + 901.6740}
= 11,601.4696
Example 3
Compute
I=
b =30
140,000
2000 ln 140,000 2100 x 9.8xdx,
a =8
using Simpson 1/3 rule (with n1 = 4), and Simpson 3/8 rule (with n2 = 3).
Solution
The segment width is
ba
h=
n
ba
=
n1 + n2
30 8
=
( 4 + 3)
= 3.1429
07.08.8
Chapter 07.08
x0 = a = 8
x1 = x0 + 1h = 8 + 3.1429 = 11.1429
x 2 = x0 + 2h = 8 + 2( 3.1429 ) = 14.2857 Simpson' s 1/3 rule
x3 = x0 + 3h = 8 + 3( 3.1429 ) = 17.4286
x 4 = 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
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 ( x2 ) = 342.3241
f ( x3 ) = 435.2749
f ( x4 ) = 536.3909
f ( x5 ) = 646.8260
f ( x6 ) = 767.9978
f ( x7 ) = 901.6740
For multiple segments ( n1 = first 4 segments) , using Simpson 1/3 rule, one obtains (See
Equation 19):
n1 1=3
n1 2 = 2
h
I 1 = f ( x0 ) + 4 f ( xi ) + 2 f ( xi ) + f x n1
3
i =1, 3,...
i = 2 ,...
()
3.1429
=
{177.2667 + 4( 256.5863 + 435.2749 ) + 2( 342.3241) + 536.3909}
3
= 4364.1197
For multiple segments ( n2 = last 3 segments) , using Simpson 3/8 rule, one obtains (See
Equation 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 x n1
8
i =1, 3,...
i = 2 ,...
i = 3, 6,...
//
()
3
= 3.1429 {177.2667 + 3( 256.5863) + 3( 342.3241) + ( no contribution ) + 435.2749}
8
= 6697.2748
The mixed (combined) Simpson 1/3 and 3/8 rules give
Simpson 3/8 Rule for Integration
07.08.9
I = I1 + I 2
= 4364.1197 + 6697.2748
= 11,061.3946
Comparing the truncated error of Simpson 1/3 rule
( b a ) 5 f ( )
(18)
Et =
2880
With Simpson 3/8 rule (See Equation 12), it 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 significantly higher than the one associated with Simpson 1/3 rule
(using 2nd order polynomial function).
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 ( x0 ) + 4 f ( x1 ) + f ( x 2 ) + f ( x 2 ) + 4 f ( x3 ) + f ( x 4 ) + ..... + f ( x n 2 ) + 4 f ( x n 1 ) + f ( x n ) }
3
n 1
n 2
h
= f ( x 0 ) + 4 f ( x i ) + 2 f ( xi ) + f ( x n )
(19)
3
i =1, 3,...
i = 2 , 4 , 6...
However, Simpson 3/8 rule can be used with the number of segments equal to 3,6,9,12,.. (can
be certain odd or even numbers that are multiples of 3).
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) would be appropriate.
Computer Algorithm for Mixed Simpson 1/3 and 3/8 Rule for Integration
Based on the earlier discussion 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 inputs information, such as
f ( x) = integrand
n1 = number of segments in conjunction with Simpson 1/3 rule (a multiple of 2 (any
even numbers)
n2 = number of segments in conjunction with Simpson 3/8 rule (a multiple of 3)
Step 2
Compute
n = n1 + n 2
ba
h=
n
07.08.10
Chapter 07.08
x0 = a
x1 = a + 1h
x 2 = a + 2h
.
.
xi = a + ih
.
.
x n = a + nh = b
Step 3
Compute result from multiple-segment Simpson 1/3 rule (See Equation 19)
n1 1
n1 2
h
I 1 = f ( x0 ) + 4 f ( xi ) + 2 f ( xi ) + f x n1
3
i =1, 3,...
i = 2 , 4 , 6...
Step 4
Compute result from multiple segment Simpson 3/8 rule (See Equation 17)
n2 2
n2 1
n2 3
3h
( x 0 ) + 3 f ( x i ) + 3 f ( x i ) + 2 f ( x i ) + f x n2
I 2 = f
8
i =1, 4 , 7...
i = 2 , 5,8...
i = 3, 6, 9 ,...
Step 5
I I1 + I 2
and print out the final approximated answer for I .
()
()
SIMPSONS 3/8 RULE FOR INTEGRATION
Topic
Simpson 3/8 Rule for Integration
Summary Textbook Chapter of Simpsons 3/8 Rule for Integration
Major
General Engineering
Authors
Duc Nguyen
Date
April 15, 2012
Web Site http://numericalmethods.eng.usf.edu
(19, repeated)
(17, repeated)
(20)