Transcript PPT

Simpson’s 1/3rd Rule of
Integration
Computer Engineering Majors
Authors: Autar Kaw, Charlie Barker
http://numericalmethods.eng.usf.edu
Transforming Numerical Methods Education for STEM
Undergraduates
7/12/2016
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1
rd
1/3
Simpson’s
Rule of
Integration
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What is Integration?
b
Integration
The process of measuring
the area under a curve.
 f ( x )dx
y
a
f(x)
b
I   f ( x )dx
a
Where:
f(x) is the integrand
a= lower limit of integration
b= upper limit of integration
3
a
b
x
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Simpson’s 1/3rd Rule
4
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Basis of Simpson’s 1/3rd Rule
Trapezoidal rule was based on approximating the integrand by a first
order polynomial, and then integrating the polynomial in the interval of
integration. Simpson’s 1/3rd rule is an extension of Trapezoidal rule
where the integrand is approximated by a second order polynomial.
Hence
b
b
a
a
I   f ( x )dx   f 2 ( x )dx
Where
f2( x )
is a second order polynomial.
f 2 ( x )  a0  a1 x  a2 x 2
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Basis of Simpson’s 1/3rd Rule
Choose
 a  b  a  b 
( a , f ( a )), 
,f
 ,
 2 
 2
and
( b , f ( b ))
as the three points of the function to evaluate a0, a1 and a2.
f ( a )  f 2 ( a )  a 0  a1 a  a 2 a 2
a  b
a  b
a  b
a  b
f
  f2 
  a0  a1 
  a2 

 2 
 2 
 2 
 2 
2
f ( b )  f 2 ( b )  a0  a1b  a 2 b 2
6
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Basis of Simpson’s 1/3rd Rule
Solving the previous equations for a0, a1 and a2 give
a  b
2
a f ( b )  abf ( b )  4abf 
  abf ( a )  b f ( a )
 2 
a0 
a 2  2ab  b 2
a  b
a  b
af ( a )  4af 
  3af ( b )  3bf ( a )  4bf 
  bf ( b )
2 
2 


a1  
a 2  2ab  b 2


a  b
2 f ( a )  2 f 

f
(
b
)


 2 


a2 
a 2  2ab  b 2
2
7
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Basis of Simpson’s 1/3rd Rule
Then
b
I   f 2 ( x )dx
a
b
  a0  a1 x  a2 x 2 dx
a
b

x
x 
 a0 x  a1  a 2 
2
3 a

2
3
b2  a2
b3  a3
 a0 ( b  a )  a1
 a2
2
3
8
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Basis of Simpson’s 1/3rd Rule
Substituting values of a0, a1, a 2 give
b
 f 2 ( x )dx 
a
ba

a  b
f
(
a
)

4
f

f
(
b
)




6 
 2 
Since for Simpson’s 1/3rd Rule, the interval [a, b] is broken
into 2 segments, the segment width
h
9
ba
2
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Basis of Simpson’s 1/3rd Rule
Hence
h

a  b
f
(
x
)
dx

f
(
a
)

4
f

f
(
b
)


 2


3
 2 
a
b
Because the above form has 1/3 in its formula, it is called Simpson’s 1/3rd Rule.
10
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Example 1
Human vision has the remarkable ability to infer 3D shapes from 2D
images. The intriguing question is: can we replicate some of these
abilities on a computer? Yes, it can be done and to do this, integration
of vector fields is required. The following integral needs to integrated.
100
I
where
 f ( x)dx
0
f(x)  0, 0  x  30
 9.1688 10 6 x 3  2.796110 3 x 2  2.8487 10 1 x  9.6778,
30  x  172
 0, 172  x  200
a) Use single segment Trapezoidal rule to find the distance covered.
b) Find the true error, E t for part (a).
c) Find the absolute relative true error, a for part (a).
11
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Solution
a)
I

ba 
ab
f
(
a
)

4
f

f
(
b
)



6 
2



a0
b  100
ab
 50
2
 100  0 

 f 0   4 f 50   f 100 
 6 
 100 

0  41.2784    0.017 
 6 
 84.947
12
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Solution (cont)
b) The exact value of the above integral is found using Maple
for calculating the true error and relative true error.
100
I
 f x dx
0
 60.793
True Error
Et  True Value  Approximate Value
 60.793  84.947
 24.154
13
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Solution (cont)
c) Absolute relative true error,
60.793  (84.947)
100
60.793
 39.732%
t 
14
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Multiple Segment Simpson’s
1/3rd Rule
15
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Multiple Segment Simpson’s 1/3rd
Rule
Just like in multiple segment Trapezoidal Rule, one can subdivide the interval
[a, b] into n segments and apply Simpson’s 1/3rd Rule repeatedly over
every two segments. Note that n needs to be even. Divide interval
[a, b] into equal segments, hence the segment width
ba
h
n
b
xn
a
x0
 f ( x )dx   f ( x )dx
where
x0  a
16
xn  b
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Multiple Segment Simpson’s 1/3rd
Rule
f(x)
b
x2
x4
a
x0
x2
 f ( x )dx   f ( x )dx   f ( x )dx  .....
. . .
xn  2
xn
xn  4
xn  2
....   f ( x )dx   f ( x )dx
x
x0
x2
xn-2
xn
Apply Simpson’s 1/3rd Rule over each interval,
 f ( x0 )  4 f ( x1 )  f ( x2 )
f
(
x
)
dx

(
x

x
)
 ...

2
0 

6


a
b
 f ( x2 )  4 f ( x3 )  f ( x4 )
 ( x4  x2 )
 ...

6


17
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Multiple Segment Simpson’s 1/3rd
Rule
 f ( xn4 )  4 f ( xn3 )  f ( xn2 )
...  ( xn2  xn4 )
 ...

6


 f ( xn2 )  4 f ( xn1 )  f ( xn )
 ( xn  xn2 )

6

Since
xi  xi  2  2 h
18
i  2, 4, ..., n
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Multiple Segment Simpson’s 1/3rd
Rule
Then
 f ( x0 )  4 f ( x1 )  f ( x2 )
 ...
 f ( x )dx  2h 

6


a
b
 f ( x2 )  4 f ( x3 )  f ( x4 )
 2h 
 ...

6


 f ( xn4 )  4 f ( xn3 )  f ( xn2 )
 2h 
 ...

6


 f ( xn2 )  4 f ( xn1 )  f ( xn )
 2h 

6

19
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Multiple Segment Simpson’s 1/3rd
Rule
b
h
 f ( x )dx  3  f ( x0 )  4 f ( x1 )  f ( x3 )  ...  f ( xn1 )  ...
a
...  2 f ( x2 )  f ( x4 )  ...  f ( xn2 )  f ( xn )}]


n

1
n

2
h
  f ( x 0 )  4  f ( xi )  2  f ( xi )  f ( x n ) 

3
i 1
i 2
i  odd
i  even




n

1
n

2
ba

f ( x 0 )  4  f ( xi )  2  f ( xi )  f ( x n ) 

3n 
i 1
i 2
i odd
i even


20
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Example 2
Human vision has the remarkable ability to infer 3D shapes from 2D
images. The intriguing question is: can we replicate some of these
abilities on a computer? Yes, it can be done and to do this, integration
of vector fields is required. The following integral needs to integrated.
100
I
f(x)  0, 0  x  30
 f ( x)dx
0
 9.1688 10 6 x 3  2.7961 10 3 x 2  2.8487 10 1 x  9.6778 ,
30  x  172
 0, 172  x  200
a) Use four segment Simpson’s 1/3rd Rule to find the approximate
value of x.
b) Find the true error, E t for part (a).
c) Find the absolute relative true error, a for part (a).
21
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Solution
a) Using n segment Simpson’s 1/3rd Rule,
h
So
100  0
4
 25
f ( x0 )  f (0)
f ( x1 )  f (0  25)  f (25)
f ( x2 )  f (25  25)  f (50)
f ( x3 )  f (50  25)  f (75)
f ( x4 )  f ( xn )  f (100)
22
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Solution (cont.)


n 1
n2
ba 
I
f x0   4  f xi   2  f xi   f xn 

3n 
i 1
i 2


i  odd
i  even


3
2
100  0 

f 0  4  f xi   2  f xi   f 100

34 
i 1
i 2


i  odd
i  even

23
100
 f (0)  4 f ( x1 )  4 f ( x3 )  2 f ( x2 )  f (100)
12
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Solution (cont.)
cont.
25
 f (0)  4 f (25)  4 f (75)  2 f (50)  f (100)

3

25
0  4(0)  4(0.17253)  2(1.2784)  (0.017000)
3
 26.917
24
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Solution (cont.)
b) In this case, the true error is
Et  60.793  26.917   33.873
c)
25
The absolute relative true
error
60.793  26.917 
t 
100%
60.793
 55.724%
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Solution (cont.)
Table 1: Values of Simpson’s 1/3rd Rule for Example 2 with
multiple segments
26
n
Approximate Value
Et
t %
2
4
6
8
10
84.947
26.917
66.606
62.318
85.820
−24.154
33.876
−5.8138
−1.5252
−25.023
39.732
55.724
9.5633
2.5088
41.169
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Error in the Multiple Segment
Simpson’s 1/3rd Rule
The true error in a single application of Simpson’s 1/3rd Rule is given as
(b  a) 5 ( 4)
Et  
f (), a    b
2880
In Multiple Segment Simpson’s 1/3rd Rule, the error is the sum of the errors
in each application of Simpson’s 1/3rd Rule. The error in n segment Simpson’s
1/3rd Rule is given by
27
h5 ( 4 )
( x2  x0 )5 ( 4 )
E1  
f ( 1 )   f ( 1 ), x0  1  x2
90
2880
h5 ( 4 )
( x4  x2 )5 ( 4 )
E2  
f (  2 )   f (  2 ), x2   2  x4
90
2880
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Error in the Multiple Segment
Simpson’s 1/3rd Rule
Ei  
( x2i  x2( i 1 ) )5
2880
f
(4)
5
h
(  i )   f ( 4 ) (  i ), x2( i 1 )   i  x2i
90
.
.
.
5

( xn 2  xn  4 )5 ( 4 ) 
h
En  
f   n    f ( 4 )   n  , x n 4   n  x n 2
1
 1 
2880
1
90
 2 1 
2
2
 2 
5
  , x   x
( xn  xn  2 )5 4  
h
( 4)
n2
n
n


En  
f n   
f   n 
2
2880
90
 2
2
 2
28
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Error in the Multiple Segment
Simpson’s 1/3rd Rule
Hence, the total error in Multiple Segment Simpson’s 1/3rd Rule is
n
2
Et   Ei  
i 1
n
5
h 2
f
( 4)
90 i 1
n
2

29
( i )
f
(b  a ) 5
( 4)
i 1
90n 4

n
(b  a) 5 2
90n 5
( 4)
f
( i )

i 1
( i )
n
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Error in the Multiple Segment
Simpson’s 1/3rd Rule
n
2
The term
f
( 4)
i 1
( i )
is an approximate average value of
n
f ( 4) ( x), a  x  b
Hence
(b  a) 5 ( 4)
Et  
f
4
90n
n
2
where
f
30
( 4)
f
 i 1
( 4)
n
( i )
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Additional Resources
For all resources on this topic such as digital audiovisual
lectures, primers, textbook chapters, multiple-choice
tests, worksheets in MATLAB, MATHEMATICA, MathCad
and MAPLE, blogs, related physical problems, please
visit
http://numericalmethods.eng.usf.edu/topics/simpsons_
13rd_rule.html
THE END
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