Transcript PPT

Simpson’s 1/3rd Rule of
Integration
Civil 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
5
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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
The concentration of benzene at a critical location is given by


c  1.75 erfc0.6560  e 32.73erfc5.758
where
erfc x  
x
e
z2
dz

So in the above formula
erfc0.6560  
Since
e
 z2
0.6560
e
z2
dz

decays rapidly as z   , we will approximate
erfc0.6560 
0.6560
z
e
 dz
2
5
a)
b)
c)
11
Use Simpson 1/3rd rule to find the approximate value of erfc(0.6560).
Find the true error, E t for part (a).
Find the absolute relative true error, a for part (a).
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Solution
a)
erfc0.6560 
0.6560
e
z2
dz
5
f ( z)  e
z2

 b  a 
 ab
erfc0.6560  
  f a   4 f 
  f b 
 6 
 2 

 0.6560  5 

 f 5  4 f 2.8280   f 0.6560 
6



  4.3440 
11
4

 1.3888 10  43.3627 10   0.65029
6



 0.47178
12
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Solution (cont)
b) The exact value of the above integral cannot be
found. We assume the value obtained by adaptive
numerical integration using Maple as the exact value
for calculating the true error and relative true error.
erfc0.6560 
0.6560
z2
e
dz
5
 0.31333
True Error
Et  True Value  Approximate Value
 0.31333   0.47178
 0.15846
13
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Solution (cont)
c) The absolute relative true error,
t 

True Error
100
True Value
0.15846
100
 0.31333
 50.573%
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
The concentration of benzene at a critical location is given by


c  1.75 erfc0.6560  e 32.73erfc5.758
where
erfc x  
x
e
z2
dz

So in the above formula
0.6560
2
erfc0.6560    e  z dz
Since
e
 z2

decays rapidly as
z  , we will approximate
erfc0.6560 
0.6560
z
e
 dz
2
5
a)
b)
c)
21
Use four segment Simpson’s 1/3rd Rule to find the approximate value of
erfc(0.6560).
Find the true error, E t for part (a).
Find the absolute relative true error, a for part (a).
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Solution
a)
Using n segment Simpson’s 1/3rd Rule,
b  a 0.6560  5
h

 1.0860
n
4
So
f ( z 0 )  f (5)
f ( z1 )  f (5  1.0860)  f (3.9140)
f ( z 2 )  f (3.9140  1.0860)  f (2.8280)
f ( z 3 )  f (2.8280  1.0860)  f (1.7420)
f ( z4 )  f (0.6560)
22
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Solution (cont.)


n 1
n2
ba 
erfc0.6560 
f z0   4  f zi   2  f zi   f zn 

3n 
i 1
i 2


i  odd
i  even


3
2
0.6560  5 

f 5  4  f  zi   2  f  zi   f 0.6560 

34  
i 1
i 2


i  odd
i  even

 4.3440
 f 5  4 f z1   4 f z3   2 f z2   f 0.6560
12

 4.3440
 f 5  4 f 3.9140  4 f 1.7420  2 f 2.8280  f 0.6560
12

 4.3440
1.3888 10 11  4 2.2226 10 7  40.048096   2 3.3627 10  4  0.65029
12





 0.30529
23
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
Solution (cont.)
b)
In this case, the true error is
Et  True Value  Approximate Value
 0.31333   0.30529
 0.0080347
c)
The absolute relative true error
t 

True Error
100
True Value
 0.0080347
100
 0.31333
 2.5643%
24
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Solution (cont.)
Table 1: Values of Simpson’s 1/3rd Rule for
Example 2 with multiple segments
2 n
4
6
8
10
25
Approximate Value
Et
−0.47178
−0.30529
−0.30678
−0.31110
−0.31248
0.15846
−0.0080347
−0.0065444
−0.0022249
−0.00084868
t
50.573%
2.5643%
2.0887%
0.71009%
0.27086%
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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
26
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
27
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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

28
( 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
29
( 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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