Simpson’s 1/3 Rule of Integration rd
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Transcript Simpson’s 1/3 Rule of Integration rd
Simpson’s 1/3rd Rule of
Integration
Chemical Engineering Majors
Authors: Autar Kaw, Charlie Barker
http://numericalmethods.eng.usf.edu
Transforming Numerical Methods Education for STEM
Undergraduates
5/27/2016
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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
ba
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
ba
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
In an attempt to understand the mechanism of the
depolarization process in a fuel cell, an electro-kinetic model for mixed
oxygen-methanol current on platinum was developed in the laboratory
at FAMU. A very simplified model of the reaction developed suggests
a functional relation in an integral form. To find the time required for
50% of the oxygen to be consumed, the time, T (s) is given by
6.73x 4.3025 10 7
T
1.2210 6
2.316 10 11 x
0.6110 6
a)
b)
c)
11
dx
Use Simpson’s 1/3rd Rule to find the time required for 50% of the
oxygen to be consumed.
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)
6.73x 4.3025 10 7
T
1.2210 6
2.316 10 11 x
0.6110 6
dx
6.73x 4.3025 10 7
f ( x)
2.316 10 11 x
b a
ab
T
f (a) 4 f
f (b)
6
2
0.61 10 6 1.22 10 6
f (1.22 10 6 ) 4 f (0.915 10 6 ) f (0.61 10 6 )
6
0.61106
3.05811011 4(3.1089 1011 ) 3.2104 1011
6
190160s
12
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Solution (cont)
b) The exact value of the above integral is
6.73x 4.3025 10 7
T
11
1.2210 6
2
.
316
10
x
0.6110 6
dx
1.90140 105 s
True Error
Et True Value Approximate Value
1.90140 105 190160
24.100
13
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Solution (cont)
c.) Absolute relative true error,
t
True Error
100
True Value
24.100
100
1.90140 105
0.012675%
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
ba
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 ( xn4 ) 4 f ( xn3 ) f ( xn2 )
... ( xn2 xn4 )
...
6
f ( xn2 ) 4 f ( xn1 ) f ( xn )
( xn xn2 )
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 ( xn4 ) 4 f ( xn3 ) f ( xn2 )
2h
...
6
f ( xn2 ) 4 f ( xn1 ) 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 ( xn1 ) ...
a
... 2 f ( x2 ) f ( x4 ) ... f ( xn2 ) 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
ba
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
In an attempt to understand the mechanism of the
depolarization process in a fuel cell, an electro-kinetic model for
mixed oxygen-methanol current on platinum was developed in
the laboratory at FAMU. A very simplified model of the reaction
developed suggests a functional relation in an integral form. To
find the time required for 50% of the oxygen to be consumed,
the time, T (s) is given by
6.73x 4.3025 10 7
T
1.2210 6
2.316 10 11 x
0.6110 6
a)
b)
c)
21
dx
Use 4-segment Simpson’s 1/3rd Rule to find the time required for 50%
of the oxygen to be consumed.
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,
ba
h
n
So
0.61 10 6 1.22 10 6
4
0.15250 10 6
f ( x0 ) f (1.22 10 6 )
f ( x1 ) f (1.22 10 6 0.15250 10 6 ) f (1.0675 10 6 )
f ( x2 ) f (1.0675 10 6 0.15250 10 6 ) f (0.915 10 6 )
f ( x3 ) f (0.915 10 6 0.15250 10 6 ) f (0.76250 10 6 )
f ( x4 ) f (0.61106 )
22
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Solution (cont.)
n 1
n2
ba
T
f ( x0 ) 4 f ( xi ) 2 f ( xi ) f ( xn )
3n
i 1
i 2
i odd
i even
3
2
0.6110 6 1.22 10 6
6
f 1.22 10 4 f xi 2 f xi f 0.6110 6
34
i 1
i 2
i odd
i even
0.61106
f 1.22 106 4 f x1 4 f x3 2 f x2 f 0.61106
12
23
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Solution (cont.)
cont.
0.61 10 6
f (1.22 10 6 ) 4 f (1.0675 10 6 ) 4 f (0.76250 10 6 ) 2 f (0.915 10 6 ) f (0.61 10 6 )
12
0.61 10 6
3.0582 1011 4(3.0799 1011 ) 4(3.1495 1011 ) 2(3.1089 1011 ) 3.2104 1011
12
190140s
24
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Solution (cont.)
b)
In this case, the true error is
Et True Value Approximate Value
1.90140 105 190140
1.9838
c)
The absolute relative true error
t
True Error
100
True Value
1.9838
100
5
1.90140 10
0.0010434%
25
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Solution (cont.)
Table. Values of Simpson’s 1/3rd Rule for
Example 2 with multiple segments
26
n
Approximate Value
Et
t %
2
4
6
8
10
190160
190140
190140
190140
190140
−24.100
−1.9838
0.42010
0.13655
0.056663
0.012675%
0.0010434%
2.2094×10−4%
7.1815×10−5%
2.9802×10−5%
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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)
n2
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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