Simpson’s 1/3rd Rule of
Integration
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
2
a
b
x
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Simpson’s 1/3rd Rule
3
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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
4
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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
5
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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
6
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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
7
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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
8
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.
9
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Example 1
The distance covered by a rocket from t=8 to t=30 is given by
140000




x    2000 ln 

9
.
8
t
dt

140000  2100t 

8
30
a) Use Simpson’s 1/3rd Rule to find the approximate value of x
b) Find the true error,
Et
c) Find the absolute relative true error,
10
t
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Solution
a)
10
x   f ( t )dt
0

 b  a 
a  b
x
f
(
a
)

4
f

f
(
b
)




 6 
 2 
 30  8 

 f ( 8 )  4 f ( 19 )  f ( 30 )
 6 
 22 
  177.2667  4( 484.7455 )  901.6740
6
 11065.72 m
11
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Solution (cont)
b) The exact value of the above integral is
140000




x    2000 ln 

9
.
8
t
dt

140000  2100t 

8
30
 11061.34 m
True Error
Et  11061.34  11065.72
 4.38 m
12
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Solution (cont)
a)c) Absolute relative true error,
11061.34  11065.72
t 
 100%
11061.34
 0.0396%
13
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Multiple Segment Simpson’s
1/3rd Rule
14
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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
15
xn  b
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Multiple Segment Simpson’s 1/3rd
Rule
f(x)
b
x2
x4
x0
x2
 f ( x )dx   f ( x )dx   f ( x )dx  .....
a
.
.
. . .
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


16
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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
17
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

18
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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


19
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Example 2
Use 4-segment Simpson’s 1/3rd Rule to approximate the distance
covered by a rocket from t= 8 to t=30 as given by
140000




x    2000 ln 

9
.
8
t
dt

140000  2100t 

8
30
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).
a)
20
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Solution
a)
Using n segment Simpson’s 1/3rd Rule,
30  8
h
 5.5
4
So
f (t 0 )  f (8)
f (t1 )  f (8  5.5)  f (13.5)
f (t 2 )  f (13.5  5.5)  f (19)
f (t 3 )  f (19  5.5)  f (24.5)
f (t4 )  f (30)
21
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Solution (cont.)


n

1
n

2
ba
x
f (t 0 )  4  f (t i )  2  f (t i )  f (t n )

3n 
i 1
i 2
i odd
i even




3
2
30  8 

f (8)  4  f (t i )  2  f (t i )  f (30)

3(4) 
i 1
i 2
i  odd
i even



22
22
 f (8)  4 f (t1 )  4 f (t 3 )  2 f (t 2 )  f (30)
12
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Solution (cont.)
cont.
11
  f (8)  4 f (13.5)  4 f (24.5)  2 f (19)  f (30)
6
11
 177.2667  4(320.2469)  4(676.0501)  2(484.7455)  901.6740
6
 11061.64 m
23
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Solution (cont.)
b)
In this case, the true error is
Et  11061.34  11061.64  0.30 m
c)
The absolute relative true error
11061.34  11061.64
t 
 100%
11061.34
 0.0027%
24
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Solution (cont.)
Table 1: Values of Simpson’s 1/3rd Rule for Example 2 with multiple segments
25
n
Approximate Value
Et
2
4
6
8
10
11065.72
11061.64
11061.40
11061.35
11061.34
4.38
0.30
0.06
0.01
0.00
|Єt |
0.0396%
0.0027%
0.0005%
0.0001%
0.0000%
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