# Simpson's 1/3rd Rule

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```					             Simpson’s 1/3rd Rule of
Integration

Major: All Engineering Majors

Authors: Autar Kaw, Charlie Barker

http://numericalmethods.eng.usf.edu
Transforming Numerical Methods Education for STEM

11/21/2012                http://numericalmethods.eng.usf.edu    1
Simpson’s      Rule of
1/3 rd

Integration

http://numericalmethods.eng.usf.edu
What is Integration?
b

Integration                  f ( x )dx
a
f(x)
y
The process of measuring
the area under a curve.

b
I   f ( x )dx
a

Where:
f(x) is the integrand
a= lower limit of integration
a                              b            x
b= upper limit of integration

3                                                 http://numericalmethods.eng.usf.edu
Simpson’s 1/3rd Rule

4                 http://numericalmethods.eng.usf.edu
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
I   f ( x )dx   f 2 ( x )dx
a               a

Where     f2( x )   is a second order polynomial.

f 2 ( x )  a0  a1 x  a2 x 2

5                                                      http://numericalmethods.eng.usf.edu
Basis of Simpson’s 1/3rd Rule
Choose
 a  b  a  b 
( a , f ( a )),       ,f        , and ( b , f ( b ))
 2       2 
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
2
f       f2        a0  a1        a2      
 2          2               2          2 

f ( b )  f 2 ( b )  a 0  a1b  a 2 b 2
6                                                http://numericalmethods.eng.usf.edu
Basis of Simpson’s 1/3rd Rule
Solving the previous equations for a0, a1 and a2 give

a  b
a f ( b )  abf ( b )  4abf 
2
  abf ( a )  b f ( a )
2

a0                                 2 
a 2  2ab  b 2
a  b                                a  b
af ( a )  4af          3af ( b )  3bf ( a )  4bf          bf ( b )
a1                    2                                    2 
a 2  2ab  b 2
              a  b           
2 f ( a )  2 f         f ( b )
               2              
a2 
a 2  2ab  b 2
7                                                      http://numericalmethods.eng.usf.edu
Basis of Simpson’s 1/3rd Rule
Then
b
I   f 2 ( x )dx
a

  a0  a1 x  a 2 x 2 dx
b

a

b
         x    x 
2       3
 a0 x  a1  a2 
         2    3 a

b2  a2      b3  a3
 a0 ( b  a )  a1          a2
2            3

8                                               http://numericalmethods.eng.usf.edu
Basis of Simpson’s 1/3rd Rule
Substituting values of a0, a1, a 2 give

b
ba           a  b          
 f 2 ( x )dx        f(a) 4 f        f ( b )
a                  6             2             

Since for Simpson’s 1/3rd Rule, the interval [a, b] is broken
into 2 segments, the segment width

ba
h
2
9                                                 http://numericalmethods.eng.usf.edu
Basis of Simpson’s 1/3rd Rule

Hence

b
h               a  b          
 f 2 ( x )dx   f ( a )  4 f        f ( b )
a              3                2             

Because the above form has 1/3 in its formula, it is called Simpson’s 1/3rd Rule.

10                                                        http://numericalmethods.eng.usf.edu
Example 1

The distance covered by a rocket from t=8 to t=30 is given by

30
            140000              
x    2000 ln                   9.8t dt
8          140000  2100t 
        

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,   t

11                                                     http://numericalmethods.eng.usf.edu
Solution
a)        30
x    f (t )dt
8
 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
12                                               http://numericalmethods.eng.usf.edu
Solution (cont)
b) The exact value of the above integral is


30
   140000              
x    2000 ln                   9.8t dt
8          140000  2100t 
        

 11061.34 m
True Error

Et  11061.34  11065.72
 4.38 m
13                                                    http://numericalmethods.eng.usf.edu
Solution (cont)
a)c) Absolute relative true error,

11061.34  11065.72
t                       100%
11061.34

 0.0396%

14                                        http://numericalmethods.eng.usf.edu
Multiple Segment Simpson’s
1/3rd Rule

15                     http://numericalmethods.eng.usf.edu
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               b             xn
h                    f ( x )dx   f ( x )dx
n                a             x0

where

x0  a              xn  b

16                                                      http://numericalmethods.eng.usf.edu
Multiple Segment Simpson’s 1/3rd
Rule
f(x)

b              x2             x4
 f ( x )dx   f ( x )dx   f ( x )dx  .....
a              x0             x2
. . .
xn  2          xn
....   f ( x )dx   f ( x )dx                                            x

xn  4         xn  2                      x0   x2       xn-2   xn

Apply Simpson’s 1/3rd Rule over each interval,
b
 f ( x0 )  4 f ( x1 )  f ( x2 )
 f ( x )dx  ( x2  x0 )                                   ...
a                                        6                
 f ( x2 )  4 f ( x3 )  f ( x4 )
 ( x4  x2 )                                   ...
                6                
17                                                             http://numericalmethods.eng.usf.edu
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  2h             i  2 , 4 , ..., n

18                                                          http://numericalmethods.eng.usf.edu
Multiple Segment Simpson’s 1/3rd
Rule
Then
b
 f ( x0 )  4 f ( x1 )  f ( x2 )
 f ( x )dx  2h                                    ...
a                                6                
 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                                                    http://numericalmethods.eng.usf.edu
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                
                                                     
ba                 n 1           n2
       f ( x 0 )  4  f ( xi )  2  f ( xi )  f ( x n ) 
3n                 i 1           i 2                  
              i  odd        i  even                
20                                                       http://numericalmethods.eng.usf.edu
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

30
            140000              
x    2000 ln                   9.8t dt
8          140000  2100t 
        

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                                                http://numericalmethods.eng.usf.edu
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 (t 4 )  f (30)

22                                           http://numericalmethods.eng.usf.edu
Solution (cont.)
                                                  
ba                n 1           n2
x       f (t 0 )  4  f (t i )  2  f (t i )  f (t n )
3n                i 1           i 2                
             i  odd        i  even              
                                             
30  8               3              2
          f (8)  4  f (t i )  2  f (t i )  f (30)
3(4)              i 1           i 2              
          i  odd        i  even            

22
       f (8)  4 f (t1 )  4 f (t 3 )  2 f (t 2 )  f (30)
12

23                                                           http://numericalmethods.eng.usf.edu
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

24                                                http://numericalmethods.eng.usf.edu
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%

25                                             http://numericalmethods.eng.usf.edu
Solution (cont.)

Table 1: Values of Simpson’s 1/3rd Rule for Example 2 with multiple segments

n     Approximate Value        Et       |Єt |
2         11065.72           4.38       0.0396%
4         11061.64           0.30       0.0027%
6         11061.40           0.06       0.0005%
8         11061.35           0.01       0.0001%
10         11061.34           0.00       0.0000%

26                                                     http://numericalmethods.eng.usf.edu
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

( x2  x0 )5 ( 4 )       h5 ( 4 )
E1                f ( 1 )   f ( 1 ), x0  1  x2
2880                 90
( x4  x2 )5 ( 4 )        h5 ( 4 )
E2               f (  2 )   f (  2 ), x2   2  x4
2880                  90
27                                                     http://numericalmethods.eng.usf.edu
Error in the Multiple Segment
Simpson’s 1/3rd Rule
( x2i  x2( i 1 ) )5                       h5 ( 4 )
Ei                             f   (4)
(  i )  f (  i ), x2( i 1 )   i  x2i
2880                                 90
.
.
.

( xn  2  xn  4 )5 ( 4 )     
En                                     h5 ( 4 ) 
f   n    f   n  , xn4   n  xn2
 1                
1         2880                 2             1         1
2                                       90        2        2

( xn  xn  2 )5 4        h 5 ( 4 )   , xn  2   n  xn
En                   f n   
          f n  
2
2880            2     90         2             2

28                                                                http://numericalmethods.eng.usf.edu
Error in the Multiple Segment
Simpson’s 1/3rd Rule
Hence, the total error in Multiple Segment Simpson’s 1/3rd Rule is

n              n                                         n
5
h 2                                (b  a) 5 2
              f ( 4 ) ( i )
2
Et   E i           f
( 4)
( i )
i 1         90 i 1                             90n 5    i 1

n
2
f
( 4)
( i )
(b  a) 5  i 1

90n 4               n

29                                                                http://numericalmethods.eng.usf.edu
Error in the Multiple Segment
Simpson’s 1/3rd Rule
n
2
f
( 4)
( i )
The term   i 1                     is an approximate average value of
n
f ( 4) ( x), a  x  b

Hence
(b  a) 5 ( 4)
Et         4
f
90n
n
where                              2
 f ( 4 ) ( i )
( 4)
f             i 1
30
n           http://numericalmethods.eng.usf.edu
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

http://numericalmethods.eng.usf.edu

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