07.03
Simpson’s 1/3 Rule for Integration-More Examples
Chemical Engineering
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

1.2210 6 
2.316  10 11 x

T  
0.6110 6

dx

a) Use Simpson’s 1/3 rule to find the time required for 50 % of the oxygen to be
consumed.
b) Find the true error, E t , for part (a).
c) Find the absolute relative true error, t , for part (a).
Solution
a)
T

ba 
ab
f (a)  4 f 
  f (b)

6 
 2 

a  1.22  10 6
b  0.61  10 6
ab
 0.91500  10 6
2
 6.73x  4.3025  10 7 
f ( x)   

2.316  10 11 x










 6.73 1.22  10 6  4.3025  10 7 
11
f 1.22  10 6  
  3.0581  10
11
6
2
.
316

10
1
.
22

10




 6.73 0.61  10 6  4.3025  10 7 
11
f 0.61  10 6  
  3.2104  10
11
6
2
.
316

10
0
.
61

10



07.03.1

07.03.2
Chapter 07.03


 6.73 0.91500  10 6  4.3025  10 7 
11
f 0.91500  10 6  
  3.1089  10
11
6
2
.
316

10
0
.
91500

10



 b  a 
ab
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


  0.61  10 6 
  3.0581  1011  4  3.1089  1011  3.2104  1011
 
6


 190160 s



b)


 



The exact value of the above integral is,
7
0.6110 6 6.73x  4.3025  10


dx
T  
6 

11
1.2210
2.316  10 x


 1.90140  10 5 s
so the true error is
Et  True Value  Approximate Value
 1.90140  10 5  190160
 24.100
c) The absolute relative true error, t , would then be
True Error
 100
True Value
 24.020

 100
1.90140  10 5
 0.012675 %
t 
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

1.2210 6 
2.316  10 11 x

T  
0.6110 6

dx

a) Use four-Simpson’s 1/3 Rule to find the time required for 50% of the oxygen to be
consumed.
Simpson’s 1/3 Rule for Integration-More Examples: Chemical Engineering
b) Find the true error, E t , for part (a).
c) Find the absolute relative true error, t , for part (a).
Solution
a)
So


n 1
n2
ba 
T
f ( x 0 )  4  f ( xi )  2  f ( xi )  f ( x n ) 


3n
i 1
i 2
i  odd
i  even


n4
a  1.22  10 6
b  0.61  10 6
ba
h
n
0.61  10 6  1.22  10 6

4
 0.15250  10 6
 6.73x  4.3025  10 7 
f ( x)   

2.316  10 11 x



f x0   f 1.22  10 6




 6.73 1.22  10 6  4.3025  10 7 
11
f 1.22  10 6  
  3.0581  10
11
6
2.316  10 1.22  10








f x1   f 1.22 10 6  0.15250 10 6  f 1.0675 10 6



 6.73 1.0675  10 6  4.3025  10 7 
11
f 1.0675  10 6  
  3.0799  10
11
6
2
.
316

10
1
.
0675

10


6
6
f x2   f 1.0675  10  0.15250  10   f 0.915  10 6 







 6.73 0.915  10 6  4.3025  10 7 
11
f 0.915  10 6  
  3.1089  10
11
6
2
.
316

10
0
.
915

10








f x3   f 0.915  10 6  0.15250  10 6  f 0.76250  10 6

f 0.76250  10
6




 6.73 0.76250  10 6  4.3025  10 7 
11
 
  3.1495  10
11
6
2.316  10 0.76250  10



f x 4   f x n   f 0.61  10 6






 6.73 0.61  10 6  4.3025  10 7 
11
f 0.61  10 6  
  3.2104  10
11
6
2.316  10 0.61  10





07.03.3
07.03.4
Chapter 07.03


n 1
n2
ba
T
f x0   4  f xi   2  f xi   f x n 


3n
i 1
i 2
i  odd
i  even


3

6
f
1
.
22

10

4
f  xi 


i 1

i  odd

2
 2
f  xi   f 0.61  10 6


i 2
 i even










0.61  10 6  1.22  10 6
34 

 0.61  10 6
f 1.22  10 6  4 f x1   4 f x3   2 f x2   f 0.61  10 6
12

6
6

 0.61  10 6  f 1.22  10  4 f 1.0675  10 

6
6
6 
12
4 f 0.76250  10  2 f 0.915  10  f 0.61  10 
 0.61  10 6

12












 

 3.0582  1011  4  3.0799  1011 


11
11
11 
4  3.1495  10  2  3.1089  10  3.2104  10 

 
 190140 s
b) The exact value of the above integral is
7
0.6110 6 6.73x  4.3025  10

T  
1.2210 6 
2.316  10 11 x


dx

 1.90140  10 5 s
so the true error is
Et  True Value  Approximate Value
 1.90140  10 5  190140
 1.9838
c) The absolute relative true error, t , would then be
True Error
 100
True Value
 1.9838

 100
1.90140  10 5
 0.0010434 %
t 




Simpson’s 1/3 Rule for Integration-More Examples: Chemical Engineering
Table 1 Values of Simpson’s 1/3 Rule for Example 2 with multiple segments.
07.03.5
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