07.03
Simpson’s 1/3 Rule for Integration-More Examples
Industrial Engineering
Example 1
A company advertises that every roll of toilet paper has at least 250 sheets. The probability
that there are 250 or more sheets in the toilet paper is given by

P( y  250)   0.3515 e 0.3881( y 252.2) dy
2
250
Approximating the above integral as
270
P( y  250)   0.3515 e 0.3881( y 252.2) dy
2
250
a) Use Simpson’s 1/3 Rule to find the probability.
b) Find the true error, E t , for part (a).
c) Find the absolute relative true error, t , for part (a).
Solution
a)
P( y  250) 

ba
ab
f (a)  4 f 
  f (b)

6 
 2 

a  250
b  270
ab
 260
2
2
f ( y)  0.3515e 0.3881( y 252.2)
f 250  0.3515e 0.3881250252.2 
 0.053721
2
f 270  0.3515e 0.3881270252.2 
 1.3888 10 54
2
f 260  0.3515e 0.3881260252.2 
 1.9560  10 11

 b  a 
ab
P( y  250)  
  f (a)  4 f 
  f (b)
 6 
 2 

 270  250 

 f 250   4 f 260   f 270
6


2
07.03.1
07.03.2
Chapter 07.03


 20 
   0.053721  41.9559  10 11   1.3888  10 54
 6 
 0.17907
b) The exact value of the above integral cannot be found. For calculating the true error and
relative true error, we assume the value obtained by adaptive numerical integration using
Maple as the exact value.
270
P( y  250)   0.3515 e 0.3881( y 252.2) dy
2
250
 0.97377
So the true error is,
Et  True Value  Approximat e Value
 0.97377  0.17907
 0.79470
Absolute Relative true error,
True Error
t 
 100 %
True Value
0.79470

 100 %
0.97377
 81.611 %
Example 2
A company advertises that every roll of toilet paper has at least 250 sheets. The probability
that there are 250 or more sheets in the toilet paper is given by

P( y  250)   0.3515 e0.3881( y252.2) dy
2
250
Approximating the above integral as
270
P( y  250)   0.3515 e 0.3881( y 252.2 ) dy
2
250
a) Use four segment Simpson’s 1/3 Rule to find the probability.
b) Find the true error, E t , for part (a).
c) Find the absolute relative true error for part (a).
Solution
a) Using n segment Simpson’s 1/3rd Rule,


n 1
n2
ba
P( y  250) 
f ( y 0 )  4  f ( y i )  2  f ( y i )  f ( y n )

3n 
i 1
i 2


i  odd
i  even
n4
a  250
b  270
ba
h
n
Simpson’s 1/3 Rule for Integration-More Examples: Industrial Engineering
270  250
4
5

f ( y)  0.3515e 0.3881( y 252.2)
2
So


n 1
n2
ba
P( y  250) 
f ( y 0 )  4  f ( y i )  2  f ( y i )  f ( y n )

3n 
i 1
i 2
i  odd
i  even




3
2
270  250 

f 250  4  f  y i   2  f  y i   f 270


34
i 1
i 2


i  odd
i  even
20
 f 250  4 f  y1   4 f  y3   2 f  y 2   f 270

12
10
  f (250)  4 f (255)  4 f (265)  2 f (260)  f (270)
6
29
10 0.053721  40.016769   48.5260  10 
 

6  21.9560  10 11   1.3888  10 54

 0.20133
Since
f  y0   f 250
 0.3515e 0.3881250252.2 
 0.053721
2
f  y1   f 250  5
 f 255
 0.3515e 0.3881255252.2 
 0.016769
2
f  y2   f 255  5
 f 260
 0.3515e 0.3881260252.2 
 1.9560  10 11
2
f  y3   f 260  5
 f 265
 0.3515e 0.3881265252.2 
 8.5260  10 29
2
f  y4   f  yn 
07.03.3
07.03.4
Chapter 07.03
 f 270
 0.3515e 0.3881270252.2 
 1.3888 10 54
2
b) The exact value of the above integral cannot be found. For calculating the true error and
relative true error, we assume the value obtained by adaptive numerical integration using
Maple as the exact value.
270
P( y  250)   0.3515 e 0.3881( y 252.2) dy
2
250
 0.97377
So the true error is
Et  True Value  Approximate Value
 0.97377  0.20133
 0.77244
Absolute Relative true error,
True Error
t 
 100 %
True Value
0.77244

 100 %
0.97377
 79.325 %
Table 1 Values of Simpson’s 1/3 Rule for Example 2 with multiple segments.
Approximate Value
n
Et
t %
2
4
6
8
10
0.17907
0.20133
1.0090
1.2042
1.0954
0.79470
0.77244
0.035226
0.23042
0.12167
81.611
79.325
3.6175
23.663
12.495