Chapter 07.06
Gauss Quadrature 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 two-point Gauss Quadrature Rule to find the probability.
b) Find the absolute relative true error.
Solution
a) First, change the limits of integration from 250, 270 to  1, 1 using
a  250
b  270
b

a
ba
ba
ba
f ( y )dy 
f
y
dy

2 1  2
2 
1
gives
270  250  270  250
270  250 
250 f ( y)dy  2 1 f  2 y  2 dy
270
1
1
 10  f 10 y  260 dy
1
Next, get weighting factors and function argument values for the two point rule,
c1  1.0000
y1  0.57735
c2  1.0000
y2  0.57735
Now we can use the Gauss Quadrature formula
1
10  f 10 y  260dy  10c1 f 10 y1  260   c 2 f 10 y 2  260
1
07.06.1
07.06.2
Chapter 07.06
 10 f 10 0.57735  260  f 100.57735  260
 10 f 254.23  f 265.77
 10 0.071407  3.1070  10 32 
 0.71408

since

f 254.23  0.3515e 0.3881254.23252.2 
2
 0.071407
f 265.77   0.3515e 0.3881265.77 252.2 
2
 3.1070  10 32
b) The absolute relative true error, t , is (Exact value = 0.97377)
True Value  Approximat e Value
 100 %
True Value
0.97377  0.71015
t 
 100 %
0.97377
 26.669 %
t 
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 three-point Gauss Quadrature to find the probability.
b) Find the absolute relative true error.
Solution
a) First, change the limits of integration from 250, 270 to  1, 1 using
a  250
b  270
b

a
ba
ba
ba
f ( y )dy 
f
y
dy

2 1  2
2 
1
gives
270  250  270  250
270  250 
250 f ( y)dy  2 1 f  2 y  2 dy
270
1
1
 10  f 10 y  260 dy
1
The weighting factors and function argument values are
Gauss Quadrature Rule for Integration-More Examples: Industrial Engineering
07.06.3
c1  0.55556
x1  0.77460
c2  0.88889
x2  0.0000
c3  0.55556
x3  0.77460
and the formula is
1
10  f 10 y  260dy  10c1 f 10 y1  260  c 2 f 10 y 2  260  c3 f 10 y 3  260
1
0.55556 f 10 0.77460  260
 10 0.88889 f 100.0000  260 
 0.55556 f 100.77460  260
0.55556 f 252.25  0.88889 f 260.00
 10

  0.55556 f 267.75

0.555560.35110  0.888891.9560  10 11 
 10

 42
 0.555566.4763  10 

 1.9509
since
f 252.25  0.3515e 0.3881252.25252.2 
2
 0.35110
f 260.00  0.3515e 0.3881260.00 252.2 
2
 1.9560  10 11
f 267.75  0.3515e 0.3881267.75 252.2 
2
 6.4763  10  42
b) The absolute relative true error, t , is (exact value = 0.97377)
True Value  Approximat e Value
 100 %
True Value
0.97377  1.9509
t 
 100 %
0.97377
 100.31 %
t 