Chapter 07.03 Simpson's 1/3 Rule for Integration-More Examples ...

Chapter 07.03
Simpson’s 1/3 Rule for Integration-More Examples
Civil Engineering
Example 1
The concentration of benzene at a critical location is given by
where
c 1.75 erfc
erfc
x
x
e
32.73
0.6560
e erfc5.758
2
z
dz
So in the above formula
Since
erfc
2
z
0.6560
2
z
0.6560
e dz
e decays rapidly as z , we will approximate
0.6560
.6560
e
5
2
z
erfc 0 dz
a) Use Simpson’s 1/3 Rule to find the value of 0.6560
b) Find the true error, E
t
, for part (a).
c) Find the absolute relative true error for part (a).
Solution
b a a b
a) erfc0
.6560
f a
4 f f b
6
a 5
b 0. 6560
a b
2.8280
2
2
z
f z
e
2
5
f 5 e
1.388810
f
0.6560
11
0.6560
e
0.65029
2
2
erfc .
07.03.1

07.03.2 Chapter 07.02
2
2.8280
f 2.8280 e
4
3.362710
b a
a b
erfc0
.6560
f a
4 f f b
6
2
0.6560 5
f
5 4 f 2.8280
f 0.6560
6
4.3440
11
4
1.388810
43.3627
10
0.65029
6
0.47178
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.
0.6560
.6560
e
5
2
z
erfc 0 dz
0.31333
so the true error is
True Value Approximate Value
E t
0.31333
0.47178
0.15846
c) The absolute relative true error,
t
, would then be
True Error
t
100
True Value
0.15846
100
0.31333
50.573 %
Example 2
The concentration of benzene at a critical location is given by
where
c 1.75 erfc
erfc
x
x
e
32.73
0.6560
e erfc5.758
2
z
dz
So in the above formula
Since
erfc
2
z
0.6560
2
z
0.6560
e dz
e decays rapidly as z , we will approximate

Simpson’s 1/3 Rule for Integration-More Examples: Civil Engineering 07.03.3
0.6560
.6560
e
z
erfc 0 dz
5
2
a) Use four segment Simpson’s 1/3 Rule to find the value of erfc 0.6560.
b) Find the true error, E
t
, for part (a).
c) Find the absolute relative true error for part (a).
Solution
b a
3n
n1
n2
0 i
2
i1
i2
iodd
ieven
a) erfc0
.6560
f z
4 f z
f z
f z
n 4
a 5
b 0. 6560
b a
h
n
0.6560
5
4
1.0860
2
z
f ( z)
e
i
n
So
z 0
f 5
2
5
5 e
f
f
f
f
f
f
f
f
1.388810
f 5 1.0860
f 3.9140
11
z1
2
3.9140
3.9140 e
7
2.222610
f 3.9140 1.0860
f 2.8280
z2
2
2.8280
2.8280 e
4
3.3627 10
2.8280 1.0860
z3 f
f 1.7420
2
1.7420
1.7420 e
0.048096

07.03.4 Chapter 07.02
z4
f z n
f
f
0.6560
2
0.6560
f (0.6560) e
0.65029
n1
n2
b a
erfc0
.6560
f z
0
4
f zi
2 f zi
f zn
3n
i1
i2
iodd
ieven
3
2
0.6560 5
f 5 4
f zi
2 f zi
f 0.6560
3 4
i1
i2
iodd
ieven
4.3440
f
5 4 f z1
4 f z3
2 f z2
f 0.6560
12
4.3440 f 5 4 f 3.9140
4 f 1.7420
12
2 f 2.8280 f 0.6560
11
7
4.3440 1.388810
42.222610
40.048096
4
12
2 3.3627 10
0.65029
0.30529
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.
0.6560
.6560
e
z
erfc 0 dz
0.31333
so the true error is
True Value Approximate Value
E t
5
0.31333
0.30529
0.0080347
c) The absolute relative true error,
t
, would then be
True Error
t
100
True Value
0.0080347
100
0.31333
2.5643 %
2

Simpson’s 1/3 Rule for Integration-More Examples: Civil Engineering 07.03.5
Table 1 Values of Simpson’s 1/3 Rule for Example 2 with multiple segments.
n Approximate Value E
t
2
4
6
8
10
0.47178
0.30529
0.30678
0.31110
0.31248
0.15846
0.0080347
0.0065444
0.0022249
0.00084868
t
%
50.573
2.5643
2.0887
0.71009
0.27086