Trapezoidal Rule of Integration
Chemical Engineering Majors
Authors: Autar Kaw, Charlie Barker
http://numericalmethods.eng.usf.edu
Transforming Numerical Methods Education for STEM
Undergraduates
7/31/2017
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1
Trapezoidal Rule of
Integration
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What is Integration
b
Integration:
 f ( x )dx
a
y
f(x)
The process of measuring
the area under a function
plotted on a graph.
b
I   f ( x )dx
a
Where:
f(x) is the integrand
a= lower limit of integration
b= upper limit of integration
3
a
b
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x
Basis of Trapezoidal Rule
Trapezoidal Rule is based on the Newton-Cotes
Formula that states if one can approximate the
integrand as an nth order polynomial…
b
I   f ( x )dx
where
f ( x )  fn( x )
a
and
4
f n ( x )  a0  a1 x  ...  an1 x n1  an x n
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Basis of Trapezoidal Rule
Then the integral of that function is approximated
by the integral of that nth order polynomial.
b
b
a
a
 f ( x )   fn( x )
Trapezoidal Rule assumes n=1, that is, the area
under the linear polynomial,
b

a
5
 f ( a )  f ( b )

(
b

a
)
f ( x )dx


2
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Derivation of the Trapezoidal Rule
6
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Method Derived From Geometry
b
The area under the
curve is a trapezoid.
The integral
 f ( x )dx
1
a
y
f(x)
b
 f ( x)dx  Area of
trapezoid
a
f1(x)
1
 ( Sum of parallel sides )( height )
2

1
 f ( b )  f ( a )( b  a )
2
 f ( a )  f ( b )
 ( b  a )

2

a
b
Figure 2: Geometric Representation
7
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x
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
7
0.6110 6  6.73x  4.3025  10


dx
T  
11
1.2210 6 
2.316  10 x


a) Use single segment Trapezoidal 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, a for part (a).
8
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Solution
 f a   f b 
I  b  a 

2

a)
a  1.22  10 6
 6.73x  4.3025  10 7 
f ( x)   

11
2
.
316

10
x


b  0.61  10 6

f 1.22 10

6
f 0.6110
9

6



 6.73 1.22 106  4.3025 107 
11
 
  3.0582 10
11
6
 2.316 10 1.22 10






 6.73 0.61106  4.3025 107 
11
 
  3.2104 10
11
6
2.316 10 0.6110



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Solution (cont)
a)

6
I  0.6110  1.22 10
6


  3.0582 1011   3.2104 1011 


2



 1.9119  10 5 s
b) The exact value of the above integral is
 6.73x  4.3025 107 
5


T  
dx

1
.
90140

10
s
11

1.22106 
2.316 10 x


0.61106
10
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Solution (cont)
b)
Et  True Value  Approximate Value
 1.90140 105  1.9119 105
 1056.2s
c)
The absolute relative true error, t
, would be
True Error
 1056.2
t 
100 
100  0.55549%
5
True Value
1.90140 10
11
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Multiple Segment Trapezoidal Rule
In Example 1, the true error using single segment trapezoidal rule was
large. We can divide the interval [8,30] into [8,19] and [19,30] intervals
and apply Trapezoidal rule over each segment.
140000


f ( t )  2000 ln
  9.8t
 140000  2100t 
30
19
30
8
8
19
 f ( t )dt   f ( t )dt   f ( t )dt
 f ( 8 )  f ( 19 )
 f ( 19 )  f ( 30 )
 ( 19  8 )
 ( 30  19 )


2
2



12
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Multiple Segment Trapezoidal Rule
With
f ( 8 )  177.27 m / s
f ( 30 )  901.67 m / s
f ( 19 )  484.75 m / s
Hence:
30

8
177.27  484.75 
 484.75  901.67 
f (t )dt  (19  8) 

(
30

19
)



2
2

 11266 m
13
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Multiple Segment Trapezoidal Rule
The true error is:
Et  11061 11266
 205 m
The true error now is reduced from -807 m to -205 m.
Extending this procedure to divide the interval into equal
segments to apply the Trapezoidal rule; the sum of the
results obtained for each segment is the approximate
value of the integral.
14
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Multiple Segment Trapezoidal Rule
y
f(x)
Divide into equal segments
as shown in Figure 4. Then
the width of each segment is:
h
ba
n
The integral I is:
b
I   f ( x )dx
a
a
ba
a
4
a2
ba
4
a3
ba
4
b
Figure 4: Multiple (n=4) Segment Trapezoidal Rule
15
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x
Multiple Segment Trapezoidal Rule
The integral I can be broken into h integrals as:
b
 f ( x )dx
a

ah
a  2h
a  ( n 1 )h
a
ah
a  ( n  2 )h
 f ( x )dx   f ( x )dx  ... 
 f ( x )dx 
b
 f ( x )dx
a  ( n 1 )h
Applying Trapezoidal rule on each segment gives:
b

a
16
ba

n 1


f
(
a
)

2
f
(
a

ih
)

f
(
b
)
f ( x )dx




2n 
 i 1

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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

T  
1.2210 6 
2.316  10 11 x

0.6110 6

dx

a) Use two-segment Trapezoidal 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, a for part (a).
17
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Solution
a) The solution using 2-segment Trapezoidal rule is
ba

n 1

I
f
(
a
)

2
f
(
a

ih
)

f
(
b
)




2n 
 i 1


n2
a  1.22  10 6
b  0.61  10 6
b  a 0.61106  1.22 106
h

 0.305 106
n
2
18
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Solution (cont)
Then
0.61106  1.22 106
I
22


 21

6
6 
 f 1.22 10  2 f a  ih   f 0.6110 
 i 1






 0.61106

f 1.22 106  2 f 0.915 106  f 0.61106
4



 
 0.6110 6

 3.0582 1011  2  3.1089 1011  3.2104 1011
4





 1.9042  10 5 s
19
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Solution (cont)
b) The exact value of the above integral is
 6.73x  4.3025 107 
5


T  
dx

1
.
90140

10
s
11

1.22106 
2.316 10 x


0.61106
so the true error is
Et  True Value  Approximate Value
 1.90140 105  1.9042 105
 282.12
20
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Solution (cont)
c)
The absolute relative true error, t
, would be
True Error
t 
 100
True Value
 282.12

 100
5
1.9014  10
 0.14838%
21
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Solution (cont)
Table 1: Multiple Segment Trapezoidal Rule Values
Table 1 gives the values
obtained using multiple
segment Trapezoidal rule
for:
 6.73x  4.3025  10 7

T  
1.2210 6 
2.316  10 11 x

0.6110 6
22

dx

t %
a %
n
Value
Et
1
191190
−1056.2
0.55549
---
2
190420
−282.12
0.14838
0.40711
3
190260
−127.31
0.066956
0.081424
4
190210
−72.017
0.037877
0.029079
5
190180
−46.216
0.024307
0.013570
6
190170
−32.142
0.016905
0.0074020
7
190160
−23.636
0.012431
0.0044740
8
190150
−18.107
0.0095231
0.0029079
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Example 3
Use Multiple Segment Trapezoidal Rule to find
the area under the curve
300 x
f(x)
1 ex
x0
from
23
300( 0 )
1 e
0
0
f (5) 
300( 5 )
1 e
x  10
10  0
h
5
2
Using two segments, we get
f(0)
to
5
 10.039
f ( 10 ) 
and
300( 10 )
1 e
10
 0.136
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Solution
Then:

n 1

ba
I
f
(
a
)

2
f
(
a

ih
)

f
(
b
)




2n 
 i 1



2 1

10  0 

f
(
0
)

2
f
(
0

5
)

f
(
10
)




2( 2 ) 
 i 1



10
 f ( 0 )  2 f ( 5 )  f ( 10 )  10 0  2( 10.039 )  0.136
4
4
 50.535
24
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Solution (cont)
So what is the true value of this integral?
10

300x
01 e
x
dx  246.59
Making the absolute relative true error:
t 
246.59  50.535
 100%
246.59
 79.506%
25
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Solution (cont)
Table 2: Values obtained using Multiple Segment
10
Trapezoidal Rule for:
300x

01 e
26
x
dx
n
Approximate
Value
Et
1
0.681
245.91
99.724%
2
50.535
196.05
79.505%
4
170.61
75.978
30.812%
8
227.04
19.546
7.927%
16
241.70
4.887
1.982%
32
245.37
1.222
0.495%
64
246.28
0.305
0.124%
t
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Error in Multiple Segment
Trapezoidal Rule
The true error for a single segment Trapezoidal rule is given by:
( b  a )3
Et 
f " (  ), a    b
12
where  is some point in a ,b
What is the error, then in the multiple segment Trapezoidal rule? It will
be simply the sum of the errors from each segment, where the error in
each segment is that of the single segment Trapezoidal rule.
The error in each segment is
E1 
27
( a  h )  a3
12
h3

f " ( 1 )
12
f " (  1 ), a   1  a  h
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Error in Multiple Segment
Trapezoidal Rule
Similarly:
Ei 
( a  ih )  ( a  ( i  1 )h )3
12
f " (  i ), a  ( i  1 )h   i  a  ih
h3

f " ( i )
12
It then follows that:
En 
b  a  ( n  1 )h3
12
f " (  n ), a  ( n  1 )h   n  b
h3

f" (n )
12
28
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Error in Multiple Segment
Trapezoidal Rule
Hence the total error in multiple segment Trapezoidal rule is
n
n
3
h3 n
(
b

a
)

f " ( i ) 

12 i 1
12n 2
Et   Ei
i 1
n
The term

i 1
 f " ( i )
i 1
n
f " (  i ) is an approximate average value of the f " ( x ), a  x  b
n
Hence:
n
Et 
29
(b  a )
12n 2
3
 f " ( i )
i 1
n
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Error in Multiple Segment
Trapezoidal Rule
Below is the table for the integral
30
140000




2000
ln

9
.
8
t

dt

140000  2100t 

8
as a function of the number of segments. You can visualize that as the
number of segments are doubled, the true error gets approximately quartered.
30
n
Value
Et
t %
a %
2
11266
-205
1.854
5.343
4
11113
-51.5
0.4655
0.3594
8
11074
-12.9
0.1165
0.03560
16
11065
-3.22
0.02913
0.00401
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Additional Resources
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
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_rule.html
THE END
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