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Trapezoidal Rule of Integration

Chemical Engineering Majors

Authors: Autar Kaw, Charlie Barker

http://numericalmethods.eng.usf.eduTransforming Numerical Methods Education for STEM

Undergraduates

http://numericalmethods.eng.usf.edu/


Trapezoidal Rule of Integration

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http://numericalmethods.eng.usf.edu3

What is IntegrationIntegration:

b

adx)x(fI

The process of measuring the area under a function plotted on a graph.

Where: f(x) is the integranda= lower limit of integrationb= upper limit of integration

f(x)

a b

b

a

dx)x(f

y

x


Basis of Trapezoidal Rule

b

adx)x(fI

Trapezoidal Rule is based on the Newton-Cotes Formula that states if one can approximate the integrand as an nth order polynomial…

where )x(f)x(f n

nn

nnn xaxa...xaa)x(f

1110and


Basis of Trapezoidal Rule

b

an

b

a)x(f)x(f

Then the integral of that function is approximated by the integral of that nth order polynomial.

Trapezoidal Rule assumes n=1, that is, the area under the linear polynomial,

2)b(f)a(f)ab(

b

adx)x(f


Derivation of the Trapezoidal Rule


Method Derived From Geometry

The area under the curve is a trapezoid. The integral

trapezoidofAreadxxfb

a

)(

)height)(sidesparallelofSum(21

)ab()a(f)b(f 21

2)b(f)a(f)ab(

Figure 2: Geometric Representation

f(x)

a b

b

a

dx)x(f1

y

x

f1(x)

Example 1In 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

a) Use single segment Trapezoidal rule to find the time required for 50% of the oxygen to be consumed.

b) Find the true error, for part (a).c) Find the absolute relative true error, for part

(a).

dxx

xT

6

6

1061.0

1022.1 11

7

10316.2103025.473.6

atE


Solution

2

bfafabIa)

61022.1 a61061.0 b

xxxf 11

7

10316.2103025.473.6)(

11611

766 100582.3

1022.110316.2103025.41022.173.61022.1

f

11611

766 102104.3

1061.010316.2103025.41061.073.61061.0

f


Solution (cont)a)

b)The exact value of the above integral is

2102104.3100582.31022.11061.0

111166I

s5109119.1

sdxx

xT 51061.0

1022.1 11

7

1090140.110316.2

103025.473.66

6


Solution (cont)b) ValueeApproximatValueTrueEt

c) The absolute relative true error, , would be t

55 109119.11090140.1

s2.1056

%55549.01001090140.12.1056100

Value TrueError True

5

t



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.

t.t

ln)t(f 892100140000

1400002000

30

19

19

8

30

8dt)t(fdt)t(fdt)t(f

2

30191930

2198

819)(f)(f

)()(f)(f

)(



With

s/m.)(f 271778

s/m.)(f 7548419

s/m.)(f 6790130

267.90175.484)1930(

275.48427.177)819()(

30

8

dttf

m11266

Hence:



1126611061 tE

m205

The true error is:

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.



f(x)

a b

y

x

4aba

42 aba

4

3 aba

Figure 4: Multiple (n=4) Segment Trapezoidal Rule

Divide into equal segments as shown in Figure 4. Then the width of each segment is:

n

abh

The integral I is:

b

adx)x(fI



b

h)n(a

h)n(a

h)n(a

ha

ha

ha

adx)x(fdx)x(f...dx)x(fdx)x(f

1

1

2

2

The integral I can be broken into h integrals as:

b

adx)x(f

Applying Trapezoidal rule on each segment gives:

b

adx)x(f

)b(f)iha(f)a(f

nab n

i

1

12

2

Example 2In 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

a) Use two-segment Trapezoidal rule to find the time required for 50% of the oxygen to be consumed.

b) Find the true error, for part (a).c) Find the absolute relative true error, for part (a).

dxx

xT

6

6

1061.0

1022.1 11

7

10316.2103025.473.6

tEa


Solutiona) The solution using 2-segment Trapezoidal rule is

)b(f)iha(f)a(f

nabI

n

i

1

12

2

2n 61022.1 a 61061.0 b

666

10305.02

1022.11061.0

n

abh


Solution (cont)Then

612

1

666

1061.021022.122

1022.11061.0 fihaffIi

6666

1061.010915.021022.14

1061.0

fff

1111116

102104.3101089.32100582.34

1061.0

s5109042.1


Solution (cont)b) The exact value of the above integral is

so the true error is

12.282109042.11090140.1 55

ValueeApproximatValueTrueEt

sdxx

xT 51061.0

1022.1 11

7

1090140.110316.2

103025.473.66

6


Solution (cont)c)The absolute relative true error, , would be t

%14838.0

100109014.112.282

100Value TrueError True

5

t


Solution (cont)Table 1 gives the

values obtained using multiple segment

Trapezoidal rule for:

Table 1: Multiple Segment Trapezoidal Rule Values

dxx

xT

6

6

1061.0

1022.1 11

7

10316.2103025.473.6

%t %an 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



Example 3Use Multiple Segment Trapezoidal Rule to find the area under the curve

xex)x(f

1300 from to 0x 10x

Using two segments, we get 52

010

h

01

03000

0

e)(

)(f 039101

53005

5.

e)(

)(f

13601

1030010

10.

e)(

)(f

and


Solution

)b(f)iha(f)a(f

nabI

n

i

1

12

2

)(f)(f)(f

)( i105020

22010 12

1

)(f)(f)(f 105204

10 13600391020

410 .).(

53550.

Then:


Solution (cont)So what is the true value of this integral?

59246130010

0.dx

exx

Making the absolute relative true error:

%.

..t 100

592465355059246

%.50679


Solution (cont)

n Approximate Value

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%

Table 2: Values obtained using Multiple Segment Trapezoidal Rule for:

10

01300 dx

exx

tE t


Error in Multiple Segment Trapezoidal Rule

The true error for a single segment Trapezoidal rule is given by:

ba),("f)ab(Et

12

3where is some point in b,a

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

haa),("fa)ha(

E

11

3

1 12

)("fh1

3

12



Similarly:

ihah)i(a),("f)h)i(a()iha(

E iii

112

1 3

)("fhi

12

3

It then follows that:

bh)n(a),("f

h)n(abE nnn

1

121 3

)("fhn

12

3



Hence the total error in multiple segment Trapezoidal rule is

n

iit EE

1

n

ii )("fh

1

3

12 n

)("f

n)ab(

n

ii

1

2

3

12

The term n

)("fn

ii

1is an approximate average value of the bxa),x("f

Hence:

n

)("f

n)ab(

E

n

ii

t

12

3

12



Below is the table for the integral

30

889

21001400001400002000 dtt.

tln

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.

tE %t %an Value2 11266 -205 1.854 5.3434 11113 -51.5 0.4655 0.35948 11074 -12.9 0.1165 0.0356

016 11065 -3.22 0.02913 0.0040

1

Additional ResourcesFor 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

http://numericalmethods.eng.usf.edu/topics/trapezoidal_rule.html



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3/20/2016 1 trapezoidal rule of integration chemical engineering majors authors: autar kaw,...

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