Simpson’s 1/3 rd Rule of Integration

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Simpson’s 1/3rd Rule of Integration

Civil Engineering Majors

Authors: Autar Kaw, Charlie Barker

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Simpson’s 1/3rd Rule of Integration

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What is Integration?Integration

b

a

dx)x(fI

The process of measuring the area under a curve.

Where:

f(x) is the integrand

a= lower limit of integration

b= upper limit of integration

f(x)

a b

y

x

b

a

dx)x(f


Simpson’s 1/3rd Rule


Basis of Simpson’s 1/3rd Rule

Trapezoidal rule was based on approximating the integrand by a firstorder polynomial, and then integrating the polynomial in the interval

ofintegration. Simpson’s 1/3rd rule is an extension of Trapezoidal rulewhere the integrand is approximated by a second order polynomial.

Hence

b

a

b

a

dx)x(fdx)x(fI 2

Where is a second order polynomial.

)x(f2

22102 xaxaa)x(f



Choose

)),a(f,a( ,ba

f,ba

22))b(f,b(an

das the three points of the function to evaluate a0, a1 and a2.

22102 aaaaa)a(f)a(f

2

2102 2222

ba

aba

aaba

fba

f

22102 babaa)b(f)b(f



Solving the previous equations for a0, a1 and a2 give

22

22

02

24

baba

)a(fb)a(abfba

abf)b(abf)b(faa

2212

2433

24

baba

)b(bfba

bf)a(bf)b(afba

af)a(afa

2222

222

baba

)b(fba

f)a(f

a



Then

b

a

dx)x(fI 2

b

a

dxxaxaa 2210

b

a

xa

xaxa

32

3

2

2

10

32

33

2

22

10

aba

aba)ab(a



Substituting values of a0, a1, a 2 give

)b(fba

f)a(fab

dx)x(fb

a 24

62

Since for Simpson’s 1/3rd Rule, the interval [a, b] is brokeninto 2 segments, the segment width

2

abh



)b(fba

f)a(fh

dx)x(fb

a 24

32

Hence

Because the above form has 1/3 in its formula, it is called Simpson’s 1/3rd Rule.


Example 1

Since decays rapidly as , we will approximate

a) Use Simpson 1/3rd rule to find the approximate value of erfc(0.6560).b) Find the true error, for part (a).c) Find the absolute relative true error, for part (a).

6560.0

5

2

6560.0 dzeerfc z

The concentration of benzene at a critical location is given by

758.56560.075.1 73.32 erfceerfcc

where dzexerfc

xz

2

So in the above formula dzeerfc z

6560.0

2

6560.0

tEa

2zez


Solution

a)

2

)( zezf

bfba

fafab

erfc2

46

6560.0

6560.08280.2456

56560.0fff

65029.0103627.34103888.16

3440.4 411

47178.0

6560.0

5

2

6560.0 dzeerfc z


Solution (cont)

b) The exact value of the above integral cannot be found. We assume the value obtained by adaptive numerical integration using Maple as the exact value for calculating the true error and relative true error.

True Error

31333.0

6560.06560.0

5

2

dzeerfc z

ValueeApproximatValueTrueEt

47178.031333.0

15846.0


Solution (cont)

c) The absolute relative true error,

%573.50

10031333.0

15846.0

100Value True

Error True

t


Multiple Segment Simpson’s 1/3rd Rule



Just like in multiple segment Trapezoidal Rule, one can subdivide the interval

[a, b] into n segments and apply Simpson’s 1/3rd Rule repeatedly overevery two segments. Note that n needs to be even. Divide interval[a, b] into equal segments, hence the segment width

n

abh

nx

x

b

a

dx)x(fdx)x(f0

where

ax 0 bxn



Apply Simpson’s 1/3rd Rule over each interval,

...)x(f)x(f)x(f

)xx(dx)x(fb

a

6

4 21002

...)x(f)x(f)x(f

)xx(

6

4 43224

f(x)

. . .

x0 x2 xn-

2

xn

x

.....dx)x(fdx)x(fdx)x(fx

x

x

x

b

a

4

2

2

0

n

n

n

n

x

x

x

x

dx)x(fdx)x(f....2

2

4



...)x(f)x(f)x(f

)xx(... nnnnn

64 234

42

6

4 122

)x(f)x(f)x(f)xx( nnn

nn

Since

hxx ii 22 n...,,,i 42



Then

...)x(f)x(f)x(f

hdx)x(fb

a

6

42 210

...)x(f)x(f)x(f

h

6

42 432

...)x(f)x(f)x(f

h nnn

6

42 234

6

42 12 )x(f)x(f)x(fh nnn



b

a

dx)x(f ...)x(f...)x(f)x(f)x(fh

n 1310 43

)}]()(...)()(2... 242 nn xfxfxfxf

)()(2)(4)(3

2

2

1

10 n

n

evenii

i

n

oddii

i xfxfxfxfh

)()(2)(4)(3

2

2

1

10 n

n

evenii

i

n

oddii

i xfxfxfxfn

ab


Example 2

Since decays rapidly as , we will approximate

a) Use four segment Simpson’s 1/3rd Rule to find the approximate value of erfc(0.6560).

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

6560.0

5

2

6560.0 dzeerfc z

The concentration of benzene at a critical location is given by

758.56560.075.1 73.32 erfceerfcc where

dzexerfcx

z

2

So in the above formula

dzeerfc z

6560.0

2

6560.0

atE

2ze z


SolutionUsing n segment Simpson’s 1/3rd

Rule,

So

a)

0860.14

56560.0

n

abh

)5()( 0 fzf

)9140.3()0860.15()( 1 ffzf

)8280.2()0860.19140.3()( 2 ffzf

)7420.1()0860.18280.2()( 3 ffzf

)6560.0()( 4 fzf


Solution (cont.)

30529.0

65029.0103627.32048096.04102226.24103888.112

3440.4

6560.08280.227420.149140.34512

3440.4

6560.0244512

3440.4

6560.024543

56560.0

4711

231

2

2

3

1

fffff

fzfzfzff

fzfzff

evenii

i

oddii

i

n

n

evenii

i

n

oddii

i zfzfzfzfn

aberfc

2

2

1

10 24

36560.0


Solution (cont.)

In this case, the true error is

The absolute relative true error

0080347.0

30529.031333.0

ValueeApproximatValueTrueEt

%5643.2

10031333.0

0080347.0

100Value True

Error True

t

c)

b)

Approximate Value

246810

−0.47178−0.30529−0.30678−0.31110−0.31248

0.15846−0.0080347−0.0065444−0.0022249

−0.00084868

50.573%2.5643%2.0887%

0.71009%0.27086%


Solution (cont.)Table 1: Values of Simpson’s 1/3rd Rule for Example 2 with multiple segments

ntE t


Error in the Multiple Segment Simpson’s 1/3rd Rule

The true error in a single application of Simpson’s 1/3rd Rule is given as

bafab

Et

),(2880

)( )4(5

In Multiple Segment Simpson’s 1/3rd Rule, the error is the sum of the errors

in each application of Simpson’s 1/3rd Rule. The error in n segment Simpson’s1/3rd Rule is given by

)(f)xx(

E )(1

45

021 2880

),(fh )(

14

5

90

)(f)xx(

E )(2

45

242 2880

),(fh )(

24

5

90

210 xx

422 xx



)(f)xx(

E i)()i(i

i

45

122

2880),(f

hi

)( 45

90

12

45

42

12 2880 n

)(nnn f

)xx(E ,f

hn

)(

12

45

90

nnn xx, 2

2

2

)4(5

90 nfh

.

.

.

ii)i( xx 212

21

2

4

nnn xx

2

45

2

22880 n

nnn f

)xx(E



Hence, the total error in Multiple Segment Simpson’s 1/3rd Rule is

2

1

n

iit EE

2

1

)4(5

)(90

n

iif

h

2

1

)4(5

5

)(90

)(n

iif

n

ab

n

f

n

ab

n

ii

2

1

)4(

4

5 )(

90

)(



The term n

f

n

ii

2

1

)4( )(is an approximate average value of

bxaxf ),()4(

Hence )4(

4

5

90

)(f

n

abEt

where

n

ff

n

ii

2

1

)4(

)4()(

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/simpsons_13rd_rule.html



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Simpson’s 1/3 rd Rule of Integration

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