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Simpson’s 1/3rd Rule of Integration

Major: All 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

adx)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 RuleTrapezoidal 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

adx)x(fdx)x(fI 2

Where is a second order polynomial. )x(f2

22102 xaxaa)x(f ++=


Basis of Simpson’s 1/3rd Rule

Choose

)),a(f,a( ,baf,ba

++

22))b(f,b(and

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

22102 aaaaa)a(f)a(f ++==

2

2102 2222

+

+

+

+=

+

=

+ baabaaabafbaf

22102 babaa)b(f)b(f ++==



Solving the previous equations for a0, a1 and a2 give

22

22

0 22

4

baba

)a(fb)a(abfbaabf)b(abf)b(faa

+−

++

+

−+=

221 22

4332

4

baba

)b(bfbabf)a(bf)b(afbaaf)a(afa

+−

+

+

−++

+

−−=

222 22

22

baba

)b(fbaf)a(fa

+−

+

+

−=



Then

∫≈b

adx)x(fI 2

( )∫ ++=b

adxxaxaa 2

210

b

a

xaxaxa

++=

32

3

2

2

10

32

33

2

22

10abaaba)ab(a −

+−

+−=



Substituting values of a0, a1, a 2 give

+

+

+−

=∫ )b(fbaf)a(fabdx)x(fb

a 24

62

Since for Simpson’s 1/3rd Rule, the interval [a, b] is broken

into 2 segments, the segment width

2abh −

=



+

+

+=∫ )b(fbaf)a(fhdx)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

a) Use Simpson’s 1/3rd Rule to find the approximate value of x

The distance covered by a rocket from t=8 to t=30 is given by

∫

−

−=

30

889

21001400001400002000 dtt.

tlnx

b) Find the true error, tE

c) Find the absolute relative true error, t∈


Solutiona)

∫=30

8

)( dttfx

+

+

+

−

= )b(fbaf)a(fabx2

46

[ ])(f)(f)(f 3019486

830++

−

=

[ ]6740901745548442667177622 .).(. ++

=

m.7211065=


Solution (cont)

b) The exact value of the above integral is

∫

−

−=

30

889

21001400001400002000 dtt.

tlnx

m.3411061=

True Error

72110653411061 ..Et −=

m.384−=


Solution (cont)a)c) Absolute relative true error,

%.

..t 100

341106172110653411061

×−

=∈

%.03960=


Multiple Segment Simpson’s 1/3rd Rule


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 over

every two segments. Note that n needs to be even. Divide interval

[a, b] into equal segments, hence the segment width

nabh −

= ∫∫ =nx

x

b

adx)x(fdx)x(f

0

where

ax =0 bxn =



Rule

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

...)x(f)x(f)x(f)xx(dx)x(fb

a+

++

−=∫ 64 210

02

...)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

xdx)x(fdx)x(f....

2

2

4



Rule

...)x(f)x(f)x(f)xx(... nnnnn +

++

−+ −−−−− 6

4 23442

++

−+ −−− 6

4 122

)x(f)x(f)x(f)xx( nnnnn

Since

hxx ii 22 =− − n...,,,i 42=



Rule

Then

...)x(f)x(f)x(fhdx)x(fb

a+

++

=∫ 642 210

...)x(f)x(f)x(fh +

++

+6

42 432

...)x(f)x(f)x(fh nnn +

++

+ −−−

642 234

++

+ −−

642 12 )x(f)x(f)x(fh nnn



Rule

∫b

adx)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 xfxfxfxfnab

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

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


Example 2Use 4-segment Simpson’s 1/3rd Rule to approximate the distance

tE

covered by a rocket from t= 8 to t=30 as given by

∫

−

−=

30

88.9

2100140000140000ln2000 dtt

tx

a∈


SolutionUsing n segment Simpson’s 1/3rd Rule,

4830 −

=h 5.5=

So )8()( 0 ftf =

)5.58()( 1 += ftf )5.13(f=

a)

)5.55.13()( 2 += ftf )19(f=

)5.519()( 3 += ftf )5.24(f=

)( 4tf )30(f=


Solution (cont.)

+++

−= ∑∑

−

==

−

==

)()(2)(4)(3

2

2

1

10 n

n

evenii

i

n

oddii

i tftftftfnabx

+++

−= ∑∑

==

==

2

2

3

1)30()(2)(4)8(

)4(3830

evenii

i

oddii

i ftftff

[ ])30()(2)(4)(4)8(1222

231 ftftftff ++++=


Solution (cont.)

[ ])30()19(2)5.24(4)5.13(4)8(6

11 fffff ++++=

[ ]6740.901)7455.484(2)0501.676(4)2469.320(42667.1776

11++++=

m64.11061=

cont.


Solution (cont.)

In this case, the true error is

64.1106134.11061 −=tE

b)

m30.0−=

The absolute relative true error

%10034.11061

64.1106134.11061×

−=∈t

c)

%0027.0=


Solution (cont.)

Table 1: Values of Simpson’s 1/3rd Rule for Example 2 with multiple segments

n Approximate Value Et |Єt |

246810

11065.7211061.6411061.4011061.3511061.34

4.380.300.060.010.00

0.0396%0.0027%0.0005%0.0001%0.0000%


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

bafabEt <ζ<ζ−

−= ),(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’s

1/3rd Rule is given by

)(f)xx(E )(1

45

021 2880

ζ−

−= ),(fh )(1

45

90ζ−=

)(f)xx(E )(2

45

242 2880

ζ−

−= ),(fh )(2

45

90ζ−=

210 xx <ζ<

422 xx <ζ<



)(f)xx(

E i)()i(i

i ζ−

−= − 45

122

2880 ),(fhi

)( ζ−= 45

90

ζ

−−=

−

−−

− 12

45

421

2 2880 n)(nn

n f)xx(E ,fhn

)(

ζ−=

−12

45

90

nnn xx, <ζ<−2

2

ζ−=

2

)4(5

90 nfh

.

.

.

ii)i( xx 212 <ζ<−

212

4 −−

− <ζ< nnn xx

ζ

−−= −

2

45

2

2 2880 nnn

n f)xx(E



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

∑=

=2

1

n

iit EE ∑

=ζ−=

2

1

)4(5

)(90

n

iifh ∑

=ζ

−−=

2

1

)4(5

5

)(90

)(n

iif

nab

n

f

nab

n

ii∑

=ζ−

−=

2

1

)4(

4

5 )(

90)(



The term n

fn

ii∑

=ζ

2

1

)4( )(is an approximate average value of

bxaxf <<),()4(

Hence )4(

4

5

90)( f

nabEt

−−=

where

n

ff

n

ii∑

=ζ

=

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

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