CSE 330 : Numerical Methods

Lecture 16: Numerical Integration - Simpson’s Rule

Dr. S. M. Lutful Kabir

Visiting Professor, BRAC University

& Professor (on leave) IICT, BUET

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Basis of Simpson’s 1/3rd Rule

Trapezoidal rule was based on approximating the integrand by a first

order polynomial, and then integrating the polynomial in the interval of

integration. Simpson’s 1/3rd rule is an extension of Trapezoidal rule

where 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

3

Basis of Simpson’s 1/3rd Rule

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

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Basis of Simpson’s 1/3rd Rule

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

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Basis of Simpson’s 1/3rd Rule

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

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Basis of Simpson’s 1/3rd Rule

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.

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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

8

892100140000

1400002000 dtt.

tlnx

b) Find the true error, tE

c) Find the absolute relative true error, t

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Solution

a)

30

8

)( dttfx

)b(fba

f)a(fab

x2

46

)(f)(f)(f 3019486

830

67409017455484426671776

22.).(.

m.7211065

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

30

8

892100140000

1400002000 dtt.

tlnx

m.3411061True Error

72110653411061 ..Et m.384

c) Absolute relative true error,

%.

..t 100

3411061

72110653411061

%.03960

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Multiple Segment Simpson’s 1/3rd Rule

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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

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Multiple Segment Simpson’s 1/3rd Rule

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

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Multiple Segment Simpson’s 1/3rd Rule

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

)xx(... nnnnn

6

4 23442

6

4 122

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

nn

Since

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

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Multiple Segment Simpson’s 1/3rd Rule

Then

6

)()(4)(2)( 210 xfxfxfhdxxf

b

a

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

h

6

42 432

6

)()(4)(2 234 nnn xfxfxfh

64

2 12 )x(f)x(f)x(fh nnn

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Multiple Segment Simpson’s 1/3rd Rule

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

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).

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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

8

8.92100140000

140000ln2000 dtt

tx

a

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

4

830 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

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Solution (cont.)

)()(2)(4)(3

2

2

1

10 n

n

evenii

i

n

oddii

i tftftftfn

abx

2

2

3

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

)4(3

830

evenii

i

oddii

i ftftff

)30()(2)(4)(4)8(12

22231 ftftftff

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Solution (cont.)

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

11fffff

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

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m64.11061

cont.

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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

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Solution (cont.)

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

n Approximate Value

Et |Єt |

2468

10

11065.7211061.6411061.4011061.3511061.34

4.380.300.060.010.00

0.0396%0.0027%0.0005%0.0001%0.0000%

Simpson’s 3/8 Rule of Integration

In a similar fashion, Simpson 3/8 rule for integration can be derived by approximating the given function f(x) with the 3rd order (cubic) polynomial f3(x)

22

3

2

1

0

32

33

22103

,,,1

)(

a

a

a

a

xxx

xaxaxaaxf

Using Lagrange interpolation, the cubic polynomial function that passes through 4 data points can be explicitly given as

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Simpson’s 3/8 Rule of Integration

3231303

2102

321202

310

1312101

3200

302010

3213

xfxxxxxx

xxxxxxxf

xxxxxx

xxxxxx

xfxxxxxx

xxxxxxxf

xxxxxx

xxxxxxxf

3210 338

3xfxfxfxf

hI

Following the same procedure of Simpson’s 1/3 rule, the expression of the Integral results,

Multi Segments for Simpson’s 3/8 Rule

Similarly the expression for the multi segment Simpson’s Rule can be derived as follows:

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nnnn xfxfxfxf

xfxfxfxfxfxfxfxfhI

123

65433210

33.....

3333

8

3

n

n

ii

n

ii

n

ii xfxfxfxfxf

h 3

,..9,6,3

1

,..8,5,2

2

,..7,4,10 233

8

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Exercise Find the distance covered by the rocket in between

t=8sec and t=30 sec using the expression of previous examples. Apply Simpson’s 3/8 rule and use six segments for finding the distance.

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Thanks

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