Simpson’s 1/3 rd Rule of Integration

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

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Simpson’s 1/3rd Rule

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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

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Basis of Simpson’s 1/3rd Rule

Choose

)),a(f,a( ,ba

f,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

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

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Basis of Simpson’s 1/3rd Rule

)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)

10

0

dt)t(fx

)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.3411061

True Error

72110653411061 ..Et

m.384

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Solution (cont)

a)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

64 234

42

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

...)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

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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

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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) 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).a

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Solution

Using 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

11

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%