Pengintegralan Numerik (lanjutan)Pertemuan 8

Matakuliah : METODE NUMERIK ITahun : 2008

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So if the table is given for

1

1

dx)x(g integrals, how does one solve

b

a

dx)x(f ? The answer lies in that any integral with limits of b,a

can be converted into an integral with limits 11, Let

cmtx

If then,ax 1t

,bx If then 1tSuch that:

2

abm

Arguments and Weighing Factors for n-point Gauss Quadrature Formulas

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Arguments and Weighing Factors for n-point Gauss Quadrature Formulas

2

abc

Then Hence

22

abt

abx

dt

abdx

2

Substituting our values of x, and dx into the integral gives us

1

1 222dx

ababx

abfdx)x(f

b

a

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Contoh 1

For an integral

derive the one-point Gaussian Quadrature

Rule.

,dx)x(fb

a

Solution

The one-point Gaussian Quadrature Rule is

11 xfcdx)x(fb

a

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Solution

Assuming the formula gives exact values for integrals

,dx

1

1

1

,xdx

1

1

and

11 cabdxb

a

11

22

2xc

abxdx

b

a

Since ,abc 1 the other equation becomes

2

22

1

abx)ab(

21

abx

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Solution (cont.)

Therefore, one-point Gauss Quadrature Rule can be expressed as

2

abf)ab(dx)x(f

b

a

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Contoh 2

Use two-point Gauss Quadrature Rule to approximate the distance

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

30

8

892100140000

1400002000 dtt.

tlnx

Find the true error, for part (a).

Also, find the absolute relative true error, for part (a).a

a)

b)

tE

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Solution

First, change the limits of integration from [8,30] to [-1,1]

by previous relations as follows

1

1

30

8 2

830

2

830

2

830dxxfdt)t(f

1

1

191111 dxxf

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Solution (cont)

.

Next, get weighting factors and function argument values from Table 1

for the two point rule,.

00000000011 .c

57735026901 .x

00000000012 .c

57735026902 .x

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Solution (cont.)

Now we can use the Gauss Quadrature formula

191111191111191111 2211

1

1

xfcxfcdxxf

1957735030111119577350301111 ).(f).(f

).(f).(f 350852511649151211

).().( 481170811831729611

m.4411058

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Solution (cont)

since

).(.).(

ln).(f 64915128964915122100140000

14000020006491512

8317296.

).(.).(

ln).(f 35085258935085252100140000

14000020003508525

4811708.

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Solution (cont)

The absolute relative true error, t, is (Exact value = 11061.34m)

%.

..t 100

3411061

44110583411061

%.02620

c)

The true error, , isb) tE

ValueeApproximatValueTrueEt

44.1105834.11061

m9000.2

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Integral Romberg

Romberg Integration is an extrapolation formula of

the Trapezoidal Rule for integration. It provides a better

approximation of the integral by reducing the True Error.

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Integral RombergRomberg integration is same as Richardson’s

extrapolation formula as given previously. However,

Romberg used a recursive algorithm for the

extrapolation. Recall 3

22

nnn

IIITV

This can alternately be written as

3

222

nnnRn

IIII

14 122

2

nn

n

III

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Determine another integral value with further halving the step size (doubling the number of segments),

3

2444

nnnRn

IIII

It follows from the two previous expressions that the true value TV can be written as

15

244

RnRnRn

IIITV

14 1324

4

RnRn

n

III

Integral Romberg

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214 1

11111

k,

IIII

k

j,kj,kj,kj,k

The index k represents the order of extrapolation. k=1 represents the values obtained from the regular

Trapezoidal rule, k=2 represents values obtained using the true estimate as O(h2). The index j represents the more and less accurate estimate of the integral.

A general expression for Romberg integration can be written as

Integral Romberg

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Contoh 2

A company advertises that every roll of toilet paper has at least 250 sheets. The probability that there are 250 or more sheets in the toilet paper is given by

250

)2.252(3881.0 2

3515.0)250( dyeyP y

Approximating the above integral as

270

250

)2.252(3881.0 2

3515.0)250( dyeyP y

Use Romberg’s rule to find the probability. Use the 1, 2, 4, and 8-segment Trapezoidal rule results as given.

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Solution

From Table 1, the needed values from original Trapezoidal rule are

where the above four values correspond to using 1, 2, 4 and 8 segment Trapezoidal rule, respectively.

53721.01,1 I 26861.02,1 I

21814.03,1 I 95767.04,1 I

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Penyelesaian (lanjutan)

To get the first order extrapolation values,

31121

2112,,

,,

IIII

Similarly,3

53721.026861.026861.0

17908.0

32,13,1

3,12,2

IIII

3

26861.021814.021814.0

20132.0

33,14,1

4,13,2

IIII

3

21814.095767.095767.0

2042.1

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For the second order extrapolation values,

Similarly,

151,22,2

2,21,3

IIII

15

17908.020132.020132.0

20280.0

152,23,2

3,22,3

IIII

15

20132.02042.12042.1

2711.1

Penyelesaian (lanjutan)

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For the third order extrapolation values,

Table 3 shows these increased correct values in a tree graph.

631,32,3

2,31,4

IIII

63

20280.02711.12711.1

2881.1

Penyelesaian (lanjutan)

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Solution (cont.)

Table 3: Improved estimates of the integral value using Romberg Integration

1-segment

2-segment

4-segment

8-segment

0.53721

0.26861

0.21814

0.95767

0.17908

0.20132

1.2042

0.20280

1.2711

1.2881

1st Order 2nd Order 3rd Order

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Soal Latihan

Hitunglah

Menggunakan a. Aturan Trapezoidal pada segmen (n) = 2,

4, 6,dan 8 b. Kuadratur Gauss untuk 2 dan 4 titikc. Integral Romberg (gunakan hasil a)

1

021 x

dx