10. Pengintegralan Numerik (lanjutan)

Pengintegralan Numerik (lanjutan)Pertemuan 10

Matakuliah : METODE NUMERIK ITahun : 2008

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Integrasi

IIntegrasi

b

a

dx)x(fI

Proses mencari luas di bawah suatu curva.

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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Aturan Kuadrat Gauss

Previously, the Trapezoidal Rule was developed by the method

of undetermined coefficients. The result of that development is

summarized below.

)b(fc)a(fcdx)x(fb

a21

)b(fab

)a(fab

22

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The four unknowns x1, x2, c1 and c2 are found by assuming that

the formula gives exact results for integrating a general third

order polynomial,

.xaxaxaa)x(f 33

2210

Hence

b

a

b

a

dxxaxaxaadx)x(f 33

2210

b

a

xa

xa

xaxa

432

4

3

3

2

2

10

432

44

3

33

2

22

10

aba

aba

abaaba


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It follows that

323

2222102

313

2121101 xaxaxaacxaxaxaacdx)x(f

b

a

Equating Equations the two previous two expressions yield

432

44

3

33

2

22

10

aba

aba

abaaba

323

2222102

313

2121101 xaxaxaacxaxaxaac

322

3113

222

211222111210 xcxcaxcxcaxcxcacca


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Since the constants a0, a1, a2, a3 are arbitrary

21 ccab 2211

22

2xcxc

ab

222

211

33

3xcxc

ab

3

223

11

44

4xcxc

ab


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The previous four simultaneous nonlinear Equations have only one acceptable solution,

21

abc

22

abc

23

1

21

ababx

23

1

22

ababx


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Hence Two-Point Gaussian Quadrature Rule

231

22231

222211abab

fababab

fab

xfcxfc

dx)x(fb

a


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)x(fc)x(fc)x(fcdx)x(fb

a332211

Dinamakan kuadrat Gauss dengan 3 titik

The coefficients c1, c2, and c3, and the functional arguments x1, x2, and x3

are calculated by assuming the formula gives exact expressions for

b

a

dxxaxaxaxaxaa 55

44

33

2210

General n-point rules would approximate the integral

)x(fc.......)x(fc)x(fcdx)x(f nn

b

a

2211

integrating a fifth order polynomial

Aturan Kuadrat Gauss dengan n titik

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Arguments and Weighing Factors for n-point Gauss Quadrature Formulas

In handbooks, coefficients and

Gauss Quadrature Rule are

1

1 1

n

iii )x(gcdx)x(g

as shown in Table 1.

Points WeightingFactors

FunctionArguments

2 c1 = 1.000000000c2 = 1.000000000

x1 = -0.577350269x2 = 0.577350269

3 c1 = 0.555555556c2 = 0.888888889c3 = 0.555555556

x1 = -0.774596669x2 = 0.000000000x3 = 0.774596669

4 c1 = 0.347854845c2 = 0.652145155c3 = 0.652145155c4 = 0.347854845

x1 = -0.861136312x2 = -0.339981044x3 = 0.861136312x4 = 0.339981044

arguments given for n-point

given for integrals

Table 1: Weighting factors c and function arguments x used in Gauss Quadrature Formulas.

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Points WeightingFactors

FunctionArguments

5 c1 = 0.236926885c2 = 0.478628670c3 = 0.568888889c4 = 0.478628670c5 = 0.236926885

x1 = -0.906179846x2 = -0.538469310x3 = 0.000000000x4 = 0.538469310x5 = 0.906179846

6 c1 = 0.171324492c2 = 0.360761573c3 = 0.467913935c4 = 0.467913935c5 = 0.360761573c6 = 0.171324492

x1 = -0.932469514x2 = -0.661209386x3 = -0.171324492x4 = 0.171324492x5 = 0.661209386x6 = 0.932469514

Table 1 (cont.) : Weighting factors c and function arguments x used in Gauss Quadrature Formulas.

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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


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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


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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


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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

10. Pengintegralan Numerik (lanjutan)

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