10. Pengintegralan Numerik (lanjutan)
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Transcript of 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
Aturan Kuadrat Gauss
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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
Aturan Kuadrat Gauss
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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
Aturan Kuadrat Gauss
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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
Aturan Kuadrat Gauss
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Hence Two-Point Gaussian Quadrature Rule
231
22231
222211abab
fababab
fab
xfcxfc
dx)x(fb
a
Aturan Kuadrat Gauss
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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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Arguments and Weighing Factors for n-point Gauss Quadrature Formulas
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
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