01 Numerical Integration
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Transcript of 01 Numerical Integration
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Numerical integration
Joko Wintoko
Matematika Teknik Kimia 2JTK/FT/UGM/2011
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Numerical integration
There are two main reasons to do numerical
integration:
analytical integration may be impossible or infeasible,
or
you may wish to integrate tabulated data rather than
known functions.
Methods:
Trapezoidal rule
Simpson's rule
Gauss quadrature
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Manual method
Manual method for determining integral by superimposing a grid on a graph of the
integrand. The boxes indicated in grey are counted.
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Newton-Cotes Integration
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Trapezoidal rule
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Trapezoidal rule
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Simpsons Rule
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Basis of Simpsons 1/3rd RuleTrapezoidal rule was based on approximating the
integrand by a first order polynomial, and then
integrating the polynomial in the interval of
integration. Simpsons 1/3rd rule is an extension of
Trapezoidal rule where the integrand is approximatedby a second order polynomial.
Hence
}!b
a
b
a
dx)x(fdx)x(fI2
Where is a second order polynomial.)x(f2
2
2102xaxaa)x(f !
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Basis of Simpsons 1/3rd Rule
}!b
a
b
a
dx)x(fdx)x(fI2
f(x)
f2(x)
2
2102xaxaa)x(f !
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Basis of Simpsons 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.
2
2102aaaaa)a(f)a(f !!
2
2102 2222
!
!
ba
a
ba
aa
ba
f
ba
f
2
2102babaa)b(f)b(f !!
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Basis of Simpsons 1/3rd Rule
Solving the previous equations for a0, a1 and a2 give
22
22
0 2
24
baba
)a(fb)a(abfba
abf)b(abf)b(fa
a
!
221
2
2433
24
baba
)b(bfba
bf)a(bf)b(afba
af)a(af
a
!
222
2
222
baba
)b(fba
f)a(f
a
!
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Basis of Simpsons 1/3rd Rule
Then
}b
a
dx)x(fI2
!b
adxxaxaa
2
210
b
a
xa
xaxa
!32
3
2
2
10
32
33
2
22
10
aba
aba)ab(a
!
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Basis of Simpsons 1/3rd Rule
Substituting values of a0, a1, a 2 give
!
)b(f
baf)a(f
abdx)x(f
b
a 2
4
62
Since for Simpsons 1/3rd Rule, the interval [a, b] isbroken
into 2 segments, the segment width
2
abh
!
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Basis of Simpsons 1/3rd Rule
! )b(f
baf)a(f
hdx)x(f
b
a 2
4
3
2
Hence
Because the above form has 1/3 in its formula,it is called Simpsons 1/3rd Rule.
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Gaussian Quadratures
Newton-Cotes Formulae
use evenly-spaced functional values
Gaussian Quadratures
select functional values at non-uniformly distributed
points to achieve higher accuracy
change of variables so that the interval of integration
is [-1,1]
Gauss-Legendre formulae
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Gaussian Quadratures
Trapezoidal rule Gaussian quadrature
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Gaussian QuadratureQuadrature onon [[--1, 1]1, 1]
Choose (c1, c2, x1, x2) such that the method
yields exact integral forf(x) = x0, x1, x2,x3
)x(fc)x(fc)x(fc)x(fcdx)x(f nn2211i1
1
n
1i
i !} ! .
)f(xc)f(xc
f(x)dx:2n
2211
1
1
!
!
x2x1-1 1
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Gaussian Quadrature onGaussian Quadrature on [[--1, 1]1, 1]
Exact integral forf = x0, x1, x2, x3
Four equations for four unknowns
)f(xc)f(xcf(x)dx:2 2211
1
1 !!
!
!
!!
!!!
!!!
!!!
!!!
31x
3
1x
1c1c
xcxc0dxxxf
xcxc
3
2dxxxf
xcxc0xdxxf
cc2dx11f
2
1
2
1
3
22
3
1
1
11
33
2
22
2
1
1
11
22
221
1
11
2
1
11
)
3
1(f)
3
1(fdx)x(fI
1
1!!
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Gaussian Quadrature onGaussian Quadrature on [[--1, 1]1, 1]
Choose (c1, c2, c3, x1, x2, x3) such that
the method yields exact integral for
f(x) = x0
, x1
, x2
,x3
,x4
, x5
)x(fc)x(fc)x(fcdx)x(f:3 3322111
1 !!
x3x1-1 1x2
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Gaussian Quadrature onGaussian Quadrature on [[--1, 1]1, 1]
5
33
5
22
5
11
1
1
55
4
33
4
22
4
11
1
1
44
3
33
3
22
3
11
1
1
33
2
33
2
22
2
11
1
1
22
332211
1
1
321
1
1
0
0
5
2
0
3
2
0
21
xcxcxcdxxxf
xcxcxcdxxxf
xcxcxcdxxxf
xcxcxcdxxxf
xcxcxcxdxxf
cccdxxf
!!!
!!!
!!!
!!!
!!!
!!!
!
!
!
!
!
!
5/3
0
5/3
9/5
9/8
9/5
3
2
1
3
2
1
x
x
x
c
c
c
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Gaussian Quadrature onGaussian Quadrature on [[--1, 1]1, 1]
Exact integral forf = x0, x1, x2, x3,x4, x5
)5
3(f
9
5)0(f
9
8)
5
3(f
9
5dx)x(fI
1
1!!
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Gaussian Quadrature onGaussian Quadrature on [a, b][a, b]
Coordinate transformation from [a,b] to [-1,1]
t2t1a b
!
!1
1
1
1
b
adx)x(gdx)
2
ab)(
2
abx
2
ab(fdt)t(f
!!
!!
!
bt1x
at1x2
abx
2
abt
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Example: Gaussian QuadratureExample: Gaussian Quadrature
Evaluate
Coordinate transformation
Two-point formula
33.34%)(543936.3477376279.3468167657324.9
e)3
44(e)
3
44()
3
1(f)
3
1(fdx)x(fI 3
44
3
441
1
!!!
!
!!
I
926477.5216dtteI4
0
t2 !!
!!!
!!
!
1
1
1
1
4x44
0
t2 dx)x(fdxe)4x4(dtteI
2dxdt;2x2
2
abx
2
abt
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Example: Gaussian QuadratureExample: Gaussian QuadratureThree-point formula
Four-point formula
4.79%)(106689.4967
)142689.8589(95)3926001.218(
98)221191545.2(
95
e)6.044(9
5e)4(
9
8e)6.044(
9
5
)6.0(f9
5)0(f
9
8)6.0(f
9
5dx)x(fI
6.0446.04
1
1
!!
!
!
!!
I
? A? A
%)37.0(54375.5197
)339981.0(f)339981.0(f652145.0
)861136.0(f)861136.0(f34785.0dx)x(fI1
1
!!
!!
I
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Contoh
Kerja yang dilakukan oleh sebuah proses termodinamis
dengan suhu tetap dapat dihitung dengan persamaan:
dengan W = kerja, p = tekanan, dan V = volume.
Hitunglah kerja yang dilakukan dengan data sebagai
berikut:
! pdVW
Tekan
an,
kPa
336 494.4 266.4 260.8 260.5 249.6 193.6 165.6
Volu
me,
m3
0.5 2 3 4 6 8 10 11
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Memakai aturan trapezoidal
P V (V Pi+1 + Pi Ii = (V/2*(Pi+1 + Pi)
336 0.5
294.4 2 1.5 630.4 472.8
266.4 3 1 560.8 280.4
260.8 4 1 527.2 263.6
260.5 6 2 521.3 521.3
249.6 8 2 510.1 510.1
193.6 10 2 443.2 443.2
165.6 11 1 359.2 179.6
I 2671
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Memakai aturan trapezoidal
Jadi kerja yang dilakukan adalah 2671 kJ
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Memakai aturan Simpson 1/3
V P (V
Faktor utk
P
(Simpson'
s rule)
Faktor * P Pi+1 + Pi Ii Metode
0.5 336Trapz
2 294.4 1.5 630.4 472.8
2 294.4 1 294.4
540.2667 Simps3 266.4 1 4 1065.6
4 260.8 1 1 260.8
4 260.8 1 260.8
1034.933 Simps6 260.5 2 4 10428 249.6 2 1 249.6
8 249.6
Trapz10 193.6 2 443.2 443.2
11 165.6 1 359.2 179.6
I 2670.8
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Memakai aturan Simpson 1/3
Jadi kerja yang dilakukan adalah 2670,8
kJ