01 Numerical Integration

8/6/2019 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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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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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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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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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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! )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


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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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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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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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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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

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01 Numerical Integration