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Numerical Integration Basic Numerical Integration

We want to find integration of functions of various forms of the equation known as the Newton Cotes integration formulas.

Trapezoidal Rule Simpson’s Rule

1/3 Rule 3/8 Rule

Midpoint Gaussian Quadrature

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Basic Numerical Integration• Weighted sum of function values

)x(fc)x(fc)x(fc

)x(fcdx)x(f

nn1100

i

n

0ii

b

a

x0 x1 xnxn-1x

f(x)

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2

4

6

8

10

12

3 5 7 9 11 13 15

Numerical IntegrationIdea is to do integral in small parts, like the way you first learned integration - a summation

Numerical methods just try to make it faster and more accurate

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Numerical Integration Newton-Cotes Closed Formulae -- Use

both end points Trapezoidal Rule : Linear Simpson’s 1/3-Rule : Quadratic Simpson’s 3/8-Rule : Cubic Boole’s Rule : Fourth-order

Newton-Cotes Open Formulae -- Use only interior points midpoint rule

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Trapezoid Rule Straight-line approximation

)x(f)x(f2

h

)x(fc)x(fc)x(fcdx)x(f

10

1100i

1

0ii

b

a

x0 x1x

f(x)

L(x)

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Trapezoid Rule Lagrange interpolation

010 1

0 1 1 0

0 1

( ) ( ) ( )

dx a x , b x , , d ;

h

0( ) (1 ) ( ) ( ) ( )

1

x xx xL x f x f x

x x x x

x alet h b a

b a

x aL f a f b

x b

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Trapezoid Rule

Integrate to obtain the rule

1

0

1 1

0 0

1 12 2

0 0

( ) ( ) ( )

( ) (1 ) ( )

( ) ( ) ( ) ( ) ( )2 2 2

b b

a af x dx L x dx h L d

f a h d f b h d

hf a h f b h f a f b

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Example:Trapezoid RuleEvaluate the integral Exact solution

Trapezoidal Rule

926477.5216)1x2(e4

1

e4

1e

2

xdxxe

1

0

x2

4

0

x2x24

0

x2

dxxe4

0

x2

%12.357926.5216

66.23847926.5216

66.23847)e40(2)4(f)0(f2

04dxxeI 84

0

x2

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Simpson’s 1/3-RuleApproximate the function by a parabola

)x(f)x(f4)x(f3

h

)x(fc)x(fc)x(fc)x(fcdx)x(f

210

221100i

2

0ii

b

a

x0 x1x

f(x)

x2h h

L(x)

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Simpson’s 1/3-Rule

1 xx

0 xx

1 xxh

dxd ,

h

xx ,

2

abh

2

ba x ,bx ,ax let

)x(f)xx)(xx(

)xx)(xx(

)x(f)xx)(xx(

)xx)(xx( )x(f

)xx)(xx(

)xx)(xx()x(L

2

1

0

1

120

21202

10

12101

200

2010

21

)x(f2

)1()x(f)1()x(f

2

)1()(L 21

20

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Simpson’s 1/3-Rule

1

1

23

2

1

1

3

1

1

1

23

0

1

12

1

0

21

1

10

1

1

b

a

)2

ξ

3

ξ(

2

h)f(x

)3

ξ(ξ)hf(x)

2

ξ

3

ξ(

2

h)f(x

)dξ1ξ(ξ2

h)f(x)dξξ1()hf(x

)dξ1ξ(ξ2

h)f(xdξ)(Lhf(x)dx

)f(x)4f(x)f(x3

hf(x)dx 210

b

a

Integrate the Lagrange interpolation

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Simpson’s 3/8-RuleApproximate by a cubic polynomial

)x(f)x(f3)x(f3)x(f8

h3

)f(xc)f(xc)f(xc)f(xc)x(fcdx)x(f

3210

33221100i

3

0ii

b

a

x0 x1x

f(x)

x2h h

L(x)

x3h

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Simpson’s 3/8-Rule

)x(f)xx)(xx)(xx(

)xx)(xx)(xx(

)x(f)xx)(xx)(xx(

)xx)(xx)(xx(

)x(f)xx)(xx)(xx(

)xx)(xx)(xx(

)x(f)xx)(xx)(xx(

)xx)(xx)(xx()x(L

3231303

210

2321202

310

1312101

320

0302010

321

)x(f)x(f3)x(f3)x(f8

h33

a-bh ; L(x)dxf(x)dx

3210

b

a

b

a

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Example: Simpson’s RulesEvaluate the integral Simpson’s 1/3-Rule

Simpson’s 3/8-Rule

dxxe4

0

x2

4 2

0

4 8

(0) 4 (2) (4)3

20 4(2 ) 4 8240.411

35216.926 8240.411

57.96%5216.926

x hI xe dx f f f

e e

4 2

0

3 4 8(0) 3 ( ) 3 ( ) (4)

8 3 3

3(4/3)0 3(19.18922) 3(552.33933) 11923.832 6819.209

85216.926 6819.209

30.71%5216.926

x hI xe dx f f f f

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Midpoint RuleNewton-Cotes Open Formula

)(f24

)ab()

2

ba(f)ab(

)x(f)ab(dx)x(f3

m

b

a

a b x

f(x)

xm

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Two-point Newton-Cotes Open Formula

Approximate by a straight line

)(f108

)ab()x(f)x(f

2

abdx)x(f

3

21

b

a

x0 x1x

f(x)

x2h h x3h

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Three-point Newton-Cotes Open Formula

Approximate by a parabola

)(f23040

)ab(7

)x(f2)x(f)x(f23

abdx)x(f

5

321

b

a

x0 x1x

f(x)

x2h h x3h h x4

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Better Numerical Integration

Composite integration Composite Trapezoidal Rule Composite Simpson’s Rule

Richardson Extrapolation Romberg integration

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Apply trapezoid rule to multiple segments over integration limits

0

1

2

3

4

5

6

7

3 5 7 9 11 13 15

Two segments

0

1

2

3

4

5

6

7

3 5 7 9 11 13 15

Four segments

0

1

2

3

4

5

6

7

3 5 7 9 11 13 15

Many segments

0

1

2

3

4

5

6

7

3 5 7 9 11 13 15

Three segments

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Composite Trapezoid Rule

)x(f)x(f2)2f(x)f(x2)f(x2

h

)f(x)f(x2

h)f(x)f(x

2

h)f(x)f(x

2

h

f(x)dxf(x)dxf(x)dxf(x)dx

n1ni10

n1n2110

x

x

x

x

x

x

b

a

n

1n

2

1

1

0

x0 x1x

f(x)

x2h h x3h h x4

n

abh

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Composite Trapezoid RuleEvaluate the integral dxxeI

4

0

x2

%66.2 95.5355

)4(f)75.3(f2)5.3(f2

)5.0(f2)25.0(f2)0(f2

hI25.0h,16n

%50.10 76.5764)4(f)5.3(f2 )3(f2)5.2(f2)2(f2)5.1(f2

)1(f2)5.0(f2)0(f2

hI5.0h,8n

%71.39 79.7288)4(f)3(f2

)2(f2)1(f2)0(f2

hI1h,4n

%75.132 23.12142)4(f)2(f2)0(f2

hI2h,2n

%12.357 66.23847)4(f)0(f2

hI4h,1n

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Composite Trapezoid Example

2

1 1

1dx

x

x f(x)1.00 0.50001.25 0.44441.50 0.40001.75 0.36362.00 0.3333

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Composite Trapezoid Rule with Unequal Segments

Evaluate the integral h1 = 2, h2 = 1, h3 = 0.5, h4 = 0.5

dxxeI4

0

x2

%45.14 58.5971 45.3 2

0.5

5.33e 2

0.532

2

120

2

2

)4()5.3(2

)5.3()3(2

)3()2(2

)2()0(2

)()()()(

87

76644

43

21

4

5.3

5.3

3

3

2

2

0

ee

eeee

ffh

ffh

ffh

ffh

dxxfdxxfdxxfdxxfI

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Composite Simpson’s Rule

x0 x2x

f(x)

x4h h xn-2h xn

n

abh

…...

Piecewise Quadratic approximations

hx3x1 xn-1

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)x(f)x(f4)x(f2

)4f(x)x(f2)f(x4

)2f(x)f(x4)2f(x)f(x4)f(x3

h

)f(x)4f(x)f(x3

h

)f(x)f(x4)f(x3

h)f(x)f(x4)f(x

3

h

f(x)dxf(x)dxf(x)dxf(x)dx

n1n2n

12ii21-2i

43210

n1n2n

432210

x

x

x

x

x

x

b

a

n

2n

4

2

2

0

Composite Simpson’s RuleMultiple applications of Simpson’s rule

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Composite Simpson’s RuleEvaluate the integral n = 2, h = 2

n = 4, h = 1

dxxeI4

0

x2

%70.8 975.5670

e4)e3(4)e2(2)e(403

1

)4(f)3(f4)2(f2)1(f4)0(f3

hI

8642

%96.57 411.8240e4)e2(403

2

)4(f)2(f4)0(f3

hI

84

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Composite Simpson’s Example

2

1 1

1dx

x

x f(x)1.00 0.50001.25 0.44441.50 0.40001.75 0.36362.00 0.3333

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Composite Simpson’s Rule with Unequal Segments

Evaluate the integral h1 = 1.5, h2 = 0.5

dxxeI4

0

x2

%76.3 23.5413

4)5.3(433

5.03)5.1(40

3

5.1

)4(2)5.3(4)3(3

)3(2)5.1(4)0(3

)()(

87663

2

1

4

3

3

0

eeeee

fffh

fffh

dxxfdxxfI

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Richardson ExtrapolationUse trapezoidal rule as an example

subintervals: n = 2j = 1, 2, 4, 8, 16, ….

j2

1jjn1n10

b

ahc)x(f)x(f2)f(x2)f(x

2

hf(x)dx

)()()()(

)()()()(

)()()()()(

)()()(

)()(

bfxf2xf2af2

hI2j

bfxf2xf2af16

hI83

bfxf2xf2xf2af8

hI42

bfxf2af4

hI21

bfaf2

hI10

Formulanj

1n1jjj

713

3212

11

0

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Richardson ExtrapolationFor trapezoidal rule

kth level of extrapolation

)h(B)2

h(B16

15

1)h(C

)2

h(b)

2

h(BA

hb)h(BA

hb)h(B h4

c)h(A)

2

h(A4

3

1A

)2

h(c)

2

h(c)

2

h(AA

hchc)h(AA

hc)h(Adx)x(fA

42

42

42

42

42

21

42

21

21

b

a

14

)h(Ch/2)(C4)h(D

k

k

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255

II256

63

II64

15

II16

3

II4I16h

II8h

III4h

IIII2h

IIIIIh

hOhOhOhOhO

4k3k2k1k0k

sBoolesSimpsonTrapezoid

3j31j2j21j1j11j0j01j

04

1303

221202

31211101

4030201000

108642

,,,,,,,,

,

,,

,,,

,,,,

,,,,,

/

/

/

/

)()()()()(

''

3, 2,1,k ;14

II4I

k

k,jk,1jk

k,j

Romberg IntegrationAccelerated Trapezoid Rule

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Romberg IntegrationAccelerated Trapezoid Rule

926477.5216dxxeI4

0

x2

%00050.0%00168.0%0053.0%0527.0%66.2

95.535525.0h

68.521976.57645.0h

20.521775.525679.72881h

01.521714.522998.56702.121422h

95.521684.522468.549941.82407.238474h

)h(O)h(O)h(O)h(O)h(O

4k3k2k1k0k

s'Booles'SimpsonTrapezoid

108642

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Romberg Integration Example

2

1 1

1dx

x

x f(x)1.00 0.50001.25 0.44441.50 0.40001.75 0.36362.00 0.3333

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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 Quadrature on [-1, 1]

Choose (c1, c2, x1, x2) such that the method yields “exact integral” for f(x) = x0, x1, x2, x3

)x(fc)x(fc)x(fc)x(fcdx)x(f nn2211i

1

1

n

1ii

)f(xc)f(xc

f(x)dx :2n

2211

1

1

x2x1-1 1

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Gaussian Quadrature on [-1, 1]

Exact integral for f = x0, x1, x2, x3

Four equations for four unknowns

)f(xc)f(xcf(x)dx :2n 2211

1

1

3

1x

3

1x

1c1c

xcxc0dxx xf

xcxc3

2dxx xf

xcxc0xdx xf

cc2dx1 1f

2

1

2

1

322

31

1

1 133

222

21

1

1 122

221

1

1 1

2

1

1 1

)3

1(f)

3

1(fdx)x(fI

1

1

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Gaussian Quadrature on [-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 :3n 332211

1

1

x3x1-1 1x2

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Gaussian Quadrature on [-1, 1]

533

522

511

1

1

55

433

422

411

1

1

44

333

322

311

1

1

33

233

222

211

1

1

22

332211

1

1

321

1

1

0

5

2

0

3

2

0

21

xcxcxcdxxxf

xcxcxcdxxxf

xcxcxcdxxxf

xcxcxcdxxxf

xcxcxcxdxxf

cccxdxf

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 on [-1, 1]

Exact integral for f = 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 on [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

bt 1xat1x2

abx

2

abt

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Example: Gaussian QuadratureEvaluate

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

926477.5216dtteI4

0

t2

1

1

1

1

4x44

0

t2 dx)x(fdxe)4x4(dtteI

2dxdt ;2x22

abx

2

abt

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Example: Gaussian QuadratureThree-point formula

Four-point formula

4.79%)( 106689.4967

)142689.8589(9

5)3926001.218(

9

8)221191545.2(

9

5

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

%)37.0( 54375.5197 )339981.0(f)339981.0(f652145.0

)861136.0(f)861136.0(f34785.0dx)x(fI1

1