Numerical Integration Lecture

7/30/2019 Numerical Integration Lecture

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1

Advanced Computational MethodsNumerical IntegrationME753


2/59

2

Introduction to NumericalIntegration

Definitions Upper and Lower Sums Trapezoid Method (Newton-Cotes Methods)Simpsons Rules Romberg Method Gauss Quadrature Examples


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3

Integration

Indefinite Integrals

Indefinite Integrals of afunction are functionsthat differ from eachother by a constant.

c

xdxx

2

2

Definite Integrals

Definite Integrals arenumbers.

2

1

2

1

0

1

0

2

xxdx


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4

Fundamental Theorem of Calculus

:forsolutionformclosedNo

:fortiveantiderivanoisThere

)f(x)(x)'F(i.e.,foftiveantiderivais

,intervalanoncontinuousisIf

2

2

b

a

x

x

b

a

dxe

e

F(a)F(b)f(x)dx

F

[a,b]f


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5

The Area Under the Curve

b

af(x)dxArea

One interpretation of the definite integral is:

Integral = area under the curve

a b

f(x)


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6

Upper and Lower Sums

a b

f(x)

3210 xxxx

iin

i

i

ii

n

i

i

iii

iii

n

xxMPfUsumUpper

xxmPfLsumLower

xxxxfM

xxxxfmDefine

bxxxxaPPartition

1

1

0

1

1

0

1

1

210

),(

),(

:)(max

:)(min

...

The interval is divided into subintervals.


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7


a b

f(x)

3210 xxxx

2

2integraltheofEstimate

),(

),(

1

1

0

1

1

0

LUError

UL

xxMPfUsumUpper

xxmPfLsumLower

ii

n

i

i

ii

n

i

i


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8

Example

14

3

2

1

4

103,2,1,0

4

1

1,

16

9,

4

1,

16

1

16

9,

4

1,

16

1,0

)intervalsequalfour(4

1,4

3,

4

2,

4

1,0:

1

3210

3210

1

0

2

iforxx

MMMM

mmmm

n

PPartition

dxx

ii


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9

Example

14

3

2

1

4

10

8

1

64

14

64

30

2

1

32

11

64

14

64

30

2

1

integraltheofEstimate

64

301

16

9

4

1

16

1

4

1),(

),(

64

14

16

9

4

1

16

10

4

1),(

),(

1

1

0

1

1

0

Error

PfU

xxMPfUsumUpper

PfL

xxmPfLsumLower

ii

n

i

i

ii

n

i

i


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10


Estimates based on Upper and LowerSums are easy to obtain for monotonicfunctions (always increasing or always

decreasing). For non-monotonic functions, finding

maximum and minimum of the function

can be difficult and other methods can bemore attractive.


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11

Newton-Cotes Methods

In Newton-Cote Methods, the functionis approximated by a polynomial oforder n.

Computing the integral of a polynomialis easy.

1

)(...

2

)()()(

...)(

1122

10

10

n

aba

abaabadxxf

dxxaxaadxxf

nn

n

b

a

b

a

n

n

b

a


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12

Newton-Cotes Methods

Trapezoid Method (First Order Polynomials are used)

Simpson 1/3 Rule (Second Order Polynomials are used)

b

a

b

a

b

a

b

a

dxxaxaadxxf

dxxaadxxf

2210

10

)(

)(


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13

Trapezoid Method

Derivation-One Interval Multiple Application Rule Estimating the Error Recursive Trapezoid Method


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14

Trapezoid Method

)()()(

)( axab

afbfaf

f(x)

ba

2

)()(

2

)()(

)()()(

)()()(

)(

)(

2

afbfab

x

ab

afbf

xab

afbfaaf

dxaxab

afbfafI

dxxfI

b

a

b

a

b

a

b

a


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15

Trapezoid MethodDerivation-One Interval

2

)()(

)()(2

)()()()()(

2

)()()()()(

)()()()()(

)()()()()(

22

2

afbfab

abab

afbfabab

afbfaaf

x

ab

afbfx

ab

afbfaaf

dxxab

afbf

ab

afbfaafI

dxaxab

afbfafdxxfI

b

a

b

a

b

a

b

a

b

a


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16

Trapezoid Method

)(bf

f(x)

ba

)()(2 bfafab

Area

)(af


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17

Trapezoid MethodMultiple Application Rule

a b

f(x)

3210 xxxx

strapezoidtheof

areastheofsum)(

...segmentsintodpartitione

isb][a,intervalThe

210

b

a

n

dxxf

bxxxxan

1212

2

)()(xx

xfxfArea

x


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18

Trapezoid MethodGeneral Formula and Special Case

)()(21

)(

...

equal)ynecessaril(notsegmentsintodividedisintervaltheIf

11

1

0

210

iiii

n

i

b

a

n

xfxfxxdxxf

bxxxxa

n

1

1

0

1

)()()(2

1)(

allfor

points)basespacedEqualiy(CaseSpecial

n

i

in

b

a

ii

xfxfxfhdxxf

ihxx


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19

Example

Given a tabulated

values of the velocity of

an object.

Obtain an estimate of

the distance traveled in

the interval [0,3].

Time (s) 0.0 1.0 2.0 3.0

Velocity (m/s) 0.0 10 12 14

Distance = integral of the velocity

3

0)(Distance dttV


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20

Example 1

Time (s) 0.0 1.0 2.0 3.0

Velocity

(m/s)

0.0 10 12 14

29)140(2

112)(101Distance

)()(2

1

)(

1

0

1

1

1

n

n

ii

ii

xfxfxfhT

xxh

MethodTrapezoid

3,2,1,0arepointsBaselssubinterva3into

dividedisintervalThe


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21

Error in estimating the integralTheorem

)(''max12

)(12

then)(eapproximat

tousedisMethodTrapezoidIf:Theorem

)(widthintervalsEqual

oncontinuousis)('':Assumption

],[

2

''2

a

xfhab

Error

[a,b]wherefhab

Error

dxxf

h

[a,b]xf

bax

b


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22

Estimating the ErrorFor Trapezoid Method

?accuracydigitdecimal5to

)sin(computetoneeded

areintervalsspacedequallymanyHow

0

dxx


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23

Example

intervals719

00437.0

)(

00437.0106

102

1

121)(''

)sin()('');cos()(';0;

)(''max12

102

1error,)sin(

52

52

],[

2

5

0

h

abn

hh

hErrorxf

xxfxxfab

xfh

ab

Error

thatsohf inddxx

bax


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24

Example

)()(2

1)(),(

,allfor:

)()(2

1),(

)(:computetomethodTrapezoidUse

0

1

1

1

11

1

0

3

1

n

n

i

i

ii

iiii

n

i

xfxfxfhPfT

ixxhCaseSpecial

xfxfxxPfTTrapezoid

dxxf

x 1.0 1.5 2.0 2.5 3.0

f(x) 2.1 3.2 3.4 2.8 2.7


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25

Example

9.5

7.21.2218.24.32.35.0

)()(2

1)()( 0

1

1

3

1

n

n

i

i xfxfxfhdxxf

x 1.0 1.5 2.0 2.5 3.0

f(x) 2.1 3.2 3.4 2.8 2.7


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ExampleUse the trapezoidal rule to estimate

8.0

0

5432 400900675200252.0 dxxxxxx

Solution

%5.89

1728.0

2

232.02.08.0

2

)8.0()0()08.0(

2

)()()(

t

ff

bfafabI


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ExampleUse the multiple-application trapezoidal rule for n = 2 toestimate

8.0

0

5432 400900675200252.0 dxxxxxx

Solution

%9.34

0688.1

4

232.0)456.2(22.08.0

4

)8.0()4.0(2)0()08.0(

)2(2

)()(2)()( 210

t

fff

xfxfxfabI


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Computer Algorithms for the

Trapezoidal Rule


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30

Recursive Trapezoid Method

f(x)

haa

)()(2

)0,0(R

:intervaloneonbasedEstimate

bfafab

abh


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f(x)

hahaa 2

)()0,0(21

)0,1(

)()(2

1)(

2)0,1(R

2

:intervals2onbasedEstimate

hafhRR

bfafhafab

abh

Based on previous estimate

Based on new point


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32


f(x)

)3()()0,1(2

1)0,2(

)()(2

1

)3()2()(

4

)0,2(R

4

hafhafhRR

bfaf

hafhafhafab

abh

hahaa 42 Based on previous estimate

Based on new points


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33

Recursive Trapezoid MethodFormulas

n

k

abh

hkafhnRnR

bfafab

n

2

)12()0,1(2

1)0,(

)()(2

)0,0(R

)1(2

1


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34


)1(

2

2

1

2

1

3

2

12

1

1

)12()0,1(2

1)0,(,

2

..................

)12()0,2(

2

1)0,3(,

2

)12()0,1(2

1)0,2(,

2

)12()0,0(2

1)0,1(,

2

)()(2

)0,0(R,

n

kn

k

k

k

hkafhnRnRab

h

hkafhRRab

h

hkafhRRab

h

hkafhRRab

h

bfafab

abh


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Example on Recursive Trapezoid

35

errortheestimatethenR(3,0)computingby)sin(

:estimatetomethodTrapezoidRecursiveUse

2/

0

dxx

n h R(n,0)

0 (b-a)=/2 (/4)[sin(0) + sin(/2)]=0.785398

1 (b-a)/2=/4 R(0,0)/2 + (/4) sin(/4) = 0.948059

2 (b-a)/4=/8 R(1,0)/2 + (/8)[sin(/8)+sin(3/8)] = 0.987116

3 (b-a)/8=/16 R(2,0)/2 + (/16)[sin(/16)+sin(3/16)+sin(5/16)+sin(7/16)] = 0.996785

Estimated Error = |R(3,0) R(2,0)| = 0.009669


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Advantages of Recursive Trapezoid

Recursive Trapezoid:

Gives the same answer as the standardTrapezoid method.

Makes use of the available information toreduce the computation time.

Useful if the number of iterations is not

known in advance.


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Simpsons Rules

Simpsons 1/3 Rule

Simpsons 3/8 Rule


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SIMPSONS RULES

More accurate estimate of an integral is obtained if a high-order polynomial is used to connect the points.

The formulas that result from taking the integrals undersuch polynomials are called Simpsons rules.


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Simpsons 1/3 Rule

This rule results when a second-order interpolatingpolynomial is used.

2

where)()(4)(

3

)())((

))((

)())((

))(()(

))((

))((

,andLet

)()(

210

2

1202

10

1

2101

200

2010

21

20

2

2

0

abhxfxfxf

hI

dxxfxxxx

xxxx

xfxxxx

xxxxxf

xxxx

xxxxI

xbxa

dxxfdxxfI

x

x

b

a

b

a

After integration,


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Simpsons 3/8 Rule

This rule results when a third-order interpolatingpolynomial is used.

3

where)()(3)(3)(8

3

,yieldsThis

)()(

3210

3

abhxfxfxfxf

hI

dxxfdxxfI

b

a

b

a


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41

Romberg Method

Motivation

Derivation of Romberg Method

Romberg Method

Example

When to stop?


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42

Motivation for Romberg Method

Trapezoid formula with a sub-interval h gives anerror of the order O(h2).

We can combine two Trapezoid estimates with

intervals h and h/2 to get a better estimate.


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43

Romberg Method

First column is obtained

using Trapezoid Method

dxf(x)ofionapproximattheimprovetocombinedare

...h/8h/4,h/2,h,sizeofintervalsmethodTrapezoidusingEstimates

b

a

R(0,0)

R(1,0) R(1,1)

R(2,0) R(2,1) R(2,2)

R(3,0) R(3,1) R(3,2) R(3,3)The other elements

are obtained using

the Romberg Method


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First ColumnRecursive Trapezoid Method

n

k

abh

hkafhnRnR

bfafab

n

2

)12()0,1(2

1)0,(

)()(2

)0,0(R

)1(2

1


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Derivation of Romberg Method

.. .)0,1()0,(43

1)(

2*41

)2(.. .64

1

16

1

4

1)0,()(

by R(n,0)obtainedisestimateaccurateMore

)1(...)0,1()(

2methodTrapezoid)()0,1()(

66

44

66

44

22

66

44

22

12

hbhbnRnRdxxf

giveseqeq

eqhahahanRdxxf

eqhahahanRdxxf

abhwithhOnRdxxf

b

a

b

a

b

a

n

b

a


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46

Romberg Method

1,1)1,1()1,(414

1),(

)12()0,1(2

1)0,(

,

2

)()(2

)0,0(R

)1(2

1

mnmnRmnRmnR

hkafhnRnR

abh

bfafab

m

m

k

n

n

R(0,0)

R(1,0) R(1,1)

R(2,0) R(2,1) R(2,2)

R(3,0) R(3,1) R(3,2) R(3,3)


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47

Property of Romberg Method

)(),()(

Theorem

22 m

b

a

hOmnRdxxf

R(0,0)

R(1,0) R(1,1)

R(2,0) R(2,1) R(2,2)

R(3,0) R(3,1) R(3,2) R(3,3)

)()()()( 8642 hOhOhOhOError Level


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Example

3

1

2

1

8

34

3

1)0,0()0,1(4

14

1)1,1(

1,1)1,1()1,(414

1),(

8

3

4

1

2

1

2

1

2

1))(()0,0(

2

1)0,1(,

2

1

5.010

2

1)()(

2

)0,0(,1

Compute

1

1

0

2

RRR

mnformnRmnRmnR

hafhRRh

bfafab

Rh

dxx

mm

0.5

3/8 1/3


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Example (cont.)

3

1

3

1

3

16

15

1)1,1()1,2(4

14

1)2,2(

3

1

8

3

32

114

3

1)0,1()0,2(4

3

1)1,2(

)1,1()1,(414

1),(

32

11

16

9

16

1

4

1

8

3

2

1))3()(()0,1(

2

1)0,2(,

4

1

2

2

RRR

RRR

mnRmnRmnR

hafhafhRRh

m

m

0.5

3/8 1/3

11/32 1/3 1/3


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50

When do we stop?

at R(4,4)STOPexample,for

steps,ofnumbergivenaAfteror

)1,(),( nnRnnR

ifSTOP


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Gauss Quadrature

Motivation General integration formula


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52

Motivation

nandih

nihcwhere

xfcdxxf

xfxfxfhdxxf

i

n

i

ii

b

a

n

i

ni

b

a

05.0

1,...,2,1

)()(

:asexpressedbecanIt

)()(2

1)()(

:MethodTrapezoid

0

1

1

0


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53

General Integration Formula

integral?theofionapproximatgoodagivesformulaethat thsoselectwedoHow

:Problem

::

)()(

0

ii

ii

n

i

ii

b

a

xandc

NodesxWeightsc

xfcdxxf


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Lagrange Interpolation

b

aii

n

i

ii

b

a

b

ai

n

i

i

b

an

b

a

n

n

b

an

b

a

dxxcwherexfcdxxf

dxxfxdxxPdxxf

xxx

xPwhere

dxxPdxxf

)()()(

)()()()(

,...,,:nodesat the

f(x)esinterpolatthatpolynomialais)(

)()(

0

0

10


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Example

Determine the Gauss Quadrature Formula of

If the nodes are given as (-1, 0 , 1)

Solution: First we need to find l0(x), l1(x), l2(x)

Then compute:

55

2

2

)( dxxf

2

2

22

2

2

11

2

2

00 )(,)(,)( dxxlcdxxlcdxxlc


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Solution

56

)1(3

8)0(

3

4)1(

3

8)(forFormulaQuadratureGaussThe

3

8

2

)1(,

3

4)1)(1(,

3

8

2

)1(

2

)1(

)12)(02(

)1)(0()(

)1)(1()21)(01(

)2)(0()(

2)1(

)20)(10()2)(1()(

2

2-

2

2

2

2

2

1

2

2

0

2

1

0

fffdxxf

dxxx

cdxxxcdxxx

c

xx

xxxx

xxxxxl

xxxxxx

xxxxxl

xxxxxx

xxxxxl


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Using the Gauss Quadrature Formula

57

answerexactsametheiswhich,3

16)1(

3

8)0(

3

4)1(

3

8

)1(3

8)0(

3

4)1(

3

8FormulaQuadratureGaussThe

3

16)(foreexact valuThe

)(Let:1Case

222

2

2-

2

2

2-

2

fff

dxxdxxf

xxf


58/59

Using the Gauss Quadrature Formula

58

answerexactsametheiswhich,0)1(3

8)0(

3

4)1(

3

8

)1(3

8)0(

3

4)1(

3

8FormulaQuadratureGaussThe

0)(foreexact valuThe

)(Let:2Case

333

2

2-

3

2

2-

3

fff

dxxdxxf

xxf


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59

Improper Integrals

1

022

12

1

1 2a

1

111:

function.newtheonmethodapply theand

)0assuming(,11)(

followingthelikemationa transforUse

)orislimitstheof(oneintegralsimpropereapproximat

todirectlyusedbecannotearlierdiscussedMethods

dt

t

tdx

xExample

abdtt

ft

dxxf a

b

b

Numerical Integration Lecture

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Transcript of Numerical Integration Lecture