CISE301_Topic7KFUPM1 SE301: Numerical Methods Topic 7 Numerical Integration Lecture 24-27 KFUPM Read Chapter 21, Section 1 Read Chapter 22, Sections 2-3

CISE301_Topic7 KFUPM 1

SE301: Numerical MethodsTopic 7

Numerical Integration Lecture 24-27

KFUPM

Read Chapter 21, Section 1Read Chapter 22, Sections 2-3


Lecture 24Introduction to

Numerical Integration

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


IntegrationIntegrationIndefinite Integrals

Indefinite Integrals of a function are functions that differ from each other by a constant.

cx

dxx2

2

Definite Integrals

Definite Integrals are numbers.

2

1

2

1

0

1

0

2

x

xdx


Fundamental Theorem of Fundamental Theorem of CalculusCalculus

:forsolution form closedNo

:for tiveantideriva no is There

) f(x)(x) 'F (i.e., f of tiveantideriva is

, intervalan on continuous is If

2

2

b

a

x

x

b

a

dxe

e

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

F

[a,b]f


The Area Under the The Area Under the CurveCurve

b

af(x)dxArea

One interpretation of the definite integral is:

Integral = area under the curve

a b

f(x)


Upper and Lower SumsUpper and Lower Sums

a b

f(x)

3210 xxxx

ii

n

ii

ii

n

ii

iii

iii

n

xxMPfUsumUpper

xxmPfLsumLower

xxxxfM

xxxxfm

Define

bxxxxaPPartition

1

1

0

1

1

0

1

1

210

),(

),(

:)(max

:)(min

...

The interval is divided into subintervals.


Upper and Lower SumsUpper and Lower Sums

a b

f(x)

3210 xxxx

2

2 integral theof Estimate

),(

),(

1

1

0

1

1

0

LUError

UL

xxMPfUsumUpper

xxmPfLsumLower

ii

n

ii

ii

n

ii


ExampleExample

14

3

2

1

4

103,2,1,0

4

1

1,16

9,

4

1,

16

116

9,

4

1,

16

1,0

)intervals equalfour (4

1,4

3,

4

2,

4

1,0 :

1

3210

3210

1

0

2

iforxx

MMMM

mmmm

n

PPartition

dxx

ii


ExampleExample

14

3

2

1

4

10

8

1

64

14

64

30

2

1

32

11

64

14

64

30

2

1integraltheofEstimate

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

ii

ii

n

ii


Upper and Lower SumsUpper and Lower Sums• Estimates based on Upper and Lower

Sums are easy to obtain for monotonic functions (always increasing or always decreasing).

• For non-monotonic functions, finding maximum and minimum of the function can be difficult and other methods can be more attractive.


Newton-Cotes Methods In Newton-Cote Methods, the function

is approximated by a polynomial of order n.

Computing the integral of a polynomial is easy.

1

)(...

2

)()()(

...)(

1122

10

10

n

aba

abaabadxxf

dxxaxaadxxf

nn

n

b

a

b

a

nn

b

a


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

)(

)(


Lecture 25Trapezoid Method

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

Read 21.1


Trapezoid MethodTrapezoid 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


Trapezoid MethodDerivation-One Interval

2

)()(

)()(2

)()()()()(

2

)()()()()(

)()()()()(

)()()(

)()(

22

2

afbfab

abab

afbfab

ab

afbfaaf

x

ab

afbfx

ab

afbfaaf

dxxab

afbf

ab

afbfaafI

dxaxab

afbfafdxxfI

b

a

b

a

b

a

b

a

b

a


Trapezoid MethodTrapezoid Method

)(bf

f(x)

ba

)()(2

bfafab

Area

)(af


Trapezoid MethodTrapezoid MethodMultiple Application RuleMultiple Application Rule

a b

f(x)

3210 xxxx

s trapezoid theof

areas theof sum)(

...

segments into dpartitione

is b][a, intervalThe

210

b

a

n

dxxf

bxxxxa

n

1212

2

)()(xx

xfxfArea

x


Trapezoid MethodTrapezoid MethodGeneral Formula and Special CaseGeneral Formula and Special Case

)()(2

1)(

...

equal)y necessaril(not segments into divided is interval theIf

11

1

0

210

iiii

n

i

b

a

n

xfxfxxdxxf

bxxxxa

n

1

10

1

)()()(2

1)(

allfor

points) base spacedEqualiy ( CaseSpecial

n

iin

b

a

ii

xfxfxfhdxxf

ihxx


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


Example 1Example 1Time (s) 0.0 1.0 2.0 3.0

Velocity (m/s)

0.0 10 12 14

29)140(2

112)(101 Distance

)()(2

1)(

1

0

1

1

1

n

n

ii

ii

xfxfxfhT

xxh

MethodTrapezoid

3,2,1,0are points Base

lssubinterva3 into

divided is interval The


Estimating the Error Estimating the Error For Trapezoid MethodFor Trapezoid Method

?accuracy digit decimal 5 to

)sin( compute toneeded

are intervals spacedequally many How

0

dxx


Error in estimating the Error in estimating the integralintegralTheoremTheorem

)(''max12

)(12

then)( eapproximat

toused is Method Trapezoid If :Theorem

) (width intervals Equal

on continuous is)('' :Assumption

],[

2

''2

a

xfhab

Error

[a,b]wherefhab

Error

dxxf

h

[a,b]xf

bax

b


ExampleExample

52

52

],[

2

5

0

106

102

1

121)(''

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

)(''max12

102

1 error ,)sin(

h

hErrorxf

xxfxxfab

xfhab

Error

thatsohfinddxx

bax


ExampleExample

)()(2

1)(),(

, allfor :

)()(2

1),(

)(:compute tomethod Trapezoid Use

0

1

1

1

11

1

0

3

1

n

n

ii

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


ExampleExample

9.5

7.21.22

18.24.32.35.0

)()(2

1)()( 0

1

1

3

1

n

n

ii xfxfxfhdxxf

x 1.0 1.5 2.0 2.5 3.0

f(x) 2.1 3.2 3.4 2.8 2.7


Recursive Trapezoid Recursive Trapezoid MethodMethod

f(x)

haa

)()(2

)0,0(R

:interval oneon based Estimate

bfafab

abh



f(x)

hahaa 2

)()0,0(2

1)0,1(

)()(2

1)(

2)0,1(R

2

:intervals 2on based Estimate

hafhRR

bfafhafab

abh

Based on previous estimate

Based on new point



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


Recursive Trapezoid Recursive Trapezoid MethodMethodFormulasFormulas

n

k

abh

hkafhnRnR

bfafab

n

2

)12()0,1(2

1)0,(

)()(2

)0,0(R

)1(2

1



)1(

2

2

1

2

13

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


Advantages of Recursive TrapezoidRecursive Trapezoid: Gives the same answer as the standard

Trapezoid method. Makes use of the available information to

reduce the computation time. Useful if the number of iterations is not

known in advance.


Lecture 26Romberg Method

Motivation Derivation of Romberg Method Romberg Method Example When to stop?

Read 22.2


Motivation for Romberg Motivation for Romberg MethodMethod

Trapezoid formula with an interval h gives an error of the order O(h2).

We can combine two Trapezoid estimates with intervals 2h and h to get a better estimate.


Romberg MethodRomberg Method

First column is obtained using Trapezoid Method

dx f(x) ofion approximat theimprove tocombined are

4h,8h,... 2h, h, size of intervals with method Trapezoid using Estimatesb

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


First Column First Column Recursive Trapezoid MethodRecursive Trapezoid Method

n

k

abh

hkafhnRnR

bfafab

n

2

)12()0,1(2

1)0,(

)()(2

)0,0(R

)1(2

1


Derivation of Romberg Derivation of Romberg MethodMethod

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

1)0,()(

2*41

)2(...64

1

16

1

4

1)0,()(

R(n,0)by obtained is estimate accurate More

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

2 method Trapezoid)()0,1()(

66

44

66

44

22

66

44

22

12

hbhbnRnRnRdxxf

giveseqeq

eqhahahanRdxxf

eqhahahanRdxxf

abhwithhOnRdxxf

b

a

b

a

b

a

n

b

a


Romberg MethodRomberg Method

1,1

)1,1()1,(14

1)1,(),(

)12()0,1(2

1)0,(

,2

)()(2

)0,0(R

)1(2

1

mnfor

mnRmnRmnRmnR

hkafhnRnR

abh

bfafab

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)


Property of Romberg Property of Romberg MethodMethod

)(),()(

Theorem

22 mb

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


Example 1Example 1

3

1

2

1

8

3

3

1

8

3)0,0()0,1(

14

1)0,1()1,1(

1,1

)1,1()1,(14

1)1,(),(

8

3

4

1

2

1

2

1

2

1))(()0,0(

2

1)0,1(,

2

1

5.0102

1)()(

2)0,0(,1

Compute

1

1

0

2

RRRR

mnfor

mnRmnRmnRmnR

hafhRRh

bfafab

Rh

dxx

m

0.5

3/8 1/3


Example 1 (cont.)Example 1 (cont.)

3

1

3

1

3

1

15

1

3

1)1,1()1,2(

14

1)1,2()2,2(

3

1

8

3

32

11

3

1

32

11)0,1()0,2(

14

1)0,1()1,2(

)1,1()1,(14

1)1,(),(

32

11

16

9

16

1

4

1

8

3

2

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

2

1)0,2(,

4

1

2

1

RRRR

RRRR

mnRmnRmnRmnR

hafhafhRRh

m

0.5

3/8 1/3

11/32 1/3 1/3


When do we stop?When do we stop?

R(4,4)at STOP example,for

steps, ofnumber given aAfter

or

)1,(),( mnRmnR

ifSTOP


Lecture 27Gauss Quadrature

Motivation General integration formula

Read 22.3


MotivationMotivation

nandih

nihcwhere

xfcdxxf

xfxfxfhdxxf

i

n

iii

b

a

n

ini

b

a

05.0

1,...,2,1

)()(

:as expressed becan It

)()(2

1)()(

:Method Trapezoid

0

1

10


General Integration General Integration FormulaFormula

integral? theofion approximat good a gives

formula e that thsoselect wedo How

:Problem

::

)()(0

ii

ii

n

iii

b

a

xandc

NodesxWeightsc

xfcdxxf


Lagrange InterpolationLagrange Interpolation

b

a ii

n

iii

b

a

b

a i

n

ii

b

a n

b

a

n

n

b

a n

b

a

dxxcwherexfcdxxf

dxxfxdxxPdxxf

xxx

xPwhere

dxxPdxxf

)()()(

)()()()(

,...,, :nodes at the

f(x) esinterpolat that polynomial a is )(

)()(

0

0

10


QuestionQuestion

What is the best way to choose the nodes and the weights?


Theorem Theorem

1.2norder of spolynomial allfor exact be willformula The

)()()(

: then)( of zeros theare,...,,,Let

00)(

:such that 1n degree of polynomial nontrivial a be Let

0

210

dxxcwherexfcdxxf

xqxxxx

nkdxxqx

q

i

b

ai

n

iii

b

a

n

kb

a


Weighted Gaussian Weighted Gaussian QuadratureQuadratureTheoremTheorem

12norder of

polynomial a isnever exact whe be willformula The

)(,)()(

:)()()(

: then)( of zeros theare,...,,,Let

00)()(

:such that 1 degree of polynomial nontrivial a be Let

0

0

210

f(x)

xx

xxxdxxwxc

wherexfcdxxwxf

xqxxxx

nkdxxwxqx

nq

n

ijj ji

jii

b

ai

n

iii

b

a

n

kb

a


Determining The Weights and Determining The Weights and NodesNodes

?5order of

spolynomial allfor exact is formula that theso

weights theand nodes select the wedo How

)()()()( 221100

1

1

xfcxfcxfcdxxf


Determining The Weights and Determining The Weights and NodesNodesSolutionSolution

5,3,0solution possible One

0)(

0)(

0)(

:satisfymust )(

)(

3120

1

1

2

1

1

1

1

33

2210

aaaa

dxxxq

xdxxq

dxxq

xq

xaxaxaaxqLet



5

30

5

3)(

use we weightsobtain the To

5

3 0,,

5

3are nodes The

5

3 0, are)(

53)(

210

1

1

210

3

fAfAfAdxxf

xxx

xqofroots

xxxqLet



5

3

9

50

9

8

5

3

9

5)(

2)(n Quadrature Gauss The

9

5 ,

9

8,

9

5are weightsThe

5

3 0,,

5

3are nodes The

1

1

210

210

fffdxxf

AAA

xxx


Gaussian QuadratureGaussian QuadratureSee more in Table 22.1 (page 626)See more in Table 22.1 (page 626)

236926.0,478628.0,568889.0,478628.0,236926.0

906179.0,538469.0,00000.0,538469.0,906179.04

34785.0,65214.0,65214.0,34785.0

86113.0,33998.0,33998.0,86113.03

555556.0,888889.0,555556.0

774596.0,000000.0,774596.02

1,1

57735.0,57735.01

)()(

43210

43210

3210

3210

210

210

10

10

0

1

1

ccccc

xxxxxn

cccc

xxxxn

ccc

xxxn

cc

xxn

xfcdxxfn

iii


Error Analysis for Gauss Error Analysis for Gauss QuadratureQuadrature

12norder of spolynomial allfor exact be willformula The

]1,1[),()!22()32(

)!1(2Error

:satisfieserror true theThen,

formula, Quadrature Gauss the

toaccording selected are

)()(

:by edapproximat be)( integral theLet

)22(3

432n

0

n

ii

n

iii

b

a

b

a

fnn

n

xandcwhere

xfcdxxf

dxxf


ExampleExample

1.n with QuadratureGaussian using1

0

2

dxeEvaluate x

b? and a arbitrary for

)(:compute toformula theuse wedo How

3

1

3

1)(

1

1

b

adxxf

ffdxxf

1

1 222)( dt

abt

abf

abdxxf

b

a


ExampleExample

22

22

5.3

15.05.

3

15.0

1

15.5.1

0

2

1

2

1

ee

dtedxe tx


Improper IntegralsImproper Integrals

1

0 221 2

1

1 2a

1

111 :

function. new on the method apply the and

)0 assuming(,11

)(

following thelikeation transforma Use

)or is limits theof (one integralsimproper eapproximat

todirectly used benot can earlier discussed Methods

dt

t

tdx

xExample

abdtt

ft

dxxf a

b

b

Documents

CISE301_Topic7KFUPM1 SE301: Numerical Methods Topic 7 Numerical Integration Lecture 24-27 KFUPM Read Chapter 21, Section 1 Read Chapter 22, Sections 2-3