cise301 : numerical methods topic 7 numerical integration lecture 24-27
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CISE301_Topic7 KFUPM 1
CISE301: Numerical Methods
Topic 7 Numerical Integration
Lecture 24-27
KFUPM
Read Chapter 21, Section 1Read Chapter 22, Sections 2-3
CISE301_Topic7 KFUPM 2
Lecture 24Introduction to
Numerical Integration Definitions Upper and Lower Sums Trapezoid Method (Newton-Cotes Methods) Romberg Method Gauss Quadrature Examples
CISE301_Topic7 KFUPM 3
IntegrationIndefinite Integrals
Indefinite Integrals of a function are functions that differ from each other by a constant.
cxdxx2
2
Definite Integrals
Definite Integrals are numbers.
21
2
1
0
1
0
2
xxdx
CISE301_Topic7 KFUPM 4
Fundamental Theorem of Calculus
: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
CISE301_Topic7 KFUPM 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)
CISE301_Topic7 KFUPM 6
Upper and Lower Sums
a b
f(x)
3210 xxxx
ii
n
ii
ii
n
ii
iii
iii
n
xxMPfUsumUpper
xxmPfLsumLower
xxxxfMxxxxfm
Define
bxxxxaPPartition
1
1
0
1
1
0
1
1
210
),(
),(
:)(max:)(min
...
The interval is divided into subintervals.
CISE301_Topic7 KFUPM 7
Upper 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
CISE301_Topic7 KFUPM 8
Example
143
21
4103,2,1,0
41
1,169,
41,
161
169,
41,
161,0
)intervals equalfour (4
1,43,
42,
41,0 :
1
3210
3210
1
0
2
iforxx
MMMM
mmmm
n
PPartition
dxx
ii
CISE301_Topic7 KFUPM 9
Example
143
21
410
81
6414
6430
21
3211
6414
6430
21integraltheofEstimate
64301
169
41
161
41),(
),(
6414
169
41
1610
41),(
),(
1
1
0
1
1
0
Error
PfU
xxMPfUsumUpper
PfL
xxmPfLsumLower
ii
n
ii
ii
n
ii
CISE301_Topic7 KFUPM 10
Upper 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.
CISE301_Topic7 KFUPM 11
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
nabaabaabadxxf
dxxaxaadxxf
nn
n
b
a
b
a
nn
b
a
CISE301_Topic7 KFUPM 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
)(
)(
CISE301_Topic7 KFUPM 13
Lecture 25Trapezoid Method
Derivation-One Interval Multiple Application Rule Estimating the Error Recursive Trapezoid Method
Read 21.1
CISE301_Topic7 KFUPM 14
Trapezoid Method
)()()()( axab
afbfaf
f(x)
ba 2
)()(
2)()(
)()()(
)()()()(
)(
2
afbfab
xab
afbf
xab
afbfaaf
dxaxab
afbfafI
dxxfI
b
a
b
a
b
a
b
a
CISE301_Topic7 KFUPM 15
Trapezoid MethodDerivation-One Interval
2
)()(
)()(2
)()()()()(
2)()()()()(
)()()()()(
)()()()()(
22
2
afbfab
ababafbfab
abafbfaaf
xab
afbfxab
afbfaaf
dxxab
afbfab
afbfaafI
dxaxab
afbfafdxxfI
b
a
b
a
b
a
b
a
b
a
CISE301_Topic7 KFUPM 16
Trapezoid Method
)(bf
f(x)
ba
)()(2
bfafabArea
)(af
CISE301_Topic7 KFUPM 17
Trapezoid MethodMultiple 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
bxxxxan
1212
2)()( xxxfxfArea
x
CISE301_Topic7 KFUPM 18
Trapezoid MethodGeneral Formula and Special Case
)()(21)(
... equal)y necessaril(not segments into divided is interval theIf
11
1
0
210
iiii
n
i
b
a
n
xfxfxxdxxf
bxxxxan
1
10
1
)()()(21)(
allfor points) base spacedEqualiy ( CaseSpecial
n
iin
b
a
ii
xfxfxfhdxxf
ihxx
CISE301_Topic7 KFUPM 19
ExampleGiven 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
CISE301_Topic7 KFUPM 20
Example 1Time (s) 0.0 1.0 2.0 3.0
Velocity (m/s)
0.0 10 12 14
29)140(2112)(101 Distance
)()(21)(
1
0
1
1
1
n
n
ii
ii
xfxfxfhT
xxhMethodTrapezoid
3,2,1,0are points Baselssubinterva3 into
divided is interval The
CISE301_Topic7 KFUPM 21
Estimating the Error For Trapezoid Method
?accuracy digit decimal 5 to
)sin( compute toneeded
are intervals spacedequally many How
0
dxx
CISE301_Topic7 KFUPM 22
Error in estimating the integralTheorem
)(''max12
)(12
then)( eapproximat
toused is Method Trapezoid If :Theorem) (width intervals Equal
on continuous is)('' :Assumption
],[
2
''2
a
xfhabError
[a,b]wherefhabError
dxxf
h[a,b]xf
bax
b
CISE301_Topic7 KFUPM 23
Example
52
52
],[
2
5
0
106
1021
121)(''
)sin()('');cos()(';0;
)(''max12
1021 error ,)sin(
h
hErrorxf
xxfxxfab
xfhabError
thatsohfinddxx
bax
CISE301_Topic7 KFUPM 24
Example
)()(21)(),(
, allfor :
)()(21),(
)(: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
CISE301_Topic7 KFUPM 25
Example
9.5
7.21.2218.24.32.35.0
)()(21)()( 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
CISE301_Topic7 KFUPM 26
Recursive Trapezoid Method
f(x)
haa
)()(2
)0,0(R
:interval oneon based Estimate
bfafababh
CISE301_Topic7 KFUPM 27
Recursive Trapezoid Method
f(x)
hahaa 2
)()0,0(21)0,1(
)()(21)(
2)0,1(R
2
:intervals 2on based Estimate
hafhRR
bfafhafab
abh
Based on previous estimate Based on new point
CISE301_Topic7 KFUPM 28
Recursive Trapezoid Method
f(x)
)3()()0,1(21)0,2(
)()(21
)3()2()(4
)0,2(R
4
hafhafhRR
bfaf
hafhafhafab
abh
hahaa 42 Based on previous estimate Based on new points
CISE301_Topic7 KFUPM 29
Recursive Trapezoid MethodFormulas
n
k
abh
hkafhnRnR
bfafab
n
2
)12()0,1(21)0,(
)()(2
)0,0(R
)1(2
1
CISE301_Topic7 KFUPM 30
Recursive Trapezoid Method
)1(
2
2
1
2
13
2
12
1
1
)12()0,1(21)0,(,
2
..................
)12()0,2(21)0,3(,
2
)12()0,1(21)0,2(,
2
)12()0,0(21)0,1(,
2
)()(2
)0,0(R,
n
kn
k
k
k
hkafhnRnRabh
hkafhRRabh
hkafhRRabh
hkafhRRabh
bfafababh
CISE301_Topic7 KFUPM 31
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.
CISE301_Topic7 KFUPM 32
Lecture 26Romberg Method
Motivation Derivation of Romberg Method Romberg Method Example When to stop?
Read 22.2
CISE301_Topic7 KFUPM 33
Motivation for Romberg Method 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.
CISE301_Topic7 KFUPM 34
Romberg 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
CISE301_Topic7 KFUPM 35
First Column Recursive Trapezoid Method
n
k
abh
hkafhnRnR
bfafab
n
2
)12()0,1(21)0,(
)()(2
)0,0(R
)1(2
1
CISE301_Topic7 KFUPM 36
Derivation of Romberg Method
...)0,1()0,(31)0,()(
2*41
)2(...641
161
41)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
CISE301_Topic7 KFUPM 37
Romberg Method
1,1
)1,1()1,(14
1)1,(),(
)12()0,1(21)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)
CISE301_Topic7 KFUPM 38
Property of Romberg Method
)(),()(
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
CISE301_Topic7 KFUPM 39
Example 1
31
21
83
31
83)0,0()0,1(
141)0,1()1,1(
1,1
)1,1()1,(14
1)1,(),(
83
41
21
21
21))(()0,0(
21)0,1(,
21
5.01021)()(
2)0,0(,1
Compute
1
1
02
RRRR
mnfor
mnRmnRmnRmnR
hafhRRh
bfafabRh
dxx
m
0.53/8 1/3
CISE301_Topic7 KFUPM 40
Example 1 (cont.)
31
31
31
151
31)1,1()1,2(
141)1,2()2,2(
31
83
3211
31
3211)0,1()0,2(
141)0,1()1,2(
)1,1()1,(14
1)1,(),(
3211
169
161
41
83
21))3()(()0,1(
21)0,2(,
41
2
1
RRRR
RRRR
mnRmnRmnRmnR
hafhafhRRh
m
0.5
3/8 1/3
11/32 1/3 1/3
CISE301_Topic7 KFUPM 41
When do we stop?
R(4,4)at STOP example,for steps, ofnumber given aAfter
or
)1,1()1,( mnRmnRifSTOP
CISE301_Topic7 KFUPM 42
Lecture 27Gauss Quadrature
Motivation General integration formula
Read 22.3
CISE301_Topic7 KFUPM 43
Motivation
nandihnih
cwhere
xfcdxxf
xfxfxfhdxxf
i
n
iii
b
a
n
ini
b
a
05.01,...,2,1
)()(
:as expressed becan It
)()(21)()(
:Method Trapezoid
0
1
10
CISE301_Topic7 KFUPM 44
General Integration Formula
integral? theofion approximat good a givesformula e that thsoselect wedo How
:Problem::
)()(0
ii
ii
n
iii
b
a
xandc
NodesxWeightsc
xfcdxxf
CISE301_Topic7 KFUPM 45
Lagrange 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
xxxxPwhere
dxxPdxxf
)()()(
)()()()(
,...,, :nodes at thef(x) esinterpolat that polynomial a is )(
)()(
0
0
10
CISE301_Topic7 KFUPM 46
Question
What is the best way to choose the nodes and the weights?
CISE301_Topic7 KFUPM 47
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
CISE301_Topic7 KFUPM 48
Weighted Gaussian QuadratureTheorem
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)
xxxx
xdxxwxc
wherexfcdxxwxf
xqxxxx
nkdxxwxqx
nq
n
ijj ji
jii
b
ai
n
iii
b
a
n
kb
a
CISE301_Topic7 KFUPM 49
Determining The Weights and Nodes
?5order of spolynomial allfor exact is formula that theso
weights theand nodes select the wedo How
)()()()( 221100
1
1
xfcxfcxfcdxxf
CISE301_Topic7 KFUPM 50
Determining The Weights and NodesSolution
5,3,0solution possible One
0)(
0)(
0)(
:satisfymust )()(
3120
1
1
2
1
1
1
1
33
2210
aaaa
dxxxq
xdxxq
dxxq
xqxaxaxaaxqLet
CISE301_Topic7 KFUPM 51
Determining The Weights and NodesSolution
3
0 1 2
1
0 1 21
( ) 3 5
3( ) are 0, 5
3 3The nodes are , 0, 5 5
To obtain the weights we use
3 3( ) 05 5
Let q x x x
roots of q x
x x x
f x dx c f c f c f
CISE301_Topic7 KFUPM 52
Determining The Weights and NodesSolution
0 1 2
0 1 2
1
1
3 3The nodes are , 0, 5 5
5 8 5The weights are , , 9 9 9
The Gauss Quadrature (n 2)
5 3 8 5 3( ) 09 5 9 9 5
x x x
c c c
f x dx f f f
CISE301_Topic7 KFUPM 53
Gaussian QuadratureSee more in Table 22.1 (page 646)
236926.0,478628.0,568889.0,478628.0,236926.0906179.0,538469.0,00000.0,538469.0,906179.04
34785.0,65214.0,65214.0,34785.086113.0,33998.0,33998.0,86113.03
555556.0,888889.0,555556.0774596.0,000000.0,774596.02
1,157735.0,57735.01
)()(
43210
43210
3210
3210
210
210
10
10
0
1
1
cccccxxxxxn
ccccxxxxn
cccxxxn
ccxxn
xfcdxxfn
iii
CISE301_Topic7 KFUPM 54
Error Analysis for Gauss Quadrature
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
CISE301_Topic7 KFUPM 55
Example1.n with QuadratureGaussian using
1
0
2
dxeEvaluate x
b? and a arbitrary for
)(:compute toformula theuse wedo How
31
31)(
1
1
b
adxxf
ffdxxf
1
1 222)( dtabtabfabdxxf
b
a
CISE301_Topic7 KFUPM 56
Example
22
22
5.315.05.
315.0
1
15.5.1
0
21
21
ee
dtedxe tx
CISE301_Topic7 KFUPM 57
Improper Integrals
1
0 221 2
1
1 2a
1111 :
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
tt
dxx
Example
abdtt
ft
dxxf a
b
b
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