lecture 10 numerical integration trapezoidal.pptx

8/9/2019 Lecture 10 Numerical Integration Trapezoidal.pptx

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What is Integration

=b

a

dx)x(fI

Integration:

1/30/15 Engineering Numerical Analysis 1

The process of measuringthe area under a functionplotted on a graph.

Where:

f(x) is the integrand

a lo!er limit ofintegration

" upper limit ofintegration


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Basis of Trapezoidal Rule

=

b

a dx)x(fI

)x(f)x(f n

Trape#oidal $ule is "ased on the Ne!ton%&otes 'ormulathat states if one can appro(imate the integrand as annthorder polynomial)

1/30/15 Engineering Numerical Analysis *

!here

nn

nnn xaxa...xaa)x(f ++++=

1

110

and


3/27

b

a

n

b

a

)x(f)x(f

Then the integral of that function is appro(imated "y theintegral of that nth order polynomial.


Trape#oidal $ule assumes n1+ that is+ the area under thelinear polynomial+

+=2

)b(f)a(f)ab(b

a

dx)x(f

Basis of Trapezoidal Rule


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Derivation of the Trapezoidal Rule

1/30/15 Engineering Numerical Analysis ,


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Method Derived From Geometry


The area under the cur-e is a trape#oid. The integral

trapezoidofAreadxxf

b

a

)(

)height)(sidesparallelofSum(2

1=

( ) )ab()a(f)b(f +=2

1

+=2

)b(f)a(f)ab(


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E(ample 1The -ertical distance co-ered "y a rocet from t to t 30 seconds is gi-en "y:

1/30/15 Engineering Numerical Analysis

=

30

8

892100140000

1400002000 dtt.

tlnx

a 2se single segment Trape#oidal rule to nd the distanceco-ered.

" 'ind the true error+ Etfor part 4a.c 'ind the a"solute relati-e true error+ 6a for part 4a.


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7olution


+

2

)b(f)a(f)ab(Ia

8=a 30=b

t.t

ln)t(f 892100140000

1400002000

=

)(.)(ln)(f 88982100140000

140000

20008

=

)(.)(

ln)(f 3089302100140000

140000200030

=

s/m.27177=

s/m.67901=

+=

2

6790127177830

..)(I m11868=


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7olution 4cont

1/30/15 Engineering Numerical Analysis

"The e(act -alue of the a"o-e integral is

=

30

8

892100140000

1400002000 dtt.

tlnx m11061=

ValueeApproximatValueTrueEt = 1186811061= m807=

c The a"solute relati-e true error+ + !ould "et

10011061

1186811061

=t %.29597=


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Multiple Segment Trapezoidal Rule


n E(ample 1+ the true error using single segment trape#oidal rule

!as large. We can di-ide the inter-al ;+30< into ;+19< and

;19+30< inter-als and apply Trape#oidal rule o-er each segment.

t.t

ln)t(f 892100140000

1400002000

=

+=30

19

19

8

30

8

dt)t(fdt)t(fdt)t(f

++

+=

2

30191930

2

198819

)(f)(f)(

)(f)(f)(


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=ultiple 7egment Trape#oidal $ule


With

s/m.)(f 271778 =

s/m.)(f 7548419 =

s/m.)(f 6790130 =

++

+= 2

67.90175.484)1930(

2

75.48427.177)819()(

30

8

dttf

m11266=

>ence:



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=ultiple 7egment Trape#oidal $ule


1126611061=tE

m205=

The true error is:

The true error no! is reduced from %08 m to %*05m.

E(tending this procedure to di-ide the inter-al intoe?ual segments to apply the Trape#oidal rule@ thesum of the results o"tained for each segment isthe appro(imate -alue of the integral.



12/271/30/15 Engineering Numerical Analysis 1*

Figure 4: Multiple (n=4) Segment Trapezoidal Rule

i-ide into e?ualsegments as sho!n in'igure ,. Then the !idthof each segment is:

nabh =

The integral is:

=b

a

dx)x(fI



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+

+

+

+

+

+

++++=b

h)n(a

h)n(a

h)n(a

ha

ha

ha

a

dx)x(fdx)x(f...dx)x(fdx)x(f1

1

2

2


The integralIcan "e "roen into

hintegralsas:

b

a

dx)x(f

plying Trape#oidal rule on each segment gi-es:

b

adx)x(f

+

++

=

= )b(f)iha(f)a(fn

ab n

i

1

122



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E(ample *

1/30/15 Engineering Numerical Analysis 1,

The -ertical distance co-ered "ya rocet from toseconds is gi-en "y:

=30

8

892100140000

1400002000 dtt.t

lnx

a 2se t!o%segment Trape#oidal rule to nd the distance

co-ered." 'ind the true error+ for part 4a.c 'ind the a"solute relati-e true error+ for part 4a.

atE


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7olution


a The solution using *%segment Trape#oidal rule is

+

++

=

= )b(f)iha(f)a(fn

ab

I

n

i

1

122

2=n 8=a 30=b

2

830 =

n

abh

= 11=


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7olution 4cont


+

++

=

=)(f)iha(f)(f

)(I

i

302822

830 12

1

[ ])(f)(f)(f 3019284

22++=

[ ]67901754842271774

22

.).(. ++=

m11266=

Then:


17/27

7olution 4cont


=30

8

892100140000

1400002000 dtt.t

lnx m11061=

" The e(act -alue of the a"o-e integral is

so the true error is

ValueeApproximatValueTrueEt =

1126611061=


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7olution 4cont


The a"solute relati-e true error+ + !ould "et

100ValueTrue

ErrorTrue =t

10011061

1126611061

=

%8534.1=


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7olution 4cont


Ta"le 1 gi-es the -alues o"tained using multiple segment Trape#oidalrule for:

n Value Et

1 11 %08 8.*9 %%%* 11* %*05 1.53 5.3,3

3 11153 %91., 0.*5 1.019

, 11113 %51.5 0.,55 0.359,

5 1109, %33.0 0.*91 0.19

110, %**.9 0.*080 0.090*

8 1108 %1. 0.15*1 0.05,*

1108, %1*.9 0.115 0.0350

=

30

8

892100140000

1400002000 dtt.

tlnx

Table 1: Multiple Segment Trapezoidal RuleValues

%t %a


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E(ample 3

1/30/15 Engineering Numerical Analysis *0

2se =ultiple 7egment Trape#oidal$ule to nd the area under the cur-e

xe

x)x(f += 1

300from to0=x 10=x


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Solution


+

++

=

=)b(f)iha(f)a(f

n

abI

n

i

1

1

22

+

++

=

=)(f)(f)(f

)( i105020

22

010 12

1

[ ])(f)(f)(f 105204

10++= [ ]13600391020

4

10.).( ++= 53550.=

Then:

sing t!o segments+ !e get 52

010=

=h

01

03000

0 =

+=

e

)()(f 03910

1

53005

5 .

e

)()(f =

+= 1360

1

1030010

10 .

e

)()(f =

+=

and


22/27

1/30/15 Engineering Numerical Analysis **

7o !hat is the true -alue of this integralB

592461

30010

0.dxe

xx =+

=aing the a"solute relati-e true error:

%.

..t 10059246

5355059246 =

%.50679=

Solution (ont!"


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n Appro(imateCalue

1 0.1 *,5.91 99.8*,D

* 50.535 19.05 89.505D

, 180.1 85.98 30.1*D **8.0, 19.5, 8.9*8D

1 *,1.80 ,.8 1.9*D

3* *,5.38 1.*** 0.,95D

, *,.* 0.305 0.1*,D

Table : Calues o"tained using =ultiple 7egment

Trape#oidal $ule for: +

10

01

300dx

e

xx

tE t

Solution (ont!"


24/27

#rror in Multiple Segment Trapezoidal Rule

1/30/15 Engineering Numerical Analysis *,

true error for a single segment Trape#oidal rule is gi-en "y:

ba),(f)ab(

Et


25/27


7imilarly:

[ ] ihah)i(a),(f)h)i(a()iha(

E iii +


26/27

1/30/15 Engineering Numerical Analysis *

nce the total error in multiple segment Trape#oidal rule is

=

=n

iit EE

1=

=n

ii )(f

h

1

3

12 n

)(f

n

)ab(

n

ii

=

= 1

2

3

12

The termn

)(fn

i i=

1 is an appro(imate a-erage -alue of the bxa),x(f ence:

n

)(f

n

)ab(

E

n

ii

t

=

= 1

2

3

12



27/27


elo! is the ta"le for theintegral

30

8

892100140000

1400002000 dtt.t

ln

as a function of the num"er of segments. Fou can -isuali#e that asthe num"er of segments are dou"led+ the true error gets

appro(imately ?uartered.

tE %t %an Value

* 11* %*05 1.5, 5.3,3

, 11113 %51.5 0.,55 0.359, 1108, %1*.9 0.115 0.035

0

1 1105 %3.** 0.0*913 0.00,01


lecture 10 numerical integration trapezoidal.pptx

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