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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
8/9/2019 Lecture 10 Numerical Integration Trapezoidal.pptx
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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
8/9/2019 Lecture 10 Numerical Integration Trapezoidal.pptx
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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.
1/30/15 Engineering Numerical Analysis 3
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
8/9/2019 Lecture 10 Numerical Integration Trapezoidal.pptx
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Derivation of the Trapezoidal Rule
1/30/15 Engineering Numerical Analysis ,
8/9/2019 Lecture 10 Numerical Integration Trapezoidal.pptx
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Method Derived From Geometry
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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
1/30/15 Engineering Numerical Analysis 8
+
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=
8/9/2019 Lecture 10 Numerical Integration Trapezoidal.pptx
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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
1/30/15 Engineering Numerical Analysis 9
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)(
8/9/2019 Lecture 10 Numerical Integration Trapezoidal.pptx
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=ultiple 7egment Trape#oidal $ule
1/30/15 Engineering Numerical Analysis 10
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:
Multiple Segment Trapezoidal Rule
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=ultiple 7egment Trape#oidal $ule
1/30/15 Engineering Numerical Analysis 11
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.
Multiple Segment Trapezoidal Rule
8/9/2019 Lecture 10 Numerical Integration Trapezoidal.pptx
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
Multiple Segment Trapezoidal Rule
8/9/2019 Lecture 10 Numerical Integration Trapezoidal.pptx
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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
1/30/15 Engineering Numerical Analysis 13
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
Multiple Segment Trapezoidal Rule
8/9/2019 Lecture 10 Numerical Integration Trapezoidal.pptx
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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
1/30/15 Engineering Numerical Analysis 15
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=
8/9/2019 Lecture 10 Numerical Integration Trapezoidal.pptx
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7olution 4cont
1/30/15 Engineering Numerical Analysis 1
+
++
=
=)(f)iha(f)(f
)(I
i
302822
830 12
1
[ ])(f)(f)(f 3019284
22++=
[ ]67901754842271774
22
.).(. ++=
m11266=
Then:
8/9/2019 Lecture 10 Numerical Integration Trapezoidal.pptx
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7olution 4cont
1/30/15 Engineering Numerical Analysis 18
=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=
8/9/2019 Lecture 10 Numerical Integration Trapezoidal.pptx
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7olution 4cont
1/30/15 Engineering Numerical Analysis 1
The a"solute relati-e true error+ + !ould "et
100ValueTrue
ErrorTrue =t
10011061
1126611061
=
%8534.1=
8/9/2019 Lecture 10 Numerical Integration Trapezoidal.pptx
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7olution 4cont
1/30/15 Engineering Numerical Analysis 19
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
8/9/2019 Lecture 10 Numerical Integration Trapezoidal.pptx
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Solution
1/30/15 Engineering Numerical Analysis *1
+
++
=
=)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
8/9/2019 Lecture 10 Numerical Integration Trapezoidal.pptx
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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!"
8/9/2019 Lecture 10 Numerical Integration Trapezoidal.pptx
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1/30/15 Engineering Numerical Analysis *3
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!"
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#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
8/9/2019 Lecture 10 Numerical Integration Trapezoidal.pptx
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1/30/15 Engineering Numerical Analysis *5
7imilarly:
[ ] ihah)i(a),(f)h)i(a()iha(
E iii +
8/9/2019 Lecture 10 Numerical Integration Trapezoidal.pptx
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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
#rror in Multiple Segment Trapezoidal Rule
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1/30/15 Engineering Numerical Analysis *8
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
#rror in Multiple Segment Trapezoidal Rule