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05/11/16 http://numericalmethods.eng.usf.edu 1
Trapezoidal Rule of
Integration
Major: ll !ngineering Majors
uthors: utar "a#$ %harlie &ar'er
http://numericalmethods.eng.usf.edu Transforming (umerical Methods !ducation for )T!M
*ndergraduates
http://numericalmethods.eng.usf.edu/http://numericalmethods.eng.usf.edu/http://numericalmethods.eng.usf.edu/http://numericalmethods.eng.usf.edu/
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Trapezoidal Rule of
Integration
http://numericalmethods.eng.usf .edu
http://numericalmethods.eng.usf.edu/http://numericalmethods.eng.usf.edu/http://numericalmethods.eng.usf.edu/http://numericalmethods.eng.usf.edu/
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http://numericalmethods.eng.usf.edu+
,hat is IntegrationIntegration:
∫ =b
a
dx ) x( f I
The process ofmeasuring the area
under a function plottedon a graph.
,here:
f(x) is the integrand
a- lo#er limit ofintegration
- upper limit ofinte ration
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http://numericalmethods.eng.usf.edu
&asis of Trapezoidal Rule
∫ =b
a
dx ) x( f I
Trapezoidal Rule is ased on the (e#ton%otesormula that states if one can appro2imate theintegrand as an nth order pol3nomial4
#here ) x( f ) x( f n≈
nn
nnn xa xa... xaa ) x( f ++++=
−−
1110and
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http://numericalmethods.eng.usf.edu5
&asis of Trapezoidal Rule
∫ ∫ ≈b
a
n
b
a
) x( f ) x( f
Then the integral of that function is appro2imated3 the integral of that nth order pol3nomial.
Trapezoidal Rule assumes n-1$ that is$ the
area under the linear pol3nomial$
+−=
2
)b( f )a( f )ab( ∫
b
a
dx ) x( f
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http://numericalmethods.eng.usf.edu6
eriation of the TrapezoidalRule
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http://numericalmethods.eng.usf.edu7
Method eried rom
8eometr3 The area under thecure is a
trapezoid. Theintegral
trapezoid of Areadx x f
b
a
≈∫ )(
)height )( sides parallel of Sum( 2
1=
( ) )ab( )a( f )b( f −+=2
1
+−=
2
)b( f )a( f )ab(
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http://numericalmethods.eng.usf.edu9
!2ample 1 The ertical distance coered 3 a roc'et from t-9to t-+0 seconds is gien 3:
∫
−
−=
30
8
892100140000
1400002000 dt t .
t ln x
a *se single segment Trapezoidal rule to ;nd the distancecoered.
ind the true error$ for part
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http://numericalmethods.eng.usf.edu=
)olution
+−≈
2
)b( f )a( f )ab( I a
8=a 30=bt .
t ln )t ( f 89
2100140000
1400002000 −
−=
)( . )( ln )( f 88982100140000
140000
20008 −
−=
)( . )(
ln )( f 3089302100140000
140000200030 −
−
=
s / m.27177=
s / m.67901=
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http://numericalmethods.eng.usf.edu10
)olution
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http://numericalmethods.eng.usf.edu11
)olution
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http://numericalmethods.eng.usf.edu1>
Multiple )egment TrapezoidalRule
In !2ample 1$ the true error using single segment trapezoidal rule#as large. ,e can diide the interal ?9$+0@ into ?9$1=@ and?1=$+0@ interals and appl3 Trapezoidal rule oer each segment.
t .t
ln )t ( f 892100140000
1400002000 −
−=
∫ ∫ ∫ +=30
19
19
8
30
8
dt )t ( f dt )t ( f dt )t ( f
+−+
+−=2
30191930
2
198819
)( f )( f )(
)( f )( f )(
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http://numericalmethods.eng.usf.edu1+
Multiple )egment Trapezoidal
Rule,ith
s / m. )( f 271778 =
s / m. )( f 7548419 =
s / m. )( f 6790130 =
+−+
+−=∫ 2
67.90175.484)1930(
2
75.48427.177)819()(
30
8
dt t f
m11266=
Aence:
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http://numericalmethods.eng.usf.edu1
Multiple )egment Trapezoidal
Rule
1126611061−=t E
m205−=
The true error is:
The true error no# is reduced from 907 m to >05m.
!2tending this procedure to diide the interal intoeBual segments to appl3 the Trapezoidal ruleC thesum of the results otained for each segment isthe appro2imate alue of the integral.
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http://numericalmethods.eng.usf.edu15
Multiple )egment Trapezoidal
Rule
Figure 4: Multiple (n=4) Segment TrapezoidalRule
iide into eBualsegments as sho#n inigure . Then the #idth
of each segment is:
n
abh −=
The integral I is:
∫ =b
a
dx ) x( f I
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∫ ∫ ∫ ∫ −+
−+
−+
+
+
+
++++=
b
h )n( a
h )n( a
h )n( a
ha
ha
ha
a
dx ) x( f dx ) x( f ...dx ) x( f dx ) x( f 1
1
2
2
http://numericalmethods.eng.usf.edu16
Multiple )egment Trapezoidal
Rule The integral I can e ro'en into h integrals
as:
∫
b
a
dx ) x( f
ppl3ing Trapezoidal rule on each segment gies:
∫ b
a
dx ) x( f
+
++−= ∑
−
= )b( f )iha( f )a( f
n
ab n
i
1
1
22
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http://numericalmethods.eng.usf.edu17
!2ample > The ertical distance coered 3
a roc'et from toseconds is gien 3:
∫ − −=30
8
892100140000
1400002000 dt t .t
ln x
a *se t#osegment Trapezoidal rule to ;nd the distance
coered. ind the true error$ for part
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http://numericalmethods.eng.usf.edu19
)olutiona The solution using >segment Trapezoidal rule is
+
++
−
= ∑
−
= )b( f )iha( f )a( f n
ab
I
n
i
1
122
2=n 8=a 30=b
2
830 −=
n
abh
−= 11=
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)olution
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http://numericalmethods.eng.usf.edu>0
)olution
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http://numericalmethods.eng.usf.edu>1
)olution
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http://numericalmethods.eng.usf.edu>>
)olution =6
> 11>66 >05 1.95+ 5.++
+ 1115+ =1. 0.9>65 1.01=
1111+ 51.5 0.655 0.+5=
5 110= ++.0 0.>=91 0.166=
6 1109 >>.= 0.>070 0.0=09>
7 11079 16.9 0.15>1 0.059>
9 1107 1>.= 0.1165 0.0+560
∫
−
−
=30
8
892100140000
1400002000 dt t .
t
ln x
Table 1: Multiple Segment Trapezoidal RuleValues
%t ∈ %a∈
!2act Dalue-11061 m
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http://numericalmethods.eng.usf.edu>+
!2ample +*se Multiple )egment Trapezoidal Rule to;nd the area under the cure
xe
x ) x( f
+=1
300from to0= x 10= x
*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
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http://numericalmethods.eng.usf.edu>
)olution
+
++−
= ∑−
=
)b( f )iha( f )a( f
n
ab I
n
i
1
1
2
2
+
++−
= ∑−
= )( f )( f )( f
)( i105020
22
010 12
1
[ ] )( f )( f )( f 10520410 ++= [ ]13600391020
410 . ).( ++=
53550.=
Then:
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)olution
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)olution 5.=1 ==.7>F
> 50.5+5 1=6.05 7=.505F
170.61 75.=79 +0.91>F9 >>7.0 1=.56 7.=>7F
16 >1.70 .997 1.=9>F
+> >5.+7 1.>>> 0.=5F
6 >6.>9 0.+05 0.1>F
Table : Dalues otained using Multiple )egment Trapezoidal Rule for:
∫ +
10
01
300dx
e
x x
t E t ∈
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!rror in Multiple )egment
Trapezoidal Rule The true error for a single segment Trapezoidal rule is gien 3:
ba ),( f )ab(
E t
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!rror in Multiple )egment
Trapezoidal Rule)imilarl3:
[ ]ihah )i( a ),( f
)h )i( a( )iha( E iii +
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!rror in Multiple )egment
Trapezoidal RuleAence the total error in multiple segment Trapezoidal rule is
∑==
n
i it
E E 1 ∑= ζ=
n
i i )( f
h
1
3
12 n
)( f
n
)ab(
n
ii∑
=ζ
−=
1
2
3
12
The term
n
)( f n
ii∑
=
ζ1
is an appro2imate aerage alue of the b xa ), x( f
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!rror in Multiple )egment
Trapezoidal Rule&elo# is the tale for theintegral ∫
−
−
30
8
892100140000
1400002000 dt t .
t ln
as a function of the numer of segments. Gou can isualize that as
the numer of segments are douled$ the true error getsappro2imatel3 Buartered.
t E %t ∈ %a∈n Value> 11>66 >05 1.95 5.++
1111+ 51.5 0.655 0.+5=9 1107 1>.= 0.1165 0.0+56
0
16 11065 +.>> 0.0>=1+ 0.0001
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dditional Resourcesor all resources on this topic such as digitalaudioisual lectures$ primers$ te2too' chapters$multiplechoice tests$ #or'sheets in MTH&$
MTA!MTI%$ Math%ad and MH!$ logs$related ph3sical prolems$ please isit
http://numericalmethods.eng.usf.edu/topics/trape
zoidalJrule.html
http://numericalmethods.eng.usf.edu/topics/trapezoidal_rule.htmlhttp://numericalmethods.eng.usf.edu/topics/trapezoidal_rule.htmlhttp://numericalmethods.eng.usf.edu/topics/trapezoidal_rule.htmlhttp://numericalmethods.eng.usf.edu/topics/trapezoidal_rule.html
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T!E E"#
http://numericalmethods.eng.usf.edu
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