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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

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Trapezoidal Rule of

Integration

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,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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&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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&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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eriation of the TrapezoidalRule


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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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!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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)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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)olution


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)olution


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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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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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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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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



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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!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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)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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)olution


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)olution


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)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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!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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)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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Trapezoidal Rule)imilarl3:

[ ]ihah )i( a ),( f

)h )i( a( )iha( E iii +


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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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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

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