Trapezoidal Rule of Integration

1

Trapezoidal Rule of Integration

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What is IntegrationIntegration:

b

adx)x(fI

The process of measuring the area under a function plotted on a graph.

Where: f(x) is the integranda= lower limit of integrationb= upper limit of integration

f(x)

a b

b

a

dx)x(f

y

x

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Basis of Trapezoidal Rule

b

adx)x(fI

Trapezoidal Rule is based on the Newton-Cotes Formula that states if one can approximate the integrand as an nth order polynomial…

where )x(f)x(f n

nn

nnn xaxa...xaa)x(f

1110and

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Basis of Trapezoidal Rule

b

an

b

a)x(f)x(f

Then the integral of that function is approximated by the integral of that nth order polynomial.

Trapezoidal Rule assumes n=1, that is, the area under the linear polynomial,

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

b

adx)x(f

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Derivation of the Trapezoidal Rule

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Method Derived From Geometry

The area under the curve is a trapezoid. The integral

trapezoidofAreadxxfb

a

)(

)height)(sidesparallelofSum(21

)ab()a(f)b(f 21

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

Figure 2: Geometric Representation

f(x)

a b

b

a

dx)x(f1

y

x

f1(x)

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Example 1The vertical distance covered by a rocket from t=8 to t=30 seconds is given by:

30

8

892100140000

1400002000 dtt.t

lnx

a) Use single segment Trapezoidal rule to find the distance covered.

b) Find the true error, for part (a).c) Find the absolute relative true error, for part (a).

tEa

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Solution

2

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

a)8a 30b

t.t

ln)t(f 892100140000

1400002000

)(.)(

ln)(f 88982100140000

14000020008

)(.)(

ln)(f 3089302100140000

140000200030

s/m.27177

s/m.67901

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Solution (cont)

2

6790127177830 ..)(I

m11868

a)

b)The exact value of the above integral is

30

8

892100140000

1400002000 dtt.t

lnx m11061

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Solution (cont)b) ValueeApproximatValueTrueEt

1186811061

m807

c) The absolute relative true error, , would be t

10011061

1186811061

t %.29597

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Multiple Segment Trapezoidal Rule

In Example 1, the true error using single segment trapezoidal rule was large. We can divide the interval [8,30] into [8,19] and [19,30] intervals and apply Trapezoidal rule over each segment.

t.t

ln)t(f 892100140000

1400002000

30

19

19

8

30

8dt)t(fdt)t(fdt)t(f

2

30191930

2198

819)(f)(f

)()(f)(f

)(

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Multiple Segment Trapezoidal Rule

With

s/m.)(f 271778

s/m.)(f 7548419

s/m.)(f 6790130

267.90175.484)1930(

275.48427.177)819()(

30

8

dttf

m11266

Hence:

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Multiple Segment Trapezoidal Rule

1126611061 tE

m205

The true error is:

The true error now is reduced from -807 m to -205 m.

Extending this procedure to divide the interval into equal segments to apply the Trapezoidal rule; the sum of the results obtained for each segment is the approximate value of the integral.

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Multiple Segment Trapezoidal Rule

f(x)

a b

y

x

4aba

42 aba

4

3 aba

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

Divide into equal segments as shown in Figure 4. Then the width of each segment is:

n

abh

The integral I is:

b

adx)x(fI

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Multiple Segment Trapezoidal Rule

b

h)n(a

h)n(a

h)n(a

ha

ha

ha

adx)x(fdx)x(f...dx)x(fdx)x(f

1

1

2

2

The integral I can be broken into h integrals as:

b

adx)x(f

Applying Trapezoidal rule on each segment gives:

b

adx)x(f

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

nab n

i

1

12

2

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Example 2The vertical distance covered by a rocket from to seconds is given by:

30

889

21001400001400002000 dtt.

tlnx

a) Use two-segment Trapezoidal rule to find the distance covered.

b) Find the true error, for part (a).c) Find the absolute relative true error, for part (a).

atE

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Solutiona) The solution using 2-segment Trapezoidal rule is

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

nabI

n

i

1

12

2

2n 8a 30b

2830

n

abh 11

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Solution (cont)

)(f)iha(f)(f

)(I

i3028

22830 12

1

)(f)(f)(f 3019284

22

6790175484227177422 .).(.

m11266

Then:

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Solution (cont)

30

8

892100140000

1400002000 dtt.t

lnx m11061

b) The exact value of the above integral is

so the true error is

ValueeApproximatValueTrueEt

1126611061

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Solution (cont)

c)The absolute relative true error, , would be t

100Value TrueError True

t

10011061

1126611061

%8534.1

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Solution (cont)Table 1 gives the values obtained using multiple segment Trapezoidal rule for:

n Value Et

1 11868 -807 7.296 ---2 11266 -205 1.853 5.3433 11153 -91.4 0.8265 1.0194 11113 -51.5 0.4655 0.35945 11094 -33.0 0.2981 0.16696 11084 -22.9 0.2070 0.0908

27 11078 -16.8 0.1521 0.0548

28 11074 -12.9 0.1165 0.0356

0

30

889

21001400001400002000 dtt.

tlnx

Table 1: Multiple Segment Trapezoidal Rule Values

%t %a

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Example 3Use Multiple Segment Trapezoidal Rule to find the area under the curve

xex)x(f

1300 from to 0x 10x

.

Using two segments, we get 52

010

h

01

03000

0

e)(

)(f 039101

53005

5.

e)(

)(f

13601

1030010

10.

e)(

)(f

and

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Solution

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

nabI

n

i

1

12

2

)(f)(f)(f

)( i105020

22010 12

1

)(f)(f)(f 10520

410

136003910204

10 .).(

53550.

Then:

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Solution (cont)So what is the true value of this integral?

59246130010

0.dx

exx

Making the absolute relative true error:

%.

..t 100

592465355059246

%.50679

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Solution (cont)

n Approximate Value

1 0.681 245.91 99.724%2 50.535 196.05 79.505%4 170.61 75.978 30.812%8 227.04 19.546 7.927%16 241.70 4.887 1.982%32 245.37 1.222 0.495%64 246.28 0.305 0.124%

Table 2: Values obtained using Multiple Segment Trapezoidal Rule for:

10

01300 dx

exx

tE t