# Trapezoidal Rule of Integration - Numerical Methods - Lecture Slides, Slides for Mathematical Methods for Numerical Analysis and Optimization. Ankit Institute of Technology and Science

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The main points are: Trapezoidal Rule of Integration, Process of Measuring Area, Lower Limit of Integration, Basis of Trapezoidal Rule, Newton-Cotes Formula, Linear Polynomial, Derivation of Trapezoidal Rule, Area of Tra...
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Trapezoidal Rule Integration

Trapezoidal Rule of Integration

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What is Integration

 b

a

dx)x(fI

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

Where:

f(x) is the integrand

a= lower limit of integration

b= upper limit of integration

f(x)

a b

b

a

dx)x(f

y

x

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Basis of Trapezoidal Rule

 b

a

dx)x(fI )x(f)x(f n

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

where

n n

n nn xaxa...xaa)x(f 

 

1 110and

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Basis of Trapezoidal Rule

  b

a

n

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

a

dx)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

b

a

 )(

)height)(sidesparallelofSum( 2

1 

  )ab()a(f)b(f  2

1



 

  

2

)b(f)a(f )ab(

Figure 2: Geometric Representation

f(x)

a b

b

a

dx)x(f 1

y

x

f1(x)

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Example 1

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

  

  

 

 

 

30

8

89 2100140000

140000 2000 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).

tE a

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Solution



 

  

2

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

8a 30b

t. t

ln)t(f 89 2100140000

140000 2000 

 

 

)(. )(

ln)(f 889 82100140000

140000 20008 

  

 

)(. )(

ln)(f 3089 302100140000

140000 200030 

  

 

s/m.27177

s/m.67901

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Solution (cont)



 

  

2

6790127177 830

.. )(I

m11868

a)

b) The exact value of the above integral is

  

  

 

 

 

30

8

89 2100140000

140000 2000 dtt.

t lnx m11061

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Solution (cont)

b) ValueeApproximatValueTrueEt 

1186811061

m807

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

100 11061

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 89 2100140000

140000 2000 

  

 

  30

19

19

8

30

8

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



 

  

 

  

2

3019 1930

2

198 819

)(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 

 

  

  

  

   2

67.90175.484 )1930(

2

75.48427.177 )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

4

ab a

 

4 2

ab a

 

4 3

ab a

 

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

ab h

 

The integral I 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(f 1

1

2

2

Multiple Segment Trapezoidal Rule

The integral I can be broken into h integrals as:

b

a

dx)x(f

Applying Trapezoidal rule on each segment gives:

b

a

dx)x(f  

  

    

  

 

  

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

ab n

i

1

1

2 2

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Example 2

The vertical distance covered bya rocket from to seconds is given by:

  

  

 

 

 

30

8

89 2100140000

140000 2000 dtt.

t lnx

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). a

tE

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Solution

a) The solution using 2-segment Trapezoidal rule is

 

  

    

  

 

  

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

ab I

n

i

1

1

2 2

2n 8a 30b

2

830  

n

ab h

  11

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Solution (cont)

 

  

    

  

 

  

)(f)iha(f)(f )(

I i

3028 22

830 12

1

 )(f)(f)(f 301928 4

22 

 6790175484227177 4

22 .).(. 

m11266

Then:

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Solution (cont)

  

  

 

 

 

30

8

89 2100140000

140000 2000 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)

The absolute relative true error, , would be t

100 Value True

Error True t

100 11061

1126611061 

 

%8534.1

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Solution (cont)

Table 1 gives the values obtained using multiple segment Trapezoidal rule for:

nValueEt

1 11868 -807 7.296 ---

2 11266 -205 1.853 5.343

3 11153 -91.4 0.8265 1.019

4 11113 -51.5 0.4655 0.3594

5 11094 -33.0 0.2981 0.1669

6 11084 -22.9 0.2070 0.09082

7 11078 -16.8 0.1521 0.05482

8 11074 -12.9 0.1165 0.03560

  

  

  

 

 

30

8

89 2100140000

140000 2000 dtt.

t lnx

Table 1: Multiple Segment Trapezoidal Rule Values

%t%a

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Example 3

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

xe

x )x(f

 1

300 from to 0x 10x

Using two segments, we get 5 2

010 

 h

0 1

0300 0

0 

 

e

)( )(f 03910

1

5300 5

5 .

e

)( )(f

  1360

1

10300 10

10 .

e

)( )(f

 

and

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Solution

 

  

    

  

 

  

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

ab I

n

i

1

1

2 2

 

  

    

  

 

  

)(f)(f)(f )( i

105020 22

010 12

1

 )(f)(f)(f 10520 4

10   13600391020

4

10 .).( 

53550.

Then:

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Solution (cont)

So what is the true value of this integral?

59246 1

30010

0

.dx e

x x

 

Making the absolute relative true error:

% .

.. t 100

59246

5355059246 

 

%.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

01

300 dx

e

x x

tE t

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