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NMSC


Unit - 4


Numerical Integration


Numerical Integration

Numerical integration is a process of evaluating or obtaining a definite integral from a set of numerical values of the integrand f(x).In case of function of single variable, the process is called quadrature.

 

Newton cotes formula-

Suppose where y takes the values

And let the integration interval (a,b) is divided into n equal sub-intervals, each of width h = b – a /n, so that,

The above formula is known as Newton’s cotes formula.

This is also known as general Quadrature formula.

 

Trapezoidal Method:

Let the interval [a,b] be divided into n equal intervals such that <<….<=b.

Here .

To find the value of .

Setting n=1, we get

Or I =

The above is known as Trapezoidal method.

 

Note: In this method second and higher difference are neglected and so f(x) is a polynomial of degree 1.

 

Geometrical Significance: The curve y=f(x),is replaced by n straight lines with the points ();() and ();…….;() and ().

The area bounded by the curve y=f(x), the ordinates ,and the x axis is approximately equivalent to the sum of the area of the n trapeziums obtained.

 

Example1: State the trapezoidal rule for finding an approximate area under the given curve. A curve is given by the points (x, y) given below:

(0, 23), (0.5, 19), (1.0, 14), (1.5, 11), (2.0, 12.5), (2.5, 16), (3.0, 19), (3.5, 20), (4.0, 20).

Estimate the area bounded by the curve, the x axis and the extreme ordinates.

We construct the data table:

 

X

0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

Y

23

19

14

11

12.5

16

19

20

20

 

Here length of interval h =0.5, initial value a = 0 and final value b = 4

By Trapezoidal method

Area of curve bounded on x axis =

 

Example2: Compute the value of  

Using the trapezoidal rule with h=0.5, 0.25 and 0.125.

Here

For h=0.5, we construct the data table:

 

X

0

0.5

1

Y

1

0.8

0.5

 

By Trapezoidal rule

For h=0.25, we construct the data table:

 

X

0

0.25

0.5

0.75

1

Y

1

0.94117

0.8

0.64

0.5

 

By Trapezoidal rule

For h = 0.125, we construct the data table:

 

X

0

0.125

0.25

0.375

0.5

0.625

0.75

0.875

1

Y

1

0.98461

0.94117

0.87671

0.8

0.71910

0.64

0.56637

0.5

 

By Trapezoidal rule

[(1+0.5)+2(0.98461+0.94117+0.87671+0.8+0.71910+0.64+0.56637)]

 

Example3: Evaluate using trapezoidal rule with five ordinates

Here

We construct the data table:

 

X

0

Y

0

0.3693161

1.195328

1.7926992

1.477265

0

 

 


Simpson’s Rule:

Let the interval [a,b] be divided into n equal intervals such that <<….<=b.

Here .

To find the value of .

Setting n = 2,

Which is known as Simpson’s 1/3- rule or Simpson’s rule.

 

Note: In this rule third and higher differences are neglected so f(x) is a polynomial of degree 2.

 

Example1: Estimate the value of the integral

By Simpson’s rule with 4 strips and 8 strips respectively.

For n=4, we have

E construct the data table:

 

X

1.0

1.5

2.0

2.5

3.0

Y=1/x

1

0.66666

0.5

0.4

0.33333

 

By Simpson’s Rule

For n = 8,  we have

 

X

1

1.25

1.50

1.75

2.0

2.25

2.50

2.75

3.0

Y=1/x

1

0.8

0.66666

0.571428

0.5

0.444444

0.4

0.3636363

0.333333

 

By Simpson’s Rule

 

Example2: Evaluate Using Simpson’s 1/3 rule with .

For , we construct the data table:

 

X

0

0

0.50874

0.707106

0.840896

0.930604

0.98281

1

 

By Simpson’s Rule

 

Example3: Using Simpson’s 1/3 rule with h = 1, evaluate

For h = 1, we construct the data table:

 

X

3

4

5

6

7

9.88751

22.108709

40.23594

64.503340

95.34959

 

By Simpson’s Rule

= 177.3853

 

Simpson’s 3/8 rule

Let the interval [a,b] be divided into n equal intervals such that <<….<=b.

Here .

To find the value of .

Setting n=3, we get

Is known as Simpson’s 3/8 rule which is not as accurate as Simpson’s rule.

 

Note: In this rule the fourth and higher differences are neglected and so f(x) is a polynomial of degree 3.

 

Example1: Evaluate

By Simpson’s 3/8 rule.

Let us divide the range of the interval [4, 5.2] into six equal parts.

For h=0.2, we construct the data table:

 

X

4.0

4.2

4.

4

4.6

4.8

5.0

5.2

Y=logx

1.3863

1.4351

1.4816

1.5261

1.5686

1.6094

1.6487

 

By Simpson’s 3/8 rule 

= 1.8278475

 

Example2: Evaluate

Let us divide the range of the interval [0,6] into six equal parts.

For h=1, we construct the data table:

 

X

0

1

2

3

4

5

6

1

0.5

0.2

0.1

0.0588

0.0385

0.027

 

By Simpson’s 3/8 rule

+3(0.0385)+0.027]

=1.3571

 

Double Integration: Trapezoidal Method

The double integration is defined by

Where.

 

Trapezoidal Rule:

The double integration is defined by

Where.

Or    

Also the double integration is defined by

Where.

Again Applying Trapezoidal rule to each term with respect to y.

 

Example1: Evaluate 

Let

Here the interval of x and y are and .

Let 

Consider the following table:

 

 

          

 

 

 

 

 

 

 

 

         

 

 

 

 

 

          

 

 

 

By Trapezoidal Rule

.

 

Example2: Evaluate 

Let

Here

Let the number of interval be .

 

 

          

 

 

 

 

 

 

 

 

 

          

 

 

 

 

 

 

 

By Trapezoidal Rule

.

 

Example3: Evaluate

Let 

And

 

 

              

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

By Trapezoidal  Rule

 

Simpson’s 1/3rd rule:

The double integration is defined by

By Simpson’s 1/3 rd Rule

Also the double integration is defined by

Where.

Again Applying Simpson’s 1/3 rd rule to each term with respect to y.

 

Example1: Evaluate 

Let

Here the interval of x and y are and .

Let 

Consider the following table:

 

 

          

 

 

 

 

 

 

 

 

 

 

 

 

 

          

 

 

 

By Simpson’s 1/3 Rule

.4444444

 

Example2: Evaluate 

Let

Here

Let the number of interval be .

 

 

          

 

 

 

 

 

 

 

 

 

          

 

 

 

 

 

 

 

By Simpson’s 1/3 Rule

 

Example3: Evaluate

Let 

And

 

 

              

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

By Simpson’s 1/3 Rule

 

Error in Integration

Error in Trapezoidal method

The total error in trapezoidal method is given by

Let is the largest value of the n quantities on the right hand side of the above equation then

 

Error in Simpson’s Rule

The error in the Simpson’s rule is given by

Where is the largest value of the fourth derivative of y(x).

 

Error in Simpson’s 3/8 Rule

The error in this rule is given by

Where is the largest value of the derivative of y(x).

 


Newton’s forward Difference formula:

This method is useful for interpolation near the beginning of a set of tabular values.

Where

Differentiating both side with respect to p, we get

h

This formula is applicable to compute the value of for non tabular values of x.

For tabular values of x , we can get formula by putting 

Therefore

In similar manner we can get the formula for higher order by differentiating the previous order formulas

Again differentiating with respect to p, we get

Hence

Also

And so on.

 

Example1: Given that

 

X

1.0

1.1

1.2

1.3

Y

0.841

0.891

0.932

0.963

 

Find at .

Here the first derivative is to be calculated at the beginning of the table, therefore forward difference  formula will be used

Forward difference  table is given below:

 

X

Y

1.0

 

1.1

 

1.2

 

1.3

0.841

 

0.891

 

0.932

 

0.962

 

0.050

 

0.041

 

0.031

 

-0.009

 

-0.010

 

-0.001

 

By Newton’s forward differentiation formula for differentiation

Here 

 

Example2: Find the first and second  derivatives of the function given below at the  point :

 

X

1

2

3

4

5

Y

0

1

5

6

8

 

Here the point of the calculation is  at the beginning of the  table,

Forward difference table is given  by:

 

X

Y

1

 

2

 

3

 

4

 

5

0

 

1

 

5

 

6

 

8

 

1

 

4

 

1

 

2

 

3

 

-3

 

1

 

-6

 

4

 

 

-10

 

 

By Newton’s forward differentiation formula for differentiation

Here  , 0.

Again 

At

 

Example3: From the following  table of values of x and y find   for 

X

1.00

1.05

1.10

1.15

1.20

1.25

1.30

Y

1.0000

1.02470

1.04881

1.07238

1.09544

1.11803

1.14017

 

Here the  value of the derivative is to be  calculated at  the beginning of the table.

Forward difference table is given by

 

X

Y

1.00

 

1.05

 

1.10

 

1.15

 

1.20

 

1.25

 

1.30

1.0000

 

1.02470

 

1.04881

 

1.07238

 

1.09544

 

1.11803

 

1.14017

 

0.02470

 

0.02411

 

0.02357

 

0.02306

 

0.02259

 

0.02214

 

-0.00059

 

-0.00054

 

-0.00051

 

-0.00047

 

-0.00045

 

0.00005

 

0.00003

 

0.00004

 

0.00002

 

-0.00002

 

0.00001

 

-0.00002

 

0.00003

 

-0.00003

 

-0.00006

 

From Newton’s forward difference formula for differentiation we get


 

Here

=0.48763

 

Newton Backward Difference Method:

This method is useful for interpolation near the ending of a set of tabular values.

Where 

Differentiating both side with respect to p, we get

This formula is applicable to compute the value of for non tabular values of x.

For tabular values of x , we can get formula by putting 

Therefore

In similar manner we can get the formula for higher order by differentiating the previous order formulas

Differentiating both side with respect to p, we get

Also

 

Example1: Given that

X

0.1

0.2

0.3

0..4

Y

1.10517

1.22140

1.34986

1.49182

 

Find ?

Backward difference table:

 

X

Y

0.1

 

0.2

 

0.3

 

0.4

1.10517

 

1.22140

 

1.34986

 

1.49182

 

0.11623

 

0.12846

 

0.14196

 

0.01223

 

0.01350

 

0.00127

 

Newton’s Backward formula for differentiation


 

Here


 

 

Example2: Given that

 

X

1.0

1.2

1.4

1.6

1.8

2.0

Y

0

0.128

0.544

1.296

2.432

4.0

 

Find the derivative of y at ?

The  difference table is  given below:

 

X

Y

1.0

 

1.2

 

1.4

 

1.6

 

1.8

 

2.0

0

 

0.128

 

0.544

 

1.296

 

2.432

 

4.0

 

0.128

 

0.416

 

0.752

 

0.136

 

1.568

 

 

0.288

 

0.336

 

0.384

 

0.432

 

0.048

 

0.048

 

0.048

 

0

 

0

 

Since the point  is at the beginning of the table therefore

From Newton’s forward difference formula for differentiation we get

Here

Since the  point is at the  end of the  table  therefore

Backward difference table is:

 

X

Y

1.0

 

1.2

 

1.4

 

1.6

 

1.8

 

2.0

0

 

0.128

 

0.544

 

1.296

 

2.432

 

4.000

 

0.128

 

0.416

 

0.752

 

0.136

 

1.568

 

0.288

 

0.336

 

0.384

 

0.432

 

0.048

 

0.048

 

0.048

 

0

 

0

 

Newton’s Backward formula for differentiation

 


This method was first introduced by Lewis Fry Richardson. The applications of Richardson extrapolation includes Romberg integration, which applies Richardson extrapolation to the trapezoidal rule.

This is a sequence accelerated method. It is used to improve approximations of integrals.

 

Romberg integration:

This is a very powerful method, which uses the method of extrapolation.

 

Romberg method for trapezium rule-

 

Step Length

Value of I

Value of I

Value of I

 

 

 

 

 

 

 

Here I denotes the exact value of integral and h, h/2, h/4 are the step lengths.

Note- Values at the end of each column are the most accurate.

 

Romberg method for Simpson’s 1/3 rule:

 

Step Length

Value of I

Value of I

Value of I

 

 

 

 

 

 

 

Example: Apply Romberg method for Simpson’s 1/3 rule to find the approximate value of the integral-

Sol.

Here we have-

The approximations using the Simpson’s 1/3 rule to the integral with various values of the step lengths can be obtained as-

We have-

 

These results can be tabulated as-

 

Step length

Value of I

Value of I

Value of I

½

 

¼

 

1/8

0.694444

 

0.693254

 

0.693155

 

0.693175

 

0.693148

 

0.693148

 

Magnitude of the error is-

|I – 0.693148| = | 0.693147 – 0.693148| = 0.000001

 

References:

  1. E. Kreyszig, “Advanced Engineering Mathematics”, John Wiley & Sons, 2006.
  2. P. G. Hoel, S. C. Port and C. J. Stone, “Introduction to Probability Theory”, Universal Book Stall, 2003.
  3. S. Ross, “A First Course in Probability”, Pearson Education India, 2002.
  4. W. Feller, “An Introduction to Probability Theory and Its Applications”, Vol. 1, Wiley, 1968.
  5. N.P. Bali and M. Goyal, “A Text Book of Engineering Mathematics”, Laxmi Publications, 2010.
  6. B.S. Grewal, “Higher Engineering Mathematics”, Khanna Publishers, 2000.

 


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