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:
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By Trapezoidal Rule
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Example2: Evaluate
Let
Here
Let the number of interval be .
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By Trapezoidal Rule
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Example3: Evaluate
Let
And
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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:
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By Simpson’s 1/3 Rule
.4444444
Example2: Evaluate
Let
Here
Let the number of interval be .
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By Simpson’s 1/3 Rule
Example3: Evaluate
Let
And
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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
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-10
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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
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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 |
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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 |
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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:
- E. Kreyszig, “Advanced Engineering Mathematics”, John Wiley & Sons, 2006.
- P. G. Hoel, S. C. Port and C. J. Stone, “Introduction to Probability Theory”, Universal Book Stall, 2003.
- S. Ross, “A First Course in Probability”, Pearson Education India, 2002.
- W. Feller, “An Introduction to Probability Theory and Its Applications”, Vol. 1, Wiley, 1968.
- N.P. Bali and M. Goyal, “A Text Book of Engineering Mathematics”, Laxmi Publications, 2010.
- B.S. Grewal, “Higher Engineering Mathematics”, Khanna Publishers, 2000.