Unit - 4

Numerical Integration

Q1) What is Newton cote’s formula.

A1)

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.

Q2) 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.

A2)

We construct the data table:

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 =

Q3) Compute the value of  

A3)

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

Here

For h=0.5, we construct the data table:

By Trapezoidal rule

For h=0.25, we construct the data table:

By Trapezoidal rule

For h = 0.125, we construct the data table:

By Trapezoidal rule

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

Q4) Evaluate using trapezoidal rule with five ordinates

A4)

Here

We construct the data table:

X	0
Y	0	0.3693161	1.195328	1.7926992	1.477265	0

Q5) Explain Simpson’s one-third rule.

A5)

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.

Q6) Estimate the value of the integral

A6)

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

For n=4, we have

E construct the data table:

By Simpson’s Rule

For n = 8,  we have

By Simpson’s Rule

Q7) Evaluate Using Simpson’s 1/3 rule with .

A7)

For , we construct the data table:

X	0
	0	0.50874	0.707106	0.840896	0.930604	0.98281	1

By Simpson’s Rule

Q8) Using Simpson’s 1/3 rule with h = 1, evaluate

A8)

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

Q9) Evaluate

A9)

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:

By Simpson’s 3/8 rule 

= 1.8278475

Q10) Evaluate

A10)

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

Q11) Evaluate 

A11)

Let

Here the interval of x and y are and .

Let 

Consider the following table:

By Trapezoidal Rule

.

Q12) Evaluate

A12)

Let 

And

By Simpson’s 1/3 Rule

Q13) Given that

Find at .

A13)

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 

Q14) Find the first and second derivatives of the function given below at the  point :

A14)

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

Q15) From the following  table of values of x and y find   for 

A15)

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

Q16) Given that

Find ?

A16)

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

Q17) Given that

Find the derivative of y at ?

A17)

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

Q18) Explain Newton’s backward difference formula.

A18)

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

Q19) Explain Newton’s forward difference formula.

A19)

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.