Generally fundamental theorem of calculus is used find the solution for definite integrals, but sometime integration becomes too hard to evaluate, numerical methods are used to find the approximated value of the integral.

Simpson’s rules are very useful in numerical integration to evaluate such integrals.

Here we will understand the concept of Simpson’s rule and evaluate integrals by using numerical technique of integration.

We find more accurate value of the integration by using Simpson’s rule than other methods

Simpson’s rule

We will study about Simpson’s one-third rule and Simpson’s three-eight rules.

But in order to get these two formulas, we should have to know about the general quadrature formula-

General quadrature formula-

The general quadrature formula is gives as-

Simpson’s one-third and three-eighth formulas are derived by putting n = 2 and n = 3 respectively in general quadrature formula.

Simpson’s one-third and three-eighth formulas are derived by putting n = 2 and n = 3 respectively in general quadrature formula.

Simpson’s one-third rule-

Put n = 2 in general quadrature formula-

We get-

Note- the given interval of integration has to be divided into an even number of sub-intervals.

Simpson’s three-eighth rule-

Put n = 3 in general quadrature formula-

We get-

Note- the given interval of integration has to be divided into sub-intervals whose number n is a multiple of 3.

Q7) Evaluate the following integral by using Simpson’s 1/3rd and 3/8th rule.

A7)

x	0	1	2	3	4	5	6
f(x)	1	0.5	0.2	0.1	1/17 = 0.05884	1/26 = 0.0385	1/37 = 0.027

A8)

A8)

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

Q9) Using Simpson’s 1/3 rule with h = 1, evaluate

A9)

A9)

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

Q10) Evaluate

A10)

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

Q11) Evaluate

A11)

Here

Using =

Also

For

For

Hence

Here

By Gauss quadrature 3-point rule

Q12) Solve by Gauss quadrature 3-point method

A12)

Given

Here

Using =

Also

For

For

Hence

Here

By Gauss quadrature 3-point rule

Hence

Q13) Define Euler’s method.

A13)

Euler’s method:

In this method the solution is in the form of a tabulated values

Integrating both side of the equation (i) we get

   

Assuming that  in  this gives Euler’s formula

In general formula

, n=0,1, 2…...

Error estimate for the Euler’s method

Q14) Use Euler’s method to find y (0.4) from the differential equation

A14)

    with h=0.1

A14)

Given equation  

Here

We break the interval in four steps.

So that

By Euler’s formula

, n=0,1,2,3   ……(i)

For n=0 in equation (i) we get

    

    

For n=1 in equation (i) we get

    

    

.01

For n=2 in equation (i) we get

    

    

For n=3 in equation (i) we get

    

    

Hence y (0.4) =1.061106.

Q15) Given  with the initial condition y=1 at   x=0. Find y for x=0.1 by Euler’s   method (five steps).

A15)

Given equation is  

Here   

No. Of steps n=5 and so that

So that

Also

By Euler’s formula

, n=0,1,2,3,4   ……(i)

For n=0 in equation (i) we get

    

    

For n=1 in equation (i) we get

    

    

For n=2 in equation (i) we get

    

    

For n=3 in equation (i) we get

    

    

For n=4 in equation (i) we get

    

    

Hence  

Q16) What is Euler’s modified method?

A16)

Modified Euler’s Method:

Instead of approximating as in Euler’s method.  In the modified Euler’s method, we have the iteration formula

Where is the nth approximation to .The iteration started with the Euler’s formula

Q17) Using modified   Euler’s method, obtain a solution of the equation

A17)

A17)

Given equation 

Here

By modified Euler’s formula the initial iteration is

 

The iteration formula by modified Euler’s method is

-----(i)

For n=0 in equation (i) we get

Where   and  as above

For n=1 in equation (i) we get

For n=2 in equation (i) we get

For n=3 in equation (i) we get

Since third and fourth approximation are equal.

Hence y=0.0952 at x=0.1

To calculate the value of  at x=0.2

By modified Euler’s formula the initial iteration is

The iteration formula by modified Euler’s method is

-----(ii)

For n=0 in equation (ii) we get

1814

For n=1 in equation (ii) we get

1814

Since first and second approximation   are equal.

Hence y = 0.1814 at x=0.2

To calculate the value of  at x=0.3

By modified Euler’s formula the initial iteration is

The iteration formula by modified Euler’s method is

-----(iii)

For n=0 in equation (iii) we get

For n=1 in equation (iii) we get

For n=2 in equation (iii) we get

For n=3 in equation (iii) we get

Since third and fourth approximation are same.

Hence y = 0.25936 at x = 0.3

Q18) Use Runge Kutta method to find y when x=1.2 in step of h=0.1 given that

A18)

A18)

Given equation

Here

Also

By Runge Kutta formula for first interval

Again

A fourth order Runge Kutta formula:

To find y at 

A fourth order Runge Kutta formula:

Q19) Using Runge Kutta method of fourth order, solve

A19)

A19)

Given equation 

Here 

Also

By Runge Kutta formula for first interval

)

A fourth order Runge Kutta formula:

Hence at x = 0.2 then y = 1.196

To find the value of y at x=0.4. In this case

A fourth order Runge Kutta formula:

Hence at x = 0.4 then y=1.37527

Q20) Using Runge-Kutta method, solve

A20)

for correct to four decimal places with initial condition

A20)

Given second order differential equation is

Let   then above equation reduces to

Or

(say)

Or .

By Runge-Kutta Method we have

A fourth order Runge-Kutta formula:

And  

.

Q21) Solve the boundary value problem defined by

A21)

By finite difference method. Compare the solution at y (0.5) by taking h=0.5 and h=0.25.

A21)

Given equation

With boundary   condition   

By finite difference method

       …. (iii)

Putting(iii) in (i) we get

       …. (iv)

For h=0.5, here for which corresponds to

For i=1 in equation (iv) we get

For h=0.25, here

Which corresponds to 

For i=1 in equation (iv) we get

For i=2 in equation (iv) we get

For i=3 in equation (iv) we get

From   equation (v), (vi) and (vii) we get

On solving above triangular   equation we get

Hence for h=0.5 we   get    y (0.5) =0.44444

And for h=0.25 we get y (0.5) =0.443674