## Simpson's Rule

If we are given odd number of tabular points,i.e. is even, then we can divide the given integral of integration in even number of sub-intervals Note that for each of these sub-intervals, we have the three tabular points and so the integrand is replaced with a quadratic interpolating polynomial. Thus using the formula (13.3.3), we get,

In view of this, we have

which gives the second quadrature formula as follows:
 (13.3.5)

This is known as SIMPSON'S RULE.

Remark 13.3.3   An estimate for the error in numerical integration using the Simpson's rule is given by

 (13.3.6)

where is the average value of the forth forward differences.

EXAMPLE 13.3.4   Using the table for the values of as is given in Example 13.3.2, compute the integral by Simpson's rule. Also estimate the error in its calculation and compare it with the error using Trapezoidal rule.
Solution: Here, thus we have odd number of nodal points. Further,

and

Thus,

To find the error estimates, we consider the forward difference table, which is given below:

 0.0 1.00000 0.01005 0.02071 0.00189 0.00149 0.1 1.01005 0.03076 0.02260 0.00338 0.00171 0.2 1.04081 0.05336 0.02598 0.00519 0.00243 0.3 1.09417 0.07934 0.03117 0.00762 0.00320 0.4 1.17351 0.11051 0.3879 0.01090 0.00459 0.5 1.28402 0.14930 0.04969 0.01549 0.00658 0.6 1.43332 0.19899 0.06518 0.02207 0.00964 0.7 1.63231 0.26417 0.08725 0.03171 0.8 1.89648 0.35142 0.11896 0.9 2.24790 0.47038 1.0 2.71828

Thus, error due to Trapezoidal rule is,

Similarly, error due to Simpson's rule is,

It shows that the error in numerical integration is much less by using Simpson's rule.

EXAMPLE 13.3.5   Compute the integral , where the table for the values of is given below:
 0.05 0.1 0.15 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0.0785 0.1564 0.2334 0.309 0.454 0.5878 0.7071 0.809 0.891 0.9511 0.9877 1

Solution: Note that here the points are not given to be equidistant, so as such we can not use any of the above two formulae. However, we notice that the tabular points and are equidistant and so are the tabular points and . Now we can divide the interval in two subinterval: and ; thus,

. The integrals then can be evaluated in each interval. We observe that the second set has odd number of points. Thus, the first integral is evaluated by using Trapezoidal rule and the second one by Simpson's rule (of course, one could have used Trapezoidal rule in both the subintervals).
For the first integral and for the second one . Thus,

which gives,

It may be mentioned here that in the above integral, and that the value of the integral is . It will be interesting for the reader to compute the two integrals using Trapezoidal rule and compare the values.

EXERCISE 13.3.6
1. Using Trapezoidal rule, compute the integral where the table for the values of is given below. Also find an error estimate for the computed value.
1.  a=1 2 3 4 5 6 7 8 9 b=10 0.09531 0.18232 0.26236 0.33647 0.40546 0.47 0.53063 0.58779 0.64185 0.69314
2.  a=1.50 1.55 1.6 1.65 1.7 1.75 b=1.80 0.40546 0.43825 0.47 0.5077 0.53063 0.55962 0.58779
3.  a = 1.0 1.5 2 2.5 3 b = 3.5 1.1752 2.1293 3.6269 6.0502 10.0179 16.5426
2. Using Simpson's rule, compute the integral Also get an error estimate of the computed integral.
1. Use the table given in Exercise 13.3.6.1b.
2.  a = 0.5 1 1.5 2 2.5 3 b = 3.5 0.493 0.946 1.325 1.605 1.778 1.849 1.833
3. Compute the integral , where the table for the values of is given below:
 0 0.5 0.7 0.9 1.1 1.2 1.3 1.4 1.5 0 0.39 0.77 1.27 1.9 2.26 2.65 3.07 3.53

A K Lal 2007-09-12