learn Simpson's rule applications: .It is extremely difficult or even impossible to evaluate a definite integral `int_a^bf(x)dx` even if f is continuous on [ a, b ]. In such cases we take a set of numerical values of the integrand f in the interval [ a, b ] and evaluate the definite integral `int_a^bf(x)dx` approximately. This process of finding the approximate value of a definite integral is called Numerical Integration. If the integrand is a function of single variable, then this process is known as Quadrature.
In Numerical Integration, first we approximate the integrand f by a polynomial p and then find `int_a^bp(x)dx` . Then `int_a^bf(x)dx` is approximately equal to `int_a^bp(x)dx` . The absolute difference `| ` `int_a^bf(x)dx` - `int_a^bp(x)dx` `|` = `|` `int_a^b[ f(x) - p(x) ]dx` `|`
Let P = { x_{0} = a, x_{1} = x_{0} + h, x_{2} = x_{0} + 2h, .................x_{n} = x_{0} + nh = b } where h = ( b - a ) / n , be a partition of [ a, b ] dividing [ a, b ] into even number n of equal parts. Let y_{0}, y_{1}, y_{2}................y_{n} be the values of y = f ( x ) at x = x_{0,} x_{1}, x_{2}, .................x_{n} respectively.
Simpson's rule is given by
`int_a^bf(x)dx` `~=` ( h / 3 ) [ ( y_{0} + y_{n} ) + 4( y_{1} + y_{3} + ..............+ y_{n-1} ) + 2( y_{2} + y_{4} +................. + y_{n-2} ) ]
= ( h / 3 ) [ Sum of the first and last ordinates + 4(Sum of the ordinates whose suffixes are odd) + 2(Sum of the ordinates whose suffixes are even excluding the first and last ordinates) ]
This is known as the Simpson's one-third rule or simply Simpson's rule which is commonly used.
Example:-
Dividing [ 0, 6 ] into 6 equal parts, evaluate `int_0^6` x^{3} dx approximately by using Simpson's rule.
Solution:- Since a = 0, b = 6 and n = 6
then h = ( b - a ) / n
h = ( 6 - 0 ) / 6
h = 1
Hence P = { x_{0} = a = 0, x_{1} = 1, x_{2} = 2, x_{3} = 3, x_{4} = 4, x_{5} = 5, x_{6} = 6 }
Now we have to find the y =f ( x )
given that f ( x ) = x^{3}
we have the following tables of values
x | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
f ( x ) | 0 | 1 | 8 | 27 | 64 | 125 | 216 |
`int_0^6` x^{3} dx = ( h / 3 ) [ ( y_{0} + y_{6} ) + 4( y_{1} + y_{3} + y_{5} ) + 2(y_{2} + y_{4}) ]
= ( 1 / 3 ) [ ( 0 + 216 ) + 4 ( 1 + 27 + 125 ) + 2 ( 8 + 64 ) ]
= ( 1 / 3 ) [ 216 + 4(153) + 2(72) ]
= ( 1 / 3 ) [ 216 + 612 + 144 ]
= 324.
`:.` `int_0^6` x^{3} dx = 324.