Boole's rule calculator for a function

SolutionHelp

Solution

Solution provided by AtoZmath.com

1. From the following table, find the area bounded by the curve and x axis from x=7.47 to x=7.52 using trapezodial, simplson's 1/3, simplson's 3/8 rule.

x	7.47	7.48	7.49	7.50	7.51	7.52
f(x)	1.93	1.95	1.98	2.01	2.03	2.06

2. Evaluate I = `int_0^1 (1)/(x+1) dx` by using simpson's rule with h=0.25 and h=0.5

Example

1. Find the approximated integral value using Boole's rule
x f(x)
1.4 4.0552
1.6 4.9530
1.8 6.0436
2.0 7.3891
2.2 9.0250

Solution:
The value of table for `x` and `f(x)`

`x`	`f(x)`
`x_0=1.4`	`f(x_(0))=4.0552`
`x_1=1.6`	`f(x_(1))=4.953`
`x_2=1.8`	`f(x_(2))=6.0436`
`x_3=2`	`f(x_(3))=7.3891`
`x_4=2.2`	`f(x_(4))=9.025`

Method-1:
Using Boole's Rule

`int f(x) dx =(2Delta x )/45 [7(f(x_(0))+f(x_(n)))+32(f(x_(1))+f(x_(3))+f(x_(5))+...)+12(f(x_(2))+f(x_(6))+f(x_(10))+...)+14(f(x_(4))+f(x_(8))+f(x_(12))+...)]`

`int f(x) dx=(2Delta x )/45 [7f(x_(0))+32f(x_(1))+12f(x_(2))+32f(x_(3))+7f(x_(4))]`

`7f(x_(0))=7*4.0552=28.3864`

`32f(x_(1))=32*4.953=158.496`

`12f(x_(2))=12*6.0436=72.5232`

`32f(x_(3))=32*7.3891=236.4512`

`7f(x_(4))=7*9.025=63.175`

`int f(x) dx=(2xx0.2)/45*(28.3864+158.496+72.5232+236.4512+63.175)`

`=(2xx0.2)/45*(559.0318)`

`=4.9692`

Solution by Boole's Rule is `4.9692`

Method-2:
Using Boole's Rule

`int f(x) dx =(2Delta x )/45 [7(f(x_(0))+f(x_(n)))+32(f(x_(1))+f(x_(3))+f(x_(5))+...)+12(f(x_(2))+f(x_(6))+f(x_(10))+...)+14(f(x_(4))+f(x_(8))+f(x_(12))+...)]`

`int f(x) dx=(2Delta x )/45 [7(f(x_(0))+f(x_(4)))+32(f(x_(1))+f(x_(3)))+12(f(x_(2)))+14()]`

`=(2xx0.2)/45 [7xx(4.0552 +9.025)+32xx(4.953+7.3891)+12xx(6.0436)+14xx()]`

`=(2xx0.2)/45 [7xx(13.0802) + 32xx(12.3421) + 12xx(6.0436) + 14xx(0)]`

`=(2xx0.2)/45 [(91.5614) + (394.9472) + (72.5232) + (0)]`

`=4.9692`

Solution by Boole's Rule is `4.9692`