Question

use simpson’s method to approximate limit 0 to 8 ∫ (x^2-x+3)dx with 8 sub-intervals.

Answer 1

if the function f(x) is continuous on [a,b], then

${\int\limits_a^b {f\left( x \right)dx} }\approx{ {\frac{{h}}{3}}\left[ {f\left( {{x_0}} \right) + 4f\left( {{x_1}} \right) }\right.}+{\left.{ 2f\left( {{x_2}} \right) + 4f\left( {{x_3}} \right) }\right.}+{\left.{ 2f\left( {{x_4}} \right) + \cdots }\right.}+{\left.{ 4f\left( {{x_{n – 1}}} \right) + f\left( {{x_n}} \right)} \right].}$

(Simpson's rule)

by the terms of the problem

f(x) = x² - x +3

a = 0; b = 8

width of each subinterval is

$h=\frac{8-0}{8} =1$

endpoints x_i have coordinates:

x_i = {0,1,2,3,4,5,6,7,8}

Calculate the function values at the points x_i

f(x₀) = f(0) = 0² - 0 + 3 = 3

f(x₁) = f(1) = 1² - 1 + 3 = 3

f(x₂) = f(2) = 2² - 2 + 3 = 5

f(x₃) = f(2) = 3² - 3 + 3 = 9

f(x₄) = f(2) = 4² - 4 + 3 = 15

f(x₅) = f(2) = 5² - 5 + 3 = 23

f(x₆) = f(2) = 6² - 6 + 3 = 33

f(x₇) = f(2) = 7² - 7 + 3 = 45

f(x₈) = f(2) = 8² - 8 + 3 = 59

Substitute all these values into the Simpson’s Rule formula:

${\int\limits_0^8 {\left( x^2 - x + 3 \right)dx} }\approx\frac{{h}}{3}$ [f(x₀)+4f(x₁) +2f(x₂)+4f(x₃) +2f(x₄)+4f(x₅) +2f(x₆)+2f(x₇)+f(x₈)] =

$\frac{{h}}{3}$ [f(x₀) +4*(f(x₁) + f(x₃) + f(x₅) + f(x₇)) + 2*(f(x₂) + f(x₄) + f(x₆) + f(x₇)) + f(x₈)] =

$\frac{{1}}{3}$ [3 + 4*(3 +9 + 23 + 45) + 2*(5 +15 +33)+59] = 488/3 $\approx$ 162.667