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Example 2 (a) Estimate by the Midpoint, Trapezoid and Simpson's

Example 2 (a) Estimate by the Midpoint, Trapezoid and Simpson's Rules using the regular partition P of the interval [ 0, /4 ] into 8 subintervals. (b) Find bounds on the errors of those approximations.

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Example 2 (a) Estimate by the Midpoint, Trapezoid and Simpson's

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  1. Example 2 (a)Estimate by the Midpoint, Trapezoid and Simpson's Rules using the regular partition P of the interval [0, /4] into 8 subintervals. (b) Find bounds on the errors of those approximations. Solution The partition P = {0, /32, /16, 3/32, /8, 5/32, 3/16, 7/32, /4} determines 8 subintervals, each of width /32. The Midpoint Rule for g(x) = tan x uses: L1=g(/64).0491 on [0,/32], L2=g(3/64)  .1483 on [/32, /16], L3=g(5/64)  .2505 on [/16, 3/32], L4=g(7/64)  .3578 on [3/32, /8], L5=g(9/64)  .4730 on [/8, 5/32], L6=g(11/64)  .5994 on [5/32, 3/16], L7=g(13/64)  .7417 on [3/16, 7/32], L8=g(15/64)  .9063 on [7/32, /4]. By the Midpoint Rule: To bound the error, we must first bound g//(x) on [0, /4]: g/(x) = sec2x, g//(x)= 2sec2x tan x and | g//(x)|  (2)(2)(1)=4 on [0, /4]. By Theorem 3.8.9(a) with a=0, b= /4, K=4, n=8:

  2. P = {0, /32, /16, 3/32, /8, 5/32, 3/16, 7/32, /4} The Trapezoid Rule for g(x) = tan x uses: L1=½[g(0)+g(/32)] .0492 on [0,/32], L2=½[g(/32)+g(/16)] .1487 on [/32, /16], L3=½[g(/16)+g(3/32)]  .2511on [/16, 3/32], L4=½[g(3/32)+g(/8)]  .3588 on [3/32, /8], L5=½[g(/8)+g(5/32)]  .4744 on [/8, 5/32], L6=½[g(5/32)+g(3/16)]  .6011 on [5/32, 3/16], L7=½[g(3/16)+g(7/32)]  .7444 on [3/16, 7/32], L8=½[g(7/32)+g(/4)]  .9103 on [7/32, /4]. By the Trapezoid Rule: By Theorem 3.8.9(b) with a=0, b= /4 , K=4, n=8:

  3. P = {0, /32, /16, 3/32, /8, 5/32, 3/16, 7/32, /4} g//(x)= 2sec2x tan x By Simpson’s Rule with x = /32 and g(x) = tan x: To bound the error on this estimate, we must first bound g(4)(x) on [0, /4]. g(3)(x) = (4sec x)(sec x tan x)(tan x) + (2sec2 x )(sec2 x) = 4sec2 x tan2 x + 2sec4 x, g(4)(x) = (8sec x)(sec x tan x)(tan2 x) + (4sec2 x)(2tan x)(sec2 x) + (8sec3 x)(sec x tan x) = 8sec2 x tan3 x + 16sec4 x tan x. Hence |g (4)(x)|  8(2)(1)+16(4)(1) = 80 on [0, /4]. By Theorem 3.8.16 with a=0, b= /4, M=80, n=8:

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