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)

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Presentation transcript:

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: L 1 =g(  /64) .0491 on [0,  /32], L 2 =g(3  /64) .1483 on [  /32,  /16], L 3 =g(5  /64) .2505 on [  /16, 3  /32], L 4 =g(7  /64) .3578 on [3  /32,  /8], L 5 =g(9  /64) .4730 on [  /8, 5  /32], L 6 =g(11  /64) .5994 on [5  /32, 3  /16], L 7 =g(13  /64) .7417 on [3  /16, 7  /32], L 8 =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) = sec 2 x, g // (x)= 2sec 2 x 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 The Trapezoid Rule for g(x) = tan x uses: L 1 =½[g(0)+g(  /32)] .0492 on [0,  /32], L 2 =½[g(  /32)+g(  /16)] .1487 on [  /32,  /16], L 3 =½[g(  /16)+g(3  /32)] .2511 on [  /16, 3  /32], L 4 =½[g(3  /32)+g(  /8)] .3588 on [3  /32,  /8], L 5 =½[g(  /8)+g(5  /32)] .4744 on [  /8, 5  /32], L 6 =½[g(5  /32)+g(3  /16)] .6011 on [5  /32, 3  /16], L 7 =½[g(3  /16)+g(7  /32)] .7444 on [3  /16, 7  /32], L 8 =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: P = {0,  /32,  /16, 3  /32,  /8, 5  /32, 3  /16, 7  /32,  /4}

3 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) + (2sec 2 x )(sec 2 x) = 4sec 2 x tan 2 x + 2sec 4 x, g (4) (x) = (8sec x)(sec x tan x)(tan 2 x) + (4sec 2 x)(2tan x)(sec 2 x) + (8sec 3 x)(sec x tan x) = 8sec 2 x tan 3 x + 16sec 4 x tan x. Hence |g (4) (x)|  8(2)(1)+16(4)(1) = 80 on [0,  /4]. By Theorem with a=0, b=  /4, M=80, n=8: g // (x)= 2sec 2 x tan x P = {0,  /32,  /16, 3  /32,  /8, 5  /32, 3  /16, 7  /32,  /4}