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Example 1 (a) Estimate by the Midpoint, Trapezoid and Simpson's

Example 1 (a) Estimate by the Midpoint, Trapezoid and Simpson's Rules using the regular partition P of the interval [ 0,2 ] into 6 subintervals. (b) Find bounds on the errors of those approximations.

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Example 1 (a) Estimate by the Midpoint, Trapezoid and Simpson's

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  1. Example 1 (a)Estimate by the Midpoint, Trapezoid and Simpson's Rules using the regular partition P of the interval [0,2] into 6 subintervals. (b) Find bounds on the errors of those approximations. Solution P is the partition {0, 1/3, 2/3, 1, 4/3, 5/3, 2} which determines six subintervals, each of width 1/3. The Midpoint Rule for uses: L1=f(1/6).027 on the subinterval [0,1/3], L2=f(1/2)=.400 on the subinterval [1/3, 2/3], L3=f(5/6)0.492 on the subinterval [2/3, 1], L4=f(7/6)0.494 on the subinterval [1, 4/3], L5=f(3/2)0.462 on the subinterval [4/3, 5/3], L6=f(11/6)0.420 on the subinterval [5/3, 2]. By the Midpoint Rule: To bound the error, we must first bound f //(x) on [0,2]. By the quotient rule: Hence |f //(x)|  2(2)(3)/(1)3 = 12on [0,2]. By Theorem 3.8.9(a) with a=0, b=2, K=12, n=6:

  2. P = {0, 1/3, 2/3, 1, 4/3, 5/3, 2} The Trapezoid Rule for uses: L1=½[f(0)+f(1/3)]  .150on the subinterval [0,1/3], L2=½[f(1/3)+f(2/3)]  .381on the subinterval [1/3, 2/3], L3=½[f(2/3)+f(1)]  .481 on the subinterval [2/3, 1], L4=½[f(1)+f(4/3)]  .490 on the subinterval [1, 4/3], L5=½[f(4/3)+f(5/3)]  .461 on the subinterval [4/3, 5/3], L6=½[f(5/3)+f(2)]  .421 on the subinterval [5/3, 2]. By the Trapezoid Rule: By Theorem 3.8.9(b) with a=0, b=2, K=12, n=6:

  3. P = {0, 1/3, 2/3, 1, 4/3, 5/3, 2} By Simpson’s Rule with x = 1/3 and To bound the error on this estimate, we must first bound f (4)(x) on [0,2]. By the quotient rule: Note that on [0,2]: D(x4-10x2+5)=4x3-20x=4x(x2-5)<0, and x4-10x2+5 is decreasing with maximum absolute value of 19 at x=2. Hence |f (4)(x)|  24(2)(19)/15 = 912 on [0,2]. By Theorem 3.8.16 with a=0, b=2, M=912, n=6:

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