1. 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: L 1 =f(1/6) .027 on the subinterval [0,1/3], L 2 =f(1/2)=.400 on the subinterval [1/3, 2/3], L 3 =f(5/6)  0.492 on the subinterval [2/3, 1], L 4 =f(7/6)  0.494 on the subinterval [1, 4/3], L 5 =f(3/2)  0.462 on the subinterval [4/3, 5/3], L 6 =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 = 12 on [0,2]. By Theorem 3.8.9(a) with a=0, b=2, K=12, n=6:

2. 2 The Trapezoid Rule for uses: L 1 =½[f(0)+f(1/3)] .150on the subinterval [0,1/3], L 2 =½[f(1/3)+f(2/3)] .381 on the subinterval [1/3, 2/3], L 3 =½[f(2/3)+f(1)] .481 on the subinterval [2/3, 1], L 4 =½[f(1)+f(4/3)] .490 on the subinterval [1, 4/3], L 5 =½[f(4/3)+f(5/3)] .461 on the subinterval [4/3, 5/3], L 6 =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: P = {0, 1/3, 2/3, 1, 4/3, 5/3, 2}

3. 3 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(x 4 -10x 2 +5)=4x 3 -20x=4x(x 2 -5)<0, and x 4 -10x 2 +5 is decreasing with maximum absolute value of 19 at x=2. Hence |f (4) (x)|  24(2)(19)/1 5 = 912 on [0,2]. By Theorem 3.8.16 with a=0, b=2, M=912, n=6: P = {0, 1/3, 2/3, 1, 4/3, 5/3, 2}