Explanation Calculate the exact value of the integral and obtain the end points. Compute T N for various values of N = 4 , 8 , 16 , 32 , 64 and compare with the previous exercise to verify the given statement Given: Observe that the error bound for T N (which has 12 in the denominator) is twice as large as the error bound for M N (which has 24 in the denominator). Calculation: The exact value of the integral is ∫ 0 π sin x d x = [ − cos x ] 0 π = 2 Δ x = π 4 and the endpoints are 0 , π 4 , 2 π 4 , 3 π 4 , π Hence, T 4 = 1 2 π 4 [ sin 0 + 2 sin ( π 4 ) + 2 sin ( 2 π 4 ) + 2 sin ( 3 π 4 ) + sin π ] ≈ 1.896119 Similarly, T 8 = 1 2 π 8 [ sin 0 + 2 sin ( π 8 ) + 2 sin ( 2 π 8 ) + ... + sin π ] ≈ 1.974232 T 16 = 1 2 π 16 [ sin 0 + 2 sin ( π 16 ) + 2 sin ( 2 π 16 ) + ... + 2 sin ( 15 π 16 ) + sin π ] ≈ 1.993570 T 32 = 1 2 π 32 [ sin 0 + 2 sin ( π 32 ) + 2 sin ( 2 π 32 ) + ...

4th Edition

ISBN: 9781319308858

Author: Rogawski

Publisher: MAC HIGHER

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Question

Chapter 7.8, Problem 57E

To determine

The actual error in $T_{N}$ for $\int_{0}^{π} \sin x d x$ for $N = 4, 8, 16, 32, 64$ and compare it with the calculations of the previous exercise. Determine if the actual error in $T_{N}$ seem to be roughly twice as large as the error in $M_{N}$ in this case

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Chapter 7.8, Problem 56E

Chapter 7.8, Problem 58E

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Chapter 7 Solutions

CALCULUS:EARLY TRANS.(LL)-W/ACCESS

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Knowledge Booster

The actual error in T N for ∫ 0 π sin x d x for N = 4 , 8 , 16 , 32 , 64 and compare it with the calculations of the previous exercise. Determine if the actual error in T N seem to be roughly twice as large as the error in M N in this case

The actual error in TNfor ∫0πsinxdx for N=4,8,16,32,64 and compare it with the calculations of the previous exercise. Determine if the actual error in TNseem to be roughly twice as large as the error in MN in this case

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Chapter 7 Solutions

The actual error in $T_{N}$ for $\int_{0}^{π} \sin x d x$ for $N = 4, 8, 16, 32, 64$ and compare it with the calculations of the previous exercise. Determine if the actual error in $T_{N}$ seem to be roughly twice as large as the error in $M_{N}$ in this case