Explanation Use the given integral to obtain the actual value by applying the limits. Then determine the midpoints and hence the value of M N for ∫ 0 π sin x d x for N = 4 , 8 , 16 , 32 , 64 . Subtract the actual value from the determined values to obtain the error. Given: The error bound for M N is proportional to 1 N 2 , so the error bound decreases by 1 4 if N is increased to 2 N . Calculation: The exact value of the integral is given by ∫ 0 π sin x d x = [ − cos x ] 0 π = 2 The value of Δ x = π 4 and hence the midpoints are π 8 , 3 π 8 , 5 π 8 , 7 π 8 Therefore, M 4 = π 4 [ sin ( π 8 ) + sin ( 3 π 8 ) + sin ( 5 π 8 ) + sin ( 7 π 8 ) ] ≈ 2.052344 Similarly, M 8 = π 8 [ sin ( π 16 ) + sin ( 3 π 16 ) + ... + sin ( 15 π 16 ) ] ≈ 2.012909 M 16 = π 16 [ sin ( π 32 ) + sin ( 3 π 32 ) + ... + sin ( 31 π 32 ) ] ≈ 2...

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Chapter 8.9, Problem 54E

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The actual error in $M_{N}$ for $\int_{0}^{π} \sin x d x$ for $N = 4, 8, 16, 32, 64$ and if the actual error seem to decrease by $\frac{1}{4}$ as $N$ is doubled

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Chapter 8.9, Problem 53E

Chapter 8.9, Problem 55E

Chapter 8 Solutions

Calculus - Standalone book

Ch. 8.1 - Prob. 1PQ Ch. 8.1 - Prob. 2PQ Ch. 8.1 - Prob. 3PQ Ch. 8.1 - Prob. 1E Ch. 8.1 - Prob. 2E Ch. 8.1 - Prob. 3E Ch. 8.1 - Prob. 4E Ch. 8.1 - Prob. 5E Ch. 8.1 - Prob. 6E Ch. 8.1 - Prob. 7E

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The actual error in M N for ∫ 0 π sin x d x for N = 4 , 8 , 16 , 32 , 64 and if the actual error seem to decrease by 1 4 as N is doubled

Solution Summary: The author explains how the error bound for M N is proportional to 1 N 2 — it decreases by 1 4 if it is increased to 2 N.

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The actual error in $M_{N}$ for $\int_{0}^{π} \sin x d x$ for $N = 4, 8, 16, 32, 64$ and if the actual error seem to decrease by $\frac{1}{4}$ as $N$ is doubled

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Chapter 8 Solutions

The actual error in M N for ∫ 0 π sin x d x for N = 4 , 8 , 16 , 32 , 64 and if the actual error seem to decrease by 1 4 as N is doubled

Solution Summary: The author explains how the error bound for M N is proportional to 1 N 2 — it decreases by 1 4 if it is increased to 2 N.

Concept explainers

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The actual error in MN for ∫0πsinxdx for N=4,8,16,32,64 and if the actual error seem to decrease by 14 as N is doubled

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Chapter 8 Solutions

The actual error in $M_{N}$ for $\int_{0}^{π} \sin x d x$ for $N = 4, 8, 16, 32, 64$ and if the actual error seem to decrease by $\frac{1}{4}$ as $N$ is doubled