Explanation Replace the value of N by 2 N in the error bound to check the correctness of the given statement. Then determine the actual value of the integral and the actual errors and observe if the actual error seems to decrease by 1 16 if N is doubled Given: ∫ 0 π sin x d x Calculation: If we replace N by 2 N the error bound becomes K 4 ( b − a ) 5 180 ( 2 N ) 4 = 1 16 ( K 4 ( b − a ) 5 180 N 4 ) Hence we see that since N is raised to the fourth power, the error bound decreases by 1 16 if N is increased to 2 N . 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 S 4 = 1 3 π 4 [ sin 0 + 4 sin ( π 4 ) + 2 sin ( 2 π 4 ) + 4 sin ( 3 π 4 ) + sin π ] ≈ 2.004560 Similarly, S 8 = 1 3 π 8 [ sin 0 + 4 sin ( π 8 ) + 2 sin ( 2 π 8 ) + ... + 4 sin ( 7 π 8 ) + sin π ] ≈ 2.0002692 S 16 = 1 3 π 16 [ sin 0 + 4 sin ( π 16 ) + 2 sin ( 2 π 16 ) + ...

4th Edition

ISBN: 9781319434526

Author: Rogawski

Publisher: MAC HIGHER

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Numerical Integration

In a mathematical investigation, numerical integration comprises a wide group of calculations for computing the mathematical estimation of definite integral, and likewise, the term is moreover in some cases used to depict the numerical solution of differ…

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Chapter 8.8, Problem 58E

To determine

To explain why the error bound for $S_{N}$ decreases by $\frac{1}{16}$ if $N$ is increased to $2 N$ and compute the actual error in $S_{N}$ for $\int_{0}^{π} \sin x d x$ for $N = 4, 8, 16, 32, 64$ . And to determine if the actual error seem to decrease by $\frac{1}{16}$ if $N$ is doubled

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Chapter 8.8, Problem 57E

Chapter 8.8, Problem 59E

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Chapter 8 Solutions

CALCULUS 4E (LL) W/ ACHIEVE ACCESS

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To explain why the error bound for S N decreases by 1 16 if N is increased to 2 N and compute the actual error in S N for ∫ 0 π sin x d x for N = 4 , 8 , 16 , 32 , 64 . And to determine if the actual error seem to decrease by 1 16 if N is doubled

Solution Summary: The author explains why the error bound for S N decreases by 1 16 if N is increased to 2 N, and determines the actual value of the integral and actual errors.

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To explain why the error bound for $S_{N}$ decreases by $\frac{1}{16}$ if $N$ is increased to $2 N$ and compute the actual error in $S_{N}$ for $\int_{0}^{π} \sin x d x$ for $N = 4, 8, 16, 32, 64$ . And to determine if the actual error seem to decrease by $\frac{1}{16}$ if $N$ is doubled

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To explain why the error bound for S N decreases by 1 16 if N is increased to 2 N and compute the actual error in S N for ∫ 0 π sin x d x for N = 4 , 8 , 16 , 32 , 64 . And to determine if the actual error seem to decrease by 1 16 if N is doubled

Solution Summary: The author explains why the error bound for S N decreases by 1 16 if N is increased to 2 N, and determines the actual value of the integral and actual errors.

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To explain why the error bound for SN decreases by 116 if N is increased to 2N and compute the actual error in SN for ∫0πsinxdx for N=4,8,16,32,64. And to determine if the actual error seem to decrease by 116 if N is doubled

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To explain why the error bound for $S_{N}$ decreases by $\frac{1}{16}$ if $N$ is increased to $2 N$ and compute the actual error in $S_{N}$ for $\int_{0}^{π} \sin x d x$ for $N = 4, 8, 16, 32, 64$ . And to determine if the actual error seem to decrease by $\frac{1}{16}$ if $N$ is doubled