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Consider the following integral: 50 i cos (10) COS = 10 sin(5)~-9.5892 0 Using an integration step size of 12.5 and the Trapezoidal method, the integral can be found to have an error of 0.582 per step. Approximately how large would you expect the per-step error to be if a step size of 6.25 is used? O a. 0.036 per step O b. 0.018 per step O c. 0.073 per step O d. 0.146 per step

Consider the following integral: 50 i cos (10) COS = 10 sin(5)~-9.5892 0 Using an integration step size of 12.5 and the Trapezoidal method, the integral can be found to have an error of 0.582 per step. Approximately how large would you expect the per-step error to be if a step size of 6.25 is used? O a. 0.036 per step O b. 0.018 per step O c. 0.073 per step O d. 0.146 per step

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Consider the following integral: 50 X COS = 10 sin(5)-9.5892 10 0 Using an integration step size of 12.5 and the Trapezoidal method, the integral can be found to have an error of 0.582 per step. Approximately how large would you expect the per-step error to be if a step size of 6.25 is used? O a. 0.036 per step O b. 0.018 per step O c. 0.073 per step O d. 0.146 per step
Transcribed Image Text:Consider the following integral: 50 X COS = 10 sin(5)-9.5892 10 0 Using an integration step size of 12.5 and the Trapezoidal method, the integral can be found to have an error of 0.582 per step. Approximately how large would you expect the per-step error to be if a step size of 6.25 is used? O a. 0.036 per step O b. 0.018 per step O c. 0.073 per step O d. 0.146 per step
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