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Below are attempts at u-substitution by students. Explain where each student went wrong, then explain how to solve the problem. (a) Jane is evaluating √ a(2x + 5)² dr. She identifies u = 2x + 5, so du = and so dx = du. Thus, ·S J x(2x+ 5)² dx = ²/2/² [x - u² du. Since the integral is 12 / 2 1 2dx, with respect to u, she treats x as a constant, and 1 3 x · u² du = 1⁄x · ²u²³ = ²√x(2x + 5)³. 2 3

Below are attempts at u-substitution by students. Explain where each student went wrong, then explain how to solve the problem. (a) Jane is evaluating √ a(2x + 5)² dr. She identifies u = 2x + 5, so du = and so dx = du. Thus, ·S J x(2x+ 5)² dx = ²/2/² [x - u² du. Since the integral is 12 / 2 1 2dx, with respect to u, she treats x as a constant, and 1 3 x · u² du = 1⁄x · ²u²³ = ²√x(2x + 5)³. 2 3

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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Below are attempts at u-substitution by students. Explain where each student went wrong, then explain how to solve the problem. (a) Jane is evaluating x(2x + 5)² dx. She identifies u = 2x + 5, so du = and so dx = du. Thus,x(2x + 5)² dx Ja = 1/2/ 1/ ST. 2 with respect to u, she treats x as a constant, and xu² du. Since the integral is 1 1 -X 2 x.u² du - 1 -U³ 2dx, 3 = x(2x- c(2x + 5)³.
Transcribed Image Text:Below are attempts at u-substitution by students. Explain where each student went wrong, then explain how to solve the problem. (a) Jane is evaluating x(2x + 5)² dx. She identifies u = 2x + 5, so du = and so dx = du. Thus,x(2x + 5)² dx Ja = 1/2/ 1/ ST. 2 with respect to u, she treats x as a constant, and xu² du. Since the integral is 1 1 -X 2 x.u² du - 1 -U³ 2dx, 3 = x(2x- c(2x + 5)³.
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