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
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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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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