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You and I think that the cube root function is the simplest approach. The student doesn't necessarily agree. And I disagree with you that the most general appro

You and I think that the cube root function is the simplest approach. The student doesn't necessarily agree. And I disagree with you that the most general approach is necessarily the simplest. > Just about any method is valid in that type of problem. Why is that not true for other types of problems? > Passing a class should mean more than I got a lot of answers correct. I agree. But you shouldn't penalise the student if the exam question is poorly framed (and we all make such mistakes). Just take a note for later and don't make the mistake again.

yequalsx 10y ago
Let's take a calc 2 example. I ask students to integrate ln(x). I want to know if they can do integration by parts when one function is 1. Some of them can memorize the answer and just write it down. I don't give them credit for this. I'm giving an easy problem because I just want to know if they know how to do parts with 1 as one of the functions. I don't want to load the test with hard problems so that I can eliminate any possibility of memorization at play. It's interesting reading all the replies I've gotten. It's nice to see other peoples' perspectives. Including yours. As you stated I would not give x^3=27 as a problem in college algebra. It's a fine line and I suspect that we mostly agree except on one part. As a grader I've given full credit for the wrong answer and no credit for the right answer.
- gizmo686 10y ago
  So, your proposed solution to integral(ln(x)) involves first noticing that d/dx (1/x) = ln(x), then proceeding to use v=1/x to integrate by parts? I think I am misunderstanding something, because as far as I can tell, integration by parts where one of the parts is 1 is literally useless.
  - yequalsx 10y ago
    u = ln(x) and dv = 1 du = 1/x and v = x integral ln(x) = x ln(x) - integral 1/x times x integral ln(x) = x ln(x) - x + C I could give integral arctan(x) but with the advent of computer algebra systems I'm mostly interested in them knowing the basic examples and to not burden them on a test with something more complicated. EDIT: The derivative of 1/x is not ln(x) as you stated. You got it backwards and my guess is that is the source of your confusion.

yequalsx 10y ago

Let's take a calc 2 example. I ask students to integrate ln(x). I want to know if they can do integration by parts when one function is 1. Some of them can memorize the answer and just write it down. I don't give them credit for this. I'm giving an easy problem because I just want to know if they know how to do parts with 1 as one of the functions. I don't want to load the test with hard problems so that I can eliminate any possibility of memorization at play. It's interesting reading all the replies I've gotten. It's nice to see other peoples' perspectives. Including yours. As you stated I would not give x^3=27 as a problem in college algebra. It's a fine line and I suspect that we mostly agree except on one part. As a grader I've given full credit for the wrong answer and no credit for the right answer.

gizmo686 10y ago
So, your proposed solution to integral(ln(x)) involves first noticing that d/dx (1/x) = ln(x), then proceeding to use v=1/x to integrate by parts? I think I am misunderstanding something, because as far as I can tell, integration by parts where one of the parts is 1 is literally useless.
- yequalsx 10y ago
  u = ln(x) and dv = 1 du = 1/x and v = x integral ln(x) = x ln(x) - integral 1/x times x integral ln(x) = x ln(x) - x + C I could give integral arctan(x) but with the advent of computer algebra systems I'm mostly interested in them knowing the basic examples and to not burden them on a test with something more complicated. EDIT: The derivative of 1/x is not ln(x) as you stated. You got it backwards and my guess is that is the source of your confusion.