Definite integration yields a strange answer. (Solved)
11-14-2015, 05:53 PM (This post was last modified: 11-14-2015 07:32 PM by pwarmuth.)
Post: #13
 pwarmuth Junior Member Posts: 35 Joined: Sep 2015
RE: Definite integration yields a strange answer. (Sufficiently Explained - Updated)
I am not an expert on US pedagogy, but this how it has been taught to me. At least at the current point that I am at in Calculus. It also seems that this is perfectly acceptable an approach for x^n, n != -1. Granted that the domain of x must be restricted to non negative numbers when the roots are even, but (-1)^(-2/3) is within the real number domain. For these fractional exponents, all you have to do is add one to the exponent, and then divide by the new value in the exponent. This will work for real numbers with the exception of n = -1 because (x^0)/0 is undefined. I realize it's a limited definition and is not a general rule as it does not take into account non-real numbers, but it works for now and is what I'm being taught.

See this example of a problem that I worked out. This is u substitution, I just didn't write out the u and du portions of it because it's trivial to do this in my head and it's not for a professor. I've been taught to rewrite the problem before moving forward with the integration.

Edit:

This is the answer key for this problem, so I am correct in doing it this way as far as the author is concerned. The book is "Calculus", by Ron Larson & Bruce Edwards, Ed. 10e.

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