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 February 3rd, 2009, 12:45 PM #1 Newbie   Joined: Feb 2009 Posts: 5 Thanks: 0 Another problem! Integrate: y = ln(x)x^(-2) The answer is supposed to be 1, but I can't figure out how to get it...
 February 3rd, 2009, 01:35 PM #2 Newbie   Joined: Jan 2009 From: Baton Rouge, LA Posts: 10 Thanks: 0 Re: Another problem! You have presented an indefinite integral; thus it is highly unlikely to get a single number as the answer. What are the bounds of integration? Either way, use integration by parts with: $u= \ln(x)$ $dv= x^{-2} = \frac{1}{x^2}$
 February 3rd, 2009, 02:18 PM #3 Newbie   Joined: Feb 2009 Posts: 5 Thanks: 0 Re: Another problem! *doh* The area to be found is within the first quadrant. So 0 to 1. That's where my problem arises. When I integrate the function, I get -ln(x)x^(-1) - x^(-1) I solve it as an improper integral using a b-substitution for the lower limit of integration and a limit as b approaches 0 from the right. However, the function ends up diverging, rather than being equal to 1. Did I perform the integration by parts incorrectly?
 February 3rd, 2009, 02:46 PM #4 Member   Joined: Dec 2008 From: Washington state, of the United States Posts: 37 Thanks: 0 Re: Another problem! $\int \frac{\ln(x)}{x^2} dx$ The anti-derivative is this (solvable through integration by parts): $C - \frac{\ln(x) + 1}{x}$ Now... Im confused about what youre looking for in terms of limits of integration. You want the area $\int_0^1\;$ ? Or do you want the area in the first quadrant? They are different. First quadrant area extends $\int_1^\infty$ Evaluating the integral from 0 to 1 is not possible... its divergent. But from 1 to infinity... $\lim_{n\to\infty}\left[-\frac{\ln(n) + 1}{n}\right] - \left[-\frac{\ln(1) + 1}{1}\right]$ $1 - \lim_{n\to\infty}\left[\frac{\ln(n) + 1}{n}\right]$ Using l'Hopitals Rule: $1 - \lim_{n\to\infty}\left[\frac{1/n}{1}\right]$ $1 - 0= 1$
 February 3rd, 2009, 05:18 PM #5 Newbie   Joined: Feb 2009 Posts: 5 Thanks: 0 Re: Another problem! But if I integrate y = ln(x)x^(-2) aren't I looking for the area bounded by that equation? I thought that the anti-derivative was a mathematical representation of the area, not the equation that bounds the area.
 February 3rd, 2009, 10:20 PM #6 Global Moderator   Joined: Dec 2006 Posts: 20,281 Thanks: 1965 The area wasn't specified originally, then you added "The area to be found is within the first quadrant." The graph of the function you are integrating lies in the first quadrant for x > 1, but a properly worded question would specify clearly what area is to be found (which hasn't been done in this case) or what integration limits to use.

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