My Math Forum integration with logarithmic values

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 March 19th, 2008, 07:12 AM #1 Newbie   Joined: Mar 2008 Posts: 3 Thanks: 0 integration with logarithmic values Hello, I have to calculate the area under curve for a part of a straight line. Of course, this is not really a problem, but both axes are logarithmic. How to handle the calculation of the area under curve? Given: x0, x1, y0, y1 (all y-values < 1.0) e.g. x0 = 1.0 x1 = 2.0 y0 = 0.046328232090399997 y1 = 0.060205390795099997 The logarithm of the y-values gives negative values. But the area under curve can't be negative?! I calculate for f(x) = m*x + n: m = (log(y1)-log(y0)) / (log(x1)-log(x0)) n = log(y0) - (m * log(x0)) and then I integrate f(x) with results in the value -2.5050021758. What am I doing wrong? Thanks in advance. Sanne
 March 19th, 2008, 08:29 AM #2 Senior Member   Joined: May 2007 Posts: 402 Thanks: 0 In mathematical sense, the area under a curve CAN be negative. This is because the curve itself is negative. Look at it this way. Acceleration is the derivative of speed. So, speed is the integral of acceleration. You say that an integral (area under a curve) cannot be negative, but if you get a negative number in the above example, it simply implies that the speed is oriented in the opposite direction! And that CAN happed. Still, if you want to calculate the absolute area, regardless of it's sign, then you must brake the integral into parts, so that the point's you brake at are your function's roots. Don't forget to change the sign of the integral in that case. Here is an example. If you want to calculate the area under f(x)=x^2-1 from 0 to t where t>=0, then you'd have: t Sf(x)dx = {t-t^3/3 for t<=1, t^3/3-t+4/3 for t>1} 0
 March 19th, 2008, 03:25 PM #3 Global Moderator   Joined: Dec 2006 Posts: 20,476 Thanks: 2039 If this was just an exercise, what was the problem's original wording? If, however, this was a "real life" problem, please explain what you were trying to accomplish, and why.
 March 19th, 2008, 11:37 PM #4 Newbie   Joined: Mar 2008 Posts: 3 Thanks: 0 @milin: Thanks for clarifying the "problem" with a negative area under curve. Now I have a better understanding of this. @skipjack: My question came from a "real life" problem. I have done experimental series and the measurement results have to be plotted in a logarithmic way. Now I want to compare the "curves" by area under curve. Every "curve" consists of an amount of line segments. I'd like to integrate for the single segments and then to sum up. I'm doing it now as follows: because the y-values are always between 0 and 1.0, I added an offset of 1.0 to them --> therefore the logarithm becomes positive. Furthermore I found an error in my code: I calculated the integral in the bounds of x0 and x1, instead of log(x0) and log(x1). Now the calculation seems to be correct. Or is there an error in reasoning? Sanne
 March 20th, 2008, 02:30 AM #5 Global Moderator   Joined: Dec 2006 Posts: 20,476 Thanks: 2039 The interesting bit is "Now I want to compare the "curves" by area under curve." Why do you want to use that means, and what do you expect the comparison to tell you?
 March 20th, 2008, 02:41 AM #6 Newbie   Joined: Mar 2008 Posts: 3 Thanks: 0 I want to know, if one curve is "better" than another one, i.e. has a higher level (over the whole x-axis and in defined parts / segments respectively).

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