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Hey sbcdave -welcome to PF!

you're generally best off posting question as new threads, gets less confusing that way, particularly for old threads.

 Quote by sbcdave I'm new to calculus, but my impression is that an integral should be a function that represents the area under the curve of the function being integrated, which f(x)=ln(x) does not do for f(x)=1/x
The reminann integral (there are other more complex defintions) is interpreted as the area under of a well behaved function

With this is mind and considering it as a definite integral (over a given integral where the function is well behaved
$$\int_a^bdx \frac{1}{x} = ln(x)|_a^bdx =ln(b)-ln(a)$$

This is equivalent to the area between the function and the horizontal axis. Notice ln(x) is a a "monotonically" increasing function (ln(b)>ln(a) for all 0<a<b), so it always gives a positive area.

 Quote by sbcdave In the graph of 1/x you can see that from 0 toward infinity the area under the curve would come on quickly and almost stop increasing. Can anyone shed light on what I'm missing.
I don't totally undertstand your question..

But, at x = 0, f(x) = 1/x is not well defined, so to calculate the area you must use limits.

Similarly to deal with taking the integral to infinity, you need to use a limiting process.

Things get subtle when you consder lmiting process, so its important to be rigrorous. First lest split the integral into 2 portions, which we can do when the function is well behaved:
$$\int_a^bdx f(x) = \int_a^cdx f(x) + \int_c^bdx f(x)$$

In this case lets choose c=1 and the integral becomes
$$\lim_{a \to 0^+} \lim_{b \to +\infty} \int_a^b dx \frac{1}{x} = \lim_{a \to 0^+} \int_a^1 dx \frac{1}{x} + \lim_{b \to +\infty}\int_1^bdx \frac{1}{x} = \lim_{a \to 0^+}(ln(1)-ln(a)) + \lim_{b \to +\infty}(ln(b)-ln(1)) = \lim_{a \to 0^+}{-ln(a)} + \lim_{b \to +\infty}ln(b)$$

As both these diverge (become infinite) we actually find there is inifinite area below 1/x in both the intervals (0,1) and (1,inf). In fact in some repsects there is saimialr amount of infinte area in each.

Though in both cases the curve compresses against the axis, it doesn't do so quickly enough to result in a finite area - things can get wierd with lmiting processes