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## Quick question about integral of (1/x)

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

Hopefully this helped answer your question and didn't confuse things more!

 Quote by sbcdave I apologize if this is elementary. I took college algebra 10 years ago, recently took a precalc class at a community college and just finished reading a text book called Brief Calculus. Love math and physics by the way and am glad I found this site.
can be a nice way to help, learn and keep up with some math skills

Mod note: fixed LaTeX

 Quote by lanedance Hey sbcdave -welcome to PF!
Thanks
 you're generally best off posting question as new threads, gets less confusing that way, particularly for old threads.
Roger
 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
But the natural log of anything less than 1 is negative, which seems to say that the area under the curve from 0 to anything less than x=1 on the graph of f(x)=1/x should be a negative area. The graph of f(x)=1/x from x=0 to 1 is above the x-axis and as you say in the next quote is infinite.

Example of what I'm struggling with: $$\int_{0.1}^1dx \frac{1}{x}$$

Should be a definable positive integer. Approximately equal to 9(0.1) increments with heights 1/0.9 + 1/0.8 + ... 1/0.1 (left hand method) or approximately equal to 2.829. However, if you do ln(1)-ln(0.1) you get approximately 2.3 (I expected a negative number and need to think about this a little more, there's clearly a flaw in my logic...haha) Sorry. Posting anyways so that you might see where I was coming from.
 I was looking at a graph of ln(x) and 1/x together on my ti-83 and assumed that because the ln(x) went negative with x<1 that that meant the area was negative, now I see that because the -y value grows as x gets smaller that your subtracting a larger negative from a negative and end up with a positive. Now I'm a little confused why the left hand method was larger than the ln(x)-ln(x) method instead of smaller though if you have any ideas on that? Thanks again for humoring me once already I'll understand if the thread dies here...lol

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 Quote by sbcdave I was looking at a graph of ln(x) and 1/x together on my ti-83 and assumed that because the ln(x) went negative with x<1 that that meant the area was negative, now I see that because the -y value grows as x gets smaller that your subtracting a larger negative from a negative and end up with a positive. Now I'm a little confused why the left hand method was larger than the ln(x)-ln(x) method instead of smaller though if you have any ideas on that? Thanks again for humoring me once already I'll understand if the thread dies here...lol
By definition,

ln(x) = $\int$$^{x}_{1}$dx/x

This integral is negative when x < 1

It is not the area under the curve when x < 1. It is minus the area.

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 Quote by sbcdave Thanks Roger But the natural log of anything less than 1 is negative, which seems to say that the area under the curve from 0 to anything less than x=1 on the graph of f(x)=1/x should be a negative area. The graph of f(x)=1/x from x=0 to 1 is above the x-axis and as you say in the next quote is infinite. Example of what I'm struggling with: $$\int_{0.1}^1dx \frac{1}{x}$$ Should be a definable positive integer. Approximately equal to 9(0.1) increments with heights 1/0.9 + 1/0.8 + ... 1/0.1 (left hand method) or approximately equal to 2.829. However, if you do ln(1)-ln(0.1) you get approximately 2.3 (I expected a negative number and need to think about this a little more, there's clearly a flaw in my logic...haha) Sorry. Posting anyways so that you might see where I was coming from.
ok so lets try the two bounding methods, dividing it into 9 blocks

Where each is greater than 1/x
= (1/0.1+1/0.2+1/0.3+1/0.4+1/0.5+1/0.6+1/0.7+1/0.8+1/0.9)*0.1~2.829

Where each is less than 1/x
= (1/0.2+1/0.3+1/0.4+1/0.5+1/0.6+1/0.7+1/0.8+1/0.9+1/1)*0.1~1.929

which both bound your exact answer of ln(1)-ln(0.1)~2.3

Note you can also use the useful property of logarithms to see that it is always positive
ln(b)-ln(a) = ln(b/a)

as b>a>0, then b/a>1 and it is always a positive number
 Thanks for the replies. Very interesting subject to me. Sorry for reviving the dead thread, hopefully some others were entertained by it.
 Im sorry but how can this: $$\int \frac{1}{x}dx = lim_{b\rightarrow -1}\frac{x^{b+1}}{b+1} = ln(x)$$ possibly be correct?

 Quote by lurflurf ...Also omiting dx is valid, but confusing to some...
That is very unintuitive however,...

$$\frac{\mbox{d}y}{\mbox{d}x}=f(x)$$
$$\mbox{d}y=f(x)\mbox{d}x$$
$$y=\int f(x)\mbox{d}x$$