Register to reply

Quick question about integral of (1/x)

by 2^Oscar
Tags: 1 or x, integral
Share this thread:
sbcdave
#19
Dec22-11, 05:45 AM
P: 8
Quote Quote by lanedance View Post
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
[tex] \int_a^bdx \frac{1}{x} = ln(x)|_a^bdx =ln(b)-ln(a) [/tex]

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.
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: [tex] \int_{0.1}^1dx \frac{1}{x} [/tex]

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.
sbcdave
#20
Dec22-11, 05:53 AM
P: 8
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
lavinia
#21
Dec22-11, 07:09 AM
Sci Advisor
P: 1,716
Quote Quote by sbcdave View Post
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) = [itex]\int[/itex][itex]^{x}_{1}[/itex]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.
lanedance
#22
Dec22-11, 11:31 AM
HW Helper
P: 3,307
Quote Quote by sbcdave View Post
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: [tex] \int_{0.1}^1dx \frac{1}{x} [/tex]

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
sbcdave
#23
Dec22-11, 09:21 PM
P: 8
Thanks for the replies. Very interesting subject to me. Sorry for reviving the dead thread, hopefully some others were entertained by it.
Poopsilon
#24
Dec25-11, 07:53 PM
P: 294
Im sorry but how can this:

[tex]\int \frac{1}{x}dx = lim_{b\rightarrow -1}\frac{x^{b+1}}{b+1} = ln(x)[/tex]

possibly be correct?
dimension10
#25
Dec28-11, 07:43 AM
P: 371
Quote Quote by lurflurf View Post
...Also omiting dx is valid, but confusing to some...
That is very unintuitive however,...

[tex] \frac{\mbox{d}y}{\mbox{d}x}=f(x) [/tex]
[tex] \mbox{d}y=f(x)\mbox{d}x [/tex]
[tex] y=\int f(x)\mbox{d}x [/tex]


Register to reply

Related Discussions
Integral help quick question Calculus & Beyond Homework 1
Integral help quick question Calculus & Beyond Homework 1
Quick integral question Calculus & Beyond Homework 1
Quick integral question Calculus 6
Quick integral question Introductory Physics Homework 6