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Quick question about integral of (1/x) 
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#19
Dec2211, 05:45 AM

P: 8

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. 


#20
Dec2211, 05:53 AM

P: 8

I was looking at a graph of ln(x) and 1/x together on my ti83 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 


#21
Dec2211, 07:09 AM

Sci Advisor
P: 1,718

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. 


#22
Dec2211, 11:31 AM

HW Helper
P: 3,307

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 


#23
Dec2211, 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.



#24
Dec2511, 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? 


#25
Dec2811, 07:43 AM

P: 371

[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] 


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