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^topic
^topic
I'll assume your limits are 1 and e.
[tex]xy=a[/tex]
[tex]y=\frac{a}{x}[/tex]
[tex]\mbox{Area under the curve}=\int\limits_{1}^{e}\frac{a}{x}\mbox{d}x[/tex]
[tex]\mbox{Area under the curve}=a\left(\mbox{ln}(e)-\mbox{ln}(1)\right)[/tex]
[tex]\mbox{Area under the curve}=a(1-0)[/tex]
[tex]\mbox{Area}=a[/tex]
Is the a you are talking about the same as the a in xy=a?Perhaps I should have specified. I knew how to find the are under the curve from a to b, where a is not equal to zero, but wanted to find the area of the curve in either the 1st or 3rd quadrant (that is, from 0 to b or from -b to 0).
Nope, sorry about that, I should have used different variables. The second a (as in from a to b) just represents that I wasn't asking about the integral from c to b where c≠0 and was concerned with the integral from 0 to b.Is the a you are talking about the same as the a in xy=a?
Ok. So,Nope, sorry about that, I should have used different variables. The second a (as in from a to b) just represents that I wasn't asking about the integral from c to b where c≠0 and was concerned with the integral from 0 to b.
Right, as I said I got this result when I tried myself. But I wanted to know what to do when c=0.Ok. So,
[tex]\mbox{Area under the curve}=\int_{c}^{b}\frac{a}{x}\mbox{d}x[/tex]
[tex] \mbox{Area under the curve}= a \left(\mbox{ln} |b|-\mbox{ln} |c| \right) [/tex]
The area is infinite because ln(0)= -∞Right, as I said I got this result when I tried myself. But I wanted to know what to do when c=0.
That's the problem, and it's what made me think I needed a new approach. Because obviously a hyperbola in the 1st quadrant doesn't have infinite area as it has asymptotes at the x and y axis. Is there perhaps another way to figure this out?The area is infinite because ln(0)= -∞
Nope. The hyperbola in the 1st quadrant does have an infinite area.That's the problem, and it's what made me think I needed a new approach. Because obviously a hyperbola in the 1st quadrant doesn't have infinite area as it has asymptotes at the x and y axis. Is there perhaps another way to figure this out?
Yeah, your right, but Wolfram was giving me a different result a minute ago, or maybe I made a typing mistake. Anyway, can you help with my other questions (sorry for so many)? There has to be some way to approximate it, especially considering that the hyperbola all but approaches the axes at an upper limit of 10*a.I'm afraid not. Your limit approaches infinity instead of 0.
Which questions?Yeah, your right, but Wolfram was giving me a different result a minute ago, or maybe I made a typing mistake. Anyway, can you help with my other questions (sorry for so many)?
These questions, you may not have noticed them because they were on the first page. Thank you for clearing up the first matter, I did not mean to imply that anyone other than myself was making a mistake.Also, are there any clever approximations for this area? Would translating the figure into say (x-3)(y-3)=a make any difference (no more singularity)? How would I find the area under a different type of hyperbola (is there a method that works for all hyperbolas that aren't in the form xy=1)? Thanks.
Clever approximations? The area is infinite. What's there to approximate?Wait, but for the second example, isn't the following true:
[itex]\int[/itex][itex]^{\infty}_{1}[/itex][itex]\frac{1}{x}[/itex]dx[itex]\equiv[/itex]-ln(1)+[itex]lim_{x->{\infty}}[/itex]ln(x)=0
?
Also, are there any clever approximations for this area? Would translating the figure into say (x-3)(y-3)=a make any difference (no more singularity)? How would I find the area under a different type of hyperbola (is there a method that works for all hyperbolas that aren't in the form xy=1)? Thanks.