Riemann integral of arcsinh (have the answer, want an explanation)

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SUMMARY

The discussion centers on converting the limit of a Riemann sum into an integral, specifically addressing the sum \lim_{n \to \infty}\Sigma^n_{k=1}\dfrac{1}{n\sqrt{1+(k/n)^2}}. The correct integral representation is \int^1_0\dfrac{1}{\sqrt{1+x^2}}dx, not \int^2_1\dfrac{1}{\sqrt{1+x^2}}dx as initially suggested. The confusion arises from the limits of integration, which should be from 0 to 1, corresponding to the substitution x = k/n.

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Homework Statement



Given the following sum, turn it into an integral:
[tex]\lim_{n \to \infty}\Sigma^n_{k=1}\dfrac{1}{n\sqrt{1+(k/n)^2}}[/tex]


Homework Equations



The answer says [tex]=\int^2_1\dfrac{1}{\sqrt{1+x^2}}[/tex]

The Attempt at a Solution



I understand how to get the equation, but why integrate from 1 to 2 and not from 0 to 1. if 1/n is the base length then the height should go from [tex]=\dfrac{1}{\sqrt{1+0}}[/tex] to [tex]=\dfrac{1}{\sqrt{1+1}}[/tex] not from [tex]=\dfrac{1}{\sqrt{1+1}}[/tex] to [tex]=\dfrac{1}{\sqrt{1+4}}[/tex]... or so i though??

Thanks
 
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No reason I can think of! Clearly to identify [itex]1/\sqrt{1+ (k/n)^2}[/itex] with [itex]1/\sqrt{1+ x^2}[/itex] you have to take x= k/n. But with x= k/n, when k= 1 you have x= 1/n, which goes to 0 as n goes to infinity, and when k= n you have x= 1. The integral is from 0 to 1. Your book must have a typo.
 
Ok cool thanks :)
 

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