Symmetric arc length of ln(x) and e^x

icesalmon
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Homework Statement


Explain why ∫(1+(1/x2)1/2dx over [1,e] = ∫(1+e2x)1/2dx over [0,1]

The Attempt at a Solution


The two original functions are ln(x) and ex and are both symmetrical about the line y = x. If I take either of the functions and translate it over the line y = x the two functions will match up completely. So it seems reasonable that the arc lengths will be the same over some region. If I plug in the bounds 1 and e into ln(x) i get 0, and 1 and if I plug the bounds 0,1 into ex I get 1, and e. I don't really know how it helps but it's something I suppose.
 
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Your argument appears sound. Are you looking for a more algebraic justification? It shouldn't be too hard to turn your argument into algebra.
 
I am looking for a more algebraic justification. I'll try and clean it up and post back when I have something, or if I have any questions. Thanks.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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