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Find the arc length of the curve over the given interval. (Round your answer to three decimal places.) y = ln x, [1, 3]
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first find \(f'(x)\)
let me know when you have that
1/x
ok good now square it
1/x^2
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ok, now add 1
okay
this is the easy part, the hard part is coming up
did you get \[\frac{x^2+1}{x^2}\]?
yes
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ok and the square root of that is \[\frac{\sqrt{x^2+1}}{x}\]
yes
so now we gotta figure out \[\int_1^3\frac{\sqrt{x^2+1}}{x}dx\]
2.3020
at this point i would just cheat and use a calculator to get the answer, finding the anti derivative is going to be a pain, require a substitution \[u=\tan(x)\] to get rid of the radical
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that is what i get too
Okay! Thank you!
yw
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