What is the arclength of $f(x)=e^(1/x)/x-e^(1/x^2)/x^2+e^(1/x^3)/x^3$ on $x in [1,2]$ ?

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Answer 1 · 2016-10-23T03:06:11

Computed by WolframAlpha
The arclength is 7.01539

Arclength along a curve

$s = int_a^b sqrt(1 + (f'(x))^2)dx$

$f'(x) = -e^(1/x)/x^3-e^(1/x)/x^2-(3 e^(1/x^3))/x^7-(2 e^(1/x^2))/x^5-(3 e^(1/x^3))/x^4-(2 e^(1/x^2))/x^3$

This is the definite integral:

$s = int_1^2 sqrt(1 + (-e^(1/x)/x^3-e^(1/x)/x^2-(3 e^(1/x^3))/x^7-(2 e^(1/x^2))/x^5-(3 e^(1/x^3))/x^4-(2 e^(1/x^2))/x^3)^2)dx$

$s = 7.01539$

What is the arclength of f(x)=e^(1/x)/x-e^(1/x^2)/x^2+e^(1/x^3)/x^3f(x)=e^(1/x)/x-e^(1/x^2)/x^2+e^(1/x^3)/x^3 on x in [1,2]x in [1,2]?