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

Question

1588 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-14T19:56:35

#approx 16.4316#

We are using the Formula

#int_a^bsqrt(1+(f'(x))^2)dx#
Calculating #f'(x)#

#f'(x)=2xe^(1/x)+x^2*e^(1/x)(-1/x^2)#
simplifying we get

#f'(x)=e^(1/x)(2x-1)#
so we have to compute

#int_1^2sqrt(1+(e^(1/x)(2x-1))^2)dx#

This integral can not expressed by the known elementary functions.

By a numerical method we get #approx 16.4316#