find the arc length of the graph of the function over the indicated interval. x^5/10+1/(6x^3) on the interval (3,6)

Answer 1

let y = x^5/10+1/(6x^3)
here is a picture of y = x^5/10+1/(6x^3) from 3 to 6
http://www.wolframalpha.com/input/?i=plot+y+%3D+x%5E5%2F10%2B1%2F%286x%5E3%29+from+3+to+6

arc length = ∫√( 1 + (dy/dx)^2) dx from 3 to 6

dy/dx = x^4/2 - 1/(2x^4)

(dy/dx)^2 = x^8/4 - 1/2 + 1/(4x^8)
= (1/4)x^8 - 1/2 + (1/4)x^-8

so arc
= ∫√(1 + (1/4)x^8 - 1/2 + (1/4)x^-8) dx from 3 to 6
nasty integral, I trusted Wolfram
http://www.wolframalpha.com/input/?i=integrate+%281+%2B+x%5E8%2F4+-+1%2F2+%2B+1%2F%284x%5E8%29+%29%5E%281%2F2%29+from+3+to+6

for appr 753.3 units of length

Answer 2

actually, the integral isn't so bad if you recognize that the integrand is just x^4 + 1/x^4

http://www.wolframalpha.com/input/?i=arc+length+of+x^5%2F10%2B1%2F%286x^3%29+from+3+to+6

Answer 3

yup, didn't see that perfect square