How do you find the lengths of the curve #y=x^3/12+1/x# for #1<=x<=3#?

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Answer 1 · 2018-05-03T18:57:14

#= 17/6 #

#s = int_1^3 \ sqrt ( 1 + (y')^2 ) \ \ dx#

#s = int_1^3 \ sqrt ( 1 + (x^2/4 - 1/x^2)^2 ) \ \ dx#

# = int_1^3 \ sqrt ( 1 + x^4/16 - 1/2 + 1/x^4) \ \ dx#

# = int_1^3 \ sqrt ( x^4/16 + 1/2 + 1/x^4) \ \ dx#

# = int_1^3 \ sqrt ( (x^2/4 + 1/x^2)^2 ) \ \ dx#

# = (\ x^3/(12) - 1/x )_1^3 #

# = ( 27/(12) - 1/3 ) - ( 1/(12) - 1 ) = 17/6 #