What is the arc length of #f(x)=(3/2)x^(2/3)# on #x in [1,8]#?

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Answer 1 · 2018-03-19T17:46:39

#L=5sqrt5-2sqrt2# units.

#f(x)=3/2x^(2/3)#

#f'(x)=x^(-1/3)#

Arc length is given by:

#L=int_1^8sqrt(1+x^(-2/3))dx#

Simplify:

#L=int_1^8sqrt(1+x^(2/3))(x^(-1/3)dx)#

Apply the substitution #x^(2/3)=u#:

#L=3/2int_1^4sqrt(1+u)du#

Integrate directly:

#L=[(1+u)^(3/2)]_1^4#

Hence

#L=5sqrt5-2sqrt2#