What is the arc length of #f(x)=6x^(3/2)+1 # on #x in [5,7]#?

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Answer 1 · 2018-03-19T17:41:15

#L=4/243(568sqrt142-203sqrt406)# units.

#f(x)=6x^(3/2)+1#

#f'(x)=9sqrtx#

Arc length is given by:

#L=int_5^7sqrt(1+81x)dx#

Integrate directly:

#L=2/243[(1+81x)^(3/2)]_5^7#

Hence

#L=4/243(568sqrt142-203sqrt406)#