What is the arc length of #f(x)=10+x^(3/2)/2# on #x in [0,2]#?

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Answer 1 · 2018-03-15T08:30:51

#L=1/27(17sqrt34-32)# units.

#f(x)=10+1/2x^(3/2)#

#f'(x)=3/4sqrtx#

Arc length is given by:

#L=int_0^2sqrt(1+9/16x)dx#

Integrate directly:

#L=32/27[(1+9/16x)^(3/2)]_0^2#

Insert the limits of integration:

#L=32/27((17/8)^(3/2)-1)#

Simplify:

#L=1/27(17sqrt34-32)#