3How do you find the lengths of the curve #y=2/3(x+2)^(3/2)# for #0<=x<=3#?

Question

9460 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-18T15:03:05

#=4 sqrt 6 - 2 sqrt 3#

Length, s, is:

#s = int_0^3 sqrt(1+ (y')^2) dx#

#y' = (x+2)^(1/2)#

#implies s = int_0^3 sqrt(3+ x) dx#

#= [2/3(3+x)^(3/2) ]_0^3#

#=2/3( 6^(3/2) - 3^(3/2) )#

#=4 sqrt 6 - 2 sqrt 3#