The arc length can be quickly derived by realizing that it's just the distance formula in combination with the derivative over some interval #Deltax#. Thus:
#s(x) = sum_(a)^(b) sqrt((Deltax)^2 + (Deltay)^2/(Deltax)^2(Deltax)^2)#
#= sum_(a)^(b) sqrt(1 + (Deltay)^2/(Deltax)^2)Deltax#
#s(x) = int_(a)^(b) sqrt(1 + ((dy)/(dx))^2)dx#
Therefore:
#(dy)/(dx) = 3/2*2/3x^"-1/3" = x^"-1/3"#
#((dy)/(dx))^2 = x^"-2/3"#
#s(x) = int_1^8 sqrt(1 + 1/x^"2/3")dx#
What normally what we would do is try to find a way to complete the square. Let's just make this look nicer for now.
#= int_1^8 1/sqrt(x^"2/3")sqrt(x^"2/3" + 1)dx#
#= int_1^8 1/x^"1/3"sqrt((x^"1/3")^2 + 1)dx#
Now let's make the substitution:
#u = x^"1/3" => u^3 = x#
#dx = 3u^2du#
#= 3int_1^8 u^2*1/usqrt(u^2 + 1)du#
#= 3int_1^8 sqrt(u^2 + 1)*udu#
The way I would proceed is with another substitution. Let:
#w = u^2 + 1#
#dw = 2udu#
#udu = (dw)/2#
#=> 3/2 int_1^8 sqrtwdw#
#= w^"3/2" = (u^2 + 1)^"3/2"#
#= [(x^"2/3" + 1)^"3/2"]|_(1)^(8)#
#= [(8^"2/3" + 1)^"3/2"] - [(1^"2/3" + 1)^"3/2"]#
#= (4 + 1)^"3/2" - (1 + 1)^"3/2"#
#= 5^"3/2" - 2^"3/2"#
#= sqrt125 - sqrt8#
#= color(blue)(5sqrt5 - 2sqrt2) ~~ 8.3519 "u"#
