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"
