Find the arc length of the graph of the function over the indicated interval.

y = 3/2x^2/3, [8, 125]

round to 3 decimal

thnxs so much

Please use parenthesis. It's very disturbing when one has to try to guess what belongs in the denominator and what doesn't. I'm going to guess that 3/2 is being multiplied by x^(2/3).

Now, let's start by taking the derivative of our function.

y = (3/2)x^(2/3)
y ' = x^(-1/3)
y ' = 1 / x^(1/3).

Now let's use the fact that L is our arc length and L = ∫ ds, where ds = sqrt( 1 + (dy/dx)^2 ) dx.

Note: Let's neglect the bounds for now, just to avoid tediousness.


∫ sqrt{ 1 + [1 / x^(1/3)]^2 } dx,
∫ sqrt[ 1 + (1 / x^(2/3)] dx,
∫ sqrt{ [x^(2/3) + 1] / x^(2/3) } dx,
∫ sqrt(x^(2/3) + 1) / sqrt(x^(2/3)) dx
∫ sqrt(x^(2/3) + 1) / x^(1/3) dx.

Now let,

u = x^(2/3) + 1
du = (2/3)x^(-1/3) dx
3du/2 = x^(-1/3) dx.

3/2 ∫sqrt(u) du
3/2 ∫ u^(1/2) du
(3/2)(2/3)u^(3/2) [ (8,125)
(x^(2/3) + 1)^(3/2)) [ (8,125).

Plugging in our bounds, gives:

(125^(2/3) + 1)^(3/2) = 132.57

(8^(2/3) + 1)^(3/2) = 11.18

132.575 - 11.18 = 121.394.


Hope this helped much.