find the arc length of the graph of the function over the indicated interval

Question

y= 3/2x^2/3 [1,8]

Answer 1

The arc length is given by the formula ∫√(1+x^-2/3)dx with limits x=1 and x=8. The integration is rather tricky. Here's how I went about it:

Begin with the substitution u=x^-1/3. Then x=u^-3 so that dx=-3u^-4du. The integral becomes ∫-3u^-4√(1+u²)du with limits u=1 and u=1/2.Note that 1/2 is the upper limit.

Now use the substitution u=tanθ. Then du=sec²θdθ. The integral becomes ∫-3sec³θ⋅(tanθ)^-4dθ with limits θ=π/4 and θ=tan^-1(1/2). Again, tan^-1(1/2) is the upper limit, despite it being less than π/4.

Rewrite the last integral using sines and cosines and you'll get ∫-3cosθ(sinθ)^-4dθ, which we can finally integrate to get (sinθ)^-3. Evaluating at the limits on θ gives the arc length as 5√5 - 2√2.

Answer 2

Unfortunately your question is not clear.

However for finding L you can use

L = ∫√1 + (dy/dx)^2 dx bounded x = 1 to x = 8

Minoo