Question

Find the arclength of:
7x^(2/3) from x=1 to x=8

Answer 1

I have gotten up to this point:
integral sq. root 9x^2/3+ 196/3 times x^-1/3 dx

but after that I'm stuck

Answer 2

f'(x)= (-1/3)7x^(-1/3)
for arclength, you have to integrate the square root of (1+f'(x)^2, so you can square the f'(x) piece by piece... (-1/3)^2 = (1/9)... (7^2)=49... [x^(-1/2)]^2 = (x^(-1)... put them all together... (1/9)(49*(x^-1)). Add 1. Square root it. Integrate it. The hardest part will be doing the busy work of integrating.
Got you started.