Answer:
6.7
Step-by-step explanation:
The function x(y) can be rewritten as
[tex]x = \frac{y^3}{15} + \frac{5}{4y}[/tex] for [tex]3 \leq y \leq 5[/tex]
The integral formula for the curve length is as the following
[tex]L = \int\limits^5_3 {\sqrt{1 + \left(\frac{dx}{dy}\right)^2}} \, dy\\ \frac{dx}{dy} = x' = 0.2y^2 - 1.25y^{-2}\\L = \int\limits^5_3 {\sqrt{1 + (0.2y^2 - 1.25y^{-2})^2}} \, dy\\\\L = \int\limits^5_3 {\sqrt{1 + 0.2^2y^4 - 2*0.2*1.25y^2y^{-2} + 1.25^2y^{-4}}} \, dy\\\\\L = \int\limits^5_3 {\sqrt{0.04y^4 + 0.5 + 1.5625y^{-4}}} \, dy\\\\[/tex]
This is an indefinite integral and can be solved numerically to get the result of 6.7
