Answer:
[tex]L=\int\limits^b_a {\sqrt{1+\frac{x^2}{64}}} \, dx[/tex]
Step-by-step explanation:
Recall the formula for the arc length of a function given a closed interval [tex][a,b][/tex]: [tex]L=\int\limits^b_a {\sqrt{1+(\frac{dy}{dx})^2}} \, dx[/tex]
Given the function [tex]y=\frac{x^2}{16}[/tex], then [tex]\frac{dy}{dx}=\frac{x}{8}[/tex] by the product rule.
Thus, the set-up integral would be:
[tex]L=\int\limits^b_a {\sqrt{1+(\frac{x}{8})^2} } \, dx\\ \\L=\int\limits^b_a {\sqrt{1+\frac{x^2}{64}}} \, dx[/tex]
