Discover the answers you need at Westonci.ca, a dynamic Q&A platform where knowledge is shared freely by a community of experts. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.
Sagot :
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]
Thank you for trusting us with your questions. We're here to help you find accurate answers quickly and efficiently. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.
