A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 one year ago
How do i find the arc length of: INT[1,x] sqrt(t^31) dt on 1<x<4?
anonymous
 one year ago
How do i find the arc length of: INT[1,x] sqrt(t^31) dt on 1<x<4?

This Question is Closed

freckles
 one year ago
Best ResponseYou've already chosen the best response.2so you want to find the arclength of \[f(x)=\int\limits _1^x \sqrt{t^31} dt \text{ on } 1<x<4 \\ \text{ recall the arclength is given by } \\ L(x)=\int\limits_a^b \sqrt{1+(f'(x))^2} dx\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0f'(x)=\[\frac{ 1 }{ 2 }x^2+\frac{ 1 }{ 2 }\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[f'(x)=\sqrt{x^31} \text{ by fundamental theorem of caluclus }\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[L=\int\limits_1^4 \sqrt{1+(\sqrt{x^31})^2} dx \\ \\ \text{ note : didn't mean to put } L(x) \text{ earlier } \\ L=\int\limits _1^4 \sqrt{1+(x^31)} dx\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2not entirely sure how you found your f'

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh thank you i understand know. I forgot to apply the first part of the fundamental theorem.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.