Find the Arc Length of a Function with Calculus 

remember this for calc 3 and during your last unit of calc 2 there should be parametrization (most important thing for calc 3 that I slacked off on because it was the end of semester)

Hehe thank you!

I'm happy that you enjoyed it!

Thank you very much! Please kindly subscribe and share! You can find all of my Calculus II videos here: ru-vid.com/group/PLeTO6OT3-FKmuCeO97iKt_Aibx-a938JA

I'm really glad to hear that Sun! Please kindly subscribe or share! I really appreciate it.

You're welcome! I'm glad you like it! You can find the rest of my Calculus videos in this playlist: ru-vid.com/group/PLeTO6OT3-FKmuCeO97iKt_Aibx-a938JA

Thanks so much Henkoegema! You can find all of my Calculus II videos in the following link (don't forget to share and kindly subscribe!): ru-vid.com/group/PLeTO6OT3-FKmuCeO97iKt_Aibx-a938JA

Hi Vincen! Please use the formula mentioned in the video and let me know what you got!

You're so welcome Stephen! Please kindly subscribe and share with classmates, and you can find the rest of my Calculus videos in the playlist here: ru-vid.com/group/PLeTO6OT3-FKmuCeO97iKt_Aibx-a938JA

Hi Nomsaphiri! I don't offer personal online tutorials, but please feel free to ask my questions in the comments. You can find the rest of my Calculus 2 videos here: ru-vid.com/group/PLeTO6OT3-FKmuCeO97iKt_Aibx-a938JA

@@QuocDatPhung thankyou this is very helpful

You're welcome Temwananimumba :)!! I'm really glad you enjoyed my video! I would really appreciate if you could share with your classmates or kindly subscribe ~ you can find all of my Calculus II videos in this link: ru-vid.com/group/PLeTO6OT3-FKmuCeO97iKt_Aibx-a938JA

Hi Zainab, can you give me an example of what you mean?

No I did not forget. I did add +1 at 12:06. Let's see why: The integrand 1+(1/4)y^6 - 1/2+1/(4y^6) = (1/4)y^6 + 1/2+1/(4y^6) and I wrote it correctly. Let me know if you see it now.