OMG thank you so much. So everytime i have to evaluate an integral as a power of series, I should try to compare it to a simple function which the representation as a power of series is known?
Thanks, Krista. I'm taking the Coursera Calculus Two: Sequences and Series from Ohio State. Dr. Fowler didn't go into much detail with how to integrate the equivalent infinite sum. Writing out several terms and finding antiderivatives was the trick that helped me see the pattern to find the generalized antiderivative.
hey, near the end wouldn't 8n+1 = (8n)+1 and not 8(n+1) as the additional t exponent was added after n was multiplied by 8? therefore A sub n+1 would be x^8n+2. the answer is the same either way
Hey Krista, I can't thank you enough! Please keep uploading. There are very few of you.
I'm so glad I've been able to help, Deepankar! I'll do my best to keep uploading new videos! :)
You broke that down so well! Thank you.
You're welcome! Glad it could help.
10 years later and its still helpful
That means so much to me, I’m so glad it helped!! ❤️
Thank you so much! This makes so much sense!!
Super awesome explanation!
I know it won't change the answer but you missed 8n+1 when n=3 at 5:17
Simply amazing!!
OMG thank you so much. So everytime i have to evaluate an integral as a power of series, I should try to compare it to a simple function which the representation as a power of series is known?
Hi, what to do when given with a difinite integral? Pls help
Good quality explanations, but a few errors. (t^8)^(1/8)=|t|, not t. In the last few steps, you've implied that the series converges for all t
Thanks, Krista. I'm taking the Coursera Calculus Two: Sequences and Series from Ohio State. Dr. Fowler didn't go into much detail with how to integrate the equivalent infinite sum. Writing out several terms and finding antiderivatives was the trick that helped me see the pattern to find the generalized antiderivative.
Oh awesome! I'm so glad that helped, sometimes it just makes it easier to see. :D
Awesome! Truly, thank you.
You're very welcome, Nova, I'm so glad it helped! :D
hey, near the end wouldn't 8n+1 = (8n)+1 and not 8(n+1) as the additional t exponent was added after n was multiplied by 8? therefore A sub n+1 would be x^8n+2. the answer is the same either way
Spooky, I just did this in class an hour ago and now boom here's the video.
Good timing! ;)
lovin this mayn
what if the exponent was an odd number?
Awesome mam nice concept
Thanks, Yogendra! :)
Fantastic
Did you forget to take the reciporical of the limit (L) at the end to get R?
What a terribly worded problem, thanks for showing me what it was asking for!
Nice help more