Comparison test for improper integrals introduction, calculus 2 tutorial
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- Опубліковано 24 гру 2024
- Learn how to use the comparison test to determine if an improper integral converges or not.
The 3 steps to use the comparison theorem for improper integral: what you know, do the check, and draw the conclusion!
Check out example 2: • Comparison test for im...
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I was honestly so confused and trying to learn math online during this pandemic has just been the worst but this was clear and straight to the point. Thank you!!
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dont think ive ever had a video help me this much. Thanks man, great job.
Thankyou Dude! It helped me alot. Pray for me because I 'm having my Calculus final terminal. :"D
Check out example 2: ua-cam.com/video/AoH58bERkiU/v-deo.html
damn you are one good teacher, keep up the good work!
This video is so informative and helpful! it really helped me on a problem I was stuck on. Thank you!
dude you are so awesome! this makes so much sense! thank u!!!!
Instead of cross-multiplying, could you not have just plugged in 1 from the beginning to show that it was true?
If you did, g(x) = 0 < f(x) = 1/2
thank you finally learned something out of your thousand videos
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Really like your videos. Instead of cross multiplying, you could also divide (the original) both numerator and denominator by x^2 which gives:
1-/x^2 for the numerator
2x^3 + 3/x + 17/x^2 for the denominator
So we see that the numerator is smaller and the denominator is bigger so therefore less than 1/2x^3
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Oh my fucking god ur a saviour. I love listening to your work. Thank you very much for your contribution to all the math noobs
I think using limit comparison test is easier than DCT .... But its really an amazing idea ❤
why we can cross-multiply? It confuses me , someone explains?
why not? everything is positive
Thank you so much. So helpful!
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Thank you it helped me a lot
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the organic chemistry tutor has good physics videos :)
thank you
What happen when the comparison will fales ?
It means you've failed intergrating the function
I enjoyed this
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So just like the direct comparison for series?
Thank you so much!!!!
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If the integral converges don't you have to find the value of it ?
he is so happy for some reason
Wouldn't p technically be -3 as 1/x^3 is the same as x^-3, therefor p < 1 and is divergent?
No, it has to have the form of 1/x^p. So p=3 in this case and would be convergent.
#riemannzeta
Does this integral have an elementary solution?
Ernest schoenmakers yes. You could split up the fraction by dividing each numerator by the denim and then solve from there using whatever method you want.
This seems like too much work for this proof. Why can't you just say that this integral is less than x^4/x^5 which converges and be done?
Remember that, comparison Theorem doesn't work for cosine, sine!
did you say, Introduction, party paragraph, conclusion? haha
Maxstar22 I did! Just like writing an essay
Maxstar22 "body" paragraph btw
blackpenredpen I prefer party paragraph.
Can someone give me that time on where I said "body" paragraph? I couldn't find it lol
11:00
:)