Evaluating an Improper integral (Type 1)

In this video, I showed how to evaluate an improper integral with an infinite boundary

[@punditgi]

Prime Newtons is the proper way to learn calculus!

[@davidgagen9856]

Great explanation.

[@jan-willemreens9010]

... A good day to you Newton, I can see you're still doing outstanding well! Regarding your clear presentation just a brief comment. I think it would clarify the problem/situation considerably, if you made just a simple sketch of the orthogonal hyperbola 2/x^e between the bounds x = 1 and + inf. and the shaded enclosed area ... a case of convergence (lol) ... perhaps it is going to far to talk about convergence/divergence in this specific presentation ... Thank you Newton for another well structured and entertaining performance on "improper(ly) behaving" integrals ... A nice day to you, Jan-W

[@PrimeNewtons]

I promise, the next video will address convergence and also come with a graph 😃. Your comments are always invaluable. Thank you.

[@jan-willemreens9010]

@@PrimeNewtons ... Just to be clear Newton, I wasn't hinting for an explanation on convergence/divergence from you, my experience with tutoring students is, that they frequently ask me what the meaning is of calculations in general, so I just simply draw a picture of the situation for them, that's all; the smile on their faces says it all for me (lol) ... I don't want to take over your excellent performances, just to be clear again! I enjoy and appreciate your presentations very much ... Jan-W p.s. By the way, talking about divergence /convergence thinking of the integrals of 1/x and 1/x^2 both from x=1 to x --> +inf. ; big unexpected outcomes when comparing the two, to be honest I was shocked when seeing the different solutions of the improper integrals (lol)... very interesting ...

[@PrimeNewtons]

You are kind. I actually planned to do that, and.......your feedback always inspires! Thank you.

[@frieszyyy3220]

how di it become -e?

[@domanicmarcus2176]

At time 5:06, how did you go from 1 over 1 to the e to just 1? I am a little lost on that. Can you please explain that?

[@ThenSaidHeUntoThem]

1^e is 1. Remember, 1 to any power is always 1.

[@eastonpeter1242]

1:00 to 1:50. I am having trouble with the reasoning behind changing the improper integral......using "t" to replace ""infinity."

[@PrimeNewtons]

Because infinity is not a number, it cannot be plugged in. You can only take limits as we approach infinity