I really am blessed to be living in a time like this in terms of math learning. The trick you used for integration by parts was nuts and I would have never known it existed without this video. Great job.
23:04 The divergence of this Integral is similar to the divergence of the Infinite GP with Common Ratio -1, if you look at the graph of cosine x, it's area till any positive number, starting from 0, ranges between -1 and 1. And, it keeps doing so in a fix pattern, never approaching anything
The last question integral can be done by factoring x² and moving it to the numerator for the integral to become x^(-2)/(1-x^(-1)). Numerator is exactly the derivative of the denominator which gives you the result of the improper integral without solving the partial fractions
I like how maths works a week ago id know what was going on in his videos and once i started my integration classes, a week later i am able to understand every single thing.
Integral converges, if limit exists and finite Integral diverges, if limit not exists or exists, but tends to infinity. Question arrises. When integral is indeterminate, then? 23:04
Im confused, he plugs in the values for u without replacing u back with its x term, doesn’t he have to change the integration bounds? Or does it not matter for infinite stuff
@@Sammie-r7d integral of 1x^-3/2 is -2x^-1/2. -1/2 multiplied by its reciprocal is 1 (-1/2 * -2/1 = 1). subtract one from exponent -1/2, you get -3/2. So the derivative of -2x^-1/2 is 1x^-3/2 and therefore that (-2x^-1/2) is the integral we are looking for
I'm confused... in 18:40 he said e to negative infinity equals zero. Then around 20:13 he was saying in x/e^infinity the e is infinity. If x and e are both infinity then yes, it equals zero. But if e is 0 then i think the lim equation is undefined. Can someone explain?
lim x -> -inf (x/e^-x) Well, lets subs that -inf/e^-(-inf) -inf/e^inf Which do you think the value "explode" faster? e^x >> x, Case similar to x/x^2 or 1/x This is an assessment without mathematical rigor but "is seen intuitively"
e^-infinity = 1/e^infinity, which you can think of as 1/infinity and it is a very small number close to 0. for the second case, x is approaching negative infinity [i will show it as (-)infinity]. x/e^(-)infinity = x/e^(-)(-)infinity = x/e^infinity. If x is approaching (-)infinity, the denominator is approaching infinity faster than how x is approaching (-)infinity because the denominator is e^infinity. So it is 0.
Très bonne vidéo, j'ai beaucoup appris de celle ci. Cependant à 18:21, je pense que la réponse est 1-e au lieu de -1+e. Vérifiez svp, car lorsqu'on inverse les bornes d'une intégrale, On multiplie l'intégrale par le signe -, chose qui n'a pas été faite dans ce cas. Depuis le CAMEROUN
The way the infinity is substituted as the limit is not correct. You have to make a definite limit and then take the limit to the infinity. If the limit diverges then you have to use Cauchy's Principal Value theorem. Just wanted to make sure everybody gets the correct way of doing it.
Leaving this here for someone if they also want to know. I think the list is just a list of common functions or examples that he uses a lot in his problems. Without the list you can use L Hopitals rule and differentiate top and bottom of fraction and get (1/x)/1 which is just 1/x and the limit of that to infinity is just 0.
Given: integral 1/sin(x) dx Strategically multiply by sin(x)/sin(x): integral sin(x)/sin(x)^2 dx Use the fundamental Pythagorean identity to rewrite sin(x)^2: integral sin(x)/(1 - cos(x)^2) dx Let u = cos(x), thus du = -sin(x). Rewrite in the u-world: integral -1/(1 - u^2) du This we can recognize a relationship to the derivative of arctanh(u): d/du arctanh(u) = 1/(1 - u^2) Thus our integrand is d/du -arctanh(u) Result: -arctanh(u) Recall u = cos(x), add +C and we have our solution: integral 1/sin(x) dx = -arctanh(cos(x)) + C This integral diverges when its bounds approach asymptotes, such as x=0 and x=pi. If we integrate it a second time, it will produce improper integrals that converge, when bounds approach the original asymptotes. I'll leave that as an exercise to you.
Check out type 2 improper integrals: ua-cam.com/video/w46sjRIkV7Y/v-deo.html
I really am blessed to be living in a time like this in terms of math learning. The trick you used for integration by parts was nuts and I would have never known it existed without this video. Great job.
Same here!
Which one?? I'm new here
@@alterez6471 Probably talking about U substitution
23:04 The divergence of this Integral is similar to the divergence of the Infinite GP with Common Ratio -1, if you look at the graph of cosine x, it's area till any positive number, starting from 0, ranges between -1 and 1. And, it keeps doing so in a fix pattern, never approaching anything
The last question integral can be done by factoring x² and moving it to the numerator for the integral to become x^(-2)/(1-x^(-1)). Numerator is exactly the derivative of the denominator which gives you the result of the improper integral without solving the partial fractions
I am Ethiopian student i watch your videos
I can gave good information .keep it up.
Thank you so much
I like how maths works a week ago id know what was going on in his videos and once i started my integration classes, a week later i am able to understand every single thing.
You are a best teacher . This lecturer very important i appreciate .
Very good. Your videos helped me improve a lot in Calculus. Thank you!
17:09 Is still find it weird that the area under the graph of a purely algebraic function, has π in it
3:43
::x sees a difficult integral::
::u comes along::
u says, "I'm in my happy place ... I'm in my happy place .. I'm in my happy place..."
Integral converges, if limit exists and finite
Integral diverges, if limit not exists or exists, but tends to infinity.
Question arrises. When integral is indeterminate, then? 23:04
The happiest place for integration 🤓
Ikr
Agreed 👍🏻
dont let me catch you disrespecting my king like that again lil bro
at 12:21 , whats 'the list'?? :0
Where can I find the list???
24:25= factoring x^2 would've left you with 1/(x^2 (1- 1/x)) and in if u= 1- 1/x then du= 1/x^2 and the whole integral becomes 1/u
Im confused, he plugs in the values for u without replacing u back with its x term, doesn’t he have to change the integration bounds? Or does it not matter for infinite stuff
He changed the integration bounds. So the change back wasn't neccessary
2:06 why is it the reciprocal? when i did it, i took the integral we usually take and got -1/2. so i got -1/(2sqrt(u)).
I got the same thing😢
@@Sammie-r7d integral of 1x^-3/2 is -2x^-1/2. -1/2 multiplied by its reciprocal is 1 (-1/2 * -2/1 = 1). subtract one from exponent -1/2, you get -3/2. So the derivative of -2x^-1/2 is 1x^-3/2 and therefore that (-2x^-1/2) is the integral we are looking for
thank you very much sir u have enabled me to understand improper integrals
I'm confused... in 18:40 he said e to negative infinity equals zero. Then around 20:13 he was saying in x/e^infinity the e is infinity. If x and e are both infinity then yes, it equals zero. But if e is 0 then i think the lim equation is undefined. Can someone explain?
lim x -> -inf (x/e^-x)
Well, lets subs that
-inf/e^-(-inf)
-inf/e^inf
Which do you think the value "explode" faster?
e^x >> x, Case similar to x/x^2 or 1/x
This is an assessment without mathematical rigor but "is seen intuitively"
e^-infinity = 1/e^infinity, which you can think of as 1/infinity and it is a very small number close to 0.
for the second case, x is approaching negative infinity [i will show it as (-)infinity].
x/e^(-)infinity = x/e^(-)(-)infinity = x/e^infinity.
If x is approaching (-)infinity, the denominator is approaching infinity faster than how x is approaching (-)infinity because the denominator is e^infinity. So it is 0.
what the hell is list , where can i find that ? help please
Just a great explanation as usual
Très bonne vidéo, j'ai beaucoup appris de celle ci. Cependant à 18:21, je pense que la réponse est 1-e au lieu de -1+e. Vérifiez svp, car lorsqu'on inverse les bornes d'une intégrale, On multiplie l'intégrale par le signe -, chose qui n'a pas été faite dans ce cas. Depuis le CAMEROUN
The way the infinity is substituted as the limit is not correct. You have to make a definite limit and then take the limit to the infinity. If the limit diverges then you have to use Cauchy's Principal Value theorem. Just wanted to make sure everybody gets the correct way of doing it.
can someone explain to me what is the list? maybe in my country we use another therms for that
Leaving this here for someone if they also want to know. I think the list is just a list of common functions or examples that he uses a lot in his problems. Without the list you can use L Hopitals rule and differentiate top and bottom of fraction and get (1/x)/1 which is just 1/x and the limit of that to infinity is just 0.
the first example id false ,you have to Either change u back to x+1 ,or to make it equals 2 in the left side
example 4 why don't you substitute 1 and infinity in lnx as the pervious example
I hate it when he smiles go fo difficult things 😅😂😂😂
Anyway he's a genius ❤️
Super helpful thank you so much!!!
10:42 but isn't it negative in the law vu-∫vdu
The negative sign from the original IBP formula, is accounted for in the signs column of the IBP table. The signs column starts on +, and alternates.
very good 😊
Man is the goat❤
This video has made me a better person 😭💔🙌🙌🙌
isn't infinity over infinity an indeterminate form?
Yes. Infinity over infinity is an indeterminate form.
Kuya, chinese ka po ba?
Diverges to DNE is the name of my black metal doom goth project
What list?
U got this guys😂😂😂
i love you math papa
I LAB U TNX FOR THIS VIDEO
What about this integral
1/sinx.
Trig sub 😅
Given: integral 1/sin(x) dx
Strategically multiply by sin(x)/sin(x):
integral sin(x)/sin(x)^2 dx
Use the fundamental Pythagorean identity to rewrite sin(x)^2:
integral sin(x)/(1 - cos(x)^2) dx
Let u = cos(x), thus du = -sin(x). Rewrite in the u-world:
integral -1/(1 - u^2) du
This we can recognize a relationship to the derivative of arctanh(u):
d/du arctanh(u) = 1/(1 - u^2)
Thus our integrand is d/du -arctanh(u)
Result:
-arctanh(u)
Recall u = cos(x), add +C and we have our solution:
integral 1/sin(x) dx = -arctanh(cos(x)) + C
This integral diverges when its bounds approach asymptotes, such as x=0 and x=pi. If we integrate it a second time, it will produce improper integrals that converge, when bounds approach the original asymptotes. I'll leave that as an exercise to you.
those are easy, on my test we'd get improper integrals that couldn't be integrated. instead we had to use different tests for convergence
I dont think this method works when integral is 1 to infinity and is 1/(2x+1)^3 dx
I get -1/100
thank you sir
dat beard tho
honestly i wish i had just ignored my professor and learnt from you from the start. he’s nice and very smart but wow he’s a bad teacher.
Superb😅
thank you daddy
*By the list??!*
He explained in this video: ua-cam.com/video/pGLOqedrk1s/v-deo.html
@@Joca-by1pd
It seems I missed that video!
Thank you so much my friend
yeah but you can simply do lhospital to do the limit
The u world... my happy place
Someone to like my comment🙏🙏🙏🙏
I wish he still had that beard.
1
i love you
SHOW ME THE SECRET WEAPONS!!!!
To do**