Type 1 improper integrals! calculus 2
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- čas přidán 26. 06. 2024
- We will solve 8 type 1 improper integrals for your calculus 2 class. A type 1 improper integral means we have to integrate over an infinite interval, such as from a to infinity, from negative infinity to b, or from negative infinity to positive infinity. An improper integral is a combination of integral and limit. So you must remember all the integration techniques and also how to take the limit.
Check out type 2 improper integrals: • Type 2 improper integr...
0:00 how do we do improper integrals (type 1 improper integral, 8 examples)
0:12 integral of 1/(x+1)^(3/2) from 0 to inf
3:17 integral of x^2/sqrt(x^3+4) from 0 to inf
5:59 integral of e^(1/x)/x^2 from 1 to inf
9:28 integral of ln(x)/x^2 from 1 to inf
13:36 integral of x/(1+x^4) from 0 to inf
17:11 integral of x*e^x from negative inf to 0
21:20 integral of cos(x) from 0 to inf
23:14 integral of 1/(x^2-x) from 2 to inf
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Check out type 2 improper integrals: czcams.com/video/w46sjRIkV7Y/video.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
Very good. Your videos helped me improve a lot in Calculus. Thank you!
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.
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
You are a best teacher . This lecturer very important i appreciate .
Just a great explanation as usual
thank you very much sir u have enabled me to understand improper integrals
Super helpful thank you so much!!!
The happiest place for integration 🤓
Ikr
solving integrals is allways fun!!!
Agreed 👍🏻
Man is the goat❤
Where can I find the list???
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
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..."
I LAB U TNX FOR THIS VIDEO
This video has made me a better person 😭💔🙌🙌🙌
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
U got this guys😂😂😂
thank you sir
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example 4 why don't you substitute 1 and infinity in lnx as the pervious example
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.
I hate it when he smiles go fo difficult things 😅😂😂😂
Anyway he's a genius ❤️
totally agree!!!
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Superb😅
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.
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😢
What list?
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.
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
Hope you passed it!
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"
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
thank you daddy
dat beard tho
isn't infinity over infinity an indeterminate form?
Yes. Infinity over infinity is an indeterminate form.
i love you
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.
I dont think this method works when integral is 1 to infinity and is 1/(2x+1)^3 dx
I get -1/100
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.
Kuya, chinese ka po ba?
*By the list??!*
He explained in this video: czcams.com/video/pGLOqedrk1s/video.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
Can you explain this in maithali ?🙄😛
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