Type 1 improper integrals! calculus 2

[@bprpcalculusbasics]

Check out type 2 improper integrals: kzbin.info/www/bejne/rWWZpJ2IftCJbbs

[@asianhero2.096]

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.

[@jcdenton3806]

Same here!

[@alterez6471]

Which one?? I'm new here

[@sin0000]

@@alterez6471 Probably talking about U substitution

[@Miki-dg1md]

I am Ethiopian student i watch your videos I can gave good information .keep it up. Thank you so much

[@dhavalsaxena9573]

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.

[@nvapisces7011]

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

[@anshumanagrawal346]

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

[@haimanotsimegn3381]

You are a best teacher . This lecturer very important i appreciate .

[@nilberthsouza]

Very good. Your videos helped me improve a lot in Calculus. Thank you!

[@user-wu8yq1rb9t]

The happiest place for integration 🤓

[@anshumanagrawal346]

Ikr

[@silentintegrals9104]

solving integrals is allways fun!!!

[@ayush_pandey_17948]

Agreed 👍🏻

[@wegojimmy]

dont let me catch you disrespecting my king like that again lil bro

[@neilgerace355]

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..."

[@fedibaklouti3239]

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

[@AmanyaOctavias]

thank you very much sir u have enabled me to understand improper integrals

[@DavideCosmaro]

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

[@anshumanagrawal346]

17:09 Is still find it weird that the area under the graph of a purely algebraic function, has π in it

[@MicklanOfficial]

This video has made me a better person 😭💔🙌🙌🙌

[@atrabilis1376]

Where can I find the list???

[@syz911]

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.

[@machoslothman]

i love you math papa

[@prudencekamara1707]

I hate it when he smiles go fo difficult things 😅😂😂😂 Anyway he's a genius ❤️

[@silentintegrals9104]

totally agree!!!

[@عليماهر-خ3ب8ك]

very good 😊

[@iñigote]

Just a great explanation as usual

[@237BrillantBABOKA]

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

[@EmpyreanLightASMR]

Diverges to DNE is the name of my black metal doom goth project

[@keeeiif]

The u world... my happy place

[@backyard282]

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

[@silentintegrals9104]

Hope you passed it!

[@DragoniteGaming]

Man is the goat❤

[@timothymuyanga1411]

U got this guys😂😂😂

[@Taro-wd5uo]

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

[@thexoxob9448]

He changed the integration bounds. So the change back wasn't neccessary

[@spencerrosenlund2779]

Super helpful thank you so much!!!

[@nelsonberm3910]

thank you daddy

[@sebbythelord567]

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.

[@michaelattafrimpong1140]

Superb😅

[@PayingPaingPhyo]

thank you sir

[@BrianKipturu-yc2kh]

Someone to like my comment🙏🙏🙏🙏

[@Emine-ri7ex]

example 4 why don't you substitute 1 and infinity in lnx as the pervious example

[@matteocurtarelli1555]

can someone explain to me what is the list? maybe in my country we use another therms for that

[@therealbigfloppa5512]

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.

[@tkbt123]

I LAB U TNX FOR THIS VIDEO

[@hansinosa6838]

Kuya, chinese ka po ba?

[@Elpsycongroo1130]

at 12:21 , whats 'the list'?? :0

[@bunkebear510]

1

[@aaronwhite556]

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)).

[@Sammie-r7d]

I got the same thing😢

[@Hello-ue2xt]

@@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

[@dipp1511]

dat beard tho

[@damjanmladenovic8890]

i love you

[@otsilediale5399]

isn't infinity over infinity an indeterminate form?

[@carultch]

Yes. Infinity over infinity is an indeterminate form.

[@dashie2580]

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?

[@matheusdossantos9252]

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"

[@Hello-ue2xt]

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.

[@thomasblackwell9507]

What list?

[@Stormnorman15]

I dont think this method works when integral is 1 to infinity and is 1/(2x+1)^3 dx

[@monsieurLDN]

I get -1/100

[@Phi1618033]

I wish he still had that beard.

[@najeebullah9658]

What about this integral 1/sinx.

[@williampeterson3498]

Trig sub 😅

[@carultch]

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.

[@user-wu8yq1rb9t]

*By the list??!*

[@Joca-by1pd]

He explained in this video: kzbin.info/www/bejne/pnivgKSbmdeeZ9U

[@user-wu8yq1rb9t]

@@Joca-by1pd It seems I missed that video! Thank you so much my friend

[@thomastran5916]

yeah but you can simply do lhospital to do the limit

[@ureal887]

10:42 but isn't it negative in the law vu-∫vdu

[@carultch]

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.

[@alexandriaarianadillonsrik4601]

SHOW ME THE SECRET WEAPONS!!!!

[@prudencekamara1707]

To do**