Introduction to improper integrals | AP Calculus BC | Khan Academy

[@Braeden123698745]

You made me unerstand this in 3:51 minutes, my teacher only confused me in 2 hours 20 minutes.

[@yansu3363]

bro u should come to our school, the teacher dont even know how to do it, and the student teach the teacher.

[@ian.ambrose]

Useless teachers, they should get executed.

[@josuequintero-td5gh]

My calc II class is asynchronous (big mistake) and the only reference material we have for each unit is a single video where the teacher does some practice problems.

[@khwajaabdurrehman1375]

for reeeeaaaaal

[@toribenita_kyo]

My calc prof asked us to calculate the area of the surface of y = sqrt(24-4x) revolved around the x-axis with an upper limit of 6 and a lower limit of 3. I was losing my patience with my calculator when it kept spitting out "math error" since 24-4(6) = 0. I ended up looking for help online to get the right answer. And then I figured out that I could have just plugged in 5.99999 instead of 6 to get a close-to-exact answer. It's a bit unorthodox, but it worked for me. I don't recommend doing it though. You want your calculations to be as true as possible. Only use shortcuts when you're absolutely desperate.

[@bichpham7089]

thank you, this makes improper integrals much easier to understand :)))

[@GothGeekAnimeFreak]

Thank you thank you thank you for helping me pass my exam last week!!!!!!!

[@liptoncheetos]

With this corona thing my teacher just told us to to go here for maths.

[@yeonhojung7185]

Thank you so much!

[@mrnosy1]

Awesome video! It would be cool if you could do some introductory video's for QM involving normalizing the wave function using improper integrals of ±∞

[@Alan157]

Thanks, very good video.

[@Juxtaroberto]

Exactly. The x value approaches a, or in this case M, but never equals that value, but the function equals its limit.

[@AshrafAli-qs6ep]

thanks sal

[@blessinggift3474]

Really helpful

[@whoknows8992]

bro come to my college here the teacher graduated from arts teach mathematics😂😂😂😂

[@zombiesalad2722]

rip

[@uzmashah7204]

Y'ALL. 1/n when n=infinity goes to zero because n can be any number like 5 or even 100. when you denominator is bigger than the numerator, it's a small number. 1/5 > 1/100. so as n gets larger, you get closer to zero. 1-0=1.

[@zombiesalad2722]

Thanks Captain Obvious

[@anteater2197]

thank you :)

[@harveythrondsen9354]

Tomorrow I have final exams.And I learned all the definite, indefinite, improper integrals in 2 days.3 days ago I knew nothing about integral.Thanks to internet.

[@nayeemrahman9358]

Thank you sir

[@pponcho8245]

This is mind blowing. How is infinity equal to exactly one. What math is this? Calculus?

[@mrnosy1]

yes calculus, but this doesn't mean infinity is exactly equal to one. We're just saying that as x approaches infinity starting from one, that since the function 1/(x^2) is getting closer and closer to a finite value as x approaches infinity that the area under the curve withing a certain boundary (1 to infinity) also approaches a finite value which can be written as (-1/n + 1). And since n approaches infinity on the bottom of the fraction the fraction will be infinitessimaly small so area = 1.

[@bhavanimadasu1548]

Thank you sir Ur telling very clearly to understand the concept

[@DrGanni13al]

0:35, already knew the ending

[@mkwarlock]

Nice video, kind sir.

[@Juxtaroberto]

No, the area is exactly 1. N approaches infinity, but A=1.

[@AvgJane19]

Juxtaroberto that's what he said...

[@Wayferrer36]

thank you

[@TheDrB0B]

Is there a Khan academy video about improper integrals when a function is discontinuous?

[@lifeDotGov]

It doesn't approach 1, it IS 1. Limit doesn't mean it's approaching, if you learn the epsilon delta (precises) definition of a limit, you will see that it IS, not "approach"

[@mTeeDev]

i think olevs right there. it approaches 1 not directly 1. but its so infinitely close to the 1 it is safe to say that it is 1. :)

[@horaciosalgado]

yes, that's exactly right type in your calculator 1/10 and 1/10000 and you will see that the bigger the denominator the closer it reaches 0.

[@kevcopo]

Lmao Bru that’s what you got out of this 😭

[@anthonybacha2521]

I like this video

[@nbme-answers]

this is the beginnings of a calculus problem known as Gabriel's Horn en.wikipedia.org/wiki/Gabriel%27s_Horn

[@Khana-Badosh-Official]

Bit confusion: How can we get exact area 1 if we are evaulting with limit infinity??? I think it should be approximate value

[@MekailTheMachine]

That is implied when he denotes it as a limit approaching infinity. After so many values of X, the change becomes so minute it becomes clear it is "approaching" 1, and that is a key part of using a limit, is it allows us to determine the area it is "approaching".

[@olev456]

i don't really know what you mean by "epsilon delta (precises) definition", maby give me source. Anyway limit does exactly that - approaching. If you think about definite integral definition then you'll see.

[@ganaramesh7278]

you are amazing i love you

[@utuberuber502]

Gana Ramesh no homo

[@lifeDotGov]

chapter 7.8... i was just doing this, lol.

[@bfc10200]

This video isn't on their Windows 8 App, they should add it

[@venzelzandrogabrielbaylan1205]

May I know what kind of app does he use on writing or making the solutions?

[@lifeDotGov]

youtube search "precise epsilon delta definition of limit"

[@christineebdalin6733]

What if the answer is π/8, is it still considered convergent?

[@olev456]

Infinity doesn't equal to one, you got some stuff wrong here. First taking definite integral gets you area under the curve 1/x^2, in this video he takes it from 1 to infinite, meaning that he finds area under the curve with span of 1 to infinite. So basically he tells you that area under the curve from 1 to inf approaches to 1. It might be something like 0.9999999999998 etc depends on how accurate u go.

[@TheZiliuz]

What happens when you take the same integrla, but this time from 0 to infinity? I evaluated it and it returns 0.

[@dfin4588]

Old comment, but Ill respond in case others have the same question: You cant start at 0 because the function is not defined at x=0. It doesnt exist. If you graph the function 1/x^2 you will notice that it has a vertical asymptote at x=0, because division by zero is not defined. So when x=0, the function isnt continuous. This only works for a continuous function. and the function 1/x^2 is continuous on the interval 1 to infinity.

[@yazzuo7213]

So I want an answer now. I hate it. How can an infinite area have a finite area??? 😂 That males no sense man 😅

[@hassaanbaiq3316]

This is not an exact value it's an approximate one. It can have an approximate value as the function is converging and after a point, the distance will become so minute that further changes won't sum up to affect the integral anymore.

[@JESalvatierra]

Da f***