Integral of Inverse Functions

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3 жыл бұрын

🎓Become a Math Master With My Intro To Proofs Course!
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Disclaimer: This video is for entertainment purposes only and should not be considered academic. Though all information is provided in good faith, no warranty of any kind, expressed or implied, is made with regards to the accuracy, validity, reliability, consistency, adequacy, or completeness of this information.
#math #brithemathguy #integral

Пікірлер: 176

@BriTheMathGuy 9 ай бұрын

🎓Become a Math Master With My Intro To Proofs Course! www.udemy.com/course/prove-it-like-a-mathematician/?referralCode=D4A14680C629BCC9D84C

@toast_recon 3 жыл бұрын

It seems crazy that you can do ANYTHING with this. The original form seems completely intractable. This is what I live for :)

@MrRyanroberson1 3 жыл бұрын

the one i always mention on these proofs is ln(x) which is usually kinda hard to integrate, but here you instantly get x ln(x) - e^ln(x) + c = x (ln(x) - 1) + c. Similarly the integral of e^x is therefore x e^x - e^x (x - 1) + c = e^x + c. nice and neat.

@lukedavis6711 3 жыл бұрын

Smart

@Johnny-tw5pr 2 жыл бұрын

Hard to integrate? It's the easiest integration by parts

@omidsedighi-mornani1636 2 жыл бұрын

i mean u just make an integration by parts and derive ln(x) and integrate 1

@MrRyanroberson1 2 жыл бұрын

y'all mentioning integration by parts when that was the proof in the video to begin with. this trick IS integration by parts in the first place. what i meant by calling it hard is, of course, that this is hard to do without integration by parts

@tinotendamandizvidza1903 3 жыл бұрын

Thnxs this is making me like math more

@BriTheMathGuy 3 жыл бұрын

Great to hear!

@aashsyed1277 3 жыл бұрын

Same for me!

@aashsyed1277 3 жыл бұрын

Damn you're so good.....

@tsehayalemu5376 3 жыл бұрын

Wish you thousands of subscribers in your career

@BriTheMathGuy 3 жыл бұрын

Thanks very much!

@maalikserebryakov Жыл бұрын

wish grantee

@mrpedrobraga 3 жыл бұрын

2:24 So accepting, I think I needed to hear that.

@energyeve2152 3 жыл бұрын

I’ve seriously been exploring more math because of inspiring people like you. Keep shining brother ☀️

@diegocoglievinadiaz5665 3 жыл бұрын

The quality of your explanation is unbelievable. You have won a subscriber :)

@atrumluminarium 3 жыл бұрын

I literally had an epiphany seeing this result. I'm definitely gonna find somewhere to use this just to flex

@AshgabatKetchumov 3 жыл бұрын

I remember that question being on one of my single-variable analysis tests last year. That was a pretty good question, not gonna lie.

@alexandrebriard9175 3 жыл бұрын

Man your videos are so cool, I bet that your KZfaq channel will gain a lot of viewers soon!

@BriTheMathGuy 3 жыл бұрын

I appreciate that!

@juandiegoparales9379 3 жыл бұрын

Great video, thanks!!!

@BriTheMathGuy 3 жыл бұрын

Glad you liked it!

@toasticide816 2 жыл бұрын

I just love how simply using a well known method (well two) allows us to integrate any inverse function, which often can seem daunting. Or at least to me they're daunting!

@foolishball9155 Жыл бұрын

Love the last line of this video

@tamazimuqeria6496 3 жыл бұрын

Am gonna use it

@BriTheMathGuy 3 жыл бұрын

Great!

@User-7986iitjee Жыл бұрын

THANK YOU VERY VERY VERY VERY MUCH.

@AstoundingJB 2 жыл бұрын

This result (sometimes referred to as a theorem) has a super nice interpretation when the integral is of the definite type, and actually requires no calculation at all

@iamtrash288 3 жыл бұрын

Beautiful presentation

@BriTheMathGuy 3 жыл бұрын

Thank you! Cheers!

@Mulkek 2 жыл бұрын

Thanks, and explain so clearly!

@BriTheMathGuy 2 жыл бұрын

Glad you think so!

@pcklop 3 жыл бұрын

I got a similar statement about definite integrals by looking at areas on a graph, and remembering that an invertible function must be monotonic. It’s more work than this method, but can be generalized to give the same result!

@bikeshike 3 жыл бұрын

well explained👍🏾

@BriTheMathGuy 3 жыл бұрын

Glad you think so!

@aashsyed1277 3 жыл бұрын

Yes! Well explained

@lukelucas5738 2 жыл бұрын

👍🏻

@hagenfarrell Жыл бұрын

I got this question on my final for calculus 1 and I was blown away

@giuseppemalaguti435 2 жыл бұрын

Bravo, sempre molto preciso

@rick4135 2 ай бұрын

Really nice way to turn into Riemann Stiljies integral after substitution then apply integration by parts!

@MrCigarro50 3 жыл бұрын

Thank you for your video.

@BriTheMathGuy 3 жыл бұрын

Thanks for watching!

@joske7804 3 жыл бұрын

Very concise, good video.

@BriTheMathGuy 3 жыл бұрын

Much appreciated!

@ExtremeAgent 3 жыл бұрын

When the video suddenly ended without any outro, my brain felt like being on the right seat while I drive on the left seat and I just brutally step on the break 😐

@Redentor92 3 жыл бұрын

Nice trick and video! Do you have any example where the integral would be hard but when considered as the inverse of a function this result makes it easy?

@lukedavis6711 3 жыл бұрын

ln(x), sin(x)

@leontsc4352 2 ай бұрын

F(f^-1(x)) can also be written as the integral of f^-1(x), which is our original function: to solve for the function itself you need to solve an equation (I represented the integral of f^-1(x) as the variable u): u = xf^-1(x) - u; 2u = xf^-1(x); u = xf^-1(x)/2; integral of f^-1(x) = xf^-1(x)/2 (If I missed anything let me know 👇)

@binodtharu8348 Жыл бұрын

Nice!

@elfwired 2 жыл бұрын

It has geometrical explanation, flip the curve so x and y axes changes, and area under curve becomes rectangle area minus areaunder the curve.

@anibalismaelfermandois6943 3 жыл бұрын

"Un Dia Vi Una Vaca sin(-) cola(∫) Vestida De Uniforme" ∫udv=uv-∫vdu. Thats the spanish way to remember integration by parts

@MusicalInquisit 3 жыл бұрын

I speak Spanish, but why does cola represent the integral? Is ther some pun I am not getting there? Does it look like a tail?

@michaeljimenez8205 3 жыл бұрын

Un día vi a una vaca vestida de uniforme

@ComicSansaMS 3 жыл бұрын

@@MusicalInquisit an integral looks like a tail, yes.

@dgr751 3 жыл бұрын

I learned It with "un día vi un valiente soldado vestido de uniforme"

@P03enix 3 жыл бұрын

actually retrieving the integration by parts formula is pretty simple, using only the derivative of uv

@SanketGarg 11 ай бұрын

If we look at the second last step, we basically have integral of f (the area under the curve with x axis) + integral of f inverse (are under the curve with y axis)= u * f(u). Which is basically the area of the overall rectangle (summing up the two areas)

@priyanshutyagi3688 3 жыл бұрын

Can u make a video on integration of implict functions

@tom-lukaslubbeke949 8 күн бұрын

I just realised that you must always be writing mirrored, you're writing on glass between you and the camera right? Honestly it's super impressive how nicely you can write mirrored

@Etothe2iPi 3 жыл бұрын

You should definitely add an example (or two).

@whatelseison8970 3 жыл бұрын

Agreed. It's jarring how abruptly that ends. I barely had a chance to see the final line.

@paulg444 2 ай бұрын

should mention that f is a one to one function, else break the integral up into monotonic domains and add.

@depressedguy9467 3 жыл бұрын

Stokes theorem on manifold plz

@justt1ice 2 жыл бұрын

There is an *intuitive* graphical explanation of this: integral(0 to t) of f(x)dx + integral(0 to f(t)) of f^(-1)(y)dy, conviently plotted on the same graph, fill up the rectangle [0,t]×[0,f(t)], so their sum equals tf(t).

@CombustibleL3mon 2 жыл бұрын

I'm a mathematics master's student and I still love watching your videos Bri

@BriTheMathGuy 2 жыл бұрын

Thanks a ton! Have a great day!

@Alaska-mk4ok 3 жыл бұрын

That’s amazing

@BriTheMathGuy 3 жыл бұрын

Glad you think so!

@WahyuHidayat-oj4ro 3 жыл бұрын

Love to watch your video....maths become easier..😊

@BriTheMathGuy 3 жыл бұрын

Thanks a lot! Glad you think so!

@Mauriciohse 3 жыл бұрын

In the end, the integral depends on F(f^-1(x)) - which is the original question, isn't it?

@irfanadlan4662 3 жыл бұрын

@cerezabay 3 жыл бұрын

@coc235 3 жыл бұрын

No, it means you first get the integral of f(x) and then plug in f^(-1)(x)

@nickbagby5313 3 жыл бұрын

@@coc235 that makes things a lot clearer, I was confused about that as well

@angelmendez-rivera351 3 жыл бұрын

F refers to the antiderivative of f, F is not the antiderivative symbol itself.

@Silly_Ah_Girl 2 жыл бұрын

You are the best!!

@BriTheMathGuy 2 жыл бұрын

You are!

@Silly_Ah_Girl 2 жыл бұрын

@@BriTheMathGuy Thanks! This helped me a lot!

@Qwerty-lq2op 2 жыл бұрын

2:57 why uf(u) is there suddenly? Where was it come from?

@ShubhamKumar-sj6dp 2 жыл бұрын

If u are generalising that f-1(f(x))=x is not that a bit wrong , what if function is sin-1(sinx) , but in the question if the domain of the function is from [pi,2pi] sin-1(sinx) comes out to be (2pi-x) not x , it will be x is domain is from [0,pi] , is not the statement said case dependent ?

@ShockerXL Жыл бұрын

bri: math can be fun! just watch also bri: *writes f(u) = x unironically*

@BramCohen 2 жыл бұрын

Can you integrate the function where f(f(x)) == e^x ?

@worldnotworld 2 жыл бұрын

Are there any good applications for this?

@babajani3569 3 жыл бұрын

Does this only work for to one functions or can you do this for other functions as well, with a restricted domain such as inverse trig functions e.g. sin^-1x

@liamhanson9538 3 жыл бұрын

The inverse of f exists if and only if f is a bijection, so yeah. Of course if you define a function on restricted domain st it's a bijection then ostensibly this would work.

@babajani3569 3 жыл бұрын

@@liamhanson9538 ok thank you so this would work for inverse trig functions right?

@idrisShiningTimes Жыл бұрын

@@babajani3569 yes in restricted domain only

@venkataramana-mb7hc 2 ай бұрын

2:51 what happened here Can anyone explain?

@SeanStephensen 3 жыл бұрын

this is easy using power rule. We know d/dx(f(x)) = 1*f^0(x) = f^0(x), and d^2/dx^2(f(x)) = d/dx(f^0(x)) = 0*f^-1(x), which we can't simplify to 0 because f^-1(x) could be 0 or infinity at some point, making this second derivative indeterminate. But this gives us an identity to help us solve the integral at hand. So the integral of f^-1(x) is simply f^0(x)/0 + c = infinity.

@fernandovictor708 2 жыл бұрын

This works for every inverse continuos function?

@willie333b Жыл бұрын

Nice

@denischen8196 2 жыл бұрын

What is the derivative of an inverse function?

@spiderjerusalem4009 2 жыл бұрын

wait, so df(u) can actually also be part of the calculation? I thought that writing d-with any variable implies integrating a function with respect to that variable i hope someone can answer this so i can at least fathom the point of writing "dx" the whole time

@taranmellacheruvu2504 2 жыл бұрын

You can think about it this way: f(u) = y. Then, you have an integral with respect to y. y is its own variable. An integral of u with respect to y doesn’t work because the integral must be in terms of only the variable y. Then, as in the video, integration by parts separates everything into digestible components. The moral of the story is that you can write anything in terms of anything else to make things easier because variables are arbitrary. That’s also the reason why u-sub works.

@KickRoozing Жыл бұрын

I love how there's not a single number in this math video :D

@absolutedesi5899 11 ай бұрын

I expected that answer. Because the area under the inverse function must be the total area - area under f(x)

@mateserie7253 3 жыл бұрын

Is there a general formula to integrate (f(x))^2?

@TheEternalVortex42 2 жыл бұрын

There is no general formula (at least as far as we know). For example, x e^(x^2) is easy to integrate, but the square of it has no known closed form.

@onemanenclave 3 жыл бұрын

Isn't f^(-1)(f(u)) unnecesary? When you let u = f^(-1)(x), you can just replace it with u in the integral.

@lotr3152 3 жыл бұрын

Ok, but... Can we do this thing after all calculations are showed in this video? In the end of the video we have: integral of f^(-1)(x) = xf^(-1)(x)-F(f^(-1)(x)) Or: F(f^(-1)(x)) = xf^(-1)(x)-F(f^(-1)(x)) But now we can add to both parts of equation F(f^(-1)(x)) and get 2*F(f^(-1)(x)) = xf^(-1)(x) After dividing by 2: F(f^(-1)(x)) = xf^(-1)(x)/2, or integral of f^(-1)(x)dx = xf^(-1)(x)/2 + C, isn't it? Ok, I know that I forgot constants in this equations, they are too many, I am just too lazy to write they, and it easily can be shown that they don't influence on result.

@jensrenders4994 3 жыл бұрын

In the beginning you say u=f^-1(x), so literally calling the entire integrand u. You can directly substitute this. No need to take the form x = f(u), plug that in and then notice f an f^-1 cancel. Good video though ;)

@andrejgrebenc3235 3 ай бұрын

I see the problem how to calculate F(f^-1(x)).

@takyc7883 3 жыл бұрын

Never knew what that circle meant!

@BriTheMathGuy 3 жыл бұрын

Now you know!

@joeeeee8738 3 жыл бұрын

You should give an example?

@tomkerruish2982 3 жыл бұрын

Couldn't this just be looked at graphically, finding the area between the curve and the y-axis rather than between the curve and the x-axis?

@johnwoods978 3 жыл бұрын

yes, it can.

@CrimS0n. Жыл бұрын

"f(u)" The Mathematical way to curse someone

@duckymomo7935 3 жыл бұрын

What’s an application

@lemniscate23 2 жыл бұрын

+c ofcourse

@jonathanbaxter4611 3 жыл бұрын

Kinda Redudant to plug in f(u) for x instead of u for f^-1(x)

@mathadventuress 3 жыл бұрын

What level of math if this Analysis

@wojciechszmyt3360 3 жыл бұрын

Isn't antiderivative an integral? You can throw the negative F to the left and simplify the equation further!

@joeeeee8738 3 жыл бұрын

Can this be used to compute e^(-x2) ?

@Bobbob-dv4hp 3 жыл бұрын

If you’re talking about e^(-2x) then yes. If you’re talking about e^-(x)^2 then no, because there are no algebraic inverse functions e^-x^2

@joeeeee8738 3 жыл бұрын

@@Bobbob-dv4hp I knew that but yeah, I misread the F(u). I guess I had my hopes high!

@medelb_w4 2 жыл бұрын

Its sqrt (-ln(x))

@jeorgealonso4823 3 жыл бұрын

I am really confused about this, is he using some fancy editing or he's actually writing reversed letters (from his perspective) on a glass blackboard?

@johnwoods978 3 жыл бұрын

he just flipped the video horizontally. as you can see, he writes with his left arm in the video.

@johanneslade2830 2 жыл бұрын

It seems like you play it a biy fast and loose with the integration by parts (IBP). Normally I would say, that you have somethings like f(x)g(x)dx and the dx is not part of the IBP. But here you just treat the df(u) as part of the IBP. I don't understand why you can do this.

@Alians0108 3 жыл бұрын

You didn't really need the +C since F(f^-1(x)) comes with that anyways :P

@Sgrunterundt 3 жыл бұрын

I always took capital F(x) to mean any antiderivative of f, not all of them

@adb012 3 жыл бұрын

F is ONE (any one) antiderivative of f, so you still need the C.

@johnwoods978 3 жыл бұрын

Ф(f^-1(x)) comes with +C, not F(f^-1(x)).

@Alians0108 7 ай бұрын

I'm back to this comment after two years, and I have no idea what I meant by this

@youneverknow5555 3 жыл бұрын

nice :)

@BriTheMathGuy 3 жыл бұрын

Thanks!

@remynettheim4918 3 жыл бұрын

It is literally an abomination this channel has so few views

@MarioRossi-sh4uk 3 жыл бұрын

Because math is normally studied on a book, with paper and pencil beside, not on social media.

@Cjendjsidj 2 жыл бұрын

@@MarioRossi-sh4uk i say you can learn maths just fine through youtube.

@mariothethird5624 Жыл бұрын

Can't I just integrate f(u)=x? So then I get F(u)=(x^2)/2

@morbidmanatee5550 3 жыл бұрын

I see what you did there haha :)

@BriTheMathGuy 3 жыл бұрын

@muhammadsindidhossain6531 2 жыл бұрын

Why don't you wear glass anymore?

@Fennaixelphox 3 жыл бұрын

It's funny because integration is, itself, also an inverse function

@ByteOfCake 3 жыл бұрын

why did you write it as d(f(u)) rather than f'(u)du?

@biblebot3947 3 жыл бұрын

They’re the same thing

@user-cr4fc3nj3i 3 жыл бұрын

@@biblebot3947 but 1+1 is the same thing as 2, why don't people write 1+1 rather than 2? sometimes we should just choose the "better" or a more "common" one when we got two same things. in my opinion d[f(u)] looks complicated, when compared to using f'(u) du, so i would upvote for writing f'(u) du rather than d[f(u)]

@biblebot3947 3 жыл бұрын

@@user-cr4fc3nj3i df(u) != f’(u) f’(u) = df(u)/du The poster was talking about f’(u)du, which is more complicated, so that actually proves my point as to why we should use df(u).

@user-cr4fc3nj3i 3 жыл бұрын

@@biblebot3947 no i was saying df(u) is f'(u) times du, not just f'(u) also why i perfer having f'(u) du is that because we usually like to have the thing after the "d" as simple as possible, for example imagine having dsin⁻¹(cos[tan(u)]), why man just do f'(u) times du, this can make the differential simple, and maybe from the f'(u) we can "cannel" something out from the original integrand too

@ByteOfCake 3 жыл бұрын

@@biblebot3947 I guess they are the same thing. It feels weird to integrate with respect to a function though

@TheRammiel 3 жыл бұрын

It totally misses the proof without words for this theorem, which can be found on Wikipedia. No need to assume f is differentiable

@BriTheMathGuy 3 жыл бұрын

Integrals Playlist! kzfaq.info/get/bejne/oJZ6gZOinay5pI0.html

@Ahahahahstayingalive 2 жыл бұрын

Um what?

@ChechoColombia1 3 жыл бұрын

f^-1(x)=1/f^(x) lol

@PASHKULI 3 жыл бұрын

Please, elaborate on how u = 1/f(x) means that f(u) = x

@sanscipher9166 3 жыл бұрын

That's not what superscript -1 means.

@apuji7555 3 жыл бұрын

f-1(x) means the inverse of f(x). the inverse of a function is when you substitute y for x and x for y: y = f(x) switch the two variables x = f(y) then solve for y y = f-1(x). the '-1' is in superscript, but it doesn't mean 1/f(x), it means the inverse. And the inverse of the non-inverse of x = x. f-1(f(x)) = x. ______________ An example to think about it: f(x) = 2x - 1 => y = 2x - 1; y = f(x) switch varibles, x = 2y - 1 solve for y, y = (x + 1) / 2 y = 1/2 * x + 1/2. That is the inverse function of f(x), denoted by f-1(x). So: if f(x) = 2x - 1 f-1(x) = 1/2 x + 1/2 u can read online about it if you want to know more.

@evolutiagames 3 жыл бұрын

Are you writing backwards?

@charliemoll5435 3 жыл бұрын

I might be stupid. But is he drawing the math backwards so it appears normal on the screen?

@johnwoods978 3 жыл бұрын

yes. he definitely wasn't able to flip the video horizontally.

@BriTheMathGuy 3 жыл бұрын

I flip the video during editing :)

@scottleung9587 2 жыл бұрын

@@BriTheMathGuy Interesting - I thought you were left-handed and very good at writing backwards!

@tomaslopez814 3 жыл бұрын

Like the math but you really be cuttin it off short. The standard bump is 5 sec but an outro wouldn't hurt, esp if you standardize it

@user-be4ib8jt6l Жыл бұрын

How are you writing by standing behind the board ...... I mean you are writing in mirror view ...... !

@BurningShipFractal Жыл бұрын

What about derivative? Edit: I found the video kzfaq.info/get/bejne/rtmFd6qKl9uYYYE.html

@andorra3862 2 жыл бұрын

haven't started the video but something tells me that the final result will include at least one instance of the gamma function or a factorial. edit: just watched the video, my disappointment is immeasurable and my day is ruined.

@muskyoxes 3 жыл бұрын

If you use this presentation style a lot, you must be asked on every video if you're really writing backwards

@schrodingerbracat2927 3 жыл бұрын

i like f(u) and F(u), what about u?

@bonbonenuranium5034 3 жыл бұрын

df(u)/du = f'(u) so df(u) = f'(u) du and it makes the intégral easier

@ASN_9320 3 жыл бұрын

If d(f(u)) is the variable..then should you not take derivative of u with respect to f(u)?

@ospreytalon8318 3 жыл бұрын

Careful! dv=f'(u) NOT df(u) (which is instead equal to f'(u)du). It should be split up in that integral you do IBP on because you cant integrate the differential operator. The logic flows and everything else is correct, but this step is wrong.

@freightmanganese4715 3 жыл бұрын

agreed