Integral of Inverse Functions
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- Опубликовано: 19 фев 2021
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#math #brithemathguy #integral
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It seems crazy that you can do ANYTHING with this. The original form seems completely intractable. This is what I live for :)
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.
Smart
Hard to integrate? It's the easiest integration by parts
i mean u just make an integration by parts and derive ln(x) and integrate 1
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
Thnxs this is making me like math more
Great to hear!
Same for me!
Damn you're so good.....
Wish you thousands of subscribers in your career
Thanks very much!
wish grantee
2:24 So accepting, I think I needed to hear that.
I’ve seriously been exploring more math because of inspiring people like you. Keep shining brother ☀️
The quality of your explanation is unbelievable. You have won a subscriber :)
Integrals Playlist!
ruclips.net/video/j0JN0q8FUtY/видео.html
I remember that question being on one of my single-variable analysis tests last year. That was a pretty good question, not gonna lie.
I literally had an epiphany seeing this result. I'm definitely gonna find somewhere to use this just to flex
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!
Man your videos are so cool, I bet that your RUclips channel will gain a lot of viewers soon!
I appreciate that!
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!
Love the last line of this video
Great video, thanks!!!
Glad you liked it!
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
THANK YOU VERY VERY VERY VERY MUCH.
Thanks, and explain so clearly!
Glad you think so!
I got this question on my final for calculus 1 and I was blown away
Really nice way to turn into Riemann Stiljies integral after substitution then apply integration by parts!
Nice!
Bravo, sempre molto preciso
well explained👍🏾
Glad you think so!
Yes! Well explained
👍🏻
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).
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)
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?
ln(x), sin(x)
Am gonna use it
Great!
"Un Dia Vi Una Vaca sin(-) cola(∫) Vestida De Uniforme" ∫udv=uv-∫vdu.
Thats the spanish way to remember integration by parts
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?
Un día vi a una vaca vestida de uniforme
@@MusicalInquisit an integral looks like a tail, yes.
I learned It with "un día vi un valiente soldado vestido de uniforme"
actually retrieving the integration by parts formula is pretty simple, using only the derivative of uv
It has geometrical explanation, flip the curve so x and y axes changes, and area under curve becomes rectangle area minus areaunder the curve.
Thank you for your video.
Thanks for watching!
Beautiful presentation
Thank you! Cheers!
Very concise, good video.
Much appreciated!
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 👇)
Can u make a video on integration of implict functions
2:57 why uf(u) is there suddenly? Where was it come from?
Can you integrate the function where f(f(x)) == e^x ?
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 😐
Stokes theorem on manifold plz
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
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.
@@liamhanson9538 ok thank you so this would work for inverse trig functions right?
@@babajani3569 yes in restricted domain only
Are there any good applications for this?
This works for every inverse continuos function?
That’s amazing
Glad you think so!
Nice
In the end, the integral depends on F(f^-1(x)) - which is the original question, isn't it?
^
+
No, it means you first get the integral of f(x) and then plug in f^(-1)(x)
@@coc235 that makes things a lot clearer, I was confused about that as well
F refers to the antiderivative of f, F is not the antiderivative symbol itself.
Love to watch your video....maths become easier..😊
Thanks a lot! Glad you think so!
2:51 what happened here Can anyone explain?
I'm a mathematics master's student and I still love watching your videos Bri
Thanks a ton! Have a great day!
should mention that f is a one to one function, else break the integral up into monotonic domains and add.
You should definitely add an example (or two).
Agreed. It's jarring how abruptly that ends. I barely had a chance to see the final line.
bri: math can be fun! just watch
also bri: *writes f(u) = x unironically*
You are the best!!
You are!
@@BriTheMathGuy Thanks! This helped me a lot!
What is the derivative of an inverse function?
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 ?
Is there a general formula to integrate (f(x))^2?
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.
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.
What’s an application
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
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
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.
You should give an example?
Can this be used to compute e^(-x2) ?
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
@@Bobbob-dv4hp I knew that but yeah, I misread the F(u). I guess I had my hopes high!
Its sqrt (-ln(x))
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 ;)
Isn't f^(-1)(f(u)) unnecesary? When you let u = f^(-1)(x), you can just replace it with u in the integral.
I see the problem how to calculate F(f^-1(x)).
I love how there's not a single number in this math video :D
I expected that answer. Because the area under the inverse function must be the total area - area under f(x)
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?
yes, it can.
What level of math if this
Analysis
Kinda Redudant to plug in f(u) for x instead of u for f^-1(x)
Never knew what that circle meant!
Now you know!
+c ofcourse
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.
nice :)
Thanks!
Can't I just integrate f(u)=x?
So then I get F(u)=(x^2)/2
why did you write it as d(f(u)) rather than f'(u)du?
They’re the same thing
@@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)]
@@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).
@@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
@@biblebot3947 I guess they are the same thing. It feels weird to integrate with respect to a function though
"f(u)" The Mathematical way to curse someone
Isn't antiderivative an integral? You can throw the negative F to the left and simplify the equation further!
It is literally an abomination this channel has so few views
Because math is normally studied on a book, with paper and pencil beside, not on social media.
@@MarioRossi-sh4uk i say you can learn maths just fine through youtube.
Are you writing backwards?
You didn't really need the +C since F(f^-1(x)) comes with that anyways :P
I always took capital F(x) to mean any antiderivative of f, not all of them
F is ONE (any one) antiderivative of f, so you still need the C.
Ф(f^-1(x)) comes with +C, not F(f^-1(x)).
I'm back to this comment after two years, and I have no idea what I meant by this
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.
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?
he just flipped the video horizontally. as you can see, he writes with his left arm in the video.
I see what you did there haha :)
:)
Why don't you wear glass anymore?
Please, elaborate on how u = 1/f(x) means that f(u) = x
That's not what superscript -1 means.
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.
f^-1(x)=1/f^(x) lol
Um what?
It's funny because integration is, itself, also an inverse function
How are you writing by standing behind the board ...... I mean you are writing in mirror view ...... !
It totally misses the proof without words for this theorem, which can be found on Wikipedia. No need to assume f is differentiable
I might be stupid. But is he drawing the math backwards so it appears normal on the screen?
yes. he definitely wasn't able to flip the video horizontally.
I flip the video during editing :)
@@BriTheMathGuy Interesting - I thought you were left-handed and very good at writing backwards!
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
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.
What about derivative?
Edit: I found the video
ruclips.net/video/xsUDGY2u41M/видео.html
If you use this presentation style a lot, you must be asked on every video if you're really writing backwards
i like f(u) and F(u), what about u?
df(u)/du = f'(u) so df(u) = f'(u) du and it makes the intégral easier
If d(f(u)) is the variable..then should you not take derivative of u with respect to f(u)?
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.
agreed
It looks quite ugly though.
True :/