Integral f inverse
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- Опубліковано 16 вер 2024
- In this video, I calculate the integral of f inverse, both by using a geometric definition, and by using a u-substitution. This problem appeared on the Math 2B final at UCI in the fall of 2018. It's a neat problem that shows that you don't always need an integration technique to calculate an integral. Enjoy!
Yo.... this is from one 🦆ing year ago...
I know 😂😂😂
Also man, now I want Peking Duck again 😋
@@drpeyam dang guys ! I Give you challenge check out this video on UA-cam .....And try to solve them. ..But you can't . Lmao 😂
🦆 ing... Nice!
Test Takers Method: Note that f(x) = x^2 has the desired properties. Also, the question implies that any function with the desired properties has gives the same answer. The inverse of x^2 is sqrt(x), so integrate that.
wait, is this legal?
3:15
bprp: *tilts camera*
Dr. Peyam: *also tilts shirt*
hahahaha i love watching your collabs :)) thank you for another great video!!
Underground School hahahahhahaha
It's not a surprise we get the same answer using integration by parts because integration by parts is also the same as subtracting the area of the graph with x axis from the area of rectangles formed my upper and lower limits.
I've seen your integration by parts intuition video and i loved it.
Keep up the good work 👏👍
I love both of these proofs. I never would have thought of the geometric proof, but what a powerful tool! Bravo!!
the second method needs f to be differentiable, but the first method doesn't. :)
If f were assumed piecewise differentiable it should be ok
If your function is not piece-wise differentiable, you got yourself a pretty ugly function. It's not something you see every day, so usually we don't care.
we can precisely substitute dy=d(f(x)). No need for diffentiable since integration by parts rule is actually |xdy+|ydx=|xy
f'(x) dx is just a way to calculate d(f(x))
Thank you for your videos, and for making math interesting, Dr Peyam. I wish the math teachers I came across during my youth had half of your enthusiasm. Cheers!
Also works is integration by parts:
∫ y dx = xy - ∫ x dy
In the future, you should watch the whole video (and read the description) before commenting on a video, so that you don't comment something that was already in the video.
5:53 LOOK AT THIS GRAAAAAPH!!! Every time I do it makes me laugh. : )
Finally, a video that was easy for me.
Maravilloso
I like that there are two ways to do the same question! Thank you Dr Peyam and bprp for shooting 😁
There is always an infinite amount of ways to solve a math problem
I really loved the first method it's so nice! 😁
I got to challenge my classmates with this 😂
Well, if you assume that the information provided is sufficient to find this integral, the only thing you have to do is to find a parametrized, invertible function that is 0 at 0 and 1 at 1, like... x^n :D you solve the first integral for n (you get 2, surprise surprise) and then you calculate the integral of x^(1/n) between 0 and 1 (and you get 2/3 ofc).
I don't know how this video has appearing in my recommended, but you really cool. Better, than most of my teachers, except maybe one. Greetings from Russia.
P.S Sorry for my mistakes, I almost don't know English.
Thank you!!! And no mistakes, impressive!
I'd argue that the first method may be more reliable, since (unless I missheard) the function f is not assumed to be differentiable, even though i'ts integrable.
I just recognised that f(x) is x^2 as this satisfies all the assumptions (f(0)=0, f(1)=1, the integral from 0 to 1 is 1/3, and taking just the +ve x values allows f(x) to be invertable) So then I just integrated sqrt(x) from 0 to 1 and obviously also got 2/3. I know this method wouldn’t work for less recognisable values of the initial integral but in this case, it seemed like a pretty quick method.
I think that there is a mistake in 2:05 . The function must be also continuous in [0,1] in order to be strictly decreasing or increasing (increasing since f(1)>f(0)).
Love from Greece anyways.
Leonidas Koumaniotis Not a mistake; in integration you assume that all functions are continuous
@@drpeyam that is simply not true.
EnNarr91 Yes it is! Of course not in Real Analysis, but in calculus you do
@@drpeyam you can assume anything you want, of course, but I wouldn't presume everyone follows the same convention.
Actually I do not agree. When we write an definite integral of a function f we assume that f is integrable (Riemann integrable for example). If f is continuous in [0,1] , then f is Riemann integrable in [0,1] for sure, but I am not sure about the opposite.
the second method is interesting in case we try to use it for nondifferentiable function. I think we can approach in two different ways. Firstly as integration by parts for Riemann-Stieltjes integral. More interestingly is to use Lebesgue theorem that says that monotone function is differentiable almost everywhere. Then we can do the substitution, but here we need to use Lebesgue integral since the Riemann integral after substitution doesn't have to exist. Damn, these fine properties of real analysis are something else.
Perhaps there's a limit method that can be used to avoid having to consider the derivative of f. Or maybe the u-sub and IBP can be done in one fell swoop, so we don't need that intermediate?
So I watch this video and there is something I still don't understand, and that is,.. come on.. don't be shy.. tell it like it is.. Ok so here it comes, I don't understand why I didn't subscribe to this channel sooner! But now I did. So we fixed that, yeeeeeay!
nice algebraic solution!
I guessed That f(x)=x^2 which is bijective fullfill the properties so the inverse function is sqrt(x)=f^-1(x) which leads to 2/3
Awesome video! Greetings from Italy ❤👋🏻
let y = f(x)
d(xy) = dxy +xdy. Integrating from x = 0 to x =1 -> xy|1,0 = Integral (ydx) + Integral(xdy).
The First one Is the Integral of f(x) dx.
The last One is the inteagral of xdy = inverseOf(y)dy. Just solve for the last integral
thank you so much
P(f-1) = ( x. f - P(f)) o (f-1) .P=Primitive . (f-1) = reciprocal .
I can't belive I got the right answer so easily(im so happy right now), I did it in the second method.
How would you solve g(x) = f((d/dx)(f(x)) for f(x) basically some function equals some other function with an input of the derivative of that other function with respect to x I got to f^-1(z)dx = df where z = g(x) but I have no idea how to integrate it
Another way: Stieltjes integral.
Int( x dy(x)) from 0 to 1. Greetings Dr.
Excellent!
A question on the second method:
How do we know that f is differentiable (I believe f might even need to be continuously differentiable) when we make the u-substitution?
Way easier method to just draw a graph and note that int f(x) dx = int f-1(x) dy, so since domain is (0,1) and to be invertible range has to be (0,1) (has to be continuously increasing over the domain so that when inverted it isn't one-to-many). Thus, over domain (0,1), int f(x) dx + int f-1(x) dx = 1×1 = 1, so int f-1(x) dx from 0 to 1 = 2/3.
Sorry for shoddy mobile formatting 😂
I discover a new formula (at least for me):
[Integral from a to b of f(x)] = bf(b)-af(a)-([Integral from f(a) to f(b) of inverse of f(x)])
You only have to generalize that... very beutifull
I get another formula that i've never seen....
[Integral of f(x)] = [Sum from n=0 to infinity of ( (-1)^n * x^(n+1) * f^("n")(x) * 1/(n+1)! )] where f^("n")(x) is the n'th derivate of f(x)
Thanks so much for the video, it's simply inspiring
Last yeari couldn't understand the algebraic method but now i could
understand it because i have taken calc 1
And ....
I have calc 1 final exam next week
So much fun for math nerds! Hooray!
Love your videos broh
Can you please explain the "method of characteristics"!!!
fav pair
Method 3: Use your intuition and recognize that f(x)=x^2.
RandomDays exactly
"Experimental math"
not necessarily, could be many other functions
@@l_szabi exactly:
for example it could be also the function
f(x)=-π/(12-3π)*sin(π/2*x)+(12-2π)/(12-3π)*x
But y=x² is the easiest one.
That makes sense if you are going to assume the answer is unchanged regardless of the form of f. But the problem did not say that. The answer was not guaranteed to be independent of f.
So fun to watch
Are there something related to the Legendre Transformation?
Nice viedo thanks D peyam السلام عليكم
Thanks so much
Summary is the following: If you plot the graph, you don't need to be clever😆
haven't touch calculus for 3 months, but still able to crack it down under 3 minutes 🤣🤣🤣
You are a great youtuber!
Every video you upload makes my feel like I want to unsub you just to be able to subscribe to you again!
Cool!!
I love u
Same Shpiel. That's make me think of someone...😂
I think the second solution is flawed, because the derivative of f doesn't necessarily exist.
Not flawed, but more that it works for differentiable functions only
I am happy to see someone caught it. I spent 2 periods last year in grade 11 explaining why continuity doesn’t imply differentiability. I had to delve into corner points using the absolute value function as an example.
Anyway, thank you very much, you made my day.
Sometimes students miss the little things: for example in grade 9 they miss that to prove a point is the midpoint of a segment, collinearity is necessary condition.
OOPS Dr Peyam . . . Got you at 7:32 !
dv = f'(x) . . . really?
Wondering why even Dr blackpenredpen didn't notice it, haha :O :P
integration by parts, dv=f'(x) would lead to v=f(x) etc.....not a typo it's right
@@harikishan5690
Nope. dv is a differential while f'(x) is a derivative. Differentials and derivatives are DIFFERENT and INCOMPARABLE. If you write dv = f'(x), then you can write ∫dv = ∫f'(x).
In ∫dv = ∫f'(x), the Left Side ∫dv becomes v
while the Right Side ∫f'(x) ≠ f(x). It is clearly wrongly written.
We could only write ∫f'(x)dx = f(x)
It’s a parabola too
This can be generalised
en.wikipedia.org/wiki/Integral_of_inverse_functions
oooooo I remember this onee!!!!!!
6:16 And same Spiel with 1
Wow!
Got i~~~~it :D
2/3
funny how no one talks about the thumbnail
lol bprp playing with the camera
NukeML lolll yup
Damn...
I think you crazy
Dr. Strange