It is so nice to see all the passion you have for math, I use to think math as just a high school subject I had to pass but after getting into the world of calculus and seeing great mathematicians such as u doctor Peyam I changed my mind and started enjoying math in a way a cant survive a day without doing a problem just for fun Keep up with the nice work
My AP calculus teacher (who retired the day after our final) always called it the "Little Fundamental Theorem of Calculus,"because it's used less often and there were ambiguities with calling which FTC 1 and which one FTC 2 in the textbook.
At 5:40, instead of mentioning the definition of the Riemann integral, you can instead draw the graph y=1 and calculate the area manually as the area is just a rectangle. In my opinion this is more rigourous
There is a slick way to get the limit by using the squeeze principle. with upper bound A_U=f(x+h)h, and lower bound A_L=f(x)h, divide by h and take the limit as h tends to zero to get that the derivative of the integral is the function itself.
i can't prove it because my understanding is /so/ visual that i could stop by 3 minutes in, as h goes to 0, the height of the rectangle approximating int(x+h)-int(x) (which is h*f(x)) approaches precisely f(x) (being of the area h*f(x) and width h), almost the same as the proof of derivatives actually.
I think this way is a bit easier. Let m and M be the min and max of f on the interval [x, x+h]. Then the integral of f over that interval is bounded between h*m and h*M. Hence the integral is equal to c*h for some c between m and M, and the derivative is g'(x) = lim_(h -> 0) (c*h)/h = lim_(h -> 0) c Note that m, M, and c depend on h and c is squeezed between m and M. One definition of continuity is that the oscillation at x is 0; in other words m and M approach x as h -> 0. Hence lim_(h -> 0) c = x
Nice theorem and nice proof doctor. You can prove this also by supposing that F is an antiderivate of f and when you get the limit of integral of f(t) from x to x+h, over h, from the beginning of your proof, it's the limit of (F(x+h)-F(x))/h, and it's in fact the derivate of F(x), so it's f(x)
You are assuming the result in this video in your proof. You are saying that you can calculate the integral as a difference in anti derivatives, but the fact that you can do that is what the fundamental theorem of calculus is saying.
Maybe this will help, although it doesn’t 100% answer your question: Covering Compactness and Uniform Continuity th-cam.com/video/xiWizwjpt8o/w-d-xo.html
The part where using the graph to see that you can change the difference of the integrals into a single integral from x to x+h seems kind of hand wavy. It makes sense logically but it doesn’t seem like you proved in a rigorous fashion that the step was allowed.
neg atory It's to prep for the epsilon-delta part; I want to show that the limit is 0, hence use an epsilon-delta argument. Once the limit of the difference is 0, the original limit is f(x), which is what we want.
Before weierstrass formalized limits, the entire field of calculus wasn't considered rigorous mathematics yet, so formalizing all the proofs in terms of the concept that makes calculus rigorous is definitely the way to go.
why? i didn't like them as student, but now i appreciate them. They are almost like a computation - in that sense you start doing something not sure what to expect, but the problem guides you. There are elegant and beautiful proofs, that teach nothing. Epsilon-delta proofs are hands-on. At least, that's my opinion.
Why do i get infront of every single video by you an ad for a netflix show called „stranger things“? Coincidence? *i think not* edit: i hope that „stranger“ means something close to „strange“ in this context. I hope it doesnt mean something like „stalker“ or such things.. My english is not the best so im very sorry if its offensive or something like that... Anyways.. nice video! :D
I prefer a more geometric proof: if look at the area g[x + h]- g[x] you can approximate it by a parallelogram ,giving h*(f[x]+f[x+h])/2. Then , for h -> 0 you get f[x] .
The expression ''parallelogram '' is not correct , but the approximation h*(f[x]+f[x+h])/2 for the area is o.k.. Hence after division by h ,and h -> 0 you find f[x] .
it purely motivates me to go further with maths when I see the undying passion of dr Peyam, this channel is such a joy, please keep going :)
Hi Dr Peyam, Id like to ask whether this can be proven with the mean value theorem and just be as valid? Mean value theorem + Squeeze theorem. Thanks!
Could you do a video about any research you have done in Mathematics and about your doctoral thesis? Thanks!
I second that
I third that
I nth that
I nth^nth that
Hi Dr Peyam, Id like to ask whether this can be proven with the mean value theorem and just be as valid? Mean value theorem + Squeeze theorem. Thanks!
It is so nice to see all the passion you have for math, I use to think math as just a high school subject I had to pass but after getting into the world of calculus and seeing great mathematicians such as u doctor Peyam I changed my mind and started enjoying math in a way a cant survive a day without doing a problem just for fun
Keep up with the nice work
euardo balint the problem of many students from Brasil is that they don't study to Understand Math...they study to get a high score...
+euardo balint
Me too😝
Explaining the obvious is the most difficult. Thank you for this great video.
I can't wait till I reach a point in my mathematics career where I can full understand these sorts of things
My AP calculus teacher (who retired the day after our final) always called it the "Little Fundamental Theorem of Calculus,"because it's used less often and there were ambiguities with calling which FTC 1 and which one FTC 2 in the textbook.
That’s so adorable!!!
I love this sort of videos, keep going please. I love maths and phisics
At 5:40, instead of mentioning the definition of the Riemann integral, you can instead draw the graph y=1 and calculate the area manually as the area is just a rectangle. In my opinion this is more rigourous
Excellent Dr Peyam ! I have seen for the the "increase" of the function cos(pi/2*t). Thank you very much.
There is a slick way to get the limit by using the squeeze principle. with upper bound A_U=f(x+h)h, and lower bound A_L=f(x)h, divide by h and take the limit as h tends to zero to get that the derivative of the integral is the function itself.
This is the best proof ever!!!
(Doesn't like epsilon delta proofs, but loves ur videos)
agree!
i can't prove it because my understanding is /so/ visual that i could stop by 3 minutes in, as h goes to 0, the height of the rectangle approximating int(x+h)-int(x) (which is h*f(x)) approaches precisely f(x) (being of the area h*f(x) and width h), almost the same as the proof of derivatives actually.
Yeah, but can you prove that it approaches precisely f(x)?
Hi Dr Peyam, Id like to ask whether this can be proven with the mean value theorem and just be as valid? Mean value theorem + Squeeze theorem. Thanks!
I think this way is a bit easier. Let m and M be the min and max of f on the interval [x, x+h]. Then the integral of f over that interval is bounded between h*m and h*M. Hence the integral is equal to c*h for some c between m and M, and the derivative is
g'(x) = lim_(h -> 0) (c*h)/h = lim_(h -> 0) c
Note that m, M, and c depend on h and c is squeezed between m and M. One definition of continuity is that the oscillation at x is 0; in other words m and M approach x as h -> 0. Hence lim_(h -> 0) c = x
This method has nice continuity with your other FTC video since it uses a different sort of mean value theorem.
Couldn't you have used the Squeeze Theorem using the local minimum and maximum of f(t) for t∈[x,x+h]?
and the mean value theorem?
The integral expression of f(x) should have the limit as h goes to zero. Then you can subtract the two integrals while subtracting f(x) from g'(x).
Can you prove the definition you mentioned at 5:40
It’s the area of a rectangle with base [x,x+h] and height 1
Nice theorem and nice proof doctor. You can prove this also by supposing that F is an antiderivate of f and when you get the limit of integral of f(t) from x to x+h, over h, from the beginning of your proof, it's the limit of (F(x+h)-F(x))/h, and it's in fact the derivate of F(x), so it's f(x)
You are assuming the result in this video in your proof.
You are saying that you can calculate the integral as a difference in anti derivatives, but the fact that you can do that is what the fundamental theorem of calculus is saying.
Amazing. Major applause
Why don't you have not use the diffrestiobality
Great proof based on the continuity of the integrand. What if f was not continuous but simply integrable?
It still holds but in a measure theory sense! Look up Lebesgue differentiation theorem!
@@drpeyam I will definitely look at the Lebesgue differential theorem to convince myself. Thank you!
Just look how happy he is while explaining the proof. I agree that math majors rejoice every time they obtain that epsilon on proofs, lol.
very nice
Thank you so much for this video! :D
Do u a video about this epsilon-delta stuff that am not really getting the idea behind !
Jihan Hamdan There are some epsilon-delta videos on my channel, and some more on blackpenredpen’s channel!
It would be interesting to know how to prove it with infinitesimals instead of limits. The structure of argument should be the same I guess.
Mr. Professor Bagenda back at UFS
can you do a video when a function is Uniform continuity, i really dont get it.
Maybe this will help, although it doesn’t 100% answer your question: Covering Compactness and Uniform Continuity th-cam.com/video/xiWizwjpt8o/w-d-xo.html
@@drpeyam Thank you!! I'll take a look!
What a great proof
f just has to be continuous not uniformly continuous, right?
Logan Reina Just continuous, because notice that we've just used continuity at x, and we didn't really care what happens away from x.
Logan Reina Oh, also a continuous function on a compact set like [a,b] is automatically uniformly continuous, so no worries about that :)
The part where using the graph to see that you can change the difference of the integrals into a single integral from x to x+h seems kind of hand wavy. It makes sense logically but it doesn’t seem like you proved in a rigorous fashion that the step was allowed.
You can prove it using integral from a to b = int a to c + int c to b
Content that independently justifies TH-cam.
No idea when epsilon and delta came out.
Ein schöner Beweis!
Danke!
At this moment, there are 5^(5-5/5) visualizations😏
During step 2, what is the purpose of equaling zero?
neg atory It's to prep for the epsilon-delta part; I want to show that the limit is 0, hence use an epsilon-delta argument. Once the limit of the difference is 0, the original limit is f(x), which is what we want.
Thank you Dr. Payem!
sorry,but when newton and leibniz discovered the fundamental theorem of calculus.the limit isn't discovered yet
Before weierstrass formalized limits, the entire field of calculus wasn't considered rigorous mathematics yet, so formalizing all the proofs in terms of the concept that makes calculus rigorous is definitely the way to go.
Yaaaaaa......got €.🏃♂️🏃♂️🏃♂️🏃♂️🏃♂️
Epsilon-delta proofs are horrible in general :(
why? i didn't like them as student, but now i appreciate them. They are almost like a computation - in that sense you start doing something not sure what to expect, but the problem guides you. There are elegant and beautiful proofs, that teach nothing. Epsilon-delta proofs are hands-on. At least, that's my opinion.
Torp Glens
Why do i get infront of every single video by you an ad for a netflix show called „stranger things“? Coincidence? *i think not*
edit: i hope that „stranger“ means something close to „strange“ in this context. I hope it doesnt mean something like „stalker“ or such things.. My english is not the best so im very sorry if its offensive or something like that...
Anyways.. nice video! :D
This comment is gonna get pinned.
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Why does this comment have 100 likes?
Good try...
I prefer a more geometric proof: if look at the area g[x + h]- g[x] you can approximate it by a parallelogram ,giving h*(f[x]+f[x+h])/2.
Then , for h -> 0 you get f[x] .
The expression ''parallelogram '' is not correct , but the approximation h*(f[x]+f[x+h])/2 for the area is o.k.. Hence after division by h ,and h -> 0 you find f[x] .