The Fundamental Theorem of Calculus - Proof

[@markkennedy9767]

This is remarkably well motivated. Something you don't get in crappy calculus classes or texts. The way he motivates the mean value theorem's role in the proof (what both parts of the proof hinge on) is lovely.

[@PopDeek23]

This proof is so clearly explained in this video it gave me goosebumps! Wonderful!

[@slcmathpc]

Great!

[@שחראוסי-פ1ב]

seen several explanations, this one us the best so far

[@Rougesteelproject]

5:40 This is when I finally understood how the Fundamental Theorem works. Thank you so much!

[@chessandmathguy]

This has got to be one of the most beautiful things I've seen in a while. Subscribed!

[@josuerios190]

Thanks, I'm a student from Colombia, and I can understand the proof. You are great teacher, thank you, very much

[@BatterflyHigh]

THIS MADE SO MUCH SENSE, THANK YOU SO MUCH

[@mathbrainius]

Clear explanation. The speed of the presentation is perfect, too.

[@charlessmith6412]

I have only seen the approach of using an area function A(x) once before. And that was in Calculus for Dummies. This exposition is much clearer. Thank you very much for an excellent video.

[@lindavadnais8093]

Like poetry. This eloquently takes the listener back to the classroom of Newton and Leibniz, or at least what I would imagine it to be. Damn good!

[@slcmathpc]

Very much appreciated, though I am not sure to be worthy of such high praise. :-)

[@roydaboii9925]

This is so much better than sal khan's proof

[@Aman_iitbh]

goosebump literally ,how easily u explained

[@ayoubdiri4553]

i wish if we had such explanation ways in moroccan classes excellent video i'm waiting your works

[@santiospina4504]

Thankyou very much!! I have been searching for this proof so many times!!

[@desrucca]

Thanks alot! This is the best and the simplest explanation i've ever seen

[@slcmathpc]

:-)

[@ObiajuluEmma-Ebere]

Excellent!! From Nigeria, thank you.

[@wagsman9999]

Nice. I always thought the FTC was a bit circular, but now I SEE THE LIGHT. Thanks.

[@slcmathpc]

Glad to hear! It is such a beautiful and powerful result!

[@paulboro5278]

King of simplicity.

[@lalalajessica3752]

Thank you so much! This is a really clear proof!

[@danny89620]

This is the greatest video on KZbin

[@wesrobertson8753]

Thank you! I get what my books have been trying to explain now. And about time to. This is going on a poster on my wall word for word!

[@kidusabebe2645]

This is What I call Feynman Technique. Thanks a lot.

[@agumondigimon]

Thank you for the explanation!

[@Santos-cz1mz]

By the way, thanks for providing material that we can download on your web page!

[@slcmathpc]

I have just posted a new version of the integral calculus (Math NYB) course pack that contains some very nice additional stuff if you're curious. :-)

[@Santos-cz1mz]

@@slcmathpc That's great, thanks!!

[@_DD_15]

Beautifully done!

[@caomoletimoloi6565]

You are the best!!! You don't deserve this number of views and subscribers. You deserve so much more than that.

[@gustavosedano294]

¡Amazing! Simple but logical

[@caomoletimoloi6565]

You are a life savior. I never really understood F.T.C but after watching this video i realised It was easy. Thank you so much

[@slcmathpc]

I am glad to hear that one more person in the world appreciates and understands this beautiful result!

[@smoothacceleration437]

Beautiful proof in beautiful handwriting.

[@DenisBencic]

Holy shit this was awesome. Thank you!

[@zakusa9891]

clear video but tsill dont get it. the independent variables are so confusing gustavo

[@ЕлизаветаЩербакова-ч9с]

It finally hit me! Thank you so much!!

[@zoekane2825]

Your handwriting is music to my eyes

[@tomdexter4996]

This is the best proof of the FTC I have ever seen! :)

[@control5586]

incredible explanation

[@mhick3333]

Great presentation thanks

[@dheerajpannem3336]

This was an awesome video, really helped me out

[@Santos-cz1mz]

Amazing video, thanks! Question: Why do we have constant "a" and variable "x"? What would happen if they were different?

[@slcmathpc]

Since "a" is left as an arbitrary constant, you can certainly think of it as a variable, but one that we do not let vary in this case; to prove the result, we only need to let "x" vary.

[@Santos-cz1mz]

@@slcmathpc Thanks a lot!! And (sorry for the dumb questions) what would happen if x didn't vary as well?

[@slcmathpc]

Well, we need to vary x since we want to show that the rate of change of the area function A(x) is f(x), so if we don't vary x, then we don't have a proof. ;-)

[@Santos-cz1mz]

@@slcmathpc 😄😄Thanks!!!

[@chewboxout]

Great explanation!

[@green-sd2nn]

this is beautiful

[@mitchellloren3568]

Fantastic video!

[@rohanrana5725]

thanks sir..u clear my concept🙏🙏

[@gustavjohansson1642]

You could have said that, formally, that you are using the mean value theorem for definite integrals.

[@matharp2653]

This is amazing. The explanation was perfect, the visual element was amazing, and you did a perfect job in making me love another element of calculus through a proof.

[@lizardgvng131]

You absolute god

[@slcmathpc]

I appreciate the sentiment and I thank you for making me laugh! :-) Good luck with your studies!

[@intheshell35ify]

Ironically that was also the proof of my stupidity.

[@louism.4980]

Thank you! :)

[@TheRamsatya]

Ever best one...

[@noone7692]

Hello, I have a simple and dumb question to ask. Why are we changing the integral f(x)dx from limit [a,x] to a dummy variable f(t)dt. To put it clear why are we using the dummy variable t u or v specifically ?. What will the consequences if we still integrste without changing it to the dummy variable?

[@slcmathpc]

It is to avoid what is commonly known as a "clash of variables". The actual variable in this instance is the upper bound of integration, which we chose to label as "x". The variable, say "v", in the integrand "f(v)dv" is what is known as a "dummy variable", since it is not a consequential variable and is completely independent of the upper bound of integration "x". Writing the integrand "f(v)dv" as "f(x)dx" seems to suggest that the "x" in "f(x)dx" has something to do with the upper bound of integration "x", which is simply not the case. Writing the integrand as "f(x)dx" and using "x" as the upper bound of integration causes a "clash" between the two expressions, which again, have nothing to do with one another. Hope this clears things up!

[@noone7692]

@@slcmathpc thank you

[@abdulhafizuddin95]

thanks you sir.. i've been thinking a way to prove FToC to my student.. but i think my explanation was too hard.. this helps me a lot.. you deserve a cookie..

[@slcmathpc]

No milk? :-)

[@zainolariffin4936]

Kau melayu ke? Ajar aku

[@abdulhafizuddin95]

@@zainolariffin4936 haah der melayu malaysia.. ajar? Mcm mn..

[@mryup6100]

@@slcmathpc lol

[@mryup6100]

@@slcmathpc 🍼 I mean 🥛

[@evanroderick91]

how is the "C = -F(a)" applicable to all cases and not just the integral between 'a' and 'a'?

[@slcmathpc]

Since the equality is true for all values of x, then it must be true for x=a, which shows that C=-F(a). There is nothing deeper going on. ;-)

[@evanroderick91]

@slcmath@pc how was it shown that it is true for all values of 'x'?

[@LucasPinheiroV8]

Uau!!!! Thanks!!! The best explanation!

[@theoyanto]

truly brilliant

[@slcmathpc]

:-)

[@sulgunrejepova3867]

Wonderfull. Thanks Sir sooooooo much

[@ChinaLostBeats]

a great video, thank you

[@ramrajmainali123]

splendid

[@slcmathpc]

:-)

[@clay0105]

very interesting and useful, however i have a doubt. from de beginning you use the statement ∫f(x)dx=F(x) + C , if F'(x)=f(x), why? Isn’t that what we want to proof or demostrate? It gives me the notion that is used as a true statement from the beginning or what is the intention to use this statement in this video? i'm so sorry if i'm misinterpreted the whole thing, please explain me please please please, thanks for a great video 😉

[@slcmathpc]

I suggest that you review the distinction between the two types of integrals: the definite integral and the indefinite integral. :-)

[@clay0105]

​@@slcmathpc thanks then, is the indefinite integral defined like an axiom already accepted? Sorry for my lack of knowledge , I actually think I know the difference between definite and indefinite integral, but I'm trying to figure out why the indefinite integral is defined in this way [∫f(x)dx=F(x) + C] thank you for your understanding and patience

[@slcmathpc]

It is nothing more than a definition, so the indefinite integral of a function is defined as the class of all functions whose derivative is equal to the original function. The definite integral of a function over a closed and bounded interval is defined as the limit of a corresponding Riemann sum. It should seem strange at first to use quite similar notation for two very seemingly different objects (indefinite vs definite integral), but they are deeply connected by the Fundamental Theorem of Calculus, which states that under the assumption of continuity, one can evaluate the definite integral using a difference of an antiderivative at the endpoints of the corresponding interval instead of taking the limit of a Riemann sum, which is a far more challenging task. I hope this helps! ;-)

[@clay0105]

@@slcmathpc thank you so much this is just what i needed ^^

[@jannesl9128]

great one

[@Gebev]

Man how good this is!

[@daniellecoutre8878]

Why have you included the +C when writing F(x)+c equals the integral from x to a of f(t)? Wouldnt the +C be cancelled out anyway due to the limits

[@slcmathpc]

If F(x) is some antiderivative of f(x), then all antiderivatives of f(x) are of the form F(x)+C. Since the area function A(x) is an antiderivative of f(x), then it must be the case that A(x)=F(x)+C.

[@daniellecoutre8878]

@@slcmathpc Isn't that only the case for indefinite integrals? in this proof we are dealing with a definite integral, so wouldnt the constant simply cancel out?

[@slcmathpc]

When stating that all antiderivatives of f(x) are of the form F(x)+C, the constant C is indefinite, which means that it can range over all real numbers. Out of these infinitely many antiderivatives, one of them must be equal to A(x), which means that there is a unique/definite value of C such that A(x)=F(x)+C. In the first case, C is an indefinite constant, but in the second case, C is a definite constant and so it has a unique value. If this double use of C still confuses you, then simply write A(x)=F(x)+c, where lower case "c" is a unique/special value of upper case "C".

[@andrewnachamkin7071]

Amazing!

[@NamTran-rz9uy]

great vid

[@johnq4841]

this is gold, omg

[@slcmathpc]

You are an individual of taste I see. :-)

[@wantsomesushi6414]

Beautiful

[@jhaokip23]

Is there any theorem on the existence of x hat?

[@slcmathpc]

The Intermediate Value Theorem for continuous functions. :-)

[@jhaokip23]

@@slcmathpc Thanks. Will look into it.

[@andreasstorvik5772]

think you made it too simple

[@blablabla12a]

Very cool

[@Ahmed-vs1ui]

Hold up I thought there was no C cuz the C from the first integral cancels the C from the second

[@slcmathpc]

Be sure not to confuse the indefinite integral from the definite integral.

[@Ahmed-vs1ui]

@@slcmathpc so am i right about defenite integrals

[@mryup6100]

So goood

[@johnjernigan5348]

It is indeed a clear video. Unfortunately the claim between 7:15 and 7:40 is wrong, but intuitively it is nice.

[@slcmathpc]

Assuming that the function y=f(x) is continuous over the range of integration, the claim is true. It follows directly from the traditional Intermediate Value Theorem.

[@ycombinator765]

@@slcmathpc I am very glad I found your video but tbh, I caught that too, the claim, as per your clarification is true only for continuous function per that considered region and wont hold true if the function were not nice. But as we all saw, intuitively it clicked for the given instance. From your following, it would sometimes be misinterpreted and Fundamental Theorem of Calculus would seem limited only to nicely linear type functions whose graph is continuous smoothly. A nicer approach would be to let h approach zero in the very first iteration. By the way, nice explanation. Keep it up bro. Am looking forward to more.

[@ggamil196]

You sound like 3blue1brown , by the way .

[@jadshmeis1389]

Wowwww

[@unknownnepali772]

This made lot of sense but it is not complete proof....

[@charlessmith6412]

Would you be so kind as to indicate what is missing? Honestly, I don't know what makes this proof incomplete.

[@unknownnepali772]

@@charlessmith6412 i mean there is much more standard proof of this theorem.

[@charlessmith6412]

Aabhash Pokharel: Do you have a recommendation for a source on a better exposition either in print or youtube? If you do I'd really appreciate it.

[@unknownnepali772]

Charles Smith still searching,whenever i will get i will surely share it with you....but i really don't think this is complete.....sorry for that.

[@charlessmith6412]

Aabhash Pokharel: Don't be sorry. I'm trying to deepen my understanding, and anything you can contribute will help. Thanks for your efforts.