Course Web Page: sites.google.com/view/slcmath... For a slight variation on the proof, you may consult the following document: drive.google.com/file/d/1uF4n...
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.
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.
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.
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. ;-)
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..
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?
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!
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.
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 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
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! ;-)
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.
@@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?
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".
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.
@@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.
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.
This proof is so clearly explained in this video it gave me goosebumps! Wonderful!
Great!
This has got to be one of the most beautiful things I've seen in a while. Subscribed!
seen several explanations, this one us the best so far
5:40 This is when I finally understood how the Fundamental Theorem works. Thank you so much!
Thanks, I'm a student from Colombia, and I can understand the proof. You are great teacher, thank you, very much
THIS MADE SO MUCH SENSE, THANK YOU SO MUCH
Thankyou very much!! I have been searching for this proof so many times!!
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.
Clear explanation. The speed of the presentation is perfect, too.
i wish if we had such explanation ways in moroccan classes
excellent video i'm waiting your works
You are the best!!! You don't deserve this number of views and subscribers. You deserve so much more than that.
Thank you so much! This is a really clear proof!
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!
This is so much better than sal khan's proof
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!
Very much appreciated, though I am not sure to be worthy of such high praise. :-)
Beautifully done!
This was an awesome video, really helped me out
goosebump literally ,how easily u explained
Excellent!! From Nigeria, thank you.
¡Amazing! Simple but logical
Holy shit this was awesome. Thank you!
Thank you for the explanation!
It finally hit me! Thank you so much!!
Thanks alot! This is the best and the simplest explanation i've ever seen
:-)
Fantastic video!
Beautiful proof in beautiful handwriting.
incredible explanation
Nice. I always thought the FTC was a bit circular, but now I SEE THE LIGHT. Thanks.
Glad to hear! It is such a beautiful and powerful result!
Great presentation thanks
thanks sir..u clear my concept🙏🙏
This is the best proof of the FTC I have ever seen! :)
Great explanation!
King of simplicity.
This is the greatest video on RUclips
This is What I call Feynman Technique. Thanks a lot.
this is beautiful
clear video but tsill dont get it. the independent variables are so confusing gustavo
Uau!!!! Thanks!!! The best explanation!
Thank you! :)
Wonderfull. Thanks Sir sooooooo much
Ever best one...
a great video, thank you
By the way, thanks for providing material that we can download on your web page!
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. :-)
@@slcmathpc That's great, thanks!!
Your handwriting is music to my eyes
truly brilliant
:-)
Amazing!
great one
great vid
Amazing video, thanks! Question: Why do we have constant "a" and variable "x"? What would happen if they were different?
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.
@@slcmathpc Thanks a lot!! And (sorry for the dumb questions) what would happen if x didn't vary as well?
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. ;-)
@@slcmathpc 😄😄Thanks!!!
Beautiful
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..
No milk? :-)
Kau melayu ke? Ajar aku
@@zainolariffin4936 haah der melayu malaysia.. ajar? Mcm mn..
@@slcmathpc lol
@@slcmathpc 🍼 I mean 🥛
You absolute god
I appreciate the sentiment and I thank you for making me laugh! :-)
Good luck with your studies!
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?
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!
@@slcmathpc thank you
Very cool
how is the "C = -F(a)" applicable to all cases and not just the integral between 'a' and 'a'?
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. ;-)
@slcmath@pc how was it shown that it is true for all values of 'x'?
splendid
:-)
So goood
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.
Is there any theorem on the existence of x hat?
The Intermediate Value Theorem for continuous functions. :-)
@@slcmathpc Thanks. Will look into it.
You could have said that, formally, that you are using the mean value theorem for definite integrals.
this is gold, omg
You are an individual of taste I see. :-)
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 😉
I suggest that you review the distinction between the two types of integrals: the definite integral and the indefinite integral. :-)
@@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
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! ;-)
@@slcmathpc thank you so much this is just what i needed ^^
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
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.
@@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?
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".
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
I am glad to hear that one more person in the world appreciates and understands this beautiful result!
think you made it too simple
Ironically that was also the proof of my stupidity.
Man how good this is!
Hold up
I thought there was no C cuz the C from the first integral cancels the C from the second
Be sure not to confuse the indefinite integral from the definite integral.
@@slcmathpc so am i right about defenite integrals
Wowwww
You sound like 3blue1brown , by the way .
It is indeed a clear video. Unfortunately the claim between 7:15 and 7:40 is wrong, but intuitively it is nice.
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.
@@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.
This made lot of sense but it is not complete proof....
Would you be so kind as to indicate what is missing? Honestly, I don't know what makes this proof incomplete.
@@charlessmith6412 i mean there is much more standard proof of this theorem.
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.
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.
Aabhash Pokharel: Don't be sorry. I'm trying to deepen my understanding, and anything you can contribute will help. Thanks for your efforts.