Great. That means we all have different rates of learning. I am by no means a math genius but I understood it the first time I watched it. I actually had to derive the proof myself to see if I really understood it. It was good.
Great vid! Very logical, really breaking the barrier that gives us students the thoughts "How could anyone figure this out? Surely one has to be a genius..." Now I feel like the inventor(s) of the integral proof are actually human and were I one of them at that time I might have figured it out! Big thanks!
So, Sal casually took the derivative of F(x) just for fun and ultimately produced a proof for the fundamental theorem of calculus? That's genius right there.
I remember watching this video back when I was in my second semester of college. I didn't understanding a thing. Now, being in my sixth semester and watching it again, everything just clicks. It's nice to see some growth for a change. (Btw, beautiful proof and splendid explanation, props!)
I have taken two semesters of calculus and have used this theorem so many times its second nature to me. However I never knew why this theorem worked until just now. I had no idea how the heck an infinite sum could be connected so directly to a derivative, and I didn't imagine it would be so simple. Now I finally understand better where this comes from and I'm so happy about that. Thank you so much Sal!
i really want to cry😂 i finally understand this TvT i have been searching for the proof for 2 days😂 (cuz i really cant accept that formula if i dont understand where it comes from ) thank u
I love your explanations! They literally got me though my engineering education at university, I didn't go to one single class - I just stayed at home watching these videos from Khan academy. Extremely helpful.
you are such a great person. i attend a community college, and out teachers are horrible. my teacher has a huge asian accent and on top of that my registration time for classes was really horrible. i always loved math, but this quarter the only calc B class left open for me was with this asian teacher at 8PM. im really sleepy, hungry, and can't understand a word the teacher is saying. thank you so much for the help khan.
the best one so far. every video I've watched before had left me with lots of questions. but this video gave me Intuitive understanding and mathematical understanding at the same time. thanks a lot and big ups
Well both t and x are placeholders for numbers that lie on the interval [a,b]. What the theorem says is that F' and f always have the same value when you evaluate them at the same number. The main reason that t is used instead of x in the integral is because there x is used as a fixed point denoting the upper bound of the integral and we must integrate with respect to a variable. Just as easily we could have written F(t) = integral from a to t of f(x)dx.
For a while in my first calculus course it's been bugging me A LOT why I was anti-differentiating when what I was writing was talking about an infinite sum.. Seriously, thank you so much for tying everything together
For all intents and purposes, x in the theorem represents any t value provided it's between some continuous region in f(t). F'(t) = f(t) would be a way you would express that if you knew the whole function beyond 'a' (in both directions) is continuous. It would be more confusing getting to that result expressing the integrals in the proof this way however. The statement is true if the whole function is continuous as it says we get f(t) from the derivative of the antiderivative (now) for all t
That is a really great video, however I did find the mean value theorem a little abrupt and thought that it would’ve been better to use the Riemann Sums, which does get you to the same result, but is more intuitive for others to understand as I’m pretty sure Riemann Sums is the bare minimum that is taught to people with respect to the various approximation methods that have been invented. But anyways, great video 👍
We use the mean value theorem for definite integrals while prooving the fundamental theorem of calculus. However when prooving m.v.t for definite integrals we also use the fundamental theorem of calculus. What exactly is going on in here?
I always thought of the point at which the line is tangent to a function as a kind of tinny little top to a trapezoid, and if we added up every little area for each trapezoid we would get the area within that interval. I guess what this is saying is that if we have an area as a function and we take the derivative, the y value f(x) is the slope of the top of our little trapezoid at x. I know it is saying more, but I am trying to picture this out loud any thoughts would be helpful.
Could we keep it simple!! As ∆x tends to 0, you may assume f is monotone, region approximates to trapezium (lower classes) ... This finishes it due to continuity.O ne may avoid mean value theorem etc. Kindly consider
@@amberheard2869 HAHA yeah, 2 years and 5 months later, you ask, and now i like your comment 7 more months after that. Where has life taken you guys, if a reply may come??
William John We used Adam & Essex - Calculus: A Complete Course for our first and second semester of real analysis. Would recommend if you’re doing real analysis.
Finally, after watching this video 10 times I think I know what the source of all (my) confusion was. In my humble opinion it is bad nomenclature or lack of explanation of the nomenclature. f(t) is a function. f(c) and f(x) are representations for the f(t) function when t=c and when t=x respectively. They are not new functions in themselves. It is pretty strange to see t as something that has undetermined (variable) values and then look at f(x) and f(c) as specific values. Am I right or am I missing the forest for the tree or viceversa?
I have a question. In your proof, you used the mean value theorem for integrals and then proved that the value of t with the mean height approaches x as delta x-> 0, but I noticed that as delta x-> 0, the size of the interval [x, x + delta x] also approaches 0. So if the size of the interval approaches 0, can't we say that the area under the curve on the interval [x, x +delta x] approaches to the area of a rectangle whose base is delta x, and height is f(x)? That path would reach the same conclusions, and would also prove the fundamental theorem of calculus, but it is faster.
I still don't intuitively understand why that integral always equals the anti derivative regardless of what the arbitrary lower bound is? Shouldn't it depend on what a equals?
the integral equals the anti-derivative evaluated at the endpoints e.g. F(b)-F(a) so yeah it does depend, but only on the actual evaluation of the area. If you are just looking for the anti derivative of a function it would be F(x).
there is an antiderivative. It's just not "elementary" in the sense that it cannot be written using polynomials, trigonometric functions, logarithms, exponentials, inverse trig functions, hyperbolic trig etc. Notice that there is a very big difference between asking "does f(x) have an antiderivative" vs "does f(x) have an elemetary/simple anti derivative". The fundamental theorem of calculus proves to you that EVERY continuous function $f$ has an antiderivative, but it says nothing about whether the result can be then expressed using such familiar functions.
Even though it seemed obvious because the derivative of an intergral of f(x) is just f(x) ,you still managed to amaze me by doing this in a mathematical way that was still helpful :D
I'm a little confused, if we can prove F'(x)=f(x) for the function f(t) does that mean when we take a definite integral normally we should change the variable, technically?
The thing is like fist of all you have a function f(t) on a closed interval [a,b]. You define a NEW function F(x)="integral from a to x "f(t)dt (imagine i've got the integral sign right there). This F(x) is related to f(t) by the fact that it represents the area under f(t) and a horizontal "t-axis" between a and x on f(t). Now this F(x) is itself a function with respect to x: for each x we choose in the interval [a,b], we get a distinct "area under curve" value out of F(x). Bear in mind that F(x) is itself a function w.r.t. x and has its own graph. We now try to find the derivative of F(x). The fundamental theorem of Cal tells us that this F'(x) equals f(x). So what is f(x)? I remember seeing a f(t), but where does this f(x) come from?Now try to recall what we first learned when we studied functions: the letters we use to represent variables does not matter. If I have a function g(x)=cosx, then this means the same thing as "g(t)=cost". It's not the x or t that makes you recognize the function. Rather, it's the "g" in front of it. You see that g, and you know it stands for cos in this case. If you put g(k), then it's cosk; if you put g(party hats), then it's "cos (party hats)"(as long as "party hats" represent a variable). Returning to the problem at hand. We know F'(x)=f(x). Also, we have an expression of f(t). Let's say f(t)=ln(arcsin t). Then what's f(x)? We know the letters do not matter. Everywhere we see t, we replace it with x. So we have f(x)=ln(arcsin x). Therefore, F'(x)=ln(arcsin x)------a nice and pretty derivative expression that we should be family with. So to directly answer your question, no, we don't change variables. The "identity" of a function is its expression. What letters we use is irrelevant. They might look different, but x, t, k, and party hats in fact stand for the same thing.
Yeh thats not the mean value theorem, the mean value theorem is f'(c) = f(b) - f(a)/ b - a, f in this case is F and u just replaced F'(c) into just f(c) using the 1st ftc while ur proving it
you used the mean value theorem for integrals to prove the FTC . problem with this is that the MVT for integrals relies on the FTC. You get caught up in a loop.
@@chappie3642 Hahaha fair at this point I was just trying to learn about integration from scratch (literally without even learning much differential calc), thankfully it's all g now
The graph is using t on the horizontal axis. F(x) = area under the curve f(t) from t=a to t=x on the horizontal axis. So instead of a definite integral from some constant 'a' to another constant 'b', we have a fixed 'a' and a variable end 'x'. f(x) = F'(x) -the derivative of F(x)
The reason the derivative of an integral of f(x) is f(x) is because of the fundamental theorem of calculus. So the claim you made is actually not obvious without proof. Math doesn't work of the basis of this is equal to that just because i say so. To really understand you need proof.
well ok...so here are we gonna define a arbitary number 0 such that after 7,8,9 tger comes 10...and so the number series can be infinitely repeated...with no end......will continue afterwords..
@@kayzero9689 ok, i'll go next, lets define the operation of addition, which takes two elements of the set of integers which you defined and combines them to form a third in such a way that the distance of each of those integers from zero is (lets just say added) to give the distance of the new element. Who's going to continue?
After watching this 20 times I finally understand!
Great 👍
i actually watched for 6 days to understand.
Great. That means we all have different rates of learning. I am by no means a math genius but I understood it the first time I watched it. I actually had to derive the proof myself to see if I really understood it. It was good.
@@No_BS_policy you my friend need a lesson on sarcasm!
@@Akshit.vats. True lol. ''omg I derived the proof myself!!''
Every-time Sal says "just for fun"
Me: " sure, just for fun."
Great vid! Very logical, really breaking the barrier that gives us students the thoughts "How could anyone figure this out? Surely one has to be a genius..."
Now I feel like the inventor(s) of the integral proof are actually human and were I one of them at that time I might have figured it out!
Big thanks!
So, Sal casually took the derivative of F(x) just for fun and ultimately produced a proof for the fundamental theorem of calculus? That's genius right there.
this video should be emmy nominated.
I remember watching this video back when I was in my second semester of college. I didn't understanding a thing. Now, being in my sixth semester and watching it again, everything just clicks. It's nice to see some growth for a change.
(Btw, beautiful proof and splendid explanation, props!)
I have taken two semesters of calculus and have used this theorem so many times its second nature to me. However I never knew why this theorem worked until just now. I had no idea how the heck an infinite sum could be connected so directly to a derivative, and I didn't imagine it would be so simple. Now I finally understand better where this comes from and I'm so happy about that. Thank you so much Sal!
i really want to cry😂 i finally understand this TvT
i have been searching for the proof for 2 days😂 (cuz i really cant accept that formula if i dont understand where it comes from )
thank u
same here pal'
has this still not won an oscar yet????
Aight I'll just watch it 20 more times
I love your explanations! They literally got me though my engineering education at university, I didn't go to one single class - I just stayed at home watching these videos from Khan academy. Extremely helpful.
Proofs for the theorems may seem monotonous but they actually give great insights into the concept.
That's the beauty of math.
if sal had a dollar for every intuition he gave us
he'd be bill gates
nah jeff bezos
@@Turnamonkey nah Elon Musk
His total views are 1.8 Billion, so sadly not Bill Gates level, but still close
@@loneranger4282organic chem tutor though
This is the innermost reasoning of Calculus, it's celestially beautiful!
i am an eight grader and yet understood everything thanks to the teaching methods thank you!
Best video on UA-cam.
you are such a great person. i attend a community college, and out teachers are horrible. my teacher has a huge asian accent and on top of that my registration time for classes was really horrible. i always loved math, but this quarter the only calc B class left open for me was with this asian teacher at 8PM. im really sleepy, hungry, and can't understand a word the teacher is saying. thank you so much for the help khan.
+Fled From Nowhere pointing out the fact that he cant understand an Asian accent isn't rasist.
the best one so far. every video I've watched before had left me with lots of questions. but this video gave me Intuitive understanding and mathematical understanding at the same time. thanks a lot
and big ups
Well both t and x are placeholders for numbers that lie on the interval [a,b]. What the theorem says is that F' and f always have the same value when you evaluate them at the same number. The main reason that t is used instead of x in the integral is because there x is used as a fixed point denoting the upper bound of the integral and we must integrate with respect to a variable. Just as easily we could have written F(t) = integral from a to t of f(x)dx.
its amazing how your new videos are always synced with what im currently doing in class...
amazing explanation, I tried understanding it from my teacher and FAILED, but here its so flowing.. thanks!!
Choice of words: "RESORT to Squeeze Theorem". That's sort of how I feel about using the Squeeze Theorem as well.
For a while in my first calculus course it's been bugging me A LOT why I was anti-differentiating when what I was writing was talking about an infinite sum.. Seriously, thank you so much for tying everything together
Really enjoyed watching like a movie. Every step is quite interesting. Thank you sir.
For all intents and purposes, x in the theorem represents any t value provided it's between some continuous region in f(t).
F'(t) = f(t) would be a way you would express that if you knew the whole function beyond 'a' (in both directions) is continuous. It would be more confusing getting to that result expressing the integrals in the proof this way however.
The statement is true if the whole function is continuous as it says we get f(t) from the derivative of the antiderivative (now) for all t
Nicely integrates (no pun/integral intended) the MVT into the explanation. Well done proof.
That is a really great video, however I did find the mean value theorem a little abrupt and thought that it would’ve been better to use the Riemann Sums, which does get you to the same result, but is more intuitive for others to understand as I’m pretty sure Riemann Sums is the bare minimum that is taught to people with respect to the various approximation methods that have been invented. But anyways, great video 👍
Thank you so much for what you do. You make a difference in many peoples' lives, and I appreciate it.
Wonderful 🤩
awesome video ,u just cleared all my doubts thank you so much
Thanks a lot sir 👍👍👍👍👍
I've watched a lot of your videos, and I have to say, this is your masterpiece. Good job and thank you Mr. Khan
Thank you so much! My math text is so hard to follow and this really helped me understand how these are connected!
Best video on UA-cam:] u made my day
We use the mean value theorem for definite integrals while prooving the fundamental theorem of calculus. However when prooving m.v.t for definite integrals we also use the fundamental theorem of calculus. What exactly is going on in here?
No fucking idea.
This is bugging me out
You can use the second fundamental theorem of calculus to prove the M.V.T and then use the M.V.T to prove the first fundamental theorem of calculus
MVT applies to any function that is differentiable
you can prove mvt without tfc
love the way he teaches
Beautiful proof. Thank you Sal.
Great ! Thank you for you work !
Very clearly explained. Thankyou.
I always thought of the point at which the line is tangent to a function as a kind of tinny little top to a trapezoid, and if we added up every little area for each trapezoid we would get the area within that interval. I guess what this is saying is that if we have an area as a function and we take the derivative, the y value f(x) is the slope of the top of our little trapezoid at x. I know it is saying more, but I am trying to picture this out loud any thoughts would be helpful.
Amazing!
Sal is a legend!
Could we keep it simple!! As ∆x tends to 0, you may assume f is monotone, region approximates to trapezium (lower classes) ... This finishes it due to continuity.O ne may avoid mean value theorem etc. Kindly consider
at 7:10 can you guys help me answer why f(c).dx = area under the curve I mean why f(c).dx I think it should added a limit when dx-->0
At last got a proper video 🧡💛💚💙💜🤎
god bless this man
Beautiful👌
Good video. Helped me out. Thanks.
is it weird that i got asmr tingles from this?
couchpotatos same here
the real mvp!!! thank you sir!
Really good video thanks for it :)
Thanks. it helped me a lot.
try taking real analysis guys. It is basically restarting calculus but with proofs. 10x harder but much more enjoyable
@@amberheard2869 HAHA yeah, 2 years and 5 months later, you ask, and now i like your comment 7 more months after that. Where has life taken you guys, if a reply may come??
@@amberheard2869 Interesting! I am in an AP Calculus class in high school. These videos are really useful, lol. Its fun to learn.
William John We used Adam & Essex - Calculus: A Complete Course for our first and second semester of real analysis. Would recommend if you’re doing real analysis.
it should be dealt T because it is x axis named as time 't'
Finally, after watching this video 10 times I think I know what the source of all (my) confusion was. In my humble opinion it is bad nomenclature or lack of explanation of the nomenclature. f(t) is a function. f(c) and f(x) are representations for the f(t) function when t=c and when t=x respectively. They are not new functions in themselves. It is pretty strange to see t as something that has undetermined (variable) values and then look at f(x) and f(c) as specific values. Am I right or am I missing the forest for the tree or viceversa?
I have a question.
In your proof, you used the mean value theorem for integrals and then proved that the value of t with the mean height approaches x as delta x-> 0, but I noticed that as delta x-> 0, the size of the interval [x, x + delta x] also approaches 0. So if the size of the interval approaches 0, can't we say that the area under the curve on the interval [x, x +delta x] approaches to the area of a rectangle whose base is delta x, and height is f(x)? That path would reach the same conclusions, and would also prove the fundamental theorem of calculus, but it is faster.
I suppose that isn't rigorous enough
Why is there a need for C? The area can be written as f(x)dx so the limit is just f(x) as dx approaches zero.
We proved that this F function gives you the area below the graph from point a to point b. How do we know that point a is 0?
Amazing video
I still don't intuitively understand why that integral always equals the anti derivative regardless of what the arbitrary lower bound is? Shouldn't it depend on what a equals?
the integral equals the anti-derivative evaluated at the endpoints e.g. F(b)-F(a) so yeah it does depend, but only on the actual evaluation of the area. If you are just looking for the anti derivative of a function it would be F(x).
but I want to ask a qustion if every continus funcction has an antaiderivtive
and that e^(-x^(2)) is a continus
why their is no antidrivtive for it
there is an antiderivative. It's just not "elementary" in the sense that it cannot be written using polynomials, trigonometric functions, logarithms, exponentials, inverse trig functions, hyperbolic trig etc. Notice that there is a very big difference between asking "does f(x) have an antiderivative" vs "does f(x) have an elemetary/simple anti derivative". The fundamental theorem of calculus proves to you that EVERY continuous function $f$ has an antiderivative, but it says nothing about whether the result can be then expressed using such familiar functions.
your explanation is amazing!! thank you very much!!!
what a wonderful video, my god, I UNDERSTAND
He proved them in this video
People used the fundemental theorem of calculus to prove the mean value theorem for integrals not the other way around.
:(
did i waste my time?
hobo doc Can you post the text of that pages in this comment section ? I would apreciate it.
The theorem also says that F(x) is a continuous function even if f(t) isn't. You need to proof that also.
I lost it at 10:00🥲🥲
who is the teacher?
does anyone have his social media accounts? or website or whatever? his awesome
What software is this video using to writing the formulas?
This is insanely beautiful
Period
ohh it was priceless video
Even though it seemed obvious because the derivative of an intergral of f(x) is just f(x) ,you still managed to amaze me by doing this in a mathematical way that was still helpful :D
are you still alive?
I'm a little confused, if we can prove F'(x)=f(x) for the function f(t) does that mean when we take a definite integral normally we should change the variable, technically?
The thing is like fist of all you have a function f(t) on a closed interval [a,b]. You define a NEW function F(x)="integral from a to x "f(t)dt (imagine i've got the integral sign right there). This F(x) is related to f(t) by the fact that it represents the area under f(t) and a horizontal "t-axis" between a and x on f(t).
Now this F(x) is itself a function with respect to x: for each x we choose in the interval [a,b], we get a distinct "area under curve" value out of F(x). Bear in mind that F(x) is itself a function w.r.t. x and has its own graph. We now try to find the derivative of F(x). The fundamental theorem of Cal tells us that this F'(x) equals f(x).
So what is f(x)? I remember seeing a f(t), but where does this f(x) come from?Now try to recall what we first learned when we studied functions: the letters we use to represent variables does not matter. If I have a function g(x)=cosx, then this means the same thing as "g(t)=cost". It's not the x or t that makes you recognize the function. Rather, it's the "g" in front of it. You see that g, and you know it stands for cos in this case. If you put g(k), then it's cosk; if you put g(party hats), then it's "cos (party hats)"(as long as "party hats" represent a variable).
Returning to the problem at hand. We know F'(x)=f(x). Also, we have an expression of f(t). Let's say f(t)=ln(arcsin t). Then what's f(x)? We know the letters do not matter. Everywhere we see t, we replace it with x. So we have f(x)=ln(arcsin x). Therefore, F'(x)=ln(arcsin x)------a nice and pretty derivative expression that we should be family with.
So to directly answer your question, no, we don't change variables. The "identity" of a function is its expression. What letters we use is irrelevant. They might look different, but x, t, k, and party hats in fact stand for the same thing.
"should be familiar with" on the second to last paragraph. Autocorrect must have changed that.
@@paulwang7229 Autocorrect looked at the context and found "nice and pretty' and went to work!
Yet another awesome math class from Salman
This is a work of art.
MIND = BLOWN
What software is he using?
Yeh thats not the mean value theorem, the mean value theorem is f'(c) = f(b) - f(a)/ b - a, f in this case is F and u just replaced F'(c) into just f(c) using the 1st ftc while ur proving it
I now associate the color magenta with Sal Khan.
SALAM!
you used the mean value theorem for integrals to prove the FTC . problem with this is that the MVT for integrals relies on the FTC. You get caught up in a loop.
hobo doc hi hobo, for MVT you need FTC2.
Or maybe his class is using Khan Academy's videos as a course guide :P
1:48 wait why
Bruh it's literally the definiton of an intrgral
@@chappie3642 Hahaha fair at this point I was just trying to learn about integration from scratch (literally without even learning much differential calc), thankfully it's all g now
@@EpiCuber7 understandable xD, I'm glad you realized your mistake
Not our calculus prof giving this to proof in final exam
I like resorting to the Sandwich Theorem. Very delicious.
HOLY I've never been so enlightened
What was that scary sound at 6:05?
dang this video rules
hi
I'm still somewhat confused... what is f(x)? How is that function defined? I see there's an f(t) but what is f(x)? I'm clearly missing something.
The graph is using t on the horizontal axis. F(x) = area under the curve f(t) from t=a to t=x on the horizontal axis. So instead of a definite integral from some constant 'a' to another constant 'b', we have a fixed 'a' and a variable end 'x'.
f(x) = F'(x) -the derivative of F(x)
Wow I was the 1.2Kth like!
estoy confundida
_nods_
The reason the derivative of an integral of f(x) is f(x) is because of the fundamental theorem of calculus. So the claim you made is actually not obvious without proof. Math doesn't work of the basis of this is equal to that just because i say so. To really understand you need proof.
My motto for the last couple of years
:( so, i wasted my time?
What do you mean? This is literally the proof of that
This business
How would integrals be solved without knowing the fundamental theorem? They always teach integrals as the inverse process of derivatives.
With infinite series, a true nightmare.
T
HANKYOU ORZ6
hello viewers
salam *sigh*
I don't like proving theories depending on other ones, if you're going to proof a theory using another theory you must proof all of them
well ok...so here are we gonna define a arbitary number 0 such that after 7,8,9 tger comes 10...and so the number series can be infinitely repeated...with no end......will continue afterwords..
@@kayzero9689 ok, i'll go next, lets define the operation of addition, which takes two elements of the set of integers which you defined and combines them to form a third in such a way that the distance of each of those integers from zero is (lets just say added) to give the distance of the new element. Who's going to continue?
In fact literally every theorem is proved, by definition, otherwise it isn't a theorem.
The mvt is proved, this is proved, every theorem is proved