Fundamental Theorem of Calculus 1 | Geometric Idea + Chain Rule Example

Derivatives are geometrically tangents to curves while definite integrals area areas under curves. How are these related? The Fundamental Theorem of Calculus parts I and II show how we can compare these two fundamental concepts in Calculus. The first derivative, the subject of this video, allows us to see geometrically how to compute the derivative of an accumulation function.
We finish the video with a computation showing how to combine chain rule and FTC 1
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Now it's your turn:
1) Summarize the big idea of this video in your own words
2) Write down anything you are unsure about to think about later
3) What questions for the future do you have? Where are we going with this content?
4) Can you come up with your own sample test problem on this material? Solve it!
Learning mathematics is best done by actually DOING mathematics. A video like this can only ever be a starting point. I might show you the basic ideas, definitions, formulas, and examples, but to truly master math means that you have to spend time - a lot of time! - sitting down and trying problems yourself, asking questions, and thinking about mathematics. So before you go on to the next video, pause and go THINK.
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[@Safwan.Hossain]

I knew how to use this theorem, its definition, how it worked, but not WHY it worked. Was desperately trying to search for videos to try to get a rock solid understanding of something as important as FTC. Most of them just showed how to use it. This was the first video that really dove into what everything meant with the helpful visuals, especially clearing up the confusion for the dummy variable t. I tried experimenting what would happen if the dummy variable function f(t) was simply replaced by f(x). It gave a completely wrong answer because it would involve changing the value of the function itself, instead of just the integral boundaries. After this video, then I'd say 3B1B's video on FTC was the next helpful resource.

[@jacksonyu1157]

You bring this to such an intuitive level. I am learning calculus and my teacher just show me the rules without explaining why. I was so confused by that f(t)dt there. But you just explains it so well and I totally understand it. I just want to say a huge thank you to your video. You are the best!

[@candelaferrari5246]

watched like 10 videos about this and this is the first time i fully understood everything

[@lambdafunctions9648]

Im your Fan, great expression of Math, please keep it up, Engineering Student, Love from INDIA

[@NeelShah0211]

This is the best explanation for fundamental theorem of calculus ive ever seen. hats off. Extremely amazing.

[@honorebelro]

I would just like to pause to acknowledge not just the educational, but also the production value of this video!

[@mikehughes6582]

The best proof I've seen. I struggled with it when I took calculus.

[@minratos6215]

thank you so much. after days of looking up videos on the topic, yours is the best and clearest explanation on the internet!!

[@Ali-mi9up]

What a commendable effort! Really well-explained and intuitive also how very astute of you to use a different input to f than the upper limit used i.e. using f(t) rather than f(x)

[@quanle9133]

So Clear again! Short but always on point! you are amazing!!!!

[@pianellrielly9803]

Makes it crystal clear, thank you so much!

[@gravygod9820]

The BEST video for this topic, thank you!

[@Dani-ge1zw]

Dr Bazett this was amazing. I have been looking for videos on the FTOC for ages and when I finally finished this one - everything clicked! Thank you so much. I loved your visual representations and enthusiasm throughout the video :)

[@DrTrefor]

I’m so glad!

[@louism.4980]

One of the best explanations out there, thank you sir! :)

[@Vegastellar]

Thank u so much I was able to get the right answer but I wasn’t really understanding what I was doing while solving it really happy that I found ur Channel

[@alisapuskala1437]

thank you, you are great at making intuitive explanations and your enthusiasm also helps!

[@ha-vz3kn]

I didn’t get it at the first time But then I watched it again and I know how it works. Thxx

[@unpopularnotion]

You are such a great teacher. Love the way you explain

[@MathMaster55]

Your explanation is wow more videos on calculus concepts thank you

[@roa1437]

Guys for better understanding the idea try to rewrite the solving question watch video solving while solving with them and come back again and you will understand every concept 💡 😉 because this video is why it worklike that Thanks you're real hero 🙌 💙 ❤ 👏

[@DrTrefor]

that's a great strategy!

[@notnow9902]

This is a solid explanation .

[@khanaltaf41]

This is my presentation Thank You❤

[@colin8923]

This was so helpful, thank you!

[@adrianarojo2104]

This helped me understand so much. Thank you!!!

[@ShivamShukla-nw6pu]

Really one of the best videos on calculus! Subscribed!

[@DrTrefor]

Glad it helped!

[@nathanhe4214]

Thank you so much, you're a hero

[@mnada72]

Amazing explanation, thanks. It seems that the lower limit (a) has no effect, is it equivalent to differentiating a constant?

[@JavierBonillaC]

I have seen the Kahn video of this 10 times. This one is the best by far! Really congratulations. I am an economics professor and I always look for the best way of making things understandable at gut level. Thank you. Note: I find it confusing to talk about f(t) -a function- and f(x) -a value-. And f(x+h) -a value-. So f(x+h) is a quantity that I can find on the Y (what we typically call) the Y axis. Let’s go back to f(x). Wouldn’t it really be f(t=x)? I know I must be wrong but where is my mistake? It's that "dummy" thing that everyone skips.

[@isavenewspapers8890]

To be precise about the terminology used here: The letter "f" refers to the function. The expression "f(t)" refers to the value that the function f returns when evaluated at the value of t that you're plugging in. The expression "f(x)" is similar, except you're evaluating for some value of x instead. In this context, a dummy variable is just a variable that we introduce to make our lives less confusing. We could speak of "the integral from a to x of f(x) dx", but the x's in "f(x) dx" are playing a different role than the x which is the upper bound of the integral. In order to make this difference clear, we tend to use "f(t) dt" instead.

[@shawncowden3909]

I'm taking Calc 1 on khan academy and for a minute I was struggling to intuitively/geometrically understand why the chain rule gets used here (I do algebraically understand why but that's not good enough for me). I think I kinda get it now, but please correct me if I'm wrong. You have a function F that is the area under a curve, and F' is the rate of change of area under that curve. In this example, e^sqrt(t) represents the rate of change of the vertical aspect of the shape under that curve, and x^3 represents the horizontal end point of the shape under the curve, so 3x^2 is the rate of change of the horizontal aspect of the shape under the curve. Now in another example if you set F(x)= integral of (t)dt from 1 to 2x, then F'(x) will = 4x. This is 4 times the rate of change of area under the curve that would occur if the upper bound was x instead of 2x. The vertical aspect of the area of the shape under the curve is accumulating twice as fast than if the upper bound was x, because 2x gets plugged into t and becomes 2x. That's the easy part. But what I had to realize is that the horizontal aspect of the area is also increasing at a different rate, and the ratio of the different rates is the derivative of 2x, which is 2. Seeing 2x in the upper bound means that you're zooming along the x-axis (technically the t-axis in this case) of the graph of y = t, at twice the rate of the normal passage of time. And since the y-value of the graph is equivalent to the t-value here, the y-value will also increase at twice the normal rate. Twice the horizonal rate, and times twice the vertical rate = 2x2 = 4 times the area rate. I imagined that for every day in May, you are to be given the number of that date in dollars, so that on May 5th you receive $5, on May 6th you get $6, etc. Let F(x) be an accumulation function to represent your total received $ after t days. If the upper bound of the integral reads "2x", that means you're passing through time at twice the normal rate of time's passage. If you could move through time twice as fast as the rest of us, you would accumulate 4 times as many dollars per your unit of time as you normally would, because 1) time is moving twice as fast [horizontal], and 2) the date's value is twice as large as it would've been if time had been passing normally [vertical]. It's a bit harder to visually grasp setting your rate of time's passage to non-linear functions, but the concept should be the same. Hopefully my analogy is at least somewhat accurate, but I'm not completely sure.

[@tellom.8375]

Very good vids , thanks for the visuals

[@NatureRandomSights]

so good! can you please explain this concept for indefinite integrals as well?

[@hurdurlemur1615]

This is absolute great explanation

[@DrTrefor]

Glad it helped!

[@Gamingzone42099]

You're the best 😀

[@hunterrees]

Thank you for this video.

[@moiseslobopvh]

Hello teacher. How can I find the shooting angle of a ball to hit a basketball hoop, if I have the speed, gravitational acceleration and vertical and horizontal distances? I can't deduce an equation to find the angle when the target is above the launch point.

[@crammingbanned2118]

Brilliant...

[@SaimKhan-xj5um]

Thanks Alot SIR ...keep it coming ...

[@Efe-pz4sc]

Great video. Thank you.

[@smoothacceleration437]

Brilliant... Tx.

[@chaoukimachreki6422]

awesome

[@tashrifa.m]

Nice

[@onion5529]

very good

[@smithcodes1243]

Excuse my language but this is so fucking good! I can't stop smiling!

[@suuuken4977]

Great explanation!

[@jamesdesantis9420]

Brilliantly explained - and this is from a math teacher. I've never seen the FTC explained quite like that. Liked and Subscribed!

[@DrTrefor]

Awesome, thank you!

[@aashsyed1277]

thanks.

[@NuclearMex]

We meet again.

[@hatmanchills]

Respect from Pakistan

[@aspiredifferent8085]

tumhe english samjh bhi aati hai iqra bhai😂, mazak kar rha hu 😊

[@JustaletterJ-ec5jy]

Uhmm how did u come up with lim as h->0 f(x)*h/h?

[@festa1999]

Well if h approaches 0 but not equal to 0 the result would approach f(x) since the hs would cancel out which makes sense, but I'm not sure how to say it mathematically as well with the limit laws and stuff. I think it's because f(x) is being treated as a constant as far as the limit is concerned but this is not explicitly stated in my textbook or his videos so I'm not sure. Sometimes I wish he was a little less concise

[@naiko1744]

So, the integral is the area under the curve of the function, in the interval between x and x+h, right? Since h is very small, a very very good estimation of that area would be f(x+h) * h What's that? That's the area of a rectangle that would estimate the value of the area under the curve there. However, we are working with exact quantities, not estimations! We want the exact amount, right? Now, notice that we are computing everything inside a limit with h -> 0 This means that, since this is a limit, the result we are giving must be within any small range of error possible. If anyone comes and asks us to compute the limit with a max error range of ± 10 or ±0.0001, the result of the limit must be the same. In turns, this means that h can be took smaller and smaller to make our answer always work for any error range, in a small interval of h around 0. A working answer that ALWAYS checks for any error range, if we can make h as small as we want, is that the area under the curve is f(x+h) * h For any error range that you may be OK with, I can take an h even smaller, and this value (area of the rectangle) would still be within approximation range. And I'm allowed to take an h as small as I need, because we are inside a limit that returns a value valid when h is around 0. Now, you may say that this value f(x+h) * h for the area is still an approximation, but you would be wrong!! That is because, if this "estimation" is always valid for any error range, THEN any other approximation greater or smaller than this one MUST be ...either the same value or WRONG! Since It's the only "estimation" possible that always works, it must be the correct value, for a certain value of h close to 0 and all of the closer ones. Inside this limit then f(x+h) * h is EXACTLY the area under the curve, if we care about it from a certain value of h close to 0 and all of the smaller ones.

[@Johnny2tc]

cheers

[@havehalkow]

Even with great explanations, some people aren’t going to get it. Unfortunately, I’m one of those :(

[@naiko1744]

Previously, we defined the anti-derivative F(x) as the function that, if we take the derivative of it, is equal to f(x) Mathematically, F ' (x) = f(x) Now we want to prove that F(x), the anti-derivative, is the same as the integral from a to x of f(x) To do that we substitute the integral to F(x) in the expression F ' (x) = f(x), and we see if we can get that it's derivative is indeed equal to f(x) Once, we substituted, we apply the definition of derivative, which is a limit, to F(x) (the integral we substituted in) After a few passages, we get to the limit of the integral from x to x+h of f(t) with h -> 0, all the expression over h Thanks to the meaning of limits, the only solution that makes sense for the integral is f(x) * h The two h cancel and we get f(x) Therefore, what we started with (the derivative of the integral from a to x of f(x) ) was really just f(x) That integral was really the anti-derivative of f(x) We now connected the concept of integral and anti-derivative

[@himm2003]

Man you are a g

[@zuckmansurov2781]

Everything was perfect until the example to explain. Could have been much better with an easier example to explain.

[@mlop2484]

thanks