# Integration and the fundamental theorem of calculus | Essence of calculus, chapter 8

This guy, Grothendieck, is somewhat of a mathematical

idol to me. And I just love this quote, don’t you? Too often in math we just dive into showing

that a certain fact is true with long series of formulas before stepping back and making

sure that it feels reasonable, and preferably obvious, at least on an intuitive level.

In this video I want to talk about integrals, and the thing that I want to become “almost

obvious” is that they are an inverse of derivatives.

Here, we’ll focus just on one example, which is kind of dual to the example of a moving

car that I talked about in chapter 2 of the series, introducing derivatives.

Then in the next video, we’ll see how the idea generalizes into some other contexts. Imagine you’re sitting in a car, and you

can’t see out the window; all you see is the speedometer. At some point, the car starts

moving, speeds up, then slows back down to a stop, all over 8 seconds.

The question is, is there a nice way to figure out how far you’ve traveled during that

time, based only on your view of the speedometer? Or better yet, find a distance function s(t)

that tells you how far you’ve traveled after any given amount of time, t, between 0 and

8 seconds. Let’s say you take note of the velocity

at each second, and make a plot over time like this… And maybe you find that a nice

function to model your velocity over time, in meters per second, is v(t)=t(8-t). You might remember, in chapter 2 of this series,

we were looking at the opposite situation, where you know a distance function, s(t),

and you want to figure out a velocity function from that.

I showed how the derivative of your distance vs. time function gives you a velocity vs.

time function, so in our current situation, where all we know is the velocity function,

it should make sense that finding a distance vs. time function s(t) comes down to asking

what function has a derivative t(8-t). This is often described as finding the anti-derivative

of a function. And indeed, that’s what we’ll end up doing,

and you could even pause and try that right now. But first, I want to spend the bulk of

this video showing how this question is related to finding an area bounded by velocity graph,

because that helps to build an intuition for a whole class of what are called “integral

problems” in math and science. This question would be much simpler if the

car was moving with a constant velocity, right? In that case, you could just multiply the

velocity, in meters per second, by the amount of time passed, in seconds, and that gives

you the number of meters traveled. Notice that you can visualize that distance

as an area, and if visualizing distance as an area seems weird, I’m right there with

you. It’s just that on this plot, where the horizontal direction has units of seconds

and the vertical direction has units of meters/second, units of area very naturally correspond to

meters. But what makes our situation hard is that

the velocity not constant, it’s incessantly changing at every instant. It would even be

a lot easier if it only ever changed at a handful of points, maybe staying static for

the first second, then suddenly discontinuously jumping to a constant 7 meters per second

for the next second, and so on, with discontinuous jumps to portions of constant velocity.

That might make it very uncomfortable for the driver, in fact, it’s physically impossible,

but it would make your calculations a lot more straightforward.

You could compute the distance traveled on each interval by multiplying the constant

velocity on that interval by the change in time. Then just add them all up.

So what we’ll do is just approximate our velocity function as if it was constant on

a bunch of different intervals. Then, as is common in calculus, we’ll see

how refining that approximation leads us to something precise. Here, let’s make this more concrete with

some numbers. Chop up the time axis between 0 and 8 into many small intervals, each with

some little width dt, like 0.25 seconds. Consider one of these intervals, like the

one between t=1, and 1.25. In reality the car speeds up from 7 m/s to

about 8.4 m/s during that time, which you can find by plugging in t=1 and 1.25 to

the equation for velocity. We want to approximate the car’s motion

as if its velocity was constant on this interval. Again, the reason for doing that is that we

don’t really know how to handle anything other than a constant velocity situations.

You could choose this constant to be anything between 7 and 8.4, it doesn’t really matter.

All that matters is that that our sequence of approximations, whatever they are, gets

better and better as dt gets smaller and smaller. That treating this car’s journey as a bunch

of discontinuous jumps in speed between small portions of constant velocity becomes a less

wrong reflection of reality as we decrease the time between those jumps.

So for convenience, let’s just approximate the speed on each interval with whatever the

true car’s velocity is at the start of the interval; the height of the graph above the

left side, which in this case is 7. So on this example interval, according to

our approximation, the car moves (7 m/s)*(0.25 s). That’s 1.75 meters, nicely visualized

as the area of this thin rectangle. This is a little under the real distance traveled,

but not by much. And the same goes for every other interval:

The approximated distance is v(t)*dt, it’s just that you plug in a different value of

t for each one, giving a different height for each rectangle. I’m going to write out an expression for

the sum of the areas of all these rectangles in kind of a funny way.

Take this symbol, which looks like a stretched “S” for sum, then put a 0 at its bottom

and an 8 at its top to indicate that we’re ranging over time steps between 0 and 8 seconds.

And as I said the amount we’re adding up at each time step is v(t)*dt.

Two things are implicit in this notation: First, the value dt plays two roles: not only

is it a factor in each quantity we’re adding up, it also indicates the spacing between

each sampled time step. So when you make dt smaller, even though it decreases the area

of each rectangle here, it increases the total number of rectangles whose areas we’re adding

up. And second, the reason we don’t use the

usual sigma notation to indicate a sum is that this expression is technically not any

particular sum for any particular choice of dt; it’s whatever that sum approaches as

dt approaches 0. As you can see, what that approaches is the

area bounded by this curve and the horizontal axis.

Remember, smaller choices of dt indicate closer approximations for our original question,

how far does the car go, right? So this limiting value for the sum, the area under this curve,

gives the precise answer to the question, in full unapproximated precision.

Now tell me that’s not surprising. We have this pretty complicated idea of approximations

that can involve adding up a huge number of very tiny things, and yet the value those

approximates approach can be described so simply, as the area under a curve.

This expression is called an “integral” of v(t), since it brings all of its values

together, it integrates them. Now you could say, “How does this help!?.

You’ve just reframed one hard question, finding how far the car has traveled, into

another equally hard problem, finding the area between this graph and the horizontal

axis?” And…you’d be right! If the velocity/distance

duo was all we cared about, most of this video with all this area under a curve nonsense

would be a waste of time. We could just skip straight ahead to figuring out an antiderivative.

But finding the area between a function’s graph and the horizontal axis is somewhat

of a common language for many disparate problems that can be broken down and approximated as

the sum of a large number of small things. You’ll see more next video, but for now

I’ll just say in the abstract that understanding how interpret and compute the area under a

graph is a very general problem-solving tool. In fact, the first video of this series already

covered the basics of how this works, but now that we have more of a background with

derivatives, we can actually take the idea to its completion. For our velocity example, think of this right

endpoint as a variable, capital T. So we’re thinking of this integral of the velocity

function between 0 and T, the area under this curve between those two inputs, as a function,

where that upper bound is the variable. That area represents the distance the car

has traveled after T seconds, right? So this is really a distance vs. time function s(T).

Now ask yourself: What is the derivative of that function?

On the one hand, a tiny change in distance over a tiny change in time is velocity; that’s

what velocity means. But there’s another way to see it purely in terms of this graph

and this area, which generalizes better to other integral problems.

A slight nudge of dT to the input causes that area to increase, some little ds represented

by the area of this sliver. The height of that sliver is the height of

the graph at that point, v(T), and its width is dT.

And for small enough dT, we can basically consider that sliver to be a rectangle. So

the area of that sliver, ds, is approximately equal to v(T)*dT.

Because this approximation gets better and better for smaller dT, the derivative of the

area function ds/dT at this point equals v(T), the value of the velocity function at whatever

time we started on. And that’s super general, the derivative

of any function giving the area under a graph like this is equal to the function for the

graph itself. So if our velocity function here is t*(8-t),

what should s be? What function of t has a derivative t*(8-t). This is where we actually

have to roll up our sleeves and do some math. It’s easier to see if we expand this out

as 8t – t2. Take each part here one at a time: What function

has a derivative 8t? Well, we know that the derivative of t2 is 2t, so if we just scale

that up by 4, we see that the derivative of 4t2 is 8t.

And for that second part, what kind of function might have -t2 as its derivative? Using the

power rule again, we know that the derivative of a cubic term, t3, gives a squared term,

3t2, so if we scale that down by a third, the derivative of (⅓)t3 is exactly t2, and

making that negative we see that -(⅓)t3 has a derivative of -t2.

Therefore, the antiderivative of 8t – t2 is 4t2 – (⅓)t3. But there’s a slight issue here: we could

add any constant to this function, and its derivative would still be 8t – t2. The derivative

of a constant is always 0. And if we graph s(t), you can think of this

in the sense that moving a graph of a distance function up and down does nothing to affect

its slope above each input. So there are actually infinitely many different

possible antiderivative functions, all of which look like 4t2 – (⅓)t3 + C for some

constant C. But there is one piece of information we haven’t

used yet that let’s us zero in on which antiderivative to use: The lower bound on

the integral. This integral must be zero when we drag that

right endpoint all the way to the left endpoint, right? The distance traveled by the car between

0 seconds and 0 seconds is…zero. So as we found, this area as a function of

capital T is an antiderivative for the stuff inside, and to choose what constant to add,

subtract off the value of that antiderivative function at the lower bound.

If you think about it for a moment, that ensures that the integral from the lower bound to

itself will indeed be 0. As it so happens, when you evaluate the function

we have here at t=0, you get zero, so in this specific case you don’t actually need to

subtract off anything. For example, the total distance traveled during

the 8 seconds is this expression evaluated at T=8, which is 85.33, minus 0.

But a more typical example would be something like this integral between 1 and 7. That’s

the area pictured here, and it represents the distance traveled between 1 second and

7 seconds. What you’d do is evaluate the antiderivative

we found at the top bound, 7, and subtract off its value at the bottom bound, 1.

Notice, it doesn’t matter what antiderivative we choose here; if for some reason it had

a constant added to it, like 5, that constant would cancel out. More generally, anytime you want to integrate

some function –and remember you think of as adding up the values f(x)*dx for inputs

in a certain range then asking what that sum approaches as dx approaches 0– the first

step is to find an antiderivative, some other function, “capital F(x)”, whose derivative

is the thing inside the integral. Then the integral equals this antiderivative

evaluated at the top bound, minus its value at the bottom bound. This fact is called the

“fundamental theorem of calculus”. Here’s what’s crazy about this fact: The

integral, the limiting value for the sum of all these thin rectangles, takes into account

every single input on the continuum from the lower bound to the upper bound, that’s why

we use the word “integrate”; it brings them together. And yet, to actually compute

it using the antiderivative, you look at only two inputs: the top and the bottom.

It almost feels like cheating! Finding the antiderivative implicitly accounts for all

the information needed to add up all the values between the lower bound and upper bound. There’s kind of a lot packed into this whole

concept, so let’s recap everything that just happened, shall we?

We wanted to figure out how far a car goes just by looking at the speedometer, and what

makes that hard is that the velocity was always changing.

If you approximate it to be constant on multiple different intervals, you can figure out how

far the car goes on each interval just with multiplication, then add them all up.

Adding up those products can be visualized as the sum of the areas of many thin rectangles

like this. Better and better approximations of the original

problem correspond to collections of rectangles whose aggregate area is closer and closer

to being the area under this curve between the start time and end time, so that area

under the curve is the precise distance traveled for the true, nowhere-constant velocity function.

If you think of this area as function, with a variable right end point, you can deduce

that the derivative of that area function must equal the height of the graph at each

point. That’s the key! So to find a function giving this area, you ask what function has

v(t) as its derivative. There are actually infinitely many antiderivatives

of a given function, since you can always just add some constant without affecting the

derivative, so you account for that by subtracting off the value of whatever antiderivative function

you choose at the bottom bound. By the way, one important thing to bring up

before leaving is the idea of negative area. What if our velocity function was negative

at some point? Meaning the car is going backwards. It’s still true that the tiny distance traveled

ds on a little time interval is about equal to the velocity times the tiny change in time,

it’s just that that the number you’d plug in for velocity would be negative, so that

tiny change in distance is negative. In terms of our thin rectangles, if the rectangle

goes below the horizontal axis like this, its area represents a bit of distance traveled

backwards, so if what you want is to find the distance between the car’s start point

and end point, you’d want to subtract it. And this is generally true of integrals: Whenever

a graph dips below the horizontal axis, that area underneath is counted as negative.

What you’ll commonly hear is that integrals measure the “signed” area between a graph

and the horizontal axis. Next up I’ll bring up more contexts where

this idea of an integral and the area under curves comes up, along with some other intuitions

for the fundamental theorem of calculus. Perhaps you remember, chapter 2 of this series,

introducing the derivative was sponsored by the Art of Problem Solving. So I think there’s

something elegant to the fact that this video, which is kind of a dual to that one, was also

supported in part by the Art of Problem Solving. I really can’t imagine a better sponsor

for the channel, because it’s a company whose books and courses I recommend to people

anyway. They were highly influential to me, when I

was a student developing a love for creative math, so if you’re a parent looking to foster

your own child’s love for the subject, or if you’re a student who wants to see what

math has to offer beyond rote school work, I cannot recommend the Art of Problem Solving

enough. Whether that’s their newest development

to build the right intuitions in elementary schools kids, called Beast academy, or their

courses on higher level topics and contest preparation.

Going to AoPS.com/3blue1brown, or clicking the link on the screen, lets them know you

came from this channel, which may encourage them to support future projects like this

one. I consider these videos a success not when

they teach people a particular bit of math, which can only ever be a drop in the ocean,

but when they encourage people to go explore the expanse of math for themselves. And the

Art of Problem Solving is among the few great places to actually do that.

School: You gotta learn this.

Me: Why?

School: You gotta learn this.

Me: But what is it good for?

School: You gotta learn this.

A few years later in university …

Me: Geee, I wonder how I can calculate the area under that curve so I can get the consumer surplus from non-linear demand and supply functions …

at 7:02, not only it a factor in each quantity that we are adding up it also indicates the spacing between each sample step. wow that is at the height of Grant's explanation. Really appreciate it.

Watches first 30 seconds.So, if it’s not almost obvious, why try and prove it. Got it.

Watches video about how 1+2+3+4+5+6…….. = -1/12Wait, what?

Now look at my teachers…i think they really don't know these things and due to such teachers students don't understand maths and science…thats why students hate…thank you very much for this knowledge…i just cant memorise formulae without understanding..i just can't….thank you very much…🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻

To the people who disliked this video….

Kill yourselves.

If you are watching from Smith's class give me a like 😀

when you said you were going to make it obvious why integral is the inverse of derivatives i was like: This video is for me!!!!!

When I first watched this video, it made no sense. Now I’m coming back after 2 months and it just sort of clicked for me.

I'm still not clear what the dx means in a definite integral. Why do we need it if finding the upper bound minus the lower bound provides us with the area already?

this is so good i'm literally crying

Sorry for stupid question but I'm still a little confused. Can anyone explain what is the relationship between infinitesimal divided parts with the formula of integration ?

V^2-u^2=2as anyone?.

It's the best explanation ever in my entire life learning calculus!

Can you do a video on double and triple integrals? I’d really like a visualisation of that kind of integrals as just computing them can get really abstract (espacially finding curl in a force field for example).

I just had a headgasm

My math teacher sent us links to these videos. Nice

This explanation is absolutely amazing! It is quintessential to understand the simplified proof, before getting into more rigorous proving. I watch the videos before going to lecture, it is a perfect combination with course materials!!

Thanks à lot sir and I wish to add also Arabic for more understanding because I have a problem with English… Thanks a lot for the great explanation

@11:10 "The derivative of any function, given the area under a graph, is equal to the function for the graph itself."

This statement seems to suggest that d[f(x)]/dx = f(x), which is incorrect. @10:30 you said 'ds' represents the sliver of change in the AREA under the graph v(t). ds/dt = v(t) suggests that the rate of change in the

cumulative areaunder v(t) is equal to v(t). Notationally, d[A(v(t))]/dt = v(t) where A(…) means 'the area under'You sir, are amazing. thank you so much for explaining integration so clearly. i could not thank you enough

One I don't understand is why aren't we multipepling by T/(dt)

Nothing but purity.

I think Grant knows who Anand Srinivas is.

Anand like grant is one of the greatest. I learnt high school physics from him.

Driving without vision on the street is dangerous. At the end he travels -100m on the y-Axis to the bottom of a cliff.

I looked at 40 different videos about this and this is the only one that makes sense.

we should have raided James Stewart's house, dragged him out onto the streets, and blew his brains out before he could come up with the 20th edition of his god forsaken textbook to confuse students all across the nation

This series is so goooood. It really makes me think intuitively rather than learning the formula, thankyou soooo much

When I first saw rectangles approximating the area under a curve in a textbook, I thought "well wouldn't it be nice is calculus was as simple as that, but clearly there's more to it than just that". But that's the thing. There's not much more to it than approximating the area under a curve by rectangles, and then letting the size of the rectangles get smaller and smaller.

Great work you are one of most influencing educators in math intuition.

Please post such kind of videos on complex analysis and it's integration.

the first time i understand clearly the integrals after long hours thank you so much

Why he writes V(t) = t(8-t) ???

15:34 “This is the fundamental theorem of calculus”

And at that moment, I paused the video, waited a second, and said to myself: “Fuck you (teacher I had a year ago). I learned something in 20 minutes, something that took you a week to sloppily teach me and hoped I’d get it. I didn’t, but now I do thanks to a YouTube video”

anyone, is it possible to integrate root of x^2 – r^2 given that r is a constant, please help, I am stuck

It's the visuals that make you understand it, very well done!

Very cognizant video

This makes 1000000% sense

The best thing about the theorem is the tiny change of time not affecting the the tiny area, the distance traveled over the tiny period of time. Velocity itself affects this distance. It is complied with the intuition about S-t, the distance function in respect of time.

Is the into/outro song an original piece, or does it have a name? I really like it.

slow learner here….how can we find the area of a curve?

Fantastic channel name📽

FEELING CALCULUS AS WRONG …BUT TRUE……

This siries is heaven like. It should get some kind of prize.

Beautiful work. Just beautiful.

Brilliant for intuition! Well done Sir.

Another clever way to obtain the total distance travelled is to multiply the average value of the velocity by the total time travelled. This still works when using curves!

Future BC Calc student here, I feel like I'm going to ace this course after watching these videos

And how do you find out the equation for any given curve??

OMG THATS WHAT dx MEANS

Thx a lot

What does the quote in the beginning mean?

If this video came out during Newton and Leibniz's time then they probably would have learned something here or there.

Thank you!

WOW!!!!! THESE ANIMATED GRAPHICS REALLY DO HELP IN TRYING TO CLEARLY "VISUALIZE" WHAT IT IS THAT YOU ARE EXPLAINING!!! "BRAVO!!!" THANK YOU!!!!!

Hi. I have a question. Is the v(t) function itself the derivative of the integral function s(T)? Is the instantaneous change in the area equals to the instantaneous speed?

This is the most beautiful series I have ever seen in my life.

Why do we need to find the integral of the velocity function? Why can't we directly plug the limits of the integral directly to the distance function?

Hi. first of all thank you .somethings is making me crazy for a long time ,and in this video you will talk about it about 15:30 pass of video and that thing is how upper bound minus lower bound will give us area ?! these are just two point which we will calculate and the shape is not something like rectangle . please, someone answer me, I've watched these video series more than 5 time ,search a lot in internet but i can not understand it ,I'm getting crazy pls someone help me about this .

2M subs !!!!!

math is becoming interesting

Can you do Lebesgue integration ?

Can you please make a video on generating function and moments of distributions , please

By any chance, could you do a video on Integration by Parts? There are no visualizations online without intense Riemann Sum work…which is pretty abstract.

8:35 Thank you for reviving the interobang.

How is v(t) = t(8-t)

I think it's just an assumption. Right?

0:18 to 0:31 this is what people should realise about math! So well said and relatable!!

Your videos are amazing. Thank you.

Dieck

At around 11:00, you said the derivative of the area function is exactly the function itself. Since when dT approaches zero, our rectangular approximation resembles dS more and more, it makes sense then that at dT=zero, the derivative of s(T) is exactly v(T). However, dT=0 invalidates the equality because a zero denominator is undefined. I think this contradiction possibly highlights a consequence of treating differentials as variables when they're actually not.

Do you think you could make a video about integration by parts? I always memorized the formula but still don't have a visual intuition.

How did you make the animation? It's so satisfied to watch them :))

integral of 0 ?

Car speedometers do not show negative speed when you are in reverse… Furthermore, if you keep looking at the speedometer while you're driving you are very likely to crash..;) So the moving car was perhaps not the best example for this demonstration.. Other than that, a very good presentation of the integral. By the way, I loved your video on the Fourier Transform. Great channel!

What is the anti derivative of the distance function s(t), and what does it mean? I mean, what would the area in the graph formed by the s(t) function mean?

Why there was v(t) =t(8-t) I didn't understand where it came from can someone explain it to me

Took me some time to get where the F(b)-F(a) came from. Suggestion: it would be more obvious if you actually plotted the F(x) on the graph and reiterated how it corresponds to the area under f(x) taken from 0 to x. Then it would be obvious that to get the area from a to b you take the area up to b, which would be F(b) and subtract the area up to a, which would be F(a). Actually, now that I've put it that way, the graph would probably not be necessary, just the notion that we subtract one area from another.

As a random geek who wants to learn calc as a 9th grader (don't worry I'm 2 years ahead), these videos are really, really awesome and I love how simple you make it all seem. Thanks for making such great videos!

1:11

Yes, there is, and they are called AP Physics equations.

Vf^2=Vi^2 + 2ax

You are a genius man.

what am I doing here? I was doing fine brushing up on my algebra on cnx.org and now I'm feeling stupid because I'm trying to understand chapter 8 of calculus lol

You are my math hero

8:35 look at the ? And the ! merged together

İ love math just because of this kind of things, you even can make impossible possible.

You have to do videos on multiple integration. Like the physical interpretation…….. That would make it soooo understandable and very helpful…. Ps: your videos increase my understanding of concepts..

wait a second…

S= d/t

St=d

8×10 = 80m

let me say this to myself

you counted that function of v = t(8-t)

ds/dt = t(8-t)

so we are calculating s (distance)

ds = (dv)(dt) which is correct

to put it in English

we have velocity and time and we want to calculate distance travelled

we know that the smaller the time frame the better guess of velocity at any point we have

the better the velocity at all points and instants the better the distance

v = s/t as change in either s or t changes

v = f(t) = t(8-t)

v at point T is v(T)

v(T) (dT) = ds

A point in velocity x small amount of time = small amount of distance

ds/dT = v(T)

distance/time = v(T) at that time

s = 4T^2 – 1/3(T)^3

People: Long videos are boring.

This guy: Hold my beer.

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And opens up the universe in a snap!

I love that the gradient of the area rectangles animated the idea of smoothness. When there are 8 large rectangles, each rectangle's color is noticeably different from the color of the adjacent rectangle. As the rectangles get smaller (approaching the area under the curve) the colors approach a smooth gradient as well.

It seems that beautiful and subtle tricks are your forté, and this one is no exception. Thank you for your very helpful videos!

You are absolutely fucking brilliant as a teacher.

nice..

So at the end, for the total distance traveled accounting for the backward direction, the solution would be either the sum of the integrals of the 2 blue areas, or the whole integral minus the red area? Or does the overall integral account for.thus? I found that part to not be explained clearly.

I've been learning how to solve derivatives and integrals two years ago. But not until watching these videos I can understand what's really happening behind the numbers

I fucking love this channel.

It doesn't seem like good Explanation You are skipping a lot, Most of your channel videos are easy to understand. But this video was not like that, it's more difficult understood. Plz Explain step by step like this https://www.youtube.com/watch?v=FsC3do74UIo #note: I'm not good at with English and I know, don't share any other youtube links in the comment box but what can I do

Gotta say. I was asleep for most of my calc class since I worked till 1am for a few months. Class at 8am. And ive watched a few of these and thanks now I know sigma? Notation and how simple they are. Preped for finals though

I feel blessed

Good video although you didn't really explain the newton-leibniz formula. You can't deduce it from what you've said at 17:05. How do you know you can just ignore all of the points in between and just do the antiderivative for the ending and starting point?

He will found father of pi

8:11

Please, make videos about surface integrals.

Thanks 🙂 great video

It always amazes me how do people actually come up with such things like integration that can calculate such complex questions so easily.