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.
