[00:12] This guy, Grothendieck, is somewhat of a mathematical idol to me,
[00:15] and I just love this quote, don't you?
[00:18] Too often in math, we dive into showing that a certain fact is true
[00:22] with a long series of formulas before stepping back and making sure it feels reasonable,
[00:27] and preferably obvious, at least at an intuitive level.
[00:31] In this video, I want to talk about integrals,
[00:33] and the thing that I want to become almost obvious is that they are an
[00:37] inverse of derivatives.
[00:39] Here we're just going to focus on one example,
[00:42] which is a kind of dual to the example of a moving car that I talked about in chapter
[00:46] 2 of the series, introducing derivatives.
[00:49] Then in the next video we're going to see how this same idea generalizes,
[00:52] but to a couple other contexts.
[00:55] Imagine you're sitting in a car, and you can't see out the window,
[00:58] all you see is the speedometer.
[01:02] At some point the car starts moving, speeds up,
[01:05] and then slows back down to a stop, all over the course of 8 seconds.
[01:11] The question is, is there a nice way to figure out how far you've
[01:15] travelled during that time based only on your view of the speedometer?
[01:19] Or better yet, can you find a distance function, s of t,
[01:23] that tells you how far you've travelled after a given amount of time, t,
[01:27] somewhere between 0 and 8 seconds?
[01:30] Let's say you take note of the velocity at every second,
[01:34] and make a plot over time that looks something like this.
[01:38] And maybe you find that a nice function to model that velocity
[01:43] over time in meters per second is v of t equals t times 8 minus t.
[01:48] You might remember, in chapter 2 of this series we were looking at
[01:51] the opposite situation, where you knew what a distance function was,
[01:55] s of t, and you wanted to figure out the velocity function from that.
[01:59] There I showed how the derivative of a distance vs.
[02:02] time function gives you a velocity vs.
[02:04] time function.
[02:06] So in our current situation, where all we know is velocity,
[02:09] it should make sense that finding a distance vs.
[02:12] time function is going to come down to asking what
[02:15] function has a derivative of t times 8 minus t.
[02:19] This is often described as finding the antiderivative of a function, and indeed,
[02:23] that's what we'll end up doing, and you could even pause right now and try that.
[02:27] But first, I want to spend the bulk of this video showing how this question is related
[02:32] to finding the area bounded by the velocity graph,
[02:35] because that helps to build an intuition for a whole class of problems,
[02:39] things called integral problems in math and science.
[02:42] To start off, notice that this question would be a lot easier
[02:45] if the car was just moving at a constant velocity, right?
[02:49] In that case, you could just multiply the velocity in meters per second times the amount
[02:54] of time that has passed in seconds, and that would give you the number of meters traveled.
[03:00] And notice, you can visualize that product, that distance, as an area.
[03:05] And if visualizing distance as area seems kind of weird, I'm right there with you.
[03:08] It's just that on this plot, where the horizontal direction has units of seconds,
[03:13] and the vertical direction has units of meters per second,
[03:17] units of area just very naturally correspond to meters.
[03:22] But what makes our situation hard is that velocity is not constant,
[03:25] it's incessantly changing at every single instant.
[03:30] It would even be a lot easier if it only ever changed at a handful of points,
[03:35] maybe staying static for the first second, and then suddenly discontinuously
[03:39] jumping to a constant 7 meters per second for the next second, and so on,
[03:43] with discontinuous jumps to portions of constant velocity.
[03:48] That would make it uncomfortable for the driver,
[03:51] in fact it's actually physically impossible, but it would make your calculations
[03:55] a lot more straightforward.
[03:57] You could just compute the distance traveled on each interval by multiplying the constant
[04:02] velocity on that interval by the change in time, and then just add all of those up.
[04:09] So what we're going to do is approximate the velocity function as if it
[04:13] was constant on a bunch of intervals, and then, as is common in calculus,
[04:17] we'll see how refining that approximation leads us to something more precise.
[04:24] Here, let's make this a little more concrete by throwing in some numbers.
[04:28] Chop up the time axis between 0 and 8 seconds into many small intervals,
[04:33] each with some little width dt, something like 0.25 seconds.
[04:38] Consider one of those intervals, like the one between t equals 1 and 1.25.
[04:45] In reality, the car speeds up from 7 m per second to about 8.4 m per
[04:49] second during that time, and you could find those numbers just by
[04:53] plugging in t equals 1 and t equals 1.25 to the equation for velocity.
[04:59] What we want to do is approximate the car's motion
[05:02] as if its velocity was constant on that interval.
[05:05] Again, the reason for doing that is we don't really know
[05:08] how to handle situations other than constant velocity ones.
[05:13] You could choose this constant to be anything between 7 and 8.4.
[05:18] It actually doesn't matter.
[05:20] All that matters is that our sequence of approximations,
[05:23] whatever they are, gets better and better as dt gets smaller and smaller.
[05:28] That treating this car's journey as a bunch of discontinuous jumps
[05:32] in speed between portions of constant velocity becomes a less-wrong
[05:36] reflection of reality as we decrease the time between those jumps.
[05:42] So for convenience, on an interval like this, let's just approximate the
[05:46] speed with whatever the true car's velocity is at the start of that interval,
[05:50] the height of the graph above the left side, which in this case is 7.
[05:55] In this example interval, according to our approximation,
[06:00] the car moves 7 m per second times 0.25 seconds.
[06:04] That's 1.75 meters, and it's nicely visualized as the area of this thin rectangle.
[06:10] In truth, that's a little under the real distance traveled, but not by much.
[06:14] The same goes for every other interval.
[06:17] The approximated distance is v of t times dt, it's just that you'd be plugging in a
[06:22] different value for t at each one of these, giving a different height for each rectangle.
[06:29] I'm going to write out an expression for the sum of
[06:32] the areas of all those rectangles in kind of a funny way.
[06:36] Take this symbol here, which looks like a stretched s for sum,
[06:40] and put a 0 at its bottom and an 8 at its top,
[06:43] to indicate that we'll be ranging over time steps between 0 and 8 seconds.
[06:48] And as I said, the amount we're adding up at each time step is v of t times dt.
[06:55] Two things are implicit in this notation.
[06:58] First of all, that value dt plays two separate roles.
[07:01] Not only is it a factor in each quantity we're adding up,
[07:05] it also indicates the spacing between each sampled time step.
[07:09] So when you make dt smaller and smaller, even though it decreases the area of
[07:13] each rectangle, it increases the total number of rectangles whose areas we're adding up,
[07:18] because if they're thinner, it takes more of them to fill that space.
[07:22] And second, the reason we don't use the usual sigma notation to indicate a sum is that
[07:28] this expression is technically not any particular sum for any particular choice of dt.
[07:33] It's meant to express whatever that sum approaches as dt approaches 0.
[07:39] And as you can see, what that approaches is the
[07:42] area bounded by this curve and the horizontal axis.
[07:46] Remember, smaller choices of dt indicate closer approximations for the original question,
[07:51] how far does the car actually go?
[07:54] So this limiting value for the sum, the area under this curve,
[07:58] gives us the precise answer to the question in full unapproximated precision.
[08:04] Now tell me that's not surprising.
[08:06] We had this pretty complicated idea of approximations that
[08:09] can involve adding up a huge number of very tiny things.
[08:13] And yet, the value that those approximations approach can be described so simply,
[08:18] it's just the area underneath this curve.
[08:22] This expression is called an integral of v of t,
[08:25] since it brings all of its values together, it integrates them.
[08:30] Now at this point, you could say, how does this help?
[08:33] You've just reframed one hard question, finding how far the car has traveled,
[08:37] into an equally hard problem, finding the area between this graph and the horizontal axis.
[08:43] And you'd be right.
[08:45] If the velocity-distance duo was the only thing we cared about, most of this video,
[08:50] with all the area under a curve nonsense, would be a waste of time.
[08:54] We could just skip straight ahead to finding an antiderivative.
[08:58] But finding the area between a function's graph and the horizontal axis
[09:02] is somewhat of a common language for many disparate problems that can be
[09:06] broken down and approximated as the sum of a large number of small things.
[09:12] You'll see more in the next video, but for now I'll just say in
[09:15] the abstract that understanding how to interpret and how to
[09:19] compute the area under a graph is a very general problem-solving tool.
[09:23] In fact, the first video of this series already covered the basics of how this works,
[09:28] but now that we have more of a background with derivatives,
[09:31] we can take this idea to its completion.
[09:34] For a velocity example, think of this right endpoint as a variable, capital T.
[09:41] So we're thinking of this integral of the velocity function between 0 and T,
[09:45] the area under this curve between those inputs,
[09:48] as a function where the upper bound is the variable.
[09:52] That area represents the distance the car has travelled after T seconds, right?
[09:57] So in reality, this is a distance vs.
[09:59] time function, s of t.
[10:01] Now ask yourself, what is the derivative of that function?
[10:07] On the one hand, a tiny change in distance over a tiny change in time is velocity,
[10:12] that is what velocity means.
[10:14] But there's another way to see this, purely in terms of this graph and this area,
[10:19] which generalizes a lot better to other integral problems.
[10:23] A slight nudge of dt to the input causes that area to increase,
[10:27] some little ds represented by the area of this sliver.
[10:32] The height of that sliver is the height of the graph at that point,
[10:37] v of t, and its width is dt.
[10:39] And for small enough dt, we can basically consider that sliver to be a rectangle,
[10:45] so this little bit of added area, ds, is approximately equal to v of t times dt.
[10:51] And because that's an approximation that gets better and better for smaller dt,
[10:56] the derivative of that area function, ds, dt, at this point equals vt,
[11:01] the value of the velocity function at whatever time we started on.
[11:06] And that right there is a super general argument.
[11:09] The derivative of any function giving the area under a
[11:12] graph like this is equal to the function for the graph itself.
[11:18] So, if our velocity function is t times 8-t, what should s be?
[11:25] What function of t has a derivative of t times 8-t?
[11:30] It's easier to see if we expand this out, writing it as 8t minus t squared,
[11:34] and then we can just take each part one at a time.
[11:37] What function has a derivative of 8t?
[11:42] We know that the derivative of t squared is 2t,
[11:45] so if we just scale that up by a factor of 4, we can see that the derivative
[11:50] of 4t squared is 8t.
[11:53] And for that second part, what kind of function do
[11:55] you think might have negative t squared as a derivative?
[12:00] Using the power rule again, we know that the derivative of a cubic term,
[12:04] t cubed, gives us a square term, 3t squared.
[12:08] So if we just scale that down by a third, the
[12:11] derivative of 1 third t cubed is exactly t squared.
[12:14] And then making that negative, we'd see that negative
[12:17] 1 third t cubed has a derivative of negative t squared.
[12:22] Therefore, the antiderivative of our function,
[12:26] 8t minus t squared, is 4t squared minus 1 third t cubed.
[12:32] But there's a slight issue here.
[12:34] We could add any constant we want to this function,
[12:37] and its derivative is still 8t minus t squared.
[12:41] The derivative of a constant always goes to zero.
[12:45] And if you were to graph s of t, you could think of this in the sense that moving a
[12:49] graph of a distance function up and down does nothing to affect its slope at every input.
[12:54] So in reality, there's actually infinitely many different
[12:58] possible antiderivative functions, and every one of them looks
[13:02] like 4t squared minus 1 third t cubed plus c, for some constant c.
[13:08] But there is one piece of information we haven't used yet that will let
[13:12] us zero in on which antiderivative to use, the lower bound of the integral.
[13:18] This integral has to be zero when we drag that right
[13:21] endpoint all the way to the left endpoint, right?
[13:24] The distance travelled by the car between 0 seconds and 0 seconds is… well, zero.
[13:31] So as we found, the area as a function of capital
[13:34] T is an antiderivative for the stuff inside.
[13:38] And to choose what constant to add to this expression,
[13:42] you subtract off the value of that antiderivative function at the lower bound.
[13:48] If you think about it for a moment, that ensures that the
[13:51] integral from the lower bound to itself will indeed be zero.
[13:57] As it so happens, when you evaluate the function we have here at t equals zero,
[14:02] you get zero.
[14:03] So in this specific case, you don't need to subtract anything off.
[14:07] For example, the total distance travelled during the full 8 seconds
[14:13] is this expression evaluated at t equals 8, which is 85.33 minus 0.
[14:18] So the answer as a whole is 85.33.
[14:23] But a more typical example would be something like the integral between 1 and 7.
[14:28] That's the area pictured here, and it represents
[14:30] the distance travelled between 1 second and 7 seconds.
[14:36] What you do is evaluate the antiderivative we found at the top bound,
[14:41] 7, and subtract off its value at the bottom bound, 1.
[14:45] Notice, by the way, it doesn't matter which antiderivative we chose here.
[14:50] If for some reason it had a constant added to it, like 5, that constant would cancel out.
[14:58] More generally, any time you want to integrate some function, and remember,
[15:03] you think of that as adding up values f of x times dx for inputs in a certain range,
[15:08] and then asking what is that sum approach as dx approaches 0.
[15:13] The first step to evaluating that integral is to find an antiderivative,
[15:18] some other function, capital F, whose derivative is the thing inside the integral.
[15:24] Then the integral equals this antiderivative evaluated
[15:28] at the top bound minus its value at the bottom bound.
[15:32] And this fact right here that you're staring at is the fundamental theorem of calculus.
[15:38] And I want you to appreciate something kind of crazy about this fact.
[15:41] The integral, the limiting value for the sum of all these thin rectangles,
[15:46] takes into account every single input on the continuum,
[15:49] from the lower bound to the upper bound.
[15:52] That's why we use the word integrate, it brings them all together.
[15:56] And yet, to actually compute it using an antiderivative,
[16:00] you only look at two inputs, the top bound and the bottom bound.
[16:05] It almost feels like cheating.
[16:06] Finding the antiderivative implicitly accounts for all the
[16:10] information needed to add up the values between those two bounds.
[16:15] That's just crazy to me.
[16:18] This idea is deep, and there's a lot packed into this whole concept,
[16:22] so let's recap everything that just happened, shall we?
[16:26] We wanted to figure out how far a car goes just by looking at the speedometer.
[16:31] And what makes that hard is that velocity is always changing.
[16:35] If you approximate velocity to be constant on multiple different intervals,
[16:39] you could figure out how far the car goes on each interval with multiplication,
[16:43] and then add all of those up.
[16:46] Better and better approximations for the original problem correspond to
[16:50] collections of rectangles whose aggregate area is closer and closer to
[16:54] being the area under this curve between the start time and the end time.
[16:58] So that area under the curve is actually the precise distance
[17:03] traveled for the true nowhere constant velocity function.
[17:08] If you think of that area as a function itself,
[17:11] with a variable right endpoint, you can deduce that the derivative
[17:15] of that area function must equal the height of the graph at every point.
[17:21] And that's really the key right there.
[17:22] It means that to find a function giving this area,
[17:26] you ask, what function has v of t as a derivative?
[17:30] There are actually infinitely many antiderivatives of a given function,
[17:34] since you can always just add some constant without affecting the derivative,
[17:38] so you account for that by subtracting off the value of whatever antiderivative function
[17:43] you choose at the bottom bound.
[17:46] By the way, one important thing to bring up before we leave is the idea of negative area.
[17:53] What if the velocity function was negative at some point, meaning the car goes backwards?
[17:58] It's still true that a tiny distance traveled ds on a little time interval is
[18:03] about equal to the velocity at that time multiplied by the tiny change in time.
[18:08] It's just that the number you'd plug in for velocity would be negative,
[18:13] so the tiny change in distance is negative.
[18:16] In terms of our thin rectangles, if a rectangle goes below the horizontal axis,
[18:21] like this, its area represents a bit of distance traveled backwards,
[18:25] so if what you want in the end is to find a distance between the car's
[18:29] start point and its end point, this is something you'll want to subtract.
[18:35] And that's generally true of integrals.
[18:37] Whenever a graph dips below the horizontal axis,
[18:40] the area between that portion of the graph and the horizontal axis is counted as negative.
[18:46] What you'll commonly hear is that integrals don't measure area per se,
[18:50] they measure the signed area between the graph and the horizontal axis.
[18:55] Next up, I'm going to bring up more context where this idea
[18:58] of an integral and area under curves comes up,
[19:01] along with some other intuitions for this fundamental theorem of calculus.
[19:06] Maybe you remember, chapter 2 of this series introducing the derivative
[19:10] was sponsored by The Art of Problem Solving, so I think there's
[19:13] something elegant to the fact that this video, which is kind of a duel to that one,
[19:18] was also supported in part by The Art of Problem Solving.
[19:22] I really can't imagine a better sponsor for this channel,
[19:25] because it's a company whose books and courses I recommend to people anyway.
[19:29] They were highly influential to me when I was a student developing a love for
[19:33] creative math, so if you're a parent looking to foster your own child's love
[19:37] for the subject, or if you're a student who wants to see what math has to offer
[19:42] beyond rote schoolwork, I cannot recommend The Art of Problem Solving enough.
[19:46] Whether that's their newest development to build the right intuitions in
[19:50] elementary school kids, called Beast Academy, or their courses in higher-level
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[20:08] I consider these videos a success not when they teach people a particular bit of math,
[20:13] which can only ever be a drop in the ocean, but when they encourage people
[20:17] to go and explore that expanse for themselves,
[20:20] and The Art of Problem Solving is among the few great places to actually do
[20:24] that exploration.