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This guy, Grothendieck, is somewhat of a mathematical idol to me,

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and I just love this quote, don't you?

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Too often in math, we dive into showing that a certain fact is true

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with a long series of formulas before stepping back and making sure it feels reasonable,

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and preferably obvious, at least at an intuitive level.

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In this video, I want to talk about integrals,

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and the thing that I want to become almost obvious is that they are an

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inverse of derivatives.

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Here we're just going to focus on one example,

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which is a kind of dual to the example of a moving car that I talked about in chapter

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2 of the series, introducing derivatives.

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Then in the next video we're going to see how this same idea generalizes,

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but to a couple other contexts.

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Imagine you're sitting in a car, and you can't see out the window,

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all you see is the speedometer.

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At some point the car starts moving, speeds up,

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and then slows back down to a stop, all over the course of 8 seconds.

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The question is, is there a nice way to figure out how far you've

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travelled during that time based only on your view of the speedometer?

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Or better yet, can you find a distance function, s of t,

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that tells you how far you've travelled after a given amount of time, t,

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somewhere between 0 and 8 seconds?

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Let's say you take note of the velocity at every second,

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and make a plot over time that looks something like this.

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And maybe you find that a nice function to model that velocity

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over time in meters per second is v of t equals t times 8 minus t.

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You might remember, in chapter 2 of this series we were looking at

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the opposite situation, where you knew what a distance function was,

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s of t, and you wanted to figure out the velocity function from that.

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There I showed how the derivative of a distance vs.

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time function gives you a velocity vs.

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time function.

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So in our current situation, where all we know is velocity,

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it should make sense that finding a distance vs.

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time function is going to come down to asking what

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function has a derivative of t times 8 minus t.

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This is often described as finding the antiderivative of a function, and indeed,

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that's what we'll end up doing, and you could even pause right now and try that.

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But first, I want to spend the bulk of this video showing how this question is related

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to finding the area bounded by the velocity graph,

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because that helps to build an intuition for a whole class of problems,

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things called integral problems in math and science.

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To start off, notice that this question would be a lot easier

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if the car was just moving at a constant velocity, right?

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In that case, you could just multiply the velocity in meters per second times the amount

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of time that has passed in seconds, and that would give you the number of meters traveled.

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And notice, you can visualize that product, that distance, as an area.

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And if visualizing distance as area seems kind of weird, I'm right there with you.

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

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and the vertical direction has units of meters per second,

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units of area just very naturally correspond to meters.

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But what makes our situation hard is that velocity is not constant,

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it's incessantly changing at every single instant.

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It would even be a lot easier if it only ever changed at a handful of points,

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maybe staying static for the first second, and then suddenly discontinuously

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jumping to a constant 7 meters per second for the next second, and so on,

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with discontinuous jumps to portions of constant velocity.

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That would make it uncomfortable for the driver,

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in fact it's actually physically impossible, but it would make your calculations

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a lot more straightforward.

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

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velocity on that interval by the change in time, and then just add all of those up.

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So what we're going to do is approximate the velocity function as if it

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was constant on a bunch of intervals, and then, as is common in calculus,

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we'll see how refining that approximation leads us to something more precise.

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Here, let's make this a little more concrete by throwing in some numbers.

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Chop up the time axis between 0 and 8 seconds into many small intervals,

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each with some little width dt, something like 0.25 seconds.

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Consider one of those intervals, like the one between t equals 1 and 1.25.

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In reality, the car speeds up from 7 m per second to about 8.4 m per

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second during that time, and you could find those numbers just by

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plugging in t equals 1 and t equals 1.25 to the equation for velocity.

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What we want to do is approximate the car's motion

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as if its velocity was constant on that interval.

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Again, the reason for doing that is we don't really know

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how to handle situations other than constant velocity ones.

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You could choose this constant to be anything between 7 and 8.4.

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It actually doesn't matter.

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All that matters is that our sequence of approximations,

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whatever they are, gets better and better as dt gets smaller and smaller.

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That treating this car's journey as a bunch of discontinuous jumps

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in speed between portions of constant velocity becomes a less-wrong

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

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So for convenience, on an interval like this, let's just approximate the

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speed with whatever the true car's velocity is at the start of that interval,

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the height of the graph above the left side, which in this case is 7.

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In this example interval, according to our approximation,

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the car moves 7 m per second times 0.25 seconds.

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That's 1.75 meters, and it's nicely visualized as the area of this thin rectangle.

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In truth, that's a little under the real distance traveled, but not by much.

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The same goes for every other interval.

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The approximated distance is v of t times dt, it's just that you'd be plugging in a

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different value for t at each one of these, giving a different height for each rectangle.

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I'm going to write out an expression for the sum of

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the areas of all those rectangles in kind of a funny way.

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Take this symbol here, which looks like a stretched s for sum,

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and put a 0 at its bottom and an 8 at its top,

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to indicate that we'll be ranging over time steps between 0 and 8 seconds.

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And as I said, the amount we're adding up at each time step is v of t times dt.

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Two things are implicit in this notation.

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First of all, that value dt plays two separate roles.

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Not only is it a factor in each quantity we're adding up,

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it also indicates the spacing between each sampled time step.

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So when you make dt smaller and smaller, even though it decreases the area of

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each rectangle, it increases the total number of rectangles whose areas we're adding up,

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because if they're thinner, it takes more of them to fill that space.

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And second, the reason we don't use the usual sigma notation to indicate a sum is that

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this expression is technically not any particular sum for any particular choice of dt.

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It's meant to express whatever that sum approaches as dt approaches 0.

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And as you can see, what that approaches is the

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area bounded by this curve and the horizontal axis.

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

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how far does the car actually go?

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So this limiting value for the sum, the area under this curve,

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

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Now tell me that's not surprising.

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We had this pretty complicated idea of approximations that

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can involve adding up a huge number of very tiny things.

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And yet, the value that those approximations approach can be described so simply,

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it's just the area underneath this curve.

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This expression is called an integral of v of t,

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since it brings all of its values together, it integrates them.

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Now at this point, you could say, how does this help?

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

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into an equally hard problem, finding the area between this graph and the horizontal axis.

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And you'd be right.

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If the velocity-distance duo was the only thing we cared about, most of this video,

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with all the area under a curve nonsense, would be a waste of time.

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We could just skip straight ahead to finding an antiderivative.

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But finding the area between a function's graph and the horizontal axis

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is somewhat of a common language for many disparate problems that can be

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broken down and approximated as the sum of a large number of small things.

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You'll see more in the next video, but for now I'll just say in

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the abstract that understanding how to interpret and how to

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compute the area under a graph is a very general problem-solving tool.

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In fact, the first video of this series already covered the basics of how this works,

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but now that we have more of a background with derivatives,

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we can take this idea to its completion.

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For a velocity example, think of this right endpoint as a variable, capital T.

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So we're thinking of this integral of the velocity function between 0 and T,

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the area under this curve between those inputs,

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as a function where the upper bound is the variable.

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That area represents the distance the car has travelled after T seconds, right?

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So in reality, this is a distance vs.

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time function, s of t.

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Now ask yourself, what is the derivative of that function?

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On the one hand, a tiny change in distance over a tiny change in time is velocity,

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that is what velocity means.

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But there's another way to see this, purely in terms of this graph and this area,

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which generalizes a lot better to other integral problems.

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A slight nudge of dt to the input causes that area to increase,

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some little ds represented by the area of this sliver.

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The height of that sliver is the height of the graph at that point,

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v of t, and its width is dt.

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And for small enough dt, we can basically consider that sliver to be a rectangle,

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so this little bit of added area, ds, is approximately equal to v of t times dt.

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And because that's an approximation that gets better and better for smaller dt,

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the derivative of that area function, ds, dt, at this point equals vt,

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the value of the velocity function at whatever time we started on.

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And that right there is a super general argument.

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The derivative of any function giving the area under a

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graph like this is equal to the function for the graph itself.

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So, if our velocity function is t times 8-t, what should s be?

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What function of t has a derivative of t times 8-t?

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It's easier to see if we expand this out, writing it as 8t minus t squared,

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and then we can just take each part one at a time.

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What function has a derivative of 8t?

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We know that the derivative of t squared is 2t,

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so if we just scale that up by a factor of 4, we can see that the derivative

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of 4t squared is 8t.

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And for that second part, what kind of function do

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you think might have negative t squared as a derivative?

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Using the power rule again, we know that the derivative of a cubic term,

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t cubed, gives us a square term, 3t squared.

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So if we just scale that down by a third, the

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derivative of 1 third t cubed is exactly t squared.

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And then making that negative, we'd see that negative

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1 third t cubed has a derivative of negative t squared.

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Therefore, the antiderivative of our function,

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8t minus t squared, is 4t squared minus 1 third t cubed.

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But there's a slight issue here.

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We could add any constant we want to this function,

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and its derivative is still 8t minus t squared.

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The derivative of a constant always goes to zero.

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And if you were to graph s of t, you could think of this in the sense that moving a

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graph of a distance function up and down does nothing to affect its slope at every input.

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So in reality, there's actually infinitely many different

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possible antiderivative functions, and every one of them looks

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like 4t squared minus 1 third t cubed plus c, for some constant c.

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But there is one piece of information we haven't used yet that will let

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us zero in on which antiderivative to use, the lower bound of the integral.

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This integral has to be zero when we drag that right

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endpoint all the way to the left endpoint, right?

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The distance travelled by the car between 0 seconds and 0 seconds is… well, zero.

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So as we found, the area as a function of capital

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T is an antiderivative for the stuff inside.

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And to choose what constant to add to this expression,

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

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If you think about it for a moment, that ensures that the

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integral from the lower bound to itself will indeed be zero.

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As it so happens, when you evaluate the function we have here at t equals zero,

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you get zero.

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So in this specific case, you don't need to subtract anything off.

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For example, the total distance travelled during the full 8 seconds

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is this expression evaluated at t equals 8, which is 85.33 minus 0.

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So the answer as a whole is 85.33.

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But a more typical example would be something like the integral between 1 and 7.

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That's the area pictured here, and it represents

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the distance travelled between 1 second and 7 seconds.

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What you do is evaluate the antiderivative we found at the top bound,

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

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Notice, by the way, it doesn't matter which antiderivative we chose here.

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If for some reason it had a constant added to it, like 5, that constant would cancel out.

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More generally, any time you want to integrate some function, and remember,

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you think of that as adding up values f of x times dx for inputs in a certain range,

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and then asking what is that sum approach as dx approaches 0.

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The first step to evaluating that integral is to find an antiderivative,

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some other function, capital F, whose derivative is the thing inside the integral.

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Then the integral equals this antiderivative evaluated

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at the top bound minus its value at the bottom bound.

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And this fact right here that you're staring at is the fundamental theorem of calculus.

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And I want you to appreciate something kind of crazy about this fact.

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The integral, the limiting value for the sum of all these thin rectangles,

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takes into account every single input on the continuum,

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from the lower bound to the upper bound.

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That's why we use the word integrate, it brings them all together.

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And yet, to actually compute it using an antiderivative,

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you only look at two inputs, the top bound and the bottom bound.

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It almost feels like cheating.

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Finding the antiderivative implicitly accounts for all the

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information needed to add up the values between those two bounds.

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That's just crazy to me.

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This idea is deep, and there's a lot packed into this whole concept,

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

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

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And what makes that hard is that velocity is always changing.

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If you approximate velocity to be constant on multiple different intervals,

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you could figure out how far the car goes on each interval with multiplication,

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and then add all of those up.

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Better and better approximations for the original problem correspond to

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collections of rectangles whose aggregate area is closer and closer to

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being the area under this curve between the start time and the end time.

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So that area under the curve is actually the precise distance

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traveled for the true nowhere constant velocity function.

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If you think of that area as a function itself,

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with a variable right endpoint, you can deduce that the derivative

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of that area function must equal the height of the graph at every point.

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And that's really the key right there.

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It means that to find a function giving this area,

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you ask, what function has v of t as a derivative?

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There are actually infinitely many antiderivatives of a given function,

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since you can always just add some constant without affecting the derivative,

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

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you choose at the bottom bound.

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By the way, one important thing to bring up before we leave is the idea of negative area.

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What if the velocity function was negative at some point, meaning the car goes backwards?

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It's still true that a tiny distance traveled ds on a little time interval is

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about equal to the velocity at that time multiplied by the tiny change in time.

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It's just that the number you'd plug in for velocity would be negative,

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so the tiny change in distance is negative.

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In terms of our thin rectangles, if a rectangle goes below the horizontal axis,

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like this, its area represents a bit of distance traveled backwards,

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so if what you want in the end is to find a distance between the car's

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start point and its end point, this is something you'll want to subtract.

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And that's generally true of integrals.

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Whenever a graph dips below the horizontal axis,

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the area between that portion of the graph and the horizontal axis is counted as negative.

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What you'll commonly hear is that integrals don't measure area per se,

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they measure the signed area between the graph and the horizontal axis.

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Next up, I'm going to bring up more context where this idea

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of an integral and area under curves comes up,

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along with some other intuitions for this fundamental theorem of calculus.

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Maybe you remember, chapter 2 of this series introducing the derivative

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was sponsored by The Art of Problem Solving, so I think there's

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something elegant to the fact that this video, which is kind of a duel to that one,

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was also supported in part by The Art of Problem Solving.

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I really can't imagine a better sponsor for this channel,

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because it's a company whose books and courses I recommend to people anyway.

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They were highly influential to me when I was a student developing a love for

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creative math, so if you're a parent looking to foster your own child's love

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for the subject, or if you're a student who wants to see what math has to offer

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beyond rote schoolwork, I cannot recommend The Art of Problem Solving enough.

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Whether that's their newest development to build the right intuitions in

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elementary school kids, called Beast Academy, or their courses in higher-level

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topics and contest preparation, going to aops.com slash 3blue1brown,

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or clicking on the link in the description, lets them know you came from this channel,

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which may encourage them to support future projects like this one.

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I consider these videos a success not when they teach people a particular bit of math,

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which can only ever be a drop in the ocean, but when they encourage people

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to go and explore that expanse for themselves,

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and The Art of Problem Solving is among the few great places to actually do

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that exploration.