What does area have to do with slope? | Chapter 9, Essence of calculus

3Blue1Brown · May 12, 2026

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Here, I want to discuss one common type of problem where integration comes up,

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finding the average of a continuous variable.

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This is a perfectly useful thing to know in its own right,

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but what's really neat is that it can give us a completely different

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perspective for why integrals and derivatives are inverses of each other.

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To start, take a look at the graph of sinx between 0 and pi, which is half of its period.

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What is the average height of this graph on that interval?

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It's not a useless question.

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All sorts of cyclic phenomena in the world are modeled using sine waves.

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For example, the number of hours the sun is up per day as a

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function of what day of the year it is follows a sine wave pattern.

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So if you wanted to predict the average effectiveness of solar panels in summer months vs.

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winter months, you'd want to be able to answer a question like this,

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what is the average value of that sine function over half of its period?

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Where as a case like this is going to have all sorts of constants mucking up the

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function, you and I are going to focus on a pure, unencumbered sinx function,

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but the substance of the approach would be totally the same in any other application.

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It's kind of a weird question to think about though, isn't it?

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The average of a continuous variable.

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Usually with averages we think of a finite number of variables,

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where you can add them all up and divide that sum by how many there are.

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But there are infinitely many values of sinx between 0 and pi,

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and it's not like we can just add up all those numbers and divide by infinity.

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This sensation comes up a lot in math, and it's worth remembering,

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where you have this vague sense that you want to add together infinitely

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many values associated with a continuum, even though that doesn't make sense.

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And almost always, when you get that sense, the key is to use an integral somehow.

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And to think through exactly how, a good first step is to

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just approximate your situation with some kind of finite sum.

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In this case, imagine sampling a finite number of points evenly spaced along this range.

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Since it's a finite sample, you can find the average by just adding up all the heights

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sinx at each one of these, and then dividing that sum by the number of points you sampled.

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And presumably, if the idea of an average height among all infinitely many

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points is going to make any sense at all, the more points we sample,

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which would involve adding up more and more heights,

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the closer the average of that sample should be to the actual average of

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the continuous variable.

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And this should feel at least somewhat related to taking an integral of sinx

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between 0 and pi, even if it might not be exactly clear how the two ideas match up.

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For that integral, remember, you also think of a sample of inputs on this continuum,

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but instead of adding the height sinx at each one and dividing by how many there are,

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you add up sinx times dx, where dx is the spacing between the samples.

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That is, you're adding up little areas, not heights.

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And technically, the integral is not quite this sum,

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it's whatever that sum approaches as dx approaches 0.

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But it is actually quite helpful to reason with respect to one of these finite

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iterations, where we're looking at a concrete size for dx and some specific number of

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rectangles.

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So what you want to do here is reframe this expression for the average,

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this sum of the heights divided by the number of sampled points,

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in terms of dx, the spacing between samples.

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And now, if I tell you that the spacing between these points is, say, 0.1,

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and you know that they range from 0 to pi, can you tell me how many there are?

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Well, you can take the length of that interval, pi,

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and divide it by the length of the space between each sample.

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If it doesn't go in perfectly evenly, you'd have to round down to the nearest integer,

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but as an approximation, this is completely fine.

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So if we write that spacing between samples as dx,

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the number of samples is pi divided by dx.

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And when we substitute that into our expression up here,

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you can rearrange it, putting that dx up top and distributing it into the sum.

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But think about what it means to distribute that dx up top.

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It means that the terms you're adding up will look like

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sinx times dx for the various inputs x that you're sampling.

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So that numerator looks exactly like an integral expression.

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And so for larger and larger samples of points,

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this average will approach the actual integral of sinx between 0 and pi,

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all divided by the length of that interval, pi.

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In other words, the average height of this graph is this area divided by its width.

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On an intuitive level, and just thinking in terms of units,

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that feels pretty reasonable, doesn't it?

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Area divided by width gives you an average height.

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So with this expression in hand, let's actually solve it.

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As we saw last video, to compute an integral, you need to find an antiderivative

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of the function inside the integral, some other function whose derivative is sinx.

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And if you're comfortable with derivatives of trig functions,

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you know that the derivative of cosine is negative sine.

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So if you just negate that, negative cosine is the function we want,

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the antiderivative of sine.

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And to gut-check yourself on that, look at this graph of negative cosine.

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At 0, the slope is 0, and then it increases up to some maximum slope at pi halves,

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and then goes back down to 0 at pi.

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And in general, its slope does indeed seem to

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match the height of the sine graph at every point.

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So what do we have to do to evaluate the integral of sine between 0 and pi?

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We evaluate this antiderivative at the upper bound,

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and subtract off its value at the lower bound.

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More visually, that's the difference in the height of

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this negative cosine graph above pi and its height at 0.

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And as you can see, that change in height is exactly 2.

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That's kind of interesting, isn't it?

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That the area under this sine graph turns out to be exactly 2?

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So the answer to our average height problem, this integral divided by the width

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of the region, evidently turns out to be 2 divided by pi, which is around 0.64.

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I promised at the start that this question of finding the average of a function offers

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an alternate perspective on why integrals and derivatives are inverses of each other,

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why the area under one graph has anything to do with the slope of another graph.

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Notice how finding this average value, 2 divided by pi,

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came down to looking at the change in the antiderivative,

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negative cosine x, over the input range, divided by the length of that range.

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And another way to think about that fraction is as the rise over run slope between

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the point of the antiderivative graph below 0 and the point of that graph above pi.

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Think about why it might make sense that this slope would

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represent an average value of sine of x on that region.

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By definition, sine of x is the derivative of this antiderivative graph,

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giving us the slope of negative cosine at every point.

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Another way to think about the average value of sine of x is

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as the average slope over all tangent lines between 0 and pi.

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And when you view things like that, doesn't it make a lot of sense

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that the average slope of a graph over all its points in a certain

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range should equal the total slope between the start and end points?

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To digest this idea, it helps to think about what it looks like for a general function.

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For any function f of x, if you want to find its average value on some interval,

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say between a and b, what you do is take the integral of f on that

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interval divided by the width of that interval, b minus a.

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You can think of this as the area under the graph divided by its width,

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or more accurately, it is the signed area of that graph,

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since any area below the x-axis is counted as negative.

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And it's worth taking a moment to remember what this area has to do with the usual notion

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of a finite average, where you add up many numbers and divide by how many there are.

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When you take some sample of points spaced out by dx,

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the number of samples is about equal to the length of the interval divided by dx.

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So if you add up the values of f of x at each sample and divide by

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the total number of samples, it's the same as adding up the product

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f of x times dx and dividing by the width of the entire interval.

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The only difference between that and the integral is that the integral asks

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what happens as dx approaches 0, but that just corresponds with samples of

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more and more points that approximate the true average increasingly well.

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Now for any integral, evaluating it comes down to finding an antiderivative of f of x,

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commonly denoted capital F of x.

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What we want is the change to this antiderivative between a and b,

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capital F of b minus capital F of a, which you can think of as

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the change in height of this new graph between the two bounds.

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I've conveniently chosen an antiderivative that passes through 0 at the lower bound here,

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but keep in mind you can freely shift this up and down adding whatever

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constant you want and it would still be a valid antiderivative.

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So the solution to the average problem is the change in the height of

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this new graph divided by the change to the x value between a and b.

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In other words, it is the slope of the antiderivative graph between the two endpoints.

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And again, when you stop to think about it, that should make a lot of sense,

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because little gives us the slope of the tangent line to this graph at each point.

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After all, it is by definition the derivative of capital F.

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So why are antiderivatives the key to solving integrals?

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My favorite intuition is still the one I showed last video,

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but a second perspective is that when you reframe the question of finding an average of

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a continuous value as instead finding the average slope of a bunch of tangent lines,

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it lets you see the answer just by comparing endpoints,

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rather than having to actually tally up all the points in between.

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In the last video I described a sensation that should bring integrals to your mind,

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namely if you feel like the problem you're solving could be approximated by

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breaking it up somehow and adding up a large number of small things.

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Here I want you to come away recognizing a second

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sensation that should also bring integrals to your mind.

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If ever there's some idea that you understand in a finite context,

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and which involves adding up multiple values, like taking the average of a

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bunch of numbers, and if you want to generalize that idea to apply to an infinite

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continuous range of values, try seeing if you can phrase things in terms of an integral.

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It's a feeling that comes up all the time, especially in probability,

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and it's definitely worth remembering.

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My thanks, as always, go to those making these videos possible.

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Thank you.

— end of transcript —

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