Here, I want to discuss one common type of problem where integration comes up: Finding
the average of a continuous variable.
This is a useful thing to know in its own right, but what’s really neat is that it
gives a completely different perspective for why integrals and derivatives are inverses
of one and other.
Take a look at the graph of sin(x) between 0 and pi, which is half its period.
What is the average height of this graph on that interval?
It’s not a useless question.
All sorts of cyclic phenomena in the world are modeled with sine waves: For example,
the number of hours the sun is up per day as a function of which-day-of-the-year-it-is
follows a sine wave pattern.
So if you wanted to predict, say, the average effectiveness of solar panels in summer months
vs. winter months, you’d want to be able to answer a question like this: What’s the
average value of that sine function over half its period.
Whereas a case like that will have all sorts of constants mucking up the function, we’ll
just focus on a pure unencumbered sin(x) function, but the substance of the approach would be
the same in any application.
It’s kind of a weird thing to think about, isn’t it, the average of a continuous variable.
Usually, with averages, we think of a finite number of values, where you add all them up
and divide that sum by how many values there are.
But there are infinitely many values of sin(x) between 0 and pi, and its not like we can
add all those numbers and divide by infinity.
This sensation actually comes up a lot in math, and is worth remembering, where you
have this vague sense that you want to add together infinitely many values associated
with a continuum like this, even though that doesn’t really make sense.
Almost always, when you get this sense, the key will be to use an integral somehow.
And to think through exactly how, a good first step is usually to approximate your situation
with some kind of finite sum.
In this case, imagine sampling a finite number of points, evenly spaced in this range.
Since it’s a finite sample, you can find the average by adding up all the heights,
sin(x), at each one, and divide that sum by the number of points you sampled, right?
And presumably, if the idea of an average height among all infinitely many points is
going to make any sense at all, the more points we sample, which would involve adding up more
heights, the closer the average of that sample should be to the actual average of the continuous
variable, don’t you think?
This should feel at least somewhat related to taking an integral of sin(x) between 0
and pi, even if it might not be clear at first exactly how the two ideas will match up.
For that integral, you also think of a sample of inputs on this continuum, but instead of
adding the height sin(x) at each one, and dividing by how many there are, you add up
sin(x)*dx where dx is the spacing between the samples; that is, you’re adding little
areas, not heights.
Technically, the integral is not quite this sum, it’s whatever that sum approaches as
dx approaches 0.
But it’s helpful to reason with respect to one of these finite iterations, where you’re
adding the areas of some specific number of rectangles.
So what you want to do is reframe this expression for the average, this sum of the heights divided
by the number of sampled points, in terms of dx, the spacing between samples.
If I tell you that the spacing between these points is 0.1, for example, and you know that
they range from 0 to pi, can you tell me how many there are?
Well, you can take that length of the interval, pi, and divide it by the length of the space
between each sample.
If it doesn’t go in evenly, you’d round down to the nearest integer, but as an approximation
this is fine.
So if we write the spacing between samples as dx, the number of samples is pi/dx.
So replacing the denominator with pi/dx here, you can rearrange, putting the dx up top and
distributing.
But, think about what it means to distribute that dx up top; it means the terms you’re
adding all look like sin(x)*dx for the various inputs x that you’re sampling, so that numerator
looks exactly like an integral expression.
And for larger and larger samples of points, the average approaches the actual integral
of sin(x) between 0 and pi, all divided by the length of that range, pi.
In other words, the average height of this graph is this area divided by its width.
On an intuitive level, and just thinking in terms of units, that feels pretty reasonable,
doesn’t it?
Area divided by with gives you average height.
So let’s actually compute this expression.
As we saw, last video, to compute an integral you need to find an antiderivative of the
function inside the integral; some function whose derivative is sin(x)
And, if you’re comfortable with trig derivatives, you know the derivative of cos(x) is -sin(x),
so if you negate that, -cos(x) is the antiderivative of sin(x).
To gut check yourself on that, look at this graph of -cos(x).
At 0, the slope is 0, then it increases to some maximum slope at pi/2, then it goes back
down to 0 at pi, and in general its slope does indeed seem to match the height of the
sine graph.
To evaluate the integral of sin(x) between 0 and pi, take that antiderivative at the
upper bound, and subtract its value at the lower bound.
More visually, that’s the difference in the height of this -cos(x) graph above pi,
and its height at 0, and as you can see, that change in height is exactly 2.
That’s kind of interesting, isn’t it?
That the area under this sine graph turns out to be exactly 2.
So the answer to our average height problem, this integral divided by the width of the
region, evidently turns out to be 2/pi, which is around 0.64.
I promised at the start that this question of finding the average value of a function
offers an alternate perspective on on why integrals and derivatives are inverses of
one and other; why the area under one graph is related to the slope of another.
Notice how finding this average value 2/pi came down to looking at the change in the
antiderivative -cos(x) over the input range, divided by the length of that input range.
Another way to think about that fraction is as the rise-over-run slope between the point
of the antiderivative graph below zero, and the point of that graph above pi.
Now think about why it might make sense that this slope represents the average value of
sin(x) on that region.
By definition, sin(x) is the derivative of this antiderivative graph; it gives the slope
of -cos(x) at every input.
So another way to think about the average value sin(x) is as the average slope over
all tangent lines here between 0 and pi.
And from that view, doesn’t it make a lot of sense that the average slope of a graph
over all its point in a certain range should equal the total slope between the start and
end point?
To digest this idea, it helps to see what it looks like for a general function.
For any function f(x), if you want to find its average value on some interval, say between
a and b, what you do is take the integral of f on that interval, divided by the width
of the interval.
You can think of this as taking the area under the graph divided by the width.
Or more accurately, it’s the signed area of that graph, since area below the x-axis
is counted as negative.
And take a moment to remember the connection between this idea of a continuous average
and the usual finite notion of an average, where you add up many numbers and divide by
how many there are.
When you take some sample of points spaced out by dx, the number of samples is about
the length of the interval divided by dx.
So if you add up the value of f(x) at each sample and divide by the total number of samples,
it’s the same as adding up the products f(x)*dx and dividing by the width of the entire
interval.
The only difference between that and the integral expression is that the integral asks what
happens as dx approaches 0, but that just corresponds with samples of more and more
points that approximate the true average increasingly well.
Like any integral, evaluating this comes down to finding an antiderivative of f(x), commonly
denoted capital F(x).
In particular, what we want is the change to this antiderivative between a and b, F(b)
- F(a), which you can think of as the change in the height of this new graph between the
two bounds.
I’ve conveniently chosen an antiderivative which passes through 0 at the lower bound
here, but keep in mind that you could freely shift this up and down, adding whatever constant
you want to it, and it would still be a valid antiderivative.
So the solution to the average problem is the change in the height of this new graph
divided by the change to its x value between a and b.
In other words, it’s the slope of the antiderivative graph between these endpoints.
And again, that should make a lot of sense, because little f(x) gives the slope of a tangent
line to this graph at each point, after all it is by definition the derivative of capital
F. So, why are antiderivative the key to solving
integrals?
Well, my favorite intuition is still the one I showed last video, but a second perspective
is that when you reframe the question of finding the average of a continuous value as finding
the average slope of bunch of tangent lines, it lets you see the answer just by the comparing
endpoints, rather than having to actually tally up all points in between.
In the last video, I described a sensation that should bring integrals to your mind.
Namely, if you feel like the problem you’re solving could be approximated by breaking
it up somehow, and adding up a large number of small things.
Here I want you to come away recognizing a second sensation that should bring integrals
to your mind.
If there’s some idea that you understand in a finite context, and which involves adding
up multiple values, like taking the average of a bunch of numbers, and if you want to
generalize that idea to apply to an infinite continuous range of values, try seeing if
you can phrase things in terms of an integral.
It’s a feeling that comes up enough that’s it’s definitely worth remembering.
My thanks, as always, to those making these videos possible.