It’s Professor Dave, let’s find some averages.
Much, much earlier in this series, we learned about different types of averages, and understood
that when we use the word average, we typically are referring to a mean value.
To get the mean value for a finite set of numbers, we just add them up, and then divide
by the number of items in the set.
But we can also talk about averages in the context of functions.
This is a little more abstract, because there are infinitely many points on any curve, and
it may seem impossible to take the mean of infinitely many values.
But that’s exactly the kind of thing we’ve been showing is possible throughout this entire
calculus series, and in fact we can do this too, using something called the mean value
theorem for integrals.
This is related to a simpler theorem, simply called the mean value theorem, and since we
haven’t discussed this yet, let’s quickly mention what that theorem says first.
The mean value theorem applies to any function that is continuous and differentiable over
some interval, A to B. Let’s write out the coordinates for these endpoints, which will
be (A, f of A), and (B, f of B).
What the theorem says is that for any such function, there must be some number C, within
this interval, whereby f prime of C, or the derivative of this function evaluated for
the input C, will give a value that is equal to the slope of the line connecting these
other two points, which we call the secant line, with a slope given by rise over run,
or f of B minus f of A, over B minus A. In other words, there must be at least one point
on the curve that has a tangent line that is parallel to the secant line.
Here we see one such point, because the curve has only one concavity.
Now look at this example, where we go from concave down to concave up before reaching
the end point.
There are now two such points where the tangent line is parallel to the secant line.
This truth has various applications.
If the function was a position function, then the derivative would be its velocity, and
the theorem would imply that there must be at least one instant over this interval where
the instantaneous velocity, given by the derivative at that point, or the slope of the tangent
line at that point, must be equal to the average velocity, given by the slope of the secant line.
Now that we’ve covered differentiation, let’s get back to integration.
The mean value theorem for integration allows us to compute the average value of a function
over some interval from A to B in the following manner.
We take one over the quantity B minus A, and multiply by the integral of the function from
A to B.
This shouldn’t be too surprising, because it means that there is some value, C, for
which the antiderivative evaluated at B minus the antiderivative evaluated at A divided
by the quantity B minus A, gives us the average value of the function.
This looks just like the mean value theorem for differentiation, it is simply that in
the previous context, the function giving us the average was the derivative of the original
function, whereas here the function in the expression is the antiderivative of the original.
So these are saying the same thing, according to the fundamental theorem of calculus.
Since we were able to interpret the mean value theorem for differentiation in a geometric
way, comparing the slope of a tangent line to the slope of a secant line, let’s also
acknowledge here that this integral, if for a positive function, represents the area under
the curve that is below a particular line, y equals f of C, such that the rectangle being
formed has an area that is the same as the area under the entire curve.
To see how this works, let’s apply the theorem to the function one plus x squared, for the
interval from negative one to two.
Well let’s plug things into this expression, with two for B and negative one for A, giving
us one third out here, and then the integral of this function will be x plus (x cubed over three).
We evaluate this at two and then negative one, finding common denominators to combine
the terms, then subtracting, and then multiplying by one third, and we end up with two.
So with this result we can say two things.
First, that the average value of this function over this interval, is two.
Second, if we find the point that corresponds to this average value by finding the x value
for which the function equals two, we get plus or minus one, so apart from this left
boundary, there is the point one, two.
If we draw the line y equals two, we also therefore know that the rectangle shown here
has the same area as the function over this interval, which tells us all kinds of interesting
things, like the fact that this portion of the function above the line has the same area
as this missing section here.
So between these two versions of the mean value theorem, we can find the average value
of a function over some interval in two different ways, one through differentiation, and one
through integration, which means that when looking at a function, we can choose the way
that will be the easiest to evaluate, depending on the complexity of the integrand.
This has all kinds of applications, like finding the average temperature over some time span,
and many other things, so for some practice putting this to use, let’s check comprehension.