We have many videos on the mean value theorem,
but I'm going to, I'm going to review it a
little bit so that we can see how this connects
the mean value theorem that we learned in Differential Calculus.
How that connects to what we learned about
the average value of a function using definite integrals.
So the mean value theorem tells us that if I have some function f, that is
continuous, continuous on the closed interval, so itâs
including the end points from a to b.
And it is differentiable, it is differentiable,
so the derivative is defined on the
open interval from a to b, so
not necessary, doesn't necessarily to be defined.
Differentiable at the boundaries as long
as it's differentiable between the boundaries.
Then we know.
Then we know that there exists, there
exists some value, or some number I guess, some number,
some number c such
that c is between the two end points of our interval.
So let me, such that a is less than c is less than b, so c is in this interval.
And, and, and this is kind of the, the meat of it.
The meat of it is that the derivative
of, the derivative of our function at that point,
the derivative of the function at that point you
could use as the slope of the tangent line.
At that point is equal to, essentially the
average, the average rate of change over the interval.
Or you can even think about it as the slope between the two endpoints.
So the, the slope between the two endpoints is
gonna be your change in y which is going to
be your change in your function value so f of b minus f of a over, over b minus a.
And once again we do this, we go into
much more depth in this when we covered this the
first time in Differential Calculus but just to give you
a visualization of it because I think it's always handy.
The mean value theorem that we learned
in Differential Calculus just tells us, hey look,
you know, if this is a, this is
b, I've got my function doing something interesting.
So this is f of a, this is f of b.
So this quantity right over here.
Where you're taking the change in the value of our function divided by.
So this right over here is f of b.
F of b minus f of a is this change in the value of our function,
divided by the change in our x axis, it's our change in y over change in x.
That gives us the slope, this right over here gives us the slope of this line.
This, this, the slope of the line that connects, that connects these two points.
That's this quantity, and this, the mean value theorem
tells us that there's some c in between a
and b where you're going to have the same slope, so it might be at least one place.
So it might be right over there, where you have the exact same slope.
There exists a c, where the slope of the tangent line at that point is going to be
the same, so this would be ac right over
there, and we actually might have a couple of c's.
That's another candidate c.
There's a least one c where the slope of the tangent
line is the same as the average slope across the interval.
And, once again, we have to assume that f is continuous and f is differentiable.
Now, when you see this, it, something might, it might
evoke some similarities with what we saw when we saw
the, the, or how we defined I guess you could
say, or the formula for the average value of a function.
Remember, what we saw for the average value of function, we said the average,
the average value of a function is going to be equal to 1 over b minus a.
Notice, 1 over b minus a, you have a b minus a in the
denominator here, times the definite integral from a to b of f of x dx.
Now this is interesting, cuz here we have a derivative, here we have an integral,
but maybe we could connect these, maybe we could connect these two things.
Well one thing that might jump out at you, is maybe we could rewrite, maybe
we could rewrite this numerator right over here in this form somehow.
And I encourage you to the pause the video and see
if you can, and I'll give you actually quite a huge hint.
Instead of it being an f of x here, what happens if there's an f prime of x there?
So I encourage you to, to re-try to do that.
So, once again, this is, let me re-write all of this.
This is going to be equal to, this over here is the exact same thing as the
definite integral from a to b of f-prime of x dx.
Think about it.
You're gonna take the anti-derivative of f-prime of x, which is gonna be f of x.
And you're going to evaluate it at b, f of b.
And then from that you're going to subtract it evaluated a minus f of a.
These two thing are identical.
And then you can, of course, divide by.
Divide by b minus a.
Now this is starting to get interesting.
One way to think about it is, there must be a, there must be a
c if we, there must be a c that takes on the average value of.
There must be a c that when you, when you evaluate the
derivative at c, it takes on the average value of the derivative.
Or another way to think about it, another way to think about it if we were to
just write, if we wrote it just right let me say g of x is equal to f prime of x.
Then we get very close to what we have over here.
Because this right over here is going to be g of c, remember f prime of c is the
same thing as g of c, is equal to 1 over is equal to 1 over b minus a.
1 over b minus a, so there exists a c, where g of c is equal to 1 over b minus
a, times the definite integral from a to b, of g of x, g of x dx.
F-prime of x is the same thing as g of x.
So, another way of thinking about it, and this is actually another form
of the mean value theorem, is called the mean value theorem for integrals.
Mean value theorem for integrals.
Let me, so this is the mean, I'll just write
the acronym, mean value theorem for, for integrals or integration.
Which essentially and, and to give you in a slightly
more formal sense, is if you have some function g.
So, if g is, let me actually go down a little bit.
Which tells us that g of x is continuous, continuous on this closed interval.
On, going from a to b, then there exists, then there exists,
a c in this interval where g of c is equal to, what is this?
This is the average value of our function.
There exists, exists a c where g of c is equal to the
average value of your function over the integral.
This was our definition of the average value of a function.
So anyway, this is just another way of
saying you might see some of the mean value
theorem of integrals, and that just to show you
that it's really closely tied is using different notation.
But it's usually, it's, it's essentially the same exact ideas the
mean value theorem you learned in
Differential Calculus, but now different notation.
And I guess you could have a slightly different interpretation.
We were thinking about it in, in Differential Calculus.
We're thinking about having a point where the slope of the tangent line
of the function of that point is the same as the average rate.
So that's when we had our kind of differential mode and we were
kind of thinking, thinking in terms of slopes and slopes of tangent lines.
And now when we're in integral mode we're thinking much
more in terms of average value, average value of the function.
So there's some c where g of c, there's, there's some c
where the function evaluated at that point is equal to the average value.
So another way of thinking about it,
if I were to draw, if I were to draw.
G of, if I were to draw a g of x, if I were to draw g of x.
So that's x, that is my y axis.
This is the graph of y is equal to g
of x, which of course was the same thing as f-prime
of x, but we core, I guess, we've just rewritten
it now to be more consistent with our average value formula.
And we're talking about the interval from a to b.
We've already seen how to calculate the average value.
We've already seen how to calculate the average value so may be the average value
is, is that right over there so that is g average.
So our average value is this.
The mean value theorem for integrals just tells there's
some c where our function must take on, where our
function must take on that value at c, whereas that
c is inside, where the c is in that interval.