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.