- [Voiceover] We have many videos on the mean value theorem,
but 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 on the closed interval,
so it's including the endpoints, from a to b,
and it is differentiable, so the derivative
is defined on the open interval, from a to b,
so it doesn't necessarily have to be differentiable
at the boundaries, as long as it's differentiable
between the boundaries, then we know that
there exists some value, or some number c,
such that c is between the two endpoints of our interval,
so such that a < c < b, so c is in this interval, AND,
and this is kind of the meat of it,
is that the derivative of our function at that point,
you could use as the slope of the tangent line
at that point, is equal to essentially the average
rate of change over the interval, or you could even
think about it as the slope between the two endpoints.
So 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 b minus a, and once again,
we go into much more depth in this
when we covered it the first time in differential calculus,
but just to give you a visualization of it,
'cause I think it's always handy, the
mean value theorem that we learned in differential calculus
just tells us, hey look, 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, so this right over here is
f of b, minus f of a, is this change in the
value of our function, divided by the change
in our x-axis, so it's a change in y over change in x,
that gives us the slope, this right over here gives us
the slope of this line, the slope of the line
that connects these two points, that's this quantity,
and the mean value theorem tell us that there's some c
in between a and b where you're gonna 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
a c right over there, and we actually might have
a couple of c's, that's another candidate c.
There's at 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 might evoke some similarities
with what we saw when we saw 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 a function,
we said 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, 'cause 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
this numerator right over here in this form somehow.
And I encourage you to 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 try to do that.
So once again, let me rewrite 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 going to 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 at a, minus f of a.
These two things are identical.
And then, you can of course divide by b minus a.
Now this is starting to get interesting.
One way to think about it is, there must be a c
that takes on the average value of,
there must be a c, that when you
evaluate the derivative at c,
it takes on the average value of the derivative.
Or another way to think about it,
if we were to just write 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 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 dx,
f prime of x is the same thing as g of x.
So another way of thinking about it, this is actually
another form of the mean value theorem,
it's called the mean value theorem for integrals.
I'll just write the acronym,
mean value theorem for integrals, or integration,
which essentially, to give it 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
if g of x is continuous on this closed interval,
going from a to b, 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 a c where g of c is equal to
the average value of your function over the interval.
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 just to show you that it's really closely tied,
it's using different notation, but it's usually,
it's essentially the same exact idea as the
mean value theorem you learned in differential calculus,
but now your different notation, and I guess
you could have a slightly different interpretation.
We were thinking about it 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 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 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 g of x,
that's x, that is my y-axis, this is the graph of
y is equal to g of x, which of course is the same thing
as f prime of x, but 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,
so maybe the average value is that right over there,
so that is g average, so our average value is this,
the mean value theorem for integrals just tells us
there's some c where our function must take on
that value at c, whereas that c is inside,
where the c is in that interval.