Mean value theorem for integrals

- [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.