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## Finding the average value of a function on an interval

Current time:0:00Total duration:9:07

# Mean value theorem for integrals

AP Calc: CHA‑4 (EU), CHA‑4.B (LO), CHA‑4.B.1 (EK)

## Video transcript

- [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.

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