The Average Value of A Function (On a Closed Interval)

Video Series: Applications of Integration (Complete In-Depth Tutorials For Calculus)

(D4M) — Here is the video transcript for this video.

00:00
in this episode you’ll learn how to find the average value of a continuous
function on a closed interval let’s do some math [Music]
hi everyone welcome back we’re going to start off with the question what is the
average value that a continuous function f takes on over an interval a b
so to do this um we’re going to start looking at what a continuous
function on a closed interval so we have something like this right here we have
some kind of continuous function and we have some kind of closed interval
right here and so i’ll just draw a generic function
and a generic closed interval it could be like 1 4 2 3 you know just any closed
interval any continuous function right here and we want to talk about the
average value now you have some kind of intuition

00:01
about what this word right here means especially if you’re in a math course or
or in some kind of class right because you may be taking some um exams or
something like that and like if you take
like three exams for example and they’re all equally weighted
then you can try to find your exam average so let’s say you take three
exams and those are your grades and then you divide by three and then you’ll see
that your average exam grade would be a 75 at that point right there
so this is you know some intuition about what the average value means
but this isn’t the case where you have a finite number of values like you have
three exams or you have even if it’s a hundred or a thousand but
a finite number of them you just add up all of them and then you divide by how
many of them you have so it’s just an elementary idea of of average but what
if you have a continuous function and now when you you look at a b here
and you look at your continuous function

00:02
right here and we’re going to be talking about infinitely many values right in
here and for each one we can go up and find a height so this would be if you
just pick an arbitrary x this would be the f of x right here that would be the
height right there and so if you look at all those heights
you how would you find the average if you have infinitely many points right if
you just have a finite number of points we can just add up all of the um f of
x’s all the outputs and then divide by however many uh of them we have
right and so that’s the idea here so if we take a continuous function
uh we’ll sample a finite number of values so let’s say i have x1 x2 x3
and let’s just say i have say in fact let’s just call them cs to
represent that we’re sampling so let’s just call this one here ck
and then this height right up here would be f of ck
and we could just add them all up and so k represents the

00:03
um the index so c1 c2 c3 and so on and so we have a finite number of them
and then in that case we can sampling of them is is the same as
dividing up the interval a b into um in sub intervals right so k is going
from you know one all the way to n and we’re going to get all these intervals
here all these little tiny tiny intervals and we can add up all the um
function heights right here um and so you know that’s exactly what
we’re going to do here is we’re going to be adding up all of these function
height the heights the outputs and then dividing by the n and this would give us
a average value here now you know each of these
sub intervals right here so we have a ck chosen right here
and we go up and we compute the height here uh f of c k
and so you know we have these um sub intervals right here and the width

00:04
of each sub interval right here the delta x here
right so so the whole a b is like way out here like here’s an a and here’s a b
right and so we’re going to have all these little sub intervals in here
and you know each each of them has width because we look at the whole thing b
minus a that’s the whole distance of the interval right here and divide it up by
an equal number of them so um so this is all the delta x’s have
the same width right here so you know we can divide by this in here
and when we put it in this form right here what we can do is we can use uh
summation notation and just say from 1 to n
and then the cks right so k is the the index right here
and then we just have one over n and now what i’m going to do is i’m
going to substitute over here and get this delta x and bring the delta x into
the problem so 1 over n right so you know
just multiply both sides by b b minus a and you get one over n

00:05
and so that’ll be delta x um and then b b minus a down here and so
now i’m going to bring the delta x in here and the reason why i do that is
bringing the delta x in here is so that now we can
uh realize that this hey this right here
looks um is is looks just like a riemann sum right here and so this would be one
over b minus a times this riemann sum right here
and so if we let n go to infinity and we’re going to be taking uh
infinitely many uh sampling points here infinitely many infinitely many of them
and past the limit if possible and we’re
kind of addressing this by assuming that the function is continuous
and you know we’re going to have the following definition right so we’re
going to integrate or we’re going to pass the limit um and then we’ll have
the average value of a function as the limit right here in the limit
right here and when we have the limit right here since this is a constant we

00:06
can bring the limit inside right here and then we have the limit of the
riemann sum and the limit of the riemann sum right there is just the um you know
the integral right there the definite integral right there and the left-hand
side that limit um we’re calling that the
average value so this is the definition of the average value
is this right here this this integral right here so
i guess i can move out of the way real quick it’s just denoted by so i’m going
to use an av of f and it’s just going to be called the
average value of a continuous function if it’s continuous then we’re going to
be good to go and so let’s look at an example now so
find the average value of a function right here
now when i do this right here when i try to find the average value of this
function right here now first i’m going to actually

00:07
use different words right here i’m going to say find and approximate
approximation find an approximation to the to the average value so i’m going to
replace the word v with find an approximation to the average value
so an approximation means you know it’s not unique you can have any kind of
approximation you want but you know how do you actually approximate the average
value first i’m going to try to address address that before we find the average
value the meaning the unique one which is of course the exact one so um yeah
let’s look at a python notebook real quick and let’s go to that so
let’s zoom in here a little bit now if you’ve never used python before
um i recommend checking out the link below in the description
where you can see how to use the python notebook and
so we’re going to use this right here python notebook and the link below is is

00:08
free it’s totally free and you don’t have to
use python you can just follow along if you want but if you are um
using the notebook and following along then i’m going to import these packages
right here numpy and matplot and scipy and actually i’m not going to actually
even use this last one here you don’t need to import this last one here if you
don’t want to now i am going to make some plots so if
you watched any of these episodes before you know that i like to customize my
plots right there by the way let me just take a moment to
say that this episode is part of the series applications of integration
complete in-depth tutorials for calculus so the link below to the full series is
below in the description all right and so yeah i’m going to
customize my axes and then i’m actually going to click up this plot function
right here real quick so let me execute that
and all right so we got that set up in there um i’m gonna um make up this plot

00:09
function here we’re gonna input a function a minimum value and a maximum value
and we’re gonna define some axes customize axes and i’m going to plot
like a thousand points for x and then input them into the
function and then show that plot right there all right so there’s those that’s
a setup right there let’s minimize that now let’s look at our first example here
so this was the example that we’re looking at right here where it says find
the average value of the function on the integral on the interval here
now let’s suppose we don’t know how to find the average value in other words we
don’t know the calculus part let’s just see how we do this by hand if we were
insistent upon trying to find the average value
so first thing i’m going to do is type in this function right here now this is
the function that’s given to us as f of x and it’s example one so i’m just going
to call the function f1 i’m defining a new function of x
and the function is just 3 and then times x squared and then minus 3. so
let’s just execute that function and now

00:10
we have f1 and it’s very easy to plug in numbers into f1 we can plug in one half
we can plug in one-third or 0.3 and then we can do
outputs you know actually get outputs now let’s go ahead and try to look at an
average value right here so i’m going to start at zero i’m going to plug in zero
into the function right here and i’m going to go all the way to 1 i’m going
to plug in 1 into the function right here
but i’m going to choose some numbers in between so i’m going to do some sampling
so 0.25 0.3 0.67 0.83 and 1. and i just chose those randomly and i’m going to
plug each of those into the function f1 and i’m going to add them all up and
then i’m going to divide by how many of them i have and so this would give us an
approximation to the average value right
here that’s not the average value that’s just an approximation
now to be more um you know to do a lot more points besides six so now i’m going
to cook up a function right here so i’m going to call this function here

00:11
approximate average value i’m going to input a function
for example here’s a function right here i’m going to input an a to b
0 1 and i’m going to input a and n and so we’ll be able to you know take as
an input however many we’re going to do and so we’re just going to say x is the
chosen and we’re going to chose them equally spaced and we’re going to choose
n of them and then we’re going to return the sum of all the outputs
and so we’re going to put all these x’s into them if n is 6 then we’ll put all
of them in we’ll add them all up and then we’ll divide by that in so let’s
execute that cell right there so this will be an approximate average value
right here so i’m going to run this on this function right here so here’s f1
that’s the function we defined here here’s 0 1 here’s 6.
and let’s run this approximate average value function right here
now you notice i got a little bit of a different answer and the reason why is

00:12
because when we execute this code right here
we’re choosing equally spaced in our sampling here i just chose random ones
and so we’re not going to get necessarily the same value right there
if you just choose random ones compared to equally spaced so
let’s just take a look at the function what it is f1 on this interval here zero
to one here it is and it looks like i’m getting average
value of something like negative 1.8 negative 1.9
which is about like right here so if we look at all the output heights
and we average them all up and we’re getting something pretty close to
negative 2 right there just for the approximations
all right so now let’s go back to the math and
let’s see here let’s see actually how to do this
using calculus now let’s find the average value the exact value
so what we’re going to do is we’re going to say a of f the average value

00:13
is equal to and so we’re going to have 1 over b
minus a so it’s going to be 1 minus and then we’re going to integrate 0 to 1
and then we’re going to integrate the function here 3x squared minus 3
and then dx and then so yeah we’re just going to
evaluate this integral right here so this integral right here
is just that’s all just 1 and so we’re integrating 0 to 1 of 3x
squared minus 3. in fact i’m just going to pull the 3 out here and just say
integral of x squared minus 1 dx and so this will be x to the third over three
minus x and then integrate this zero to one
and then now let’s come in here with the one so we’re gonna get three times
uh one third minus one and then when i substitute in zero it all vanishes
and so it looks like we’re getting here this is think of this as three over

00:14
three so this is minus minus two thirds and so the 3s cancel we end up with
minus 2. so the average value of this function right here over this interval was
exactly minus 2. so here’s the calculus to find the average value and this is
not sampling this is not doing an approximation this is doing infinitely
many of them and so yeah calculus is really powerful
there really quick and easy to find the average value
all right so let’s look at another example example two here and let’s go here
and look at this function right here now f of x is three x squared minus two x
and we’re gonna be looking over the interval one four
um and so yeah we can find the average value real quick
but i really want to drive home that um that um of how you approximate it
and so i want to find the average value of this function over this interval but

00:15
i actually want to approximate it i want to just get an approximate value
and you know the reason why this is valuable is because even though you may
technically be able to say that an integral exists it may be a different
thing entirely to actually evaluate it so this is still has um lots of value in
here um but i’m going to define the function
so 3x squared and then i’m going to say minus 2x
and then i’m going to just you know do some inputs and outputs i’m going to add
up all the all these right here and so i’m going from one to four so i’m just
choosing one 1.25 and then a 2.3 3.67 so i’m just sampling between one and four
and i’m just adding them all up so let’s execute these cells here
let’s uh shift enter in there and then let’s go down here and hit shift enter
very good so we get an approximation to the average value right here it’s about
20.645 and so yeah we can do this again here and

00:16
let’s do say 600 you know what how long would it take to do 600 of them
right so if you had to approximate and you know it’s very easy
to do lots of them really fast uh okay yeah so here we did did it in
again right here all right so let’s take a look at the
plot of this function right here f2 on interval one to four and so we can see
the uh approximate the average value is close to 16. so 16 is about right here
so that’s about the average value of the outputs all right so
um yeah so then now let’s go find the actual um exact value right there
so find the average value of this function over the interval right here
so the average value will be and we’re going to go
it’s going to be 1 over b minus a so 4 minus 3 sorry 4 minus 1

00:17
and then integral from 1 to 4 and then we’re going to integrate
integrate this function right here and yeah so now we can get 1 3
and we can integrate this right here this will be
3 times x to the third over three and then minus two and then this will be
x squared over two and i’m going to go one to four
and so yeah without boring you you get 48 over three or 16
and that is the exact value right there so very good so
there’s the exact average value right there all right so let’s just ask this
question now if this is the average value according to our tuition which we
believe it is if the average value really is a typical value of the integral
of the integrable function then the average value should have the

00:18
same integral over a b that f does and so let’s actually show that that’s
actually true and so the way i’m going to do that is
i’m going to say the average value of f which is 1 over b minus a integral from
a to b of f of x and i’m going to call that a constant
just to make this a little bit easier to write out now
why is this a constant well of course it’s a constant the average value is a
number right so you integrate with respect to x you do
um you know the anti-derivative at b minus you know a
at a and then you know you multiply by this number right here so this is a
number right here so i’m just going to call this k
so that’s a constant right there so let’s see if we can start with the left
and get to the right and let’s see how to do that here so let’s go a to b
and average value of f and now like i said this is just a k

00:19
it’s just a constant after you go find it
it may be hard to find or maybe easy to find whatever it is to find you found it
it’s a number let’s move that number outside of the integral
and then this is just dx here now the integral from a to b of dx is
just well it’s just x a to b or in other words b minus a here
all right so we can find this integral right here it’s just this it’s just k
times it looks like i used a capital k so i should be consistent there
um all right and so this right here uh k here
um let’s substitute back in now what the k is remember k is all of this
so let’s write that for k so instead of using k here i’m going to
write all this out again so 1 over b minus a integral a to b f of x dx
so that’s just the k right there and then all that times b minus a so

00:20
i’ll just put big brackets there just to make it look clear
and so the b minus a’s here actually cancel
and so we just get the integral from a to b of f of x so if you
integrate the average value it’s just the same thing as integrating the
function right there f of x all right so that’s just kind of a
maybe you want to call it a sanity check or something i don’t know
all right so how can we use this how can we use this average value here so let’s
use it to do something interesting here so
let’s say we’re taking the temperature um in a room or somewhere and we want to
take the temperature so let’s look at temperature and we want to do this for
24 hours so we’re going to take the temperature every hour on the hour

00:21
and so what we can do is we can say that we have f of x one star
plus f of x two star and we’re gonna do this here
um n times and n is twenty four so we’re gonna so basically every hour on the
hours think of the x1 star as let’s say you started at midnight and then that
would be one o’clock you’d and then the function f is take the temperature right
so that means at one o’clock you take the temperature so that right there is
representing the temperature at one o’clock and then here’s the temperature at
two o’clock and then so on for 24 hours and then
we’re going to divide by this in here and so we can get the average
temperature right here and so what we can do is we can write
this as 1 1 over b minus a and then integral of or not integral but sum

00:22
of the f of x let’s use a k k [Music] k equals 1 to n and get the delta x here
so just as we did before this is going to be an approximation to the average
value this is a the average value and if you’re if you’re discrete a
finite number of choices write every hour on the hour then we can write it
like this right here and so yeah if we take the limit as n
goes to infinity then we’ll get the um exact value and we’ll get the
integral over here for example what if you wanted to take
the temperature not every hour but what if you wanted to take it every 30
minutes right so then then this would be
many more and you would have a different larger end right and so if you want to
go every um every minute or actually every second
so how would you actually find the right the limiting value right there
right so you would have the average value of f
and so this would be the same as taking the temperature
at every instant in time whatever that would mean

00:23
and that would be this average value of the function right there
all right and so um um let’s say we’re going to try to do
something with it mathematically and we can
um you know instead of thinking about the temperature we can ask a question like
um so remember at the beginning when i talked about what if you have three
three exam grades and they’re all equally weighted and you want to try to
find your exam average and so in this case right here the exam average is what
225 over 3 or in other words 75 and so here’s an example where one of
the exam grades was actually equal to the exam average so
if you have a finite number of of choices though um that that may be

00:24
not happening right so what if you have say a 51 and a 76 or how about a 74
we would still have the 225 over 3 we would still have 75 but now it happened
that you didn’t actually make a 75 on the exam
so you know if you have a continuous function though if you’re doing
infinitely many of these and trying to find the average value of
a continuous function so then you can try to ask the question
is there always an input where the output is equal to the average value
um and so that’s the question right there is um and the answer to that is
called the mean value theorem for integrals
and it states that if f is continuous on a closed interval a b
so that’s exactly the the condition that we have to define the average value

00:25
so then there exists so this is an existence theorem then there exists
a number c in a b in this inside the interval here um so b is you know
greater than or equal to a less than or equal to b such that
f of c is equal to the average value so we’ll just say the average value
average value of f which we know is one minus one over b minus a of integral
from a to b of f of x so it’s saying that there is an input in
which the average value so the examples that we’re looking at the average
temperature or the average grade um the average value
there was some input that hit that average value
so that doesn’t necessarily happen to hap have to happen for the discontinuous

00:26
or the discrete case but for continuous functions there has to exist a c
that is equal to the average value now sometimes you might want to rewrite
it sometimes you might want to say integral a to b f of x dx
is equal to so i’m just going to multiply both sides by b minus a so just
move the b minus a up here and we’ll get f of c times b minus a
and so that’s a kind of a nice slick way there
saying that if f is continuous on a closed interval right here a b
then there exists a c such that the output times b minus a the length of
the interval here that’s actually really nice to know because
what that tells us is that something that may be difficult to compute
remember this is the limit of riemann sum
you know for a large number of functions that you can integrate and you may be
able to actually find an anti-derivative and actually come up with this number
right here pretty easily but for a large number of functions it may be difficult

00:27
but this right hand side looks like you just substitute in c into the function
right so this is actually very handy not
only just because it looks so pretty and nice but actually you can use this in a
lot of different theorems to prove more interesting results all right so
let’s look at an example of this right here how can we
use this an example right here so let’s look at a quick example before we go
here so um we’re going to say f of x is one plus x squared
and we’re looking on the interval minus one to two and so let’s just actually
find the average value of this rule really quick so what’s the average value
of this function right here on minus one to two
so it’s going to be one over and then it’s going to be two minus a negative
one and then we’re going to integrate from minus one to two of one plus x
squared dx and so yeah this is just what 1 3

00:28
and so we’re going to integrate here from minus 1 to 2 1 plus x squared right
so we’re going to be integrating the one we’re just going to get an x and we’re
going to integrate the x squared so x to the third over 3
and then we’re going to go from -1 to 2 and long story short when you integrate
this out plug in all the numbers simplify you get 2. so the average value of f
on this interval right here is just 2. now what is the mean value theorem for
integral say it says that the integral from -1 to two of f of x dx is equal to
f of c times two minus the negative one which we know is three so it says that
let’s just write a three right we already talked about that that’s three
so three times f of c so we can actually find the c so the mean

00:29
value theorem says there exists a c where this equation is true but actually
because we have a concrete function and we’re able to find the actual
average value so now we can actually go and find out what this is right here so
the average value right here is is one third integral of all this we found the
average value is just a two and what happens when you plug in the f of c
so the f of c is one plus c squared and so yeah just looking at this you can
see what the c is so this implies that c is equal to now
just because we’re solving this equation you have to take into account the over
the uh starting interval right here because we’re only looking on this
interval right here but both both c’s actually work right minus 1 and 1
are both in here so c is plus or minus 1. so the mean value theorem says there
has to exist a c for this to hold but actually sometimes you can actually go
and find those c’s and this in this example here there are actually two c’s

00:30
that actually make the mean value theorem for integrals work
all right so i hope you enjoyed this video and uh have a great day and i’ll
see you in the next episode if you enjoyed this video please like
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About The Author
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David A. Smith (Dave)

Mathematics Educator

David A. Smith is the CEO and founder of Dave4Math. His background is in mathematics (B.S. & M.S. in Mathematics), computer science, and undergraduate teaching (15+ years). With extensive experience in higher education and a passion for learning, his professional and academic careers revolve around advancing knowledge for himself and others. His work helps others learn about subjects that can help them in their personal and professional lives.

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