- Let's say that we have the function
F of X is equal to X squared
plus one and what we want to do
is we want to figure out the average value
of our function F on the interval,
on the closed interval between zero
and let's say between zero and three.
I encourage you to pause this video
especially if you've seen the other videos
on introducing the idea of an average
value of a function and figure out
what this is.
What is the average value of our function F
over this interval?
So, I'm assuming you've had a go at it.
Let's just visualize what's going on
and then we can actually find the average.
So that's my Y axis.
This is my X axis.
Now over the interval between
zero and three, so let's say
this is the zero, this is one, two, three.
It's a closed interval.
When X is zero F of zero
is going to be one.
So, we're going to be right over here.
F of one is two.
So it's one, two, three.
Actually, let me make my scale
a little bit smaller on that.
I have to go all the way up to 10.
So this is going to be 10.
This is going to be five.
And then one, two, three.
This is the hardest part is making
this even.
So see this is going to be in the middle.
Pretty good, and then let's see in the middle.
Then we have that.
Good enough. All right.
So, we're going to be there.
We're going to be there.
I have obviously different scales
for X and Y axis.
Two squared plus one is five.
Three squared plus one is 10.
So it's going to look something like this.
This is what our function is going
to look like.
So, that's the graph of Y is
equal to F of X.
And we care about the average value
on the closed interval between zero and three.
Between zero and three.
So, one way to think about it,
you could apply the formula,
but it's very important to think about
what does that formula actually mean?
Once again, you shouldn't memorize
this formula because it actually kind
of falls out out of what it actually means.
So the average of our function
is going to be equal to the definite
integral over this interval.
So, essentially the area under this curve.
So, it's going to be the definite intergral
from zero to three of F of X which is
X squared plus one DX.
Then we're going to take this area.
We're going to take this area
right over here and we're going
to divide it by the width of our interval
to essentially come up with the average height,
or the average value of our function.
So, we're going to divide it by B minus A,
or three minus zero, which is just
going to be three.
And so now we just have to evaluate this.
So, this is going to be equal to
one third times --
Let's see the antiderivative of X squared
is X to the third over three.
Antiderivative of one is X,
and we're going to evaluate it from zero to three.
So, this is going to be equal to one third
times when we evaluate it at three.
Let me use another color here.
When we evaluate it at three it's
going to be three to the third
divided by three.
Well, that's just going to be 27 divided
by three.
That's nine plus three and then
when we evaluated zero, minus zero
minus zero.
So, it's just minus.
When you evaluated zero it's
just going to be zero.
And so, we are left with --
I'm going to make the brackets
that same color.
This is going to be one third
times 12.
One third times 12, which is equal
to four.
Which is equal to four.
So this is the average value
of our function.
The average value of our function
over this interval is equal to four.
Notice, our function actually hits
that value at some point in the interval.
At some point in the interval,
something lower than two but greater than one.
We can maybe call that C.
It looks like our function hits that value.
This is actually a generally true thing.
This is a mean value theorem
for integrals and we'll go into more depth there.
But you can see this kind of does look
like it's average value.
That if you imagine the box, if you
multiplied this height, this average value
times this width you would have this
area right over here, and this
area right over here is the same,
this area that I'm highlighting
in yellow right over here is the same
as the area under the curve
because we have the average height
times the width is the same
thing as the area under the curve.
So, anyway hopefully you found
that interesting.