BAM!!! Mr. Tarrou. In this video I am going to introduce two theorems and tie them together
with one example that has two parts. Now when we first started learning about the Definition
of Area we find the area bound by a function and an axis by estimating it's area with a
bunch of rectangles. Then developed that eventually into the definition of the definite integral
of Riemann Sums. Well you saw that when we inscribed rectangles and used as those as
our partitions, that if we used inscribed rectangles those areas of those rectangles
underestimated that actual area bound the function and the axis. And when we used circumscribed
rectangles we over estimated that. We called them the lower and the upper sums. What we
are going to talk about in the first theorem today is that there is a rectangle that will
perfectly, or exactly estimate... or be equal to, that area of that rectangle will be equal
to the actual area bound between the function and an axis on a closed interval. We see that
here in our third diagram. Again the inscribed rectangle. I have drawn this lower value here
as a relative minimum of our function just for convenience. It does not have to be on
the right endpoint of our closed interval. But that inscribed rectangle is going to underestimate
the actual area. And our circumscribed rectangle is going to over estimate the area. Lower
case m is the minimum value of our function on that closed interval. And capital M is
going to be identifying the maximum value for what.. you know... what independent variable
you can plug in and get that maximum value for the function. That would be our circumscribed
rectangle. Well somewhere in that closed interval there is guaranteed, by the the Mean Value
Theorem for Integral, a value of c in that closed interval such that when you take the
value of c and you plug into the function you get a height. And that height times width,
the area of that rectangle, is going to be equal to the bound area that we discussed.
So here we have the first theorem The Mean Value Theorem for Integrals. If function f
is continuous on a closed interval [a,b], then there exists some number c in that closed
interval [a,b] such that the definite integral of f(x) on that closed interval... or with
a lower bound of a and an upper bound of b is equal to f(c) times b-a. Well that c value
that we plug in and get a y value out of the function is going to be the height of our
rectangle, then b-a is of course going to be the width along the x or it could be the
y axis. Now this theorem just simply guarantees that there is a value of c such that the height
times width, that area of that rectangle, is going to be equal to the area again bound
by the function and an axis on a closed interval. It does not tell us where that value of c
is. So we are not going to know from this theorem, and we will not know, until we finish
the second theorem the height of that rectangle. If you.. you know.. need that for your particular
problem. The height of the rectangle, ok, this one here... this blue one... is going
to be equal to the average value of the function one [a,b]. That is going to be the next theorem
that we take a look at. Once I state that theorem I want to tie together again just
the basic... you know... where can you see within that theorem the basic area of a rectangle
which is just simply length times width. And again, using the definite integral in the
purpose of finding this area. Area is equal to again length times width and we are drawing
that rectangle. That is all there is to it. Let's get to the next theorem. BAM! Here we
go. The Average Value of a Function. If f is integrable on a closed interval of [a,b],
then the average value of that function f on the interval is given by the definite integral
from a to b of f(x)dx multiplied by 1 over b-a. Now this is how my textbook gives you
the average value of a function. That is the way that is is written. But if I add a little
bit to it and say that this is going to be equal, like I said in the previous screen,
the height of the rectangle then this is equal to f(c). So you can see all they did with
our previous theorem is divide both sides by b-a and voila you have the Average Value
of a Function. So what we are going to is, we are going to get this board erased, I am
going to put up a problem and we are going to... first let's see here... We are going
to first find the average value of a function and then use that average value, the height
of that rectangle, to then go back and find what is that value of c that is guaranteed
such that f(c) times b-a does actually equal the definite integral from a to b. Let's get
to it! There we go, there is my remote. Find the average value of f(x) is equal to negative
x squared plus 8x minus 7 on the closed interval of one to five. So whenever you are learning
a brand new concept it is a very good idea, especially when it has a formula attached
to it, that you need to memorize to write that down as you do the problem and then fill
in the spots... fill in the blanks.. and then continue on with your problem. So again the
average value of a function is equal to 1 over b-a times the definite integral of f(x)dx
from a to b. So when... Remember when they give us our closed interval, the lower limit
is a... or the lower value... and the upper value is b. So we have this is going to be
equal to 1 over 5-1 times the definite integral... and again this is a and b... so 1 to 5 of
f(x). Now of course f(x) is -x^2+8x-7 and don't forget the dx just for good notation.
You know what value are we integration with respect to. So I am going to pause the video
and come back with the solution written up. If you want to try it on your own, you might
want to pause the video now and make sure that you remember how to do these definite
integrals. We will get the answer in just a second. I hope this is what you got. The
average value of our function on the closed interval of [1,5] came out to be 20/3. How
did we get there? Well we went again through the definite integration process and used
the power rule. We took this power of 2 and increased it by 1 and divided by 3. I just
pulled that negative one and the division of 3 out as a fraction, -1/3. Took that exponent
of 1 and increased it by 1 and divided, so we had that 8 divided by 2. Then finally our
-7x. Then remember we don't need to write the plus c that we get from indefinite integration
because as we go through the First Fundamental Theorem of Calculus and find this definite
integral they are going to cancel out anyway. So clouds are coming through. Sorry about
the white balance change there. And then we took the... Then remember this is a and b.
So we take f(b) and we subtract f(a) along with that 1/4 now that is sitting out front
from the 1 over b-a. And basically just went through the simplification process making
sure to distribute that negative sign through the second group, through the minus f(a) part
of our expression here. Very long expression. And then common denominators. Ultimately get
1/4 of 80/3. Now 80/3 is our value from the definite integral of f(x) dx. So the area
on that closed interval was 80/3. I finished by multiplying it by 1/4 and getting that
average value of 20/3. Let me adjust my white balance and get to the second part of this
example. Now the question comes in... now that we know what the average value is, let's
find that value of c on that closed interval where f(c) is equal to the average value.
I am writing this exactly how it appears in our book which will take me an extra step.
Because remember if we take this equation and divide both sides by b-a, we are going
to have f(c). We are going to have that f(c), the height of that rectangle is the average
value. So I am going to plug in the numbers just how it looks here, but that is actually
going to add one step to my work that you might catch on to... remember. Bring this
over and we have the Average Value. So at any rate, the value of the definite integral
of f(x) dx from a to b, or the area on that closed interval was 80/3. f(c) is um... well
let's see here. We don't know what c is right. So we are just going to basically... f(c)...
we are going to look at our function and just write in c everywhere there is an x. We are
going to have -c^2+8c-7 times b-a. b-a, b was 5 and a was 1 so that would be 4. I am
going to write the 5-1 just to show all my work, but it is 4. When we divide both sides
by 4, we are going to have 80/12 is equal to -c^2+8c-7. We know that the 80/12 as we
have already reduced is equal to 20/3. Ok, well we are getting into a situation where
we have a quadratic. And we need to solve for that. I am going to multiply everything
by 3. We get 20 is equal to -3c^2+24c minus 3 times 7 is 21. We are going to subtract
this 20 over to the right hand side of the equation. We get... Actually for convenience
sake let's move everything to the left so that my leading term is positive. We have
bringing this over we have 3c^2. Bring the 24c over and we have negative 24c. And if
we add 21 to both sides we get positive 41. Let me just check my notes to make sure that
this is good. Ok. Now at this point we have a quadratic. That means that there might not
only be one value of c on that closed interval. There might be two, and that is possible.
There could even be more than that if the question is really complicated. So when we
find that c on the closed interval, it may not be a unique solution. There may be two
or more. So what we have now is a quadratic that we might solve by completing the square,
we might try and factor it, but if we do 3 times 41 and look at the product. There are
no factors of that which add up to be -24. In other words this is not a factorable quadratic.
So what we are going to do is, well plug it into the quadratic formula. We have c is equal
to negative b, so negative... or the opposite of b. So the opposite of negative twenty-four...
plus or minus the square root of b squared minus 4ac. So 3 times 41... all over 2a. And
when you finish solving this, right off the top of my head, we get... Oh.. Just checking
the signs of my answers. What we end up with is c is equal to 2.472 and 5.528. So I am
going to go ahead and solve that quadratic with the quadratic formula. We do get two
answer. So there is possibly two places, to values of c, where this equation is true,
where the definite integral is equal to f(c) times b-a. But our definite integral, or closed
interval, where the question was concerned with was the closed interval between 1 and
5. So while it is possible to get two answers, or more than one answer... more than one value
of c on that closed interval, my second value of 5.528 is not within the closed interval
that was asked for in the original problem. So we did get a unique solution. That is the
end of my video. So, you know what that means? Time for you to go do your homework! BAM!