BAM!
Mr. Tarrou.
Hey, do you remember many many years ago in math, when you first learned estimation, it
would take some irregular shaped plot of land and over that, they would put a grid and they
would tell you usually that each of those little boxes in the grid was one square mile
or square feet or whatever.
And, you had to count those boxes that were inside, and then go around the boxes and count
partial boxes and maybe if just a piece of a box was inside that irregular shape, you
would count as maybe half a unit, and you would do that to estimate the area of that
irregular shape, meaning it's not a triangle or a rectangle or pentagon, we don't really
find areas of pentagons, maybe parallelogram or trapezoid.
Well, we are going to start now the second main sort of classic type problem that we
have in the beginning of calculus.
First one being, of course, derivative to find the area of a curved line along any given
value of x or almost all of them that worked.
And the second is going to be finding the area of a plane region, specifically in this
video, we're going to be finding the areas bound by a function, the x axis, and two points
along the x axis.
Here's what our problem is going to sound like, or is going to be, excuse me.
Use 3 rectangles to find two approximations of the area of the region between the graph,
that graph is going to be y=-x^2+6, and the x axis between the x values of 0 and 2.
So we have this boundary along the x axis between 0 and 2.
So we're looking for, I drew sort of the entire parabola here, the parabola that opens up
downward in quadrant I, but we're not going to be looking for the area over here in this
last little bit because we're not going to go past 2.
Okay and we're gonna do that, like I said, with just old school area of rectangles.
So the point of this of course, is we just did a video studying or reviewing our skills
with the sigma notation and incorporating it with limits.
So ultimately we're gonna do this estimation with an infinite number of rectangles, not
just 3.
But I want to build it up so you understand the basic concept when we are done.
Okay, so what I did was I took and marked off 0 and 2, and we were asked to estimate
the area with three rectangles.
So I simply took that horizontal distance of two and divided it by three.
Now I have three rectangles that all have a width of 2/3rds so if I have the width,
then all I need is to find the height of those rectangles, I can find the three areas and
add them together.
Well, what I have done is I've inscribed these rectangles, so they go up to very very edge
of the function y=-x^2+6.
That means, that the function is going to define the height of each of these three rectangles,
based on the right hand side where that rectangle just happens to touch the function.
It doesn't matter really that it's the right hand side, the fact that I inscribed these
rectangles which means the corner of the rectangle happens to hit that function to find the height.
So, let's find the area of the first rectangle, the area of the first rectangle is gonna be,
I'm gonna do height times width because ultimately, I want to get back to that sigma notation
we talked about in the last video.
So we're gonna do the height of this rectangle which we're gonna find by taking 2/3rds and
plugging it into this function.
I'm going to leave a little space because I want to go back to sigma notation here,
so we'll say maybe around here somewhere, I'll need two lines for this work.
We're gonna take, let's see here, square bracket, I want to put this function in here to identify
the height, so it's gonna be parenthesis, negative parenthesis, because you should always
use those when you do substitution.
We're gonna plug in that 2/3rds raised to the second power, plus 6.
So there's the height of the rectangle, the width is 2/3rds.
And I'm using these square brackets to just really emphasize where we're finding the area
of each of the three rectangles.
They're kind of overkill a little bit, but that's why I'm doing it.
Second rectangle, we have the height there, well now the height again, looking at the
right hand side, we have 4/5ths needs to get plugged into this function.
Now I'm not going to write in 4/5ths as x, what I really want to do in this is highlight
the pattern that I'm using as the 2/3rds here.
Because again, I want to get back to that sigma notation and we need an index, a counting
value.
So for this second rectangle, from this original left bound, lower bound.
I need to move 2/3rds units twice, so we're gonna do plus another square bracket, parenthesis,
now we're gonna plug in that function for the height, negative x, which is 2/3rds times
2, plus the width.
So, I'm highlighting here that again we moved 2/3rds twice, that means this is 2/3rds time
1.
So height times width.
And now the last rectangle, okay well again, one movement, two movement, three movements
of 2/3rds.
Again, taking that x value and plugging it into that original function here that I have,
y=-x^2+6.
And times the width of the rectangle.
Now, this area happens to come out, right off the top of my head, of 7.852.
Now, you know, why did I teach you summation notation and limits, which we don't need limits
for this problem, but summation notation.
Let's look at that pattern.
What I've done is, I needed to take the summation as i starts at 1, and we had three rectangles,
so we have three additions here, three terms that we're adding, so the upper limits is
3.
And what did we plug in?
Well, we plugged in, let's see.... and then what do we plug into that function?
Well, we plugged in, 2/3rds, but it was 2/3rds times 1, 2/3rds times 2, 2/3rds times 3, so
what we did basically was, again we took the overall width that we needed to find the area
between, we divided it by 3, cause we wanted to use 3 rectangles, and we got this width
of the rectangles of 2/3rds.
Then, to find the height, I had to start from my lower limit and move by 2/3rds once, by
twice, by three times.
So the heights were found by taking 2/3rds and the very first height, we had to move
2/3rds once.
Okay, so the lower index starts off as being 1, so the first thing I did that plugged into
my function is take 2/3rds and multiply it by one, to get back to that sigma notation,
we multiply it by i.
We have, f of 2/3rds i, and the function was -x^2+6, and then we multiplied that by the
width of the rectangle, which was 2/3rds, and that sigma notation is what we just walked
through here to get a area of 7.852.
Now, this area as we can see, is not gonna be correct, it's gonna be an estimation, but
it's not going to be correct, and it's actually going to underestimate the area between the
curve and the x axis bound by the values of 0 and 2.
We can see there's empty spaces here, so what if instead of inscribing the rectangles, we
circumscribed them?
So we're gonna start here, basically we're going to let the left side of the rectangle
be defined or touch that function, and then draw the rectangles like this.
And we're going to develop this idea of lower and upper limit.
We're going to underestimate and over estimate the area underneath the function.
Now, when you use the right corner, you're not always going to underestimate the area
and as you can see now, I'm using the left corner of my rectangles to define their height,
and I have some extra area, so I'm going to overestimate the area.
It's not always the right side underestimates and the left side overestimates, it's just
because my graph is falling to the right.
If it was rising, then just the opposite as far as what side corner of the rectangle would
over or under estimate the area.
I'm getting a little ahead of myself here.
So now we have three rectangles and let's write this out in multiplication long-hand
form, and then go back to the sigma notation.
Okay, so this first height of the rectangle is defined as the active function, is still
defined by the function, but it's defined by plugging in value of 0 because it's 0 that
left corner that's touching the function now, not the right hand corner.
So we need to do negative, now I want to do this a little bit differently, I want to focus
on the fact that we're moving or breaking the x axis by values of 2/3rds.
So we need to do 2/3rds times 0, okay?
So we're gonna do 2/3rds times 0, squared, plus 6.
Did you catch why I'm doing that?
Because I need the value of 0 to go into this function, I'm looking at x=0.
So I need my expression to show that I'm plugging in the value of 0 into the function and not
2/3rds.
Why am I doing this times 0, 2/3rds times 0, why am I not just plugging in 0?
Because I want to show you the difference between the first problem when the rectangles
are inscribed, and the second example where they are circumscribed, and ultimately we're
not going to do this long handed arithmetic for every single problem.
We need that sigma notation because we're going to bring back the idea of limits here
in a few minutes.
So 2/3rds times 0, squared, plus 6, times 2/3rds, still the width of each rectangle.
Kind of taking up too much space here so I'll drop my second line now.
The second rectangle, its hieght is defined by 2/3rds, so we're gonna do negative 2/3rds
times 1, squared, plus 6, times 2/3rds, (its width) plus negative 2/3rds times 2, squared,
plus 6, times 2/3rds.
And I'm for some reason really writing quite large.
This area comes out to be, right off the top of my head...
I'm putting my negative in the wrong spot here, started talking, started writing faster
and messed up a little bit.
Okay, so that comes out to be 10.519.
Okay, now the pattern for this is the summation as i starts at 1 and goes to 3.
Now I know the lower index of sigma notation, or the summation notation, doesn't have to
start at 1, but it will need to start at 1 so we can use those 3 formulas, the summation
for when we're taking the summation of i, i^2, and i^3, because again, we're not gonna
do all these problems long hand, we're going to ultimately put in a infinite number of
rectangles to get an infinitely accurate estimation of the area of the curve, which means we have
to bring in the idea of limits, which means again, for how many times now we need that
sigma notation.
So we do need to start that at 1, which means that we're gonna take the idea of what we
did with the function.
We took 2/3rds and we did not just take 2/3rds and multiply it by i, do you see here this
in the original problem we did 2/3rds times 1, 2/3rds times 2, and 2/3rds times 3, well
now all those values that we're multiplying 2/3rds by are 0, 1, and 2.
They've shrank by a value of 1.
So we need to do f of 2/3rds time i, minus 1, close the parenthesis, times 2/3rds.
Now, we know that the actual area is between 7.85 and 10.519.
Let's see what happens, little bit of a spread here, that's really not a great summation.
Let's see if on your own, you can do the same problem with six rectangles.
For the second example, I'm just going to step out, well I'll rewrite the question,
and step out and sort of reveal the solution in small sections and not really talk because
it's the exact same process, it would be a good idea for you to stop the video, and change
the 3 to a 6, and do this estimation on your own.
It's going to be the exact same process, of course the numbers are going to be little
bit different, and we are going to get a tighter interval of the area, this will go a lot quicker
because I'm not going to talk this next example, however, I really think it's strongly suggested
that you attempt the next section, or example, on your own.
We're gonna do 6 rectangles instead of 3, and then do a little bit of definitions, defining
some processes and some limits, and then do not a rough estimate like we're doing in the
3 and the 6 rectangles, but the exact estimate that we're going to do by letting the number
of rectangles go to infinity.
Okay, let me get out of the way and get that other problem for you.
Okay, so what I've done is simply change the 3 to a 6, and you can already tell by the
picture that we're not gonna underestimate as badly as we did before, nor is our overestimation,
our upper limits, going to be as off as it was because these bases are all now smaller,
using 6 rectangles instead of just 3.
Now as you attempt the problem, when I reveal it on the board, let's not forget that we
were looking for a boundary between 0 and 2, that's a length of 2 and we want to estimate
it with 6 rectangles now, not just 3.
So I've marked off intervals of 1/3rd as opposed to 2/3rds.
Yes, 2/3rds.
And sorry about that.
And also note, as you try and do the problem, finding lower and the upper limit, or boundary
or whatever you want to call it, that, you know, we're marking off not only our rectangles
a width of 1/3rd now as opposed to 2/3rd, but when we go in steps to find the heights
of these rectangles, finding values to plug into our function of y=-x^2+6, that we're
gonna be stepping in intervals, again, of 1/3rd is wrong, not 2/3rds.
So let's get out of the way again, and reveal that answer to you as you attempt it on your
own.
Okay, I hope you got it.
We have a lower sum of 8.63 and an upper sum of 9.96, getting closer to the actual estimated,
or it's going to be estimated indefinitely accurate, so let's just say it's getting closer
to the actual area underneath the curve and the x axis between the values of 0 and 2,
tighter now.
So, I want to emphasize a couple of things.
As you saw, I was highlighting that we are finding the heights based on the either the
right or the left upper corner of the rectangle.
I don't want you to always think that if you're working with the upper lefthand corner, you're
going to find the sum where you're overestimating the actual area.
And if you're using the upper right corner, you're finding the lower sum.
I just want to emphasize again that is only the situation we have here because our graph
is falling down to the right.
If it was rising up, you would have just the opposite relationship there.
So don't get that memorized in your head and think that's the end all, be all.
The 1/3rd that we see here, we're about to get to the definition so that 1/3rd, that
width of that rectangle, verbally said it a couple of times, is gonna be identified
though with the variable of delta x, the change in x, and the change in x or the width of
those rectangle are gonna be found by taking the upper boundary that we have on the x axis,
could be the y axis, we're gonna do one of those in the next video.
The lower limit, we're gonna subtract those again and divide by n, the number of rectangles
you'd like to use to estimate the area.
Okay, so we've seen that with 3 rectangles, we have a certain estimation, lower and upper
sum.
With 6 rectangles, we're getting closer, but still a little bit of a fair gap there of,
whatever that is, .03 units.
So now we're going to let the number of rectangles expand to infinity.
So that means we need to get some definitions down that I have not yet given you, and then
finish up this example after those definitions.
And I got one more example for you to try on your own as well, but like we did here.
Okay, let me get this board erased and get those definitions and theorems and whatever
you want to call it up there for you.
BAM!