PROFESSOR: Welcome back to recitation.
Today what I'd like to do is work on using Riemann sums to
approximate definite integrals.
So, what we're going to do first is I'm going to let you
work on the problem.
So, let me state the problem.
I'll give you a little time to work on it.
And then I'll come back.
So the problem is the following.
We want to use four subintervals and left
endpoints to approximate the following definite integral:
the integral from minus 1 to 3 of the function x cubed.
And I've drawn a rough sketch of what x cubed, y equals x
cubed looks like to help you out as the starting point.
So why don't you use the four subintervals and left
endpoints to approximate the definite integral, then I'll
come back and show you how I do it.
OK, welcome back.
So what we want to do again, is use Riemann sums to
approximate this defnite integral.
And I've given you the specifications of four
subintervals and left endpoints.
So I'm going to draw what these four subintervals and
their left endpoints are going to give us on the graph.
Then we're going to calculate what the actual value is of
this estimate.
And then I'll show you a way you can write it in some
different notation.
So let me actually find some blue chalk.
Excuse me.
So, using blue chalk I'm going to show you where the four
subintervals and the left endpoints.
So it's minus 1 to 3, so I've conveniently done it so we
have every unit length is one of the seven intervals.
They're all unit length.
And if I want left endpoints, then my rectangle's heights,
if you remember, are going to be the output at the left side
of the interval.
So I'm going to designate the left interval's outputs.
And what do we have here?
I'm going to now draw rectangles of those, using
those with the one side on x-axis and one side at the
height of the left endpoint output.
So there's one rectangle.
Oh, I almost did something wrong here.
This left endpoint's rectangle has no height.
Because the left endpoint's value is 0.
The output is 0.
There's another rectangle.
And there's the last rectangle.
So, to designate everything carefully, I will call this
value y0, the output value here y0.
This value will be y1.
The output value here will be y2.
And to show exactly what do I mean by output value, here I
have enough room to write the height is really
what it is, is y3.
So y0, y1, y2 and y3 designate the heights of these four
rectangles.
And the length of each rectangle, if you remember, is
really what we designate as delta x.
So this right here is delta x.
And in this case delta x is each time equal to 1.
So, remember formally, what we want to do is, if we want to
find this definite integral, we want to find the area under
the curve between the x-axis and the function x cubed.
And remember this is sined area.
So anything below the x-axis has a negative sign
associated to it.
Anything above the x-axis has a positive sign
associated to it.
So, when I'm trying to determine this, when I'm
trying to do an estimate, I'm finding the areas of these
rectangles.
So my first approximation that I've given here, is we want
the four subintervals and the left endpoints.
What we really need is this delta x, which is the base,
times the height, which is y0, y1, y2 and y3 of each of these
rectangles.
So, to write it carefully--
well I guess this won't be the most careful way we write it--
but I have the base, which is delta x in each case, times y0
plus y1 plus y2 plus y3.
And the only thing I have to be careful of is that the area
of the rectangle here will be negative.
But how do I pick that up?
I pick that up because this y0 value as an
actual output is negative.
So that actually will be picked up automatically by the
value of the output on the function.
So let's evaluate these things. y0 should be the value
of the function at x equal negative 1.
When x is negative 1, the function, remember, is x
cubed, so I get negative 1 cubed, which is
just negative 1.
So delta x is 1.
y0 is negative 1.
y1 is the value the function at x equals 0, because my left
endpoints are minus 1, 0, 1 and 2.
So the second one is 0 is the input.
at x equals 0 I get 0 as an output.
The y2 is going to be this third left endpoint.
That's at x equal 1.
1 cubed is 1.
So I get a 1 there.
And then, the third left endpoint, I had
minus 1, 0, 1, 2.
So the third left endpoint is 2.
2 cubed is 8.
And so y3 is 8.
Hopefully the subscripts here didn't confuse you because I
actually was very close to having the subscripts
represent the x value.
But that's not in fact what the subscripts are doing.
The subscripts are just representing what interval we
are looking at.
So the four intervals are designated by 0, 1, 2 and 3.
So I get, when I put this all together I get 1 times--
well, minus 1 plus 1 is 0--
1 times 8.
So I get 8.
And if I wanted to look at the picture, how is that
represented?
Well this rectangle has base 1 and height 8.
So that has area 8.
This rectangle is actually a square.
It has base 1, height 1.
And this one is also a square.
It has base 1, height 1.
But because it's below, it has a minus sign associated to it,
when I think about it in terms of area.
So I get minus 1 plus 1 plus 8.
And that's where the 8 comes from.
Now if I wanted to write this in terms of a Riemann sum, you
see them sometimes written in this way.
You can think about it in a formal sum in this way.
You've probably seen this in the lecture
videos more like this.
The sum from i equals 0 to 3 of f of x sub i delta x.
And now, in this case, we could even be more deliberate
if we wanted to.
And we could let x sub i be designated by some value.
We could say, starting at negative 1 and adding some
number to it each time based on the interval.
But this is sort of the easiest way to write out what
we're interested in.
And then the x sub i's are going to be designated
separately.
We can write x sub 0 is equal to minus 1 x sub
1 is equal to 0.
x sub 2 is equal to-- sorry, not minus 1-- is
equal to plus 1.
An x sub 3 is equal to 2.
So in that case, that's what we would get there.
And we could write it in that way and then
get the same values.
So I think I will stop there.