WEBVTT

00:00:07.250 --> 00:00:08.960
PROFESSOR: Welcome
back to recitation.

00:00:08.960 --> 00:00:11.090
Today what I'd
like to do is work

00:00:11.090 --> 00:00:14.360
on using Riemann sums to
approximate definite integrals.

00:00:14.360 --> 00:00:16.632
So, what we're
going to do first is

00:00:16.632 --> 00:00:18.340
I'm going to let you
work on the problem.

00:00:18.340 --> 00:00:19.400
So, let me state the problem.

00:00:19.400 --> 00:00:20.900
I'll give you a little
time to work on it.

00:00:20.900 --> 00:00:21.940
And then I'll come back.

00:00:21.940 --> 00:00:23.620
So the problem is the following.

00:00:23.620 --> 00:00:28.010
We want to use four
subintervals and left endpoints

00:00:28.010 --> 00:00:30.600
to approximate the
following definite integral:

00:00:30.600 --> 00:00:34.150
the integral from minus 1 to
3 of the function x cubed.

00:00:34.150 --> 00:00:36.420
And I've drawn a rough
sketch of what x cubed,

00:00:36.420 --> 00:00:38.790
y equals x cubed looks
like to help you out

00:00:38.790 --> 00:00:40.410
as the starting point.

00:00:40.410 --> 00:00:43.630
So why don't you use the four
subintervals and left endpoints

00:00:43.630 --> 00:00:45.790
to approximate the
definite integral, then

00:00:45.790 --> 00:00:47.750
I'll come back and
show you how I do it.

00:00:56.360 --> 00:00:57.600
OK, welcome back.

00:00:57.600 --> 00:01:00.496
So what we want to
do, again, is use

00:01:00.496 --> 00:01:02.620
Riemann sums to approximate
this definite integral.

00:01:02.620 --> 00:01:05.980
And I've given you
the specifications

00:01:05.980 --> 00:01:08.070
of four subintervals
and left endpoints.

00:01:08.070 --> 00:01:10.229
So I'm going to
draw what these four

00:01:10.229 --> 00:01:11.770
subintervals and
their left endpoints

00:01:11.770 --> 00:01:13.540
are going to give
us on the graph.

00:01:13.540 --> 00:01:16.120
Then we're going to calculate
what the actual value is

00:01:16.120 --> 00:01:16.930
of this estimate.

00:01:16.930 --> 00:01:18.805
And then I'll show you
a way you can write it

00:01:18.805 --> 00:01:20.010
in some different notation.

00:01:20.010 --> 00:01:22.456
So let me actually
find some blue chalk.

00:01:22.456 --> 00:01:22.956
Excuse me.

00:01:22.956 --> 00:01:26.785
So, using blue chalk
I'm going to show you

00:01:26.785 --> 00:01:28.910
where the four subintervals
and the left endpoints.

00:01:28.910 --> 00:01:31.630
So it's minus 1 to 3,
so I've conveniently

00:01:31.630 --> 00:01:35.220
done it so we have
every unit length is

00:01:35.220 --> 00:01:36.220
one of the subintervals.

00:01:36.220 --> 00:01:38.120
They're all unit length.

00:01:38.120 --> 00:01:41.990
And if I want left endpoints,
then my rectangle's heights,

00:01:41.990 --> 00:01:44.664
if you remember, are going to
be the output at the left side

00:01:44.664 --> 00:01:45.330
of the interval.

00:01:45.330 --> 00:01:48.820
So I'm going to designate
the left interval as output.

00:01:53.859 --> 00:01:54.900
And what do we have here?

00:01:54.900 --> 00:01:57.550
I'm going to now draw
rectangles of those, using those

00:01:57.550 --> 00:02:02.590
with the one side on
x-axis and one side

00:02:02.590 --> 00:02:07.200
at the height of the
left endpoint output.

00:02:07.200 --> 00:02:09.940
So there's one rectangle.

00:02:09.940 --> 00:02:12.180
Oh, I almost did
something wrong here.

00:02:12.180 --> 00:02:16.080
This left endpoint's
rectangle has no height.

00:02:16.080 --> 00:02:18.370
Because the left
endpoint's value is 0.

00:02:18.370 --> 00:02:19.220
The output is 0.

00:02:22.680 --> 00:02:25.030
There's another rectangle.

00:02:25.030 --> 00:02:26.433
And there's the last rectangle.

00:02:30.060 --> 00:02:32.460
So, to designate
everything carefully,

00:02:32.460 --> 00:02:38.120
I will call this value y_0,
the output value here y_0.

00:02:38.120 --> 00:02:39.190
This value will be y_1.

00:02:41.594 --> 00:02:43.010
The output value
here will be y_2.

00:02:46.480 --> 00:02:48.770
And to show exactly what
do I mean by output value,

00:02:48.770 --> 00:02:50.850
here I have enough room
to write the height

00:02:50.850 --> 00:02:53.860
is really what it is, is y_3.

00:02:53.860 --> 00:02:56.800
So y_0, y_1, y_2 and y_3
designate the heights

00:02:56.800 --> 00:02:58.570
of these four rectangles.

00:02:58.570 --> 00:03:02.620
And the length of each
rectangle, if you remember,

00:03:02.620 --> 00:03:05.150
is really what we
designate as delta x.

00:03:05.150 --> 00:03:09.610
So this right here is delta x.

00:03:09.610 --> 00:03:13.610
And in this case delta x
is each time equal to 1.

00:03:13.610 --> 00:03:15.974
So, remember, formally,
what we want to do

00:03:15.974 --> 00:03:17.890
is, if we want to find
this definite integral,

00:03:17.890 --> 00:03:22.280
we want to find the area under
the curve between the x-axis

00:03:22.280 --> 00:03:23.560
and the function x cubed.

00:03:23.560 --> 00:03:25.080
And remember this
is signed area.

00:03:25.080 --> 00:03:28.350
So anything below the
x-axis has a negative sign

00:03:28.350 --> 00:03:29.170
associated to it.

00:03:29.170 --> 00:03:33.830
Anything above the x-axis has a
positive sign associated to it.

00:03:33.830 --> 00:03:35.420
So, when I'm trying
to determine this,

00:03:35.420 --> 00:03:36.836
when I'm trying
to do an estimate,

00:03:36.836 --> 00:03:39.200
I'm finding the areas
of these rectangles.

00:03:39.200 --> 00:03:41.720
So my first approximation
that I've given here,

00:03:41.720 --> 00:03:44.390
is we want the four subintervals
and the left endpoints.

00:03:44.390 --> 00:03:47.220
What we really need
is this delta x,

00:03:47.220 --> 00:03:49.520
which is the base,
times the height, which

00:03:49.520 --> 00:03:52.950
is y_0, 1, 2 and 3, of
each of these rectangles.

00:03:52.950 --> 00:03:56.000
So, to write it
carefully-- well I

00:03:56.000 --> 00:03:58.600
guess this won't be the most
careful way we write it--

00:03:58.600 --> 00:04:01.510
but I have the base, which
is delta x in each case,

00:04:01.510 --> 00:04:08.490
times y_0 plus y_1
plus y_2 plus y_3.

00:04:08.490 --> 00:04:10.240
And the only thing I
have to be careful of

00:04:10.240 --> 00:04:12.940
is that the area
of the rectangle

00:04:12.940 --> 00:04:14.190
here will be negative.

00:04:14.190 --> 00:04:15.680
But how do I pick that up?

00:04:15.680 --> 00:04:20.040
I pick that up because this
y_0 value, as an actual output,

00:04:20.040 --> 00:04:21.450
is negative.

00:04:21.450 --> 00:04:23.450
So that actually will be
picked up automatically

00:04:23.450 --> 00:04:26.870
by the value of the
output on the function.

00:04:26.870 --> 00:04:29.470
So let's evaluate
these things. y_0

00:04:29.470 --> 00:04:32.520
should be the value of the
function at x equal negative 1.

00:04:32.520 --> 00:04:34.990
When x is negative 1-- the
function, remember, is x cubed,

00:04:34.990 --> 00:04:38.690
so I get negative 1 cubed,
which is just negative 1.

00:04:38.690 --> 00:04:40.790
So delta x is 1.

00:04:40.790 --> 00:04:42.900
y_0 is negative 1.

00:04:42.900 --> 00:04:46.620
y_1 is the value the
function at x equals 0,

00:04:46.620 --> 00:04:51.450
because my left endpoints
are minus 1, 0, 1 and 2.

00:04:51.450 --> 00:04:55.000
So the second one
is 0 is the input.

00:04:55.000 --> 00:04:57.190
At x equals 0 I
get 0 as an output.

00:04:59.890 --> 00:05:03.950
The y_2 is going to be
this third left endpoint.

00:05:03.950 --> 00:05:05.790
That's at x equal 1.

00:05:05.790 --> 00:05:08.900
1 cubed is 1.

00:05:08.900 --> 00:05:11.180
So I get a 1 there.

00:05:11.180 --> 00:05:13.850
And then, the third left
endpoint, I had minus 1,

00:05:13.850 --> 00:05:16.000
0, 1, 2.

00:05:16.000 --> 00:05:17.950
So the third left endpoint is 2.

00:05:17.950 --> 00:05:19.480
2 cubed is 8.

00:05:19.480 --> 00:05:23.190
And so y_3 is 8.

00:05:23.190 --> 00:05:25.300
Hopefully the subscripts
here didn't confuse you

00:05:25.300 --> 00:05:28.500
because I actually was very
close to having the subscripts

00:05:28.500 --> 00:05:29.630
represent the x-value.

00:05:29.630 --> 00:05:31.838
But that's not in fact what
the subscripts are doing.

00:05:31.838 --> 00:05:33.710
The subscripts are
just representing what

00:05:33.710 --> 00:05:35.130
interval we are looking at.

00:05:35.130 --> 00:05:38.820
So the four intervals are
designated by 0, 1, 2 and 3.

00:05:38.820 --> 00:05:42.790
So I get, when I put this all
together I get 1 times-- well,

00:05:42.790 --> 00:05:45.790
minus 1 plus 1 is 0-- 1 times 8.

00:05:45.790 --> 00:05:48.000
So I get 8.

00:05:48.000 --> 00:05:49.730
And if I wanted to
look at the picture,

00:05:49.730 --> 00:05:50.870
how is that represented?

00:05:50.870 --> 00:05:54.770
Well this rectangle has
base 1 and height 8.

00:05:54.770 --> 00:05:56.460
So that has area 8.

00:05:56.460 --> 00:05:57.960
This rectangle is
actually a square.

00:05:57.960 --> 00:05:59.670
It has base 1, height 1.

00:05:59.670 --> 00:06:00.920
And this one is also a square.

00:06:00.920 --> 00:06:02.030
It has base 1, height 1.

00:06:02.030 --> 00:06:04.290
But because it's below,
it has a minus sign

00:06:04.290 --> 00:06:07.370
associated to it, when I think
about it in terms of area.

00:06:07.370 --> 00:06:10.740
So I get minus 1 plus 1 plus 8.

00:06:10.740 --> 00:06:12.680
And that's where
the 8 comes from.

00:06:12.680 --> 00:06:15.760
Now if I wanted to write this
in terms of a Riemann sum,

00:06:15.760 --> 00:06:17.960
you see them sometimes
written in this way.

00:06:17.960 --> 00:06:24.232
You can think about it in
a formal sum in this way.

00:06:24.232 --> 00:06:29.360
You've probably seen this in the
lecture videos more like this.

00:06:29.360 --> 00:06:38.160
The sum from i equals 0 to
3 of f of x sub i delta x.

00:06:38.160 --> 00:06:40.020
And now, in this
case, we could even

00:06:40.020 --> 00:06:41.970
be more deliberate
if we wanted to.

00:06:41.970 --> 00:06:45.910
And we could let x sub i be
designated by some value.

00:06:45.910 --> 00:06:47.510
We could say,
starting at negative 1

00:06:47.510 --> 00:06:50.180
and adding some
number to it each time

00:06:50.180 --> 00:06:51.280
based on the interval.

00:06:51.280 --> 00:06:54.730
But this is sort
of the easiest way

00:06:54.730 --> 00:06:56.430
to write out what
we're interested in.

00:06:56.430 --> 00:07:00.330
And then the x sub i's are going
to be designated separately.

00:07:00.330 --> 00:07:07.280
We can write x sub 0 is equal to
minus 1, x sub 1 is equal to 0,

00:07:07.280 --> 00:07:12.510
x sub 2 is equal to-- sorry, not
minus 1-- is equal to plus 1.

00:07:12.510 --> 00:07:16.031
And x sub 3 is equal to 2.

00:07:16.031 --> 00:07:18.030
So in that case, that's
what we would get there.

00:07:18.030 --> 00:07:19.450
And we could write
it in that way

00:07:19.450 --> 00:07:22.590
and then get the same values.

00:07:22.590 --> 00:07:24.878
So I think I will stop there.