1
00:00:07,250 --> 00:00:08,960
PROFESSOR: Welcome
back to recitation.

2
00:00:08,960 --> 00:00:11,090
Today what I'd
like to do is work

3
00:00:11,090 --> 00:00:14,360
on using Riemann sums to
approximate definite integrals.

4
00:00:14,360 --> 00:00:16,632
So, what we're
going to do first is

5
00:00:16,632 --> 00:00:18,340
I'm going to let you
work on the problem.

6
00:00:18,340 --> 00:00:19,400
So, let me state the problem.

7
00:00:19,400 --> 00:00:20,900
I'll give you a little
time to work on it.

8
00:00:20,900 --> 00:00:21,940
And then I'll come back.

9
00:00:21,940 --> 00:00:23,620
So the problem is the following.

10
00:00:23,620 --> 00:00:28,010
We want to use four
subintervals and left endpoints

11
00:00:28,010 --> 00:00:30,600
to approximate the
following definite integral:

12
00:00:30,600 --> 00:00:34,150
the integral from minus 1 to
3 of the function x cubed.

13
00:00:34,150 --> 00:00:36,420
And I've drawn a rough
sketch of what x cubed,

14
00:00:36,420 --> 00:00:38,790
y equals x cubed looks
like to help you out

15
00:00:38,790 --> 00:00:40,410
as the starting point.

16
00:00:40,410 --> 00:00:43,630
So why don't you use the four
subintervals and left endpoints

17
00:00:43,630 --> 00:00:45,790
to approximate the
definite integral, then

18
00:00:45,790 --> 00:00:47,750
I'll come back and
show you how I do it.

19
00:00:56,360 --> 00:00:57,600
OK, welcome back.

20
00:00:57,600 --> 00:01:00,496
So what we want to
do, again, is use

21
00:01:00,496 --> 00:01:02,620
Riemann sums to approximate
this definite integral.

22
00:01:02,620 --> 00:01:05,980
And I've given you
the specifications

23
00:01:05,980 --> 00:01:08,070
of four subintervals
and left endpoints.

24
00:01:08,070 --> 00:01:10,229
So I'm going to
draw what these four

25
00:01:10,229 --> 00:01:11,770
subintervals and
their left endpoints

26
00:01:11,770 --> 00:01:13,540
are going to give
us on the graph.

27
00:01:13,540 --> 00:01:16,120
Then we're going to calculate
what the actual value is

28
00:01:16,120 --> 00:01:16,930
of this estimate.

29
00:01:16,930 --> 00:01:18,805
And then I'll show you
a way you can write it

30
00:01:18,805 --> 00:01:20,010
in some different notation.

31
00:01:20,010 --> 00:01:22,456
So let me actually
find some blue chalk.

32
00:01:22,456 --> 00:01:22,956
Excuse me.

33
00:01:22,956 --> 00:01:26,785
So, using blue chalk
I'm going to show you

34
00:01:26,785 --> 00:01:28,910
where the four subintervals
and the left endpoints.

35
00:01:28,910 --> 00:01:31,630
So it's minus 1 to 3,
so I've conveniently

36
00:01:31,630 --> 00:01:35,220
done it so we have
every unit length is

37
00:01:35,220 --> 00:01:36,220
one of the subintervals.

38
00:01:36,220 --> 00:01:38,120
They're all unit length.

39
00:01:38,120 --> 00:01:41,990
And if I want left endpoints,
then my rectangle's heights,

40
00:01:41,990 --> 00:01:44,664
if you remember, are going to
be the output at the left side

41
00:01:44,664 --> 00:01:45,330
of the interval.

42
00:01:45,330 --> 00:01:48,820
So I'm going to designate
the left interval as output.

43
00:01:53,859 --> 00:01:54,900
And what do we have here?

44
00:01:54,900 --> 00:01:57,550
I'm going to now draw
rectangles of those, using those

45
00:01:57,550 --> 00:02:02,590
with the one side on
x-axis and one side

46
00:02:02,590 --> 00:02:07,200
at the height of the
left endpoint output.

47
00:02:07,200 --> 00:02:09,940
So there's one rectangle.

48
00:02:09,940 --> 00:02:12,180
Oh, I almost did
something wrong here.

49
00:02:12,180 --> 00:02:16,080
This left endpoint's
rectangle has no height.

50
00:02:16,080 --> 00:02:18,370
Because the left
endpoint's value is 0.

51
00:02:18,370 --> 00:02:19,220
The output is 0.

52
00:02:22,680 --> 00:02:25,030
There's another rectangle.

53
00:02:25,030 --> 00:02:26,433
And there's the last rectangle.

54
00:02:30,060 --> 00:02:32,460
So, to designate
everything carefully,

55
00:02:32,460 --> 00:02:38,120
I will call this value y_0,
the output value here y_0.

56
00:02:38,120 --> 00:02:39,190
This value will be y_1.

57
00:02:41,594 --> 00:02:43,010
The output value
here will be y_2.

58
00:02:46,480 --> 00:02:48,770
And to show exactly what
do I mean by output value,

59
00:02:48,770 --> 00:02:50,850
here I have enough room
to write the height

60
00:02:50,850 --> 00:02:53,860
is really what it is, is y_3.

61
00:02:53,860 --> 00:02:56,800
So y_0, y_1, y_2 and y_3
designate the heights

62
00:02:56,800 --> 00:02:58,570
of these four rectangles.

63
00:02:58,570 --> 00:03:02,620
And the length of each
rectangle, if you remember,

64
00:03:02,620 --> 00:03:05,150
is really what we
designate as delta x.

65
00:03:05,150 --> 00:03:09,610
So this right here is delta x.

66
00:03:09,610 --> 00:03:13,610
And in this case delta x
is each time equal to 1.

67
00:03:13,610 --> 00:03:15,974
So, remember, formally,
what we want to do

68
00:03:15,974 --> 00:03:17,890
is, if we want to find
this definite integral,

69
00:03:17,890 --> 00:03:22,280
we want to find the area under
the curve between the x-axis

70
00:03:22,280 --> 00:03:23,560
and the function x cubed.

71
00:03:23,560 --> 00:03:25,080
And remember this
is signed area.

72
00:03:25,080 --> 00:03:28,350
So anything below the
x-axis has a negative sign

73
00:03:28,350 --> 00:03:29,170
associated to it.

74
00:03:29,170 --> 00:03:33,830
Anything above the x-axis has a
positive sign associated to it.

75
00:03:33,830 --> 00:03:35,420
So, when I'm trying
to determine this,

76
00:03:35,420 --> 00:03:36,836
when I'm trying
to do an estimate,

77
00:03:36,836 --> 00:03:39,200
I'm finding the areas
of these rectangles.

78
00:03:39,200 --> 00:03:41,720
So my first approximation
that I've given here,

79
00:03:41,720 --> 00:03:44,390
is we want the four subintervals
and the left endpoints.

80
00:03:44,390 --> 00:03:47,220
What we really need
is this delta x,

81
00:03:47,220 --> 00:03:49,520
which is the base,
times the height, which

82
00:03:49,520 --> 00:03:52,950
is y_0, 1, 2 and 3, of
each of these rectangles.

83
00:03:52,950 --> 00:03:56,000
So, to write it
carefully-- well I

84
00:03:56,000 --> 00:03:58,600
guess this won't be the most
careful way we write it--

85
00:03:58,600 --> 00:04:01,510
but I have the base, which
is delta x in each case,

86
00:04:01,510 --> 00:04:08,490
times y_0 plus y_1
plus y_2 plus y_3.

87
00:04:08,490 --> 00:04:10,240
And the only thing I
have to be careful of

88
00:04:10,240 --> 00:04:12,940
is that the area
of the rectangle

89
00:04:12,940 --> 00:04:14,190
here will be negative.

90
00:04:14,190 --> 00:04:15,680
But how do I pick that up?

91
00:04:15,680 --> 00:04:20,040
I pick that up because this
y_0 value, as an actual output,

92
00:04:20,040 --> 00:04:21,450
is negative.

93
00:04:21,450 --> 00:04:23,450
So that actually will be
picked up automatically

94
00:04:23,450 --> 00:04:26,870
by the value of the
output on the function.

95
00:04:26,870 --> 00:04:29,470
So let's evaluate
these things. y_0

96
00:04:29,470 --> 00:04:32,520
should be the value of the
function at x equal negative 1.

97
00:04:32,520 --> 00:04:34,990
When x is negative 1-- the
function, remember, is x cubed,

98
00:04:34,990 --> 00:04:38,690
so I get negative 1 cubed,
which is just negative 1.

99
00:04:38,690 --> 00:04:40,790
So delta x is 1.

100
00:04:40,790 --> 00:04:42,900
y_0 is negative 1.

101
00:04:42,900 --> 00:04:46,620
y_1 is the value the
function at x equals 0,

102
00:04:46,620 --> 00:04:51,450
because my left endpoints
are minus 1, 0, 1 and 2.

103
00:04:51,450 --> 00:04:55,000
So the second one
is 0 is the input.

104
00:04:55,000 --> 00:04:57,190
At x equals 0 I
get 0 as an output.

105
00:04:59,890 --> 00:05:03,950
The y_2 is going to be
this third left endpoint.

106
00:05:03,950 --> 00:05:05,790
That's at x equal 1.

107
00:05:05,790 --> 00:05:08,900
1 cubed is 1.

108
00:05:08,900 --> 00:05:11,180
So I get a 1 there.

109
00:05:11,180 --> 00:05:13,850
And then, the third left
endpoint, I had minus 1,

110
00:05:13,850 --> 00:05:16,000
0, 1, 2.

111
00:05:16,000 --> 00:05:17,950
So the third left endpoint is 2.

112
00:05:17,950 --> 00:05:19,480
2 cubed is 8.

113
00:05:19,480 --> 00:05:23,190
And so y_3 is 8.

114
00:05:23,190 --> 00:05:25,300
Hopefully the subscripts
here didn't confuse you

115
00:05:25,300 --> 00:05:28,500
because I actually was very
close to having the subscripts

116
00:05:28,500 --> 00:05:29,630
represent the x-value.

117
00:05:29,630 --> 00:05:31,838
But that's not in fact what
the subscripts are doing.

118
00:05:31,838 --> 00:05:33,710
The subscripts are
just representing what

119
00:05:33,710 --> 00:05:35,130
interval we are looking at.

120
00:05:35,130 --> 00:05:38,820
So the four intervals are
designated by 0, 1, 2 and 3.

121
00:05:38,820 --> 00:05:42,790
So I get, when I put this all
together I get 1 times-- well,

122
00:05:42,790 --> 00:05:45,790
minus 1 plus 1 is 0-- 1 times 8.

123
00:05:45,790 --> 00:05:48,000
So I get 8.

124
00:05:48,000 --> 00:05:49,730
And if I wanted to
look at the picture,

125
00:05:49,730 --> 00:05:50,870
how is that represented?

126
00:05:50,870 --> 00:05:54,770
Well this rectangle has
base 1 and height 8.

127
00:05:54,770 --> 00:05:56,460
So that has area 8.

128
00:05:56,460 --> 00:05:57,960
This rectangle is
actually a square.

129
00:05:57,960 --> 00:05:59,670
It has base 1, height 1.

130
00:05:59,670 --> 00:06:00,920
And this one is also a square.

131
00:06:00,920 --> 00:06:02,030
It has base 1, height 1.

132
00:06:02,030 --> 00:06:04,290
But because it's below,
it has a minus sign

133
00:06:04,290 --> 00:06:07,370
associated to it, when I think
about it in terms of area.

134
00:06:07,370 --> 00:06:10,740
So I get minus 1 plus 1 plus 8.

135
00:06:10,740 --> 00:06:12,680
And that's where
the 8 comes from.

136
00:06:12,680 --> 00:06:15,760
Now if I wanted to write this
in terms of a Riemann sum,

137
00:06:15,760 --> 00:06:17,960
you see them sometimes
written in this way.

138
00:06:17,960 --> 00:06:24,232
You can think about it in
a formal sum in this way.

139
00:06:24,232 --> 00:06:29,360
You've probably seen this in the
lecture videos more like this.

140
00:06:29,360 --> 00:06:38,160
The sum from i equals 0 to
3 of f of x sub i delta x.

141
00:06:38,160 --> 00:06:40,020
And now, in this
case, we could even

142
00:06:40,020 --> 00:06:41,970
be more deliberate
if we wanted to.

143
00:06:41,970 --> 00:06:45,910
And we could let x sub i be
designated by some value.

144
00:06:45,910 --> 00:06:47,510
We could say,
starting at negative 1

145
00:06:47,510 --> 00:06:50,180
and adding some
number to it each time

146
00:06:50,180 --> 00:06:51,280
based on the interval.

147
00:06:51,280 --> 00:06:54,730
But this is sort
of the easiest way

148
00:06:54,730 --> 00:06:56,430
to write out what
we're interested in.

149
00:06:56,430 --> 00:07:00,330
And then the x sub i's are going
to be designated separately.

150
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,

151
00:07:07,280 --> 00:07:12,510
x sub 2 is equal to-- sorry, not
minus 1-- is equal to plus 1.

152
00:07:12,510 --> 00:07:16,031
And x sub 3 is equal to 2.

153
00:07:16,031 --> 00:07:18,030
So in that case, that's
what we would get there.

154
00:07:18,030 --> 00:07:19,450
And we could write
it in that way

155
00:07:19,450 --> 00:07:22,590
and then get the same values.

156
00:07:22,590 --> 00:07:24,878
So I think I will stop there.