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PROFESSOR: Hi.
11
00:00:35,110 --> 00:00:39,070
I thought that today what I'd
like to do is to in a sense
12
00:00:39,070 --> 00:00:42,020
clean up a few little things
that were implied by our last
13
00:00:42,020 --> 00:00:44,630
lecture, things that bothered
me a little bit and which I
14
00:00:44,630 --> 00:00:46,380
thought were worthy
of discussing
15
00:00:46,380 --> 00:00:47,810
in more detail today.
16
00:00:47,810 --> 00:00:51,240
You may recall that last time
we discussed the inverse of
17
00:00:51,240 --> 00:00:52,710
differentiation.
18
00:00:52,710 --> 00:00:55,720
And towards the end of the
lecture, we pointed out that
19
00:00:55,720 --> 00:00:58,920
traditionally this operation
was called
20
00:00:58,920 --> 00:01:00,670
the 'indefinite integral'.
21
00:01:00,670 --> 00:01:03,250
Now, you see, two points
bothered me here.
22
00:01:03,250 --> 00:01:05,709
One is why the word integral?
23
00:01:05,709 --> 00:01:08,770
In other words, why not just
'inverse derivative'?
24
00:01:08,770 --> 00:01:10,940
And we're going to talk about
that in more detail later in
25
00:01:10,940 --> 00:01:12,070
the course.
26
00:01:12,070 --> 00:01:14,860
The other part that bothered me
was the inference of what
27
00:01:14,860 --> 00:01:17,860
do you mean by the 'indefinite'
integral, say, as
28
00:01:17,860 --> 00:01:20,310
opposed to the 'definite'
integral?
29
00:01:20,310 --> 00:01:22,760
In essence then, I think that
when you say indefinite
30
00:01:22,760 --> 00:01:25,690
integral, one probably assumes
that there was something
31
00:01:25,690 --> 00:01:28,260
called the definite integral
that exists.
32
00:01:28,260 --> 00:01:30,930
And in order not to prejudice
anything and not to use the
33
00:01:30,930 --> 00:01:33,660
word definite integral in a
context that I don't really
34
00:01:33,660 --> 00:01:37,940
want it in, I have entitled
today's lecture The "Definite"
35
00:01:37,940 --> 00:01:39,480
Indefinite Integral.
36
00:01:39,480 --> 00:01:41,840
And let's see what this
really means now.
37
00:01:41,840 --> 00:01:46,650
Recall that last time we used
the notation that 'D inverse'
38
00:01:46,650 --> 00:01:48,720
'f of x' meant--
39
00:01:48,720 --> 00:01:49,660
and that was what?
40
00:01:49,660 --> 00:01:52,440
This is notation we used, and
we said in the book but this
41
00:01:52,440 --> 00:01:56,490
was denoted by integral ''f of
x' dx', and that this was
42
00:01:56,490 --> 00:01:58,320
called the indefinite
integral.
43
00:01:58,320 --> 00:02:02,840
And what it meant was it was
a name for all function 'G'
44
00:02:02,840 --> 00:02:06,640
whose derivative with respect
to 'x' was 'f'.
45
00:02:06,640 --> 00:02:09,590
And by the mean value theorem,
we also showed what?
46
00:02:09,590 --> 00:02:14,250
That another name for that
set was simply what?
47
00:02:14,250 --> 00:02:17,080
If we knew one function whose
derivative was 'f', call that
48
00:02:17,080 --> 00:02:18,050
capital 'F'.
49
00:02:18,050 --> 00:02:21,750
Then the family capital 'F of
x' plus 'c' was another name
50
00:02:21,750 --> 00:02:23,600
for this set of functions.
51
00:02:23,600 --> 00:02:27,270
And by way of a review, let's
do a typical problem that we
52
00:02:27,270 --> 00:02:28,950
could tackle last time.
53
00:02:28,950 --> 00:02:32,010
Let's see where the word
'indefinite' comes from.
54
00:02:32,010 --> 00:02:34,500
You see, in a sense, the
indefiniteness comes from the
55
00:02:34,500 --> 00:02:36,380
arbitrary constant 'c'.
56
00:02:36,380 --> 00:02:38,920
You see, let's suppose that we
know that 'f prime of x' is 'x
57
00:02:38,920 --> 00:02:41,440
squared', and we now want
to find out what
58
00:02:41,440 --> 00:02:42,670
'f of x' looks like.
59
00:02:42,670 --> 00:02:44,770
You see the inverse of
differentiation again.
60
00:02:44,770 --> 00:02:46,260
We have the derivative.
61
00:02:46,260 --> 00:02:48,580
Well, we know that one function
whose derivative is
62
00:02:48,580 --> 00:02:51,440
'x squared' is '1/3 x cubed'.
63
00:02:51,440 --> 00:02:54,180
Therefore, any other function
whose derivative is 'x
64
00:02:54,180 --> 00:02:58,260
squared' must differ from '1/3
x cubed' by a constant.
65
00:02:58,260 --> 00:03:00,890
Hence, we know that
'f of x' is '1/3 x
66
00:03:00,890 --> 00:03:03,300
cubed' plus a constant.
67
00:03:03,300 --> 00:03:06,610
And it's this 'c' that makes
this thing called the
68
00:03:06,610 --> 00:03:07,450
'indefinite' integral.
69
00:03:07,450 --> 00:03:09,850
You see, at this particular
stage, 'c' could be any
70
00:03:09,850 --> 00:03:11,810
constant whatsoever, and
this would still
71
00:03:11,810 --> 00:03:13,230
be a correct answer.
72
00:03:13,230 --> 00:03:16,930
We have only got a correct
answer up to a family of
73
00:03:16,930 --> 00:03:19,970
functions which differ
only by a constant.
74
00:03:19,970 --> 00:03:24,680
Now, if I specify one piece of
information, which is called
75
00:03:24,680 --> 00:03:26,100
an 'initial condition',
shall we say.
76
00:03:26,100 --> 00:03:30,130
In other words, let's suppose
I specify that when 'x' is
77
00:03:30,130 --> 00:03:34,680
equal to 'x0', 'f of x' is
equal to 'y sub 0', the
78
00:03:34,680 --> 00:03:37,670
subscript 0 here just being
the abbreviation for a
79
00:03:37,670 --> 00:03:38,360
beginning point.
80
00:03:38,360 --> 00:03:41,540
In other words, I now specify
one additional piece of
81
00:03:41,540 --> 00:03:42,640
information.
82
00:03:42,640 --> 00:03:46,910
Well, if I plug this back into
the equation before, I get
83
00:03:46,910 --> 00:03:53,570
that 'f of x0', which is 'y0',
equals '1/3 'x sub 0' cubed
84
00:03:53,570 --> 00:03:57,670
plus c', from which I can
determine that 'c' is equal to
85
00:03:57,670 --> 00:04:01,240
''y0' minus '1/3 x0 cubed''.
86
00:04:01,240 --> 00:04:04,270
Now, notice that 'x0' and
'y0' are given numbers.
87
00:04:04,270 --> 00:04:07,400
Consequently, 'c' here is
a particular number.
88
00:04:07,400 --> 00:04:10,980
And 'c' being a constant, it
means that whatever 'c' is for
89
00:04:10,980 --> 00:04:14,220
one given value of 'x', it
must be that for all.
90
00:04:14,220 --> 00:04:18,240
So now what I can do is replace
the indefinite 'c' by
91
00:04:18,240 --> 00:04:23,460
the definite specific value
''y0' minus '1/3 x0 cubed''
92
00:04:23,460 --> 00:04:27,380
and arrive at the fact that 'f
of x' is that unique function,
93
00:04:27,380 --> 00:04:31,570
the only function that there
is: '1/3 x cubed' plus 'y0'
94
00:04:31,570 --> 00:04:33,590
minus '1/3 x0 cubed'.
95
00:04:33,590 --> 00:04:36,140
In other words, no more
arbitrary constant in here.
96
00:04:36,140 --> 00:04:37,890
What is this function
'f of x'?
97
00:04:37,890 --> 00:04:42,720
Well, it's the function whose
derivative is 'x squared' and
98
00:04:42,720 --> 00:04:47,370
satisfies the condition that
when 'x' is equal to 'x0', 'y'
99
00:04:47,370 --> 00:04:49,540
is equal to 'y0'.
100
00:04:49,540 --> 00:04:53,260
And if you want to see that in
terms of our usual analogy of
101
00:04:53,260 --> 00:04:56,720
resorting to curve plotting,
observe that what we can say
102
00:04:56,720 --> 00:04:57,790
here is what?
103
00:04:57,790 --> 00:04:59,790
The slope is 'x squared'.
104
00:04:59,790 --> 00:05:02,960
Therefore, the curve has the
form 'y' equals '1/3
105
00:05:02,960 --> 00:05:04,610
x cubed plus c'.
106
00:05:04,610 --> 00:05:07,300
We now specify that the
curve passes through
107
00:05:07,300 --> 00:05:10,400
the point (x0, y0).
108
00:05:10,400 --> 00:05:13,410
That helps us determine
that 'c' is 'y0'
109
00:05:13,410 --> 00:05:15,680
minus '1/3 x0 cubed'.
110
00:05:15,680 --> 00:05:19,710
We put that value 'c' back into
our original equation,
111
00:05:19,710 --> 00:05:24,470
and we wind up with 'y' equals
'1/3 x cubed' plus 'y0' minus
112
00:05:24,470 --> 00:05:26,660
'1/3 x0 cubed'.
113
00:05:26,660 --> 00:05:29,260
And what is that particular
curve?
114
00:05:29,260 --> 00:05:32,550
It's the curve that's
characterized by the fact that
115
00:05:32,550 --> 00:05:38,150
its slope at any point (x , y)
is given by 'x squared' and it
116
00:05:38,150 --> 00:05:41,270
passes through the
point (x0, y0).
117
00:05:41,270 --> 00:05:45,080
In other words, notice then that
if we know the slope of a
118
00:05:45,080 --> 00:05:48,180
curve and we know a point on the
curve, in other words, we
119
00:05:48,180 --> 00:05:50,870
know the slope everywhere and we
know a point that the curve
120
00:05:50,870 --> 00:05:54,180
passes through, it shouldn't
seem too surprising that we
121
00:05:54,180 --> 00:05:58,230
can uniquely locate where the
curve is at any time.
122
00:05:58,230 --> 00:06:00,050
And that's exactly
what this says.
123
00:06:00,050 --> 00:06:04,830
We have a unique expression
for this particular curve.
124
00:06:04,830 --> 00:06:07,620
Let's see if we can't generalize
this result.
125
00:06:07,620 --> 00:06:11,580
And instead of talking about 'f
of x' equals 'x squared',
126
00:06:11,580 --> 00:06:13,120
let's consider the following.
127
00:06:13,120 --> 00:06:15,500
Suppose we just generalize it.
128
00:06:15,500 --> 00:06:20,010
We have that we're given that
'dy dx' is 'f of x' and the
129
00:06:20,010 --> 00:06:24,280
domain of 'f' is the closed
interval from 'a' to 'b'.
130
00:06:24,280 --> 00:06:27,010
And now what we'd like to do is
to find out what function
131
00:06:27,010 --> 00:06:28,840
'y' is specifically.
132
00:06:28,840 --> 00:06:31,590
Well, assume that 'g prime'
is any function whose
133
00:06:31,590 --> 00:06:32,960
derivative is 'f'.
134
00:06:32,960 --> 00:06:35,630
And by the way, this is a
pretty big assumption.
135
00:06:35,630 --> 00:06:37,320
In other words, it's one thing
to say let's assume that we
136
00:06:37,320 --> 00:06:39,240
have a function whose
derivative is 'f'.
137
00:06:39,240 --> 00:06:42,940
But for a given 'f', it may not
be quite that simple to
138
00:06:42,940 --> 00:06:45,070
find what 'G' must be.
139
00:06:45,070 --> 00:06:47,480
And we'll talk about
that in more detail
140
00:06:47,480 --> 00:06:49,130
as our course unfolds.
141
00:06:49,130 --> 00:06:52,530
But it's not always that simple
to find a function
142
00:06:52,530 --> 00:06:53,900
which has a given derivative.
143
00:06:53,900 --> 00:06:55,970
We mentioned that in the
last lecture, too.
144
00:06:55,970 --> 00:06:57,950
At any rate, all we're assuming
now though is that by
145
00:06:57,950 --> 00:07:01,540
hook or by crook, somehow or
other, we know a function 'G'
146
00:07:01,540 --> 00:07:03,440
whose derivative is 'f'.
147
00:07:03,440 --> 00:07:07,820
Then therefore, what we conclude
is that since 'G' has
148
00:07:07,820 --> 00:07:12,540
the same derivative as 'y', that
'y' must be 'G of x' plus
149
00:07:12,540 --> 00:07:14,320
a constant.
150
00:07:14,320 --> 00:07:18,260
Now, here's where the word
'initial condition' comes in.
151
00:07:18,260 --> 00:07:22,040
We're starting this problem
at 'x' equals 'a'.
152
00:07:22,040 --> 00:07:23,420
See, 'x' equals 'a'.
153
00:07:23,420 --> 00:07:26,570
What we assume is the curve has
to be someplace when 'x'
154
00:07:26,570 --> 00:07:27,480
equals 'a'.
155
00:07:27,480 --> 00:07:30,460
Let's call that y-coordinate
'y' of 'a'.
156
00:07:30,460 --> 00:07:34,390
In other words, 'y' of 'a' is
equal to 'G of a' plus 'c',
157
00:07:34,390 --> 00:07:37,810
from which we determine that
'c' is equal to 'y of
158
00:07:37,810 --> 00:07:39,400
a' minus 'G of a'.
159
00:07:39,400 --> 00:07:42,950
You see, the only indefinite
thing in here is the 'y of a'.
160
00:07:42,950 --> 00:07:45,730
Because, you see, all we're
saying is we know the
161
00:07:45,730 --> 00:07:50,520
x-coordinate when 'x' is 'a',
and the curve can then pass
162
00:07:50,520 --> 00:07:53,910
through any point whose
x-coordinate is 'a'.
163
00:07:53,910 --> 00:07:56,300
And all we're saying is let's
call the y-coordinate of that
164
00:07:56,300 --> 00:07:57,930
point 'y of a'.
165
00:07:57,930 --> 00:08:01,740
At any rate, knowing what 'c'
is, we now have that 'y of x'
166
00:08:01,740 --> 00:08:06,580
is 'G of x' plus 'y of
a' minus 'G of a'.
167
00:08:06,580 --> 00:08:10,510
Now, in particular, from this,
you see, notice from this, we
168
00:08:10,510 --> 00:08:12,860
can now find out what
'y of b' is.
169
00:08:12,860 --> 00:08:16,420
Namely, we just replace 'x' by
'b', and we get that 'y of b'
170
00:08:16,420 --> 00:08:20,290
is 'G of b' plus 'y of
a' minus 'G of a'.
171
00:08:20,290 --> 00:08:22,850
And again, the only thing
that's indefinite over
172
00:08:22,850 --> 00:08:24,930
here is 'y of a'.
173
00:08:24,930 --> 00:08:27,800
Now, to get rid of the only
indefinite part, it seems
174
00:08:27,800 --> 00:08:31,190
rather clever that maybe all we
should do is subtract 'y of
175
00:08:31,190 --> 00:08:32,860
a' from 'y to b'.
176
00:08:32,860 --> 00:08:37,230
If we do that, we get that 'y of
b' minus 'y of a' is 'G of
177
00:08:37,230 --> 00:08:40,270
b' minus 'G of a'.
178
00:08:40,270 --> 00:08:44,400
Notice that name for 'y of b'
minus 'y of a' is just the
179
00:08:44,400 --> 00:08:48,280
change in 'y' as 'x' moved
from 'a' to 'b'.
180
00:08:48,280 --> 00:08:53,370
In other words, if we know the
derivative of 'y' with respect
181
00:08:53,370 --> 00:08:58,380
to 'x' and we want to find out
how much 'y' changed by as we
182
00:08:58,380 --> 00:09:02,120
moved from 'x' equals 'a' to 'x'
equals 'b', all we have to
183
00:09:02,120 --> 00:09:07,280
do is to compute 'G of b' minus
'G of a' where 'G' is
184
00:09:07,280 --> 00:09:10,090
'a' function whose derivative
is 'f'.
185
00:09:10,090 --> 00:09:12,930
And by the way, that's all this
notation means over here.
186
00:09:12,930 --> 00:09:17,260
This is read this evaluated at
the upper value minus the
187
00:09:17,260 --> 00:09:21,290
value of this evaluated
at the lower value.
188
00:09:21,290 --> 00:09:23,470
In fact, now I suppose the only
problem that's dangling,
189
00:09:23,470 --> 00:09:26,910
as we said, you know, let 'G'
be the function whose
190
00:09:26,910 --> 00:09:28,090
derivative is 'f'.
191
00:09:28,090 --> 00:09:31,890
There are many functions whose
derivative is 'f' once we know
192
00:09:31,890 --> 00:09:34,110
one function whose derivative
is 'f'.
193
00:09:34,110 --> 00:09:36,420
Which function should
we use over here?
194
00:09:36,420 --> 00:09:39,170
And what I intend to show next
is that it really doesn't make
195
00:09:39,170 --> 00:09:39,800
any difference.
196
00:09:39,800 --> 00:09:42,240
That's how definite
this really is.
197
00:09:42,240 --> 00:09:48,080
Namely, let's suppose that 'H'
is any other function whose
198
00:09:48,080 --> 00:09:49,980
derivative is 'f'.
199
00:09:49,980 --> 00:09:54,920
Well, if we use this particular
expression, my
200
00:09:54,920 --> 00:09:59,330
claim is that the change in 'y'
would be a 'H of b' minus
201
00:09:59,330 --> 00:10:01,510
'H of a', the same way
as we did with 'G'.
202
00:10:01,510 --> 00:10:04,760
And the reason for this is that
if 'H' and 'G' have the
203
00:10:04,760 --> 00:10:07,920
same derivative, then they
differ by a constant.
204
00:10:07,920 --> 00:10:11,570
That means that 'H of x' is 'G
of x' plus some constant,
205
00:10:11,570 --> 00:10:15,090
which I denote by 'c1' to
indicate that this is no
206
00:10:15,090 --> 00:10:16,870
longer an arbitrary constant.
207
00:10:16,870 --> 00:10:19,810
What I'm saying is I've picked
a particular 'H'.
208
00:10:19,810 --> 00:10:22,170
I have the 'G' of the
previous problem.
209
00:10:22,170 --> 00:10:24,100
The difference is some
specific constant
210
00:10:24,100 --> 00:10:25,470
which I call 'c1'.
211
00:10:25,470 --> 00:10:29,240
Well, at any rate, I can now
compute 'H to b', which is
212
00:10:29,240 --> 00:10:33,430
just 'G of b' plus 'c1',
'H of a', which is
213
00:10:33,430 --> 00:10:35,670
'G of a' plus 'c1'.
214
00:10:35,670 --> 00:10:40,320
And if I now subtract, I get
'H of b' minus 'H of a' is
215
00:10:40,320 --> 00:10:44,660
equal to 'G of b' minus 'G of
a' because the constant 'c1'
216
00:10:44,660 --> 00:10:47,040
drops out when I form
the subtraction.
217
00:10:47,040 --> 00:10:52,730
In other words, notice that 'G
of b' minus 'G of a' is a
218
00:10:52,730 --> 00:10:56,510
well-determined number
independent of what function
219
00:10:56,510 --> 00:11:00,170
we choose whose derivative
is 'f'.
220
00:11:00,170 --> 00:11:02,970
And to see what this thing
means pictorially, just
221
00:11:02,970 --> 00:11:04,260
observe the following.
222
00:11:04,260 --> 00:11:06,440
Our geometric interpretation
is this.
223
00:11:06,440 --> 00:11:10,300
Let's suppose that we have the
curve 'y' equals 'G of x' on
224
00:11:10,300 --> 00:11:12,240
the closed interval
from 'a' to 'b'.
225
00:11:12,240 --> 00:11:13,930
And what's the property
of 'G'?
226
00:11:13,930 --> 00:11:17,320
It's derivative is 'f'.
227
00:11:17,320 --> 00:11:19,570
And now we look at 'H of x'.
228
00:11:19,570 --> 00:11:21,230
Now, what is 'H of x'?
229
00:11:21,230 --> 00:11:24,020
'H of x' is 'G of x'
plus a constant.
230
00:11:24,020 --> 00:11:26,570
Graphically, what happens
to a curve when
231
00:11:26,570 --> 00:11:28,340
you add on a constant?
232
00:11:28,340 --> 00:11:32,670
You see, adding on a constant
just raises the curve.
233
00:11:32,670 --> 00:11:37,300
In other words, it displaces the
curve parallel to itself
234
00:11:37,300 --> 00:11:40,380
vertically, in other words, with
respect to the y-axis,
235
00:11:40,380 --> 00:11:43,160
that the constant just
lifts the curve.
236
00:11:43,160 --> 00:11:45,450
In other words then, what we're
saying is that when
237
00:11:45,450 --> 00:11:49,160
you're measuring the change in
'y', observe that it really
238
00:11:49,160 --> 00:11:52,830
makes no difference whether
you're talking about this or
239
00:11:52,830 --> 00:11:54,630
whether you're talking
about this.
240
00:11:54,630 --> 00:11:59,030
Because since these two curves
are parallel, you see, the
241
00:11:59,030 --> 00:12:01,560
displacements over these
intervals are the same.
242
00:12:01,560 --> 00:12:03,740
In other words, all we're really
saying, I guess, is
243
00:12:03,740 --> 00:12:07,160
that these two regions
here are congruent.
244
00:12:07,160 --> 00:12:12,130
In fact, it might be wise in
a sense to pick a new axis
245
00:12:12,130 --> 00:12:20,590
either way here, shift this
thing over, and label this the
246
00:12:20,590 --> 00:12:22,220
'delta y' axis.
247
00:12:22,220 --> 00:12:24,820
And what you're really saying
is the arbitrary constant
248
00:12:24,820 --> 00:12:26,480
doesn't really make
any difference.
249
00:12:26,480 --> 00:12:28,980
Because what you've raised the
curve by at this end, you
250
00:12:28,980 --> 00:12:31,900
raised it at this end, and
consequently, the arbitrary
251
00:12:31,900 --> 00:12:35,520
constant drops out in
determining the change in 'y'.
252
00:12:35,520 --> 00:12:38,790
Well, again, this may
still seem abstract.
253
00:12:38,790 --> 00:12:42,370
Let's give a physical
interpretation to what we're
254
00:12:42,370 --> 00:12:43,990
talking about right now.
255
00:12:43,990 --> 00:12:47,000
Let's suppose now that
we have a particle
256
00:12:47,000 --> 00:12:49,785
moving along the x-axis.
257
00:12:49,785 --> 00:12:53,080
A particle is moving
along the x-axis.
258
00:12:53,080 --> 00:12:56,920
We know that its speed at any
time 't' is given by 't
259
00:12:56,920 --> 00:12:59,120
squared', and we know
that the particles
260
00:12:59,120 --> 00:13:00,540
moves for one second.
261
00:13:00,540 --> 00:13:04,990
In other words, the domain
of 't' here is the closed
262
00:13:04,990 --> 00:13:07,580
interval from 0 to 1.
263
00:13:07,580 --> 00:13:10,720
Now, we know that
'v' is 'dx dt'.
264
00:13:10,720 --> 00:13:16,070
Therefore, since '1/3 t cubed'
has its derivative equal to 't
265
00:13:16,070 --> 00:13:20,840
squared', 'x' must be '1/3
t cubed' plus a constant.
266
00:13:20,840 --> 00:13:21,990
And here's where the
word 'initial
267
00:13:21,990 --> 00:13:23,270
condition' comes in again.
268
00:13:23,270 --> 00:13:26,710
We're starting this problem
when 't' equals 0.
269
00:13:26,710 --> 00:13:30,180
When 't' equals 0, the particle
has to be someplace.
270
00:13:30,180 --> 00:13:33,140
Let's say it's at 'x'
equals 'x sub 0'.
271
00:13:33,140 --> 00:13:35,850
That's 'x of 0', you see.
272
00:13:35,850 --> 00:13:38,920
And if we do this, notice
that when 't' is 0,
273
00:13:38,920 --> 00:13:41,060
'x' is 'x sub 0'.
274
00:13:41,060 --> 00:13:43,860
That says that 'c'
is 'x sub 0'.
275
00:13:43,860 --> 00:13:49,180
Replacing 'c' by 'x sub 0', we
find that the particle moves
276
00:13:49,180 --> 00:13:52,720
according to the rule
'x' equals '1/3 t
277
00:13:52,720 --> 00:13:55,440
cubed' plus 'x sub 0'.
278
00:13:55,440 --> 00:13:59,890
Now, we can find out where the
particle is when 't' equals 1.
279
00:13:59,890 --> 00:14:04,180
When 't' equals 1, 'x'
is '1/3 plus 'x0''.
280
00:14:04,180 --> 00:14:07,960
We know that when the particle
started at 't' equals 0, 'x'
281
00:14:07,960 --> 00:14:10,590
of 0 was 'x sub 0'.
282
00:14:10,590 --> 00:14:13,490
And therefore, the change in
'x', in other words, 'x'
283
00:14:13,490 --> 00:14:17,680
evaluated when 't' equals '1
minus x' evaluated when 't'
284
00:14:17,680 --> 00:14:25,160
equals 0 is simply what? '1/3
plus 'x0' minus 'x0'', or 1/3.
285
00:14:25,160 --> 00:14:27,810
In other words, in this
particular problem, if we
286
00:14:27,810 --> 00:14:31,540
choose as our unit, say, feet
and seconds, if a particle
287
00:14:31,540 --> 00:14:35,670
moves for 1 second according to
the rule that speed in feet
288
00:14:35,670 --> 00:14:39,920
per second is the square of the
time, in 1 second it has
289
00:14:39,920 --> 00:14:43,940
been displaced by a distance
of 1/3 of a foot.
290
00:14:43,940 --> 00:14:46,200
And if I want to draw that for
you, let me show you what's
291
00:14:46,200 --> 00:14:47,350
happening over here.
292
00:14:47,350 --> 00:14:49,860
What we're saying is
here is a particle
293
00:14:49,860 --> 00:14:52,160
moving along the x-axis.
294
00:14:52,160 --> 00:14:56,020
when we start this problem
at time 't' equals 0, the
295
00:14:56,020 --> 00:14:59,510
particle is at some point
'x' equals 'x0'.
296
00:14:59,510 --> 00:15:02,360
We observe that the particle
is moving because of the
297
00:15:02,360 --> 00:15:05,690
velocity 't squared', and that
means it's always moving from
298
00:15:05,690 --> 00:15:07,080
left to right.
299
00:15:07,080 --> 00:15:11,610
And what we find is that when
the time is 1, the particle is
300
00:15:11,610 --> 00:15:15,430
now at 'x0 + 1/3'.
301
00:15:15,430 --> 00:15:19,200
In other words, in this 1
second, the particle has moved
302
00:15:19,200 --> 00:15:21,140
1/3 of a foot.
303
00:15:21,140 --> 00:15:23,670
By the way, to make a little
interesting aside while we're
304
00:15:23,670 --> 00:15:27,080
here, and we'll come back to
this in more detail also into
305
00:15:27,080 --> 00:15:30,610
future lectures, notice that
since 'v' equals 't squared',
306
00:15:30,610 --> 00:15:34,760
when 't' is 0, the
velocity is 0.
307
00:15:34,760 --> 00:15:37,890
And when 't' is 1, since 't
squared' is then 1, the
308
00:15:37,890 --> 00:15:40,140
velocity is 1.
309
00:15:40,140 --> 00:15:43,210
Notice that the particle starts
with 0 velocity, ends
310
00:15:43,210 --> 00:15:47,870
up with a velocity of 1, but the
average velocity is not a
311
00:15:47,870 --> 00:15:50,000
half a foot per second.
312
00:15:50,000 --> 00:15:52,190
In other words, notice that the
particle traveled a total
313
00:15:52,190 --> 00:15:55,460
distance of 1/3 of a
foot in 1 second.
314
00:15:55,460 --> 00:15:58,610
In other words, the average
speed was just 1/3 of a foot
315
00:15:58,610 --> 00:15:59,780
per second.
316
00:15:59,780 --> 00:16:02,880
And you see, notice that in this
particular problem, since
317
00:16:02,880 --> 00:16:05,200
'v' is equal to 't
squared', the
318
00:16:05,200 --> 00:16:07,260
acceleration is not a constant.
319
00:16:07,260 --> 00:16:09,040
The acceleration is '2t'.
320
00:16:09,040 --> 00:16:10,060
It's a variable.
321
00:16:10,060 --> 00:16:15,080
Notice, for example, that the
time is approximately 7/10 of
322
00:16:15,080 --> 00:16:19,500
a second before the particle
would move a half a foot.
323
00:16:19,500 --> 00:16:25,160
See, in other words, notice that
the time has to be 7/10
324
00:16:25,160 --> 00:16:28,250
of a second before the
velocity is 1/2.
325
00:16:28,250 --> 00:16:31,630
Namely, when you square
0.7, you get 0.49.
326
00:16:31,630 --> 00:16:34,870
And notice again the
nonlinearity of this, which,
327
00:16:34,870 --> 00:16:37,080
as I say, we'll come back to
later, but I thought it was
328
00:16:37,080 --> 00:16:39,500
worth mentioning in
passing over here.
329
00:16:39,500 --> 00:16:42,900
At any rate, let's generalize
this velocity problem because
330
00:16:42,900 --> 00:16:45,650
I think it's much easier to
think in terms of distance,
331
00:16:45,650 --> 00:16:48,790
rate, and time than in terms
of slopes of curves.
332
00:16:48,790 --> 00:16:50,360
But let's just take
a look here.
333
00:16:50,360 --> 00:16:51,790
The generalization is this.
334
00:16:51,790 --> 00:16:53,490
Let's oppose that we
have a particle
335
00:16:53,490 --> 00:16:55,870
moving along the x-axis.
336
00:16:55,870 --> 00:16:59,780
Its speed at any time is given
by 'v' equals 'f of t', and
337
00:16:59,780 --> 00:17:03,580
the time interval is from 't'
equals 'a' to 't' equals 'b'.
338
00:17:03,580 --> 00:17:07,640
We assume that capital 'G' is
any function whose derivative
339
00:17:07,640 --> 00:17:11,780
is 'f', and we then can conclude
that therefore 'x'
340
00:17:11,780 --> 00:17:15,440
must be 'G of t' plus
a constant.
341
00:17:15,440 --> 00:17:18,750
We also say OK, initially
that's what?
342
00:17:18,750 --> 00:17:19,960
Not necessarily 0.
343
00:17:19,960 --> 00:17:21,960
Initially is when we
start the problem.
344
00:17:21,960 --> 00:17:26,109
When 't' equals 'a', 'x of
a' is 'G of a' plus 'c'.
345
00:17:26,109 --> 00:17:29,340
'x of t', therefore,
is 'G of t'.
346
00:17:29,340 --> 00:17:34,420
Notice that 'c' is 'x of a'
minus 'G of a', so 'x of t' is
347
00:17:34,420 --> 00:17:38,710
'G of t' plus 'x of
a' minus 'G of a'.
348
00:17:38,710 --> 00:17:43,170
From this, just replacing 't'
by 'b', I find that 'x of b'
349
00:17:43,170 --> 00:17:47,050
is 'G of b' plus 'x of
a' minus 'G of a'.
350
00:17:47,050 --> 00:17:50,140
Again, this is the only
indefinite part here.
351
00:17:50,140 --> 00:17:52,990
Notice, by the way, that this
is mimicking what I did
352
00:17:52,990 --> 00:17:56,310
earlier for 'dy dx', but I
wanted to go through this one
353
00:17:56,310 --> 00:17:59,660
more time in terms of a more
tangible physical example.
354
00:17:59,660 --> 00:18:03,630
At any rate, if I now subtract
'x of a' from this result, I
355
00:18:03,630 --> 00:18:08,140
wind up with that 'x of b' minus
'x of a' is 'G of b'
356
00:18:08,140 --> 00:18:09,480
minus 'G of a'.
357
00:18:09,480 --> 00:18:13,310
In other words, the change in
'x' on this time interval from
358
00:18:13,310 --> 00:18:19,020
'a' to 'b' is determined just
by evaluating 'G of b' minus
359
00:18:19,020 --> 00:18:21,630
'G of a' where 'G'
is any function
360
00:18:21,630 --> 00:18:23,550
whose derivative changes.
361
00:18:23,550 --> 00:18:25,860
See, a rather easy
way of evaluating
362
00:18:25,860 --> 00:18:27,780
this particular thing.
363
00:18:27,780 --> 00:18:31,530
Now, again I should point out
as an interesting physical
364
00:18:31,530 --> 00:18:36,590
aside that this 'delta x' is
a displacement, not a total
365
00:18:36,590 --> 00:18:39,370
distance, and I'll explain
what that means
366
00:18:39,370 --> 00:18:40,850
in terms of an example.
367
00:18:40,850 --> 00:18:44,230
You see, let's observe that in
the problem that I did over
368
00:18:44,230 --> 00:18:47,480
here, in this particular problem
over here, notice that
369
00:18:47,480 --> 00:18:50,200
the velocity was always
positive.
370
00:18:50,200 --> 00:18:53,100
In given problems, the velocity
can oscillate between
371
00:18:53,100 --> 00:18:56,440
positive and negative, meaning
in terms of moving along the
372
00:18:56,440 --> 00:18:58,940
x-axis, the particle
can sometimes move
373
00:18:58,940 --> 00:19:00,080
from left to right.
374
00:19:00,080 --> 00:19:02,740
Other times, it can from
right to left.
375
00:19:02,740 --> 00:19:08,400
But at any rate, what I want
to point out is that the
376
00:19:08,400 --> 00:19:12,350
change in 'x' is a displacement,
a net change.
377
00:19:12,350 --> 00:19:16,060
378
00:19:16,060 --> 00:19:17,430
Let me give you an example.
379
00:19:17,430 --> 00:19:19,500
Let's suppose again
that a particle is
380
00:19:19,500 --> 00:19:21,330
moving along the x-axis.
381
00:19:21,330 --> 00:19:25,710
Now its speed 'v' is given
by 't - 1' where 't' is
382
00:19:25,710 --> 00:19:28,320
between 0 and 2.
383
00:19:28,320 --> 00:19:33,070
Now, the function I'm looking
for, 'x', must differ from a
384
00:19:33,070 --> 00:19:37,260
constant by a constant from any
function whose derivative
385
00:19:37,260 --> 00:19:38,470
is 't - 1'.
386
00:19:38,470 --> 00:19:42,370
In particular, one function
whose derivative is 't - 1' is
387
00:19:42,370 --> 00:19:45,120
'1/2 t squared' minus 't'.
388
00:19:45,120 --> 00:19:48,590
So whatever 'x' is, it
must have what form?
389
00:19:48,590 --> 00:19:53,210
It's '1/2 t squared' minus
't' plus a constant.
390
00:19:53,210 --> 00:19:59,410
Now, I want to compute 'x'
evaluated when 't' is 2, 'x'
391
00:19:59,410 --> 00:20:02,900
evaluated when 't' is 0, and
take this difference because
392
00:20:02,900 --> 00:20:05,910
that will be the change in
'x', you see, as 't'
393
00:20:05,910 --> 00:20:07,880
goes from 0 to 2.
394
00:20:07,880 --> 00:20:11,480
Well, at any rate, you see 'x of
2', just putting 2 in here,
395
00:20:11,480 --> 00:20:15,200
gives me 1/2 of 4, which
is 2, minus 2 plus
396
00:20:15,200 --> 00:20:16,810
'c', which is 'c'.
397
00:20:16,810 --> 00:20:21,440
'x of 0' is 'c', and therefore,
'x of 2' minus 'x
398
00:20:21,440 --> 00:20:23,390
of 0' is 0.
399
00:20:23,390 --> 00:20:28,120
Or written more symbolically,
'delta x' as 't' goes
400
00:20:28,120 --> 00:20:32,150
from 0 to 2 is 0.
401
00:20:32,150 --> 00:20:34,540
In other words, there
is no change in 'x'.
402
00:20:34,540 --> 00:20:36,680
Now, obviously, we don't mean
the particle didn't
403
00:20:36,680 --> 00:20:38,760
move in this case.
404
00:20:38,760 --> 00:20:41,900
For no motion, the velocity
should be 0, but the velocity
405
00:20:41,900 --> 00:20:44,470
is 't - 1'.
406
00:20:44,470 --> 00:20:48,240
Let's see what really happened
in this particular problem.
407
00:20:48,240 --> 00:20:52,140
To do this, I have drawn some
separate graphs over here.
408
00:20:52,140 --> 00:20:55,280
The first one shows 'v'
in terms of 't'.
409
00:20:55,280 --> 00:20:56,870
And what this shows is what?
410
00:20:56,870 --> 00:21:01,230
That when 't' is between 0
and 1, 'v' is negative.
411
00:21:01,230 --> 00:21:05,120
Now, there's nothing imaginary,
again recall, about
412
00:21:05,120 --> 00:21:06,990
a negative velocity.
413
00:21:06,990 --> 00:21:09,090
Remember, that since the
particle is moving along the
414
00:21:09,090 --> 00:21:13,810
x-axis and the positive sense of
the x-axis is from left to
415
00:21:13,810 --> 00:21:16,440
right, a negative velocity means
that the particle is
416
00:21:16,440 --> 00:21:18,540
just moving from
right to left.
417
00:21:18,540 --> 00:21:21,220
So what this says is the
particle was moving from right
418
00:21:21,220 --> 00:21:25,050
to left for the first second
and then from left to right
419
00:21:25,050 --> 00:21:27,010
for the second second.
420
00:21:27,010 --> 00:21:33,590
Also, if we now want to plot the
distance versus the time,
421
00:21:33,590 --> 00:21:37,240
you see, what we see is that
'x' is equal to '1/2 t
422
00:21:37,240 --> 00:21:38,990
squared' minus 't'.
423
00:21:38,990 --> 00:21:41,160
That's this particular shape.
424
00:21:41,160 --> 00:21:45,160
And the plus 'x sub 0' is
our initial condition.
425
00:21:45,160 --> 00:21:47,520
In other words, this tells us
lookit, when we started this
426
00:21:47,520 --> 00:21:51,860
problem at 't' equals 0, the
particle was positioned at 'x'
427
00:21:51,860 --> 00:21:53,350
equals 'x sub 0'.
428
00:21:53,350 --> 00:21:55,820
And notice what this
thing seems to say.
429
00:21:55,820 --> 00:21:58,320
This seems to say that what?
430
00:21:58,320 --> 00:21:59,650
You started here.
431
00:21:59,650 --> 00:22:03,790
Then for the first second, the
displacement was decreasing.
432
00:22:03,790 --> 00:22:07,060
That means you were moving in
the negative direction.
433
00:22:07,060 --> 00:22:10,270
And then you reached a certain
peak here, and when you got
434
00:22:10,270 --> 00:22:13,940
down to this particular point,
you started to come back.
435
00:22:13,940 --> 00:22:17,240
And at the end of the second
second, your displacement was
436
00:22:17,240 --> 00:22:20,090
exactly what it was when
you first started here.
437
00:22:20,090 --> 00:22:24,600
And by the way, let me again
mention, this is not the graph
438
00:22:24,600 --> 00:22:26,740
of how the particle moved.
439
00:22:26,740 --> 00:22:31,210
The particle is moving
horizontally along the x-axis.
440
00:22:31,210 --> 00:22:35,000
This graph is just how the
displacement looks as a
441
00:22:35,000 --> 00:22:37,730
function of time.
442
00:22:37,730 --> 00:22:40,280
In fact, by the way, to see how
we can eliminate the 'x
443
00:22:40,280 --> 00:22:43,000
sub 0' here, I have
drawn one more.
444
00:22:43,000 --> 00:22:46,040
You'll notice that in our
recipe, we said let's compute
445
00:22:46,040 --> 00:22:50,650
'1/2 t squared' minus 't'
as 't' goes from 0 to 2.
446
00:22:50,650 --> 00:22:55,920
That was the special case where
I replaced the x-axis by
447
00:22:55,920 --> 00:22:58,030
the 'delta x' axis.
448
00:22:58,030 --> 00:23:02,230
In other words, the curve 'x'
equals '1/2 t squared' minus
449
00:23:02,230 --> 00:23:05,180
't' would be this curve here.
450
00:23:05,180 --> 00:23:08,610
In other words, when
't' is 0, 'x' is 0.
451
00:23:08,610 --> 00:23:09,550
This says what?
452
00:23:09,550 --> 00:23:12,510
The change in 'x' when you start
the problem is 0 because
453
00:23:12,510 --> 00:23:14,210
the particle hasn't moved yet.
454
00:23:14,210 --> 00:23:14,950
And this says what?
455
00:23:14,950 --> 00:23:18,890
The particle moves, reaches
a minimum value
456
00:23:18,890 --> 00:23:21,400
when 't' equals 1.
457
00:23:21,400 --> 00:23:25,200
And when 't' equals 1, 'x'
equals minus 1/2, and then
458
00:23:25,200 --> 00:23:28,650
comes back to 0 when
't' equals 2.
459
00:23:28,650 --> 00:23:29,990
In other words, physically what
460
00:23:29,990 --> 00:23:31,930
happened here is the following.
461
00:23:31,930 --> 00:23:34,770
We started this problem
at 't' equals 0.
462
00:23:34,770 --> 00:23:38,760
At 't' equals 0, the particle
had a certain displacement, a
463
00:23:38,760 --> 00:23:42,490
certain initial position which
we'll call 'x sub 0'.
464
00:23:42,490 --> 00:23:45,260
And what happened in this
problem is that for the first
465
00:23:45,260 --> 00:23:50,030
second, the particle moved
from right to left.
466
00:23:50,030 --> 00:23:53,130
And when 't' was equal to 1,
the particle was at the
467
00:23:53,130 --> 00:23:57,250
position 'x sub 0' minus 1/2,
which is just another way of
468
00:23:57,250 --> 00:24:01,460
saying it was 1/2 of a unit, or
in this case, 1/2 a foot to
469
00:24:01,460 --> 00:24:03,300
the left of its starting
point.
470
00:24:03,300 --> 00:24:06,910
Then during the next second,
the velocity is positive.
471
00:24:06,910 --> 00:24:11,420
The particle doubles back,
returns to the point 'x0' when
472
00:24:11,420 --> 00:24:16,440
't' equals 2, so that the
displacement, meaning the net
473
00:24:16,440 --> 00:24:20,450
change in position during the
time interval of this problem,
474
00:24:20,450 --> 00:24:24,300
the net distance
traveled is 0.
475
00:24:24,300 --> 00:24:26,880
It started and finished
at the same point.
476
00:24:26,880 --> 00:24:30,460
On the other hand, the total
distance traveled is 1/2 a
477
00:24:30,460 --> 00:24:34,190
foot in this direction, 1/2 a
foot in this direction, so
478
00:24:34,190 --> 00:24:37,140
therefore, a total of 1 foot.
479
00:24:37,140 --> 00:24:40,760
Well, you see, I thought this
was a worthwhile discussion to
480
00:24:40,760 --> 00:24:42,970
have on distance,
rate, and time.
481
00:24:42,970 --> 00:24:46,070
But I also thought another thing
was rather important.
482
00:24:46,070 --> 00:24:51,160
Did you notice that except for
my mentioning at the beginning
483
00:24:51,160 --> 00:24:55,010
of our lesson that the inverse
derivative is called the
484
00:24:55,010 --> 00:24:58,830
indefinite integral, that the
so-called integral sign has
485
00:24:58,830 --> 00:25:01,600
never appeared in anything
that I've done.
486
00:25:01,600 --> 00:25:03,550
In other words, to emphasize
what I was talking about in
487
00:25:03,550 --> 00:25:06,960
the last lecture, observe that
I could get by wonderfully
488
00:25:06,960 --> 00:25:09,720
without ever having heard of
an integral sign, without
489
00:25:09,720 --> 00:25:13,280
having heard of the phrase
indefinite integral.
490
00:25:13,280 --> 00:25:16,980
At any rate, though, let me ask
the following query, so to
491
00:25:16,980 --> 00:25:21,040
speak, to sort of lead into
a summation point of view.
492
00:25:21,040 --> 00:25:27,070
Once we invent the notation
integral ''f of x' dx' to
493
00:25:27,070 --> 00:25:31,050
denote the set of all functions
'G' whose derivative
494
00:25:31,050 --> 00:25:37,650
is 'f', why not invent a new
symbol, namely, what?
495
00:25:37,650 --> 00:25:41,260
We'll still use this integral
sign, but now put that lower
496
00:25:41,260 --> 00:25:43,420
and upper bound on this thing.
497
00:25:43,420 --> 00:25:46,530
We'll call this the definite
integral ''f of x' dx', or the
498
00:25:46,530 --> 00:25:48,590
definite indefinite integral--
499
00:25:48,590 --> 00:25:50,910
I put these limits
on, 'a' to 'b'--
500
00:25:50,910 --> 00:25:55,800
to denote 'G of b' minus 'G of
a' where 'G prime' is any
501
00:25:55,800 --> 00:25:57,710
function whose derivative
is 'f'.
502
00:25:57,710 --> 00:26:00,630
In other words, to put this in
still another perspective,
503
00:26:00,630 --> 00:26:06,310
remember the 'G' was a member
of the family described by
504
00:26:06,310 --> 00:26:08,250
interval ''f of x' dx'.
505
00:26:08,250 --> 00:26:11,110
To indicate that I'm going to
compute 'G of b', why not put
506
00:26:11,110 --> 00:26:13,690
a 'b' on top of this
integral sign?
507
00:26:13,690 --> 00:26:16,800
And then to indicate that I'm
going to subtract from that 'G
508
00:26:16,800 --> 00:26:21,020
of a', why not denote 'G of a'
by the same integral sign with
509
00:26:21,020 --> 00:26:22,770
an a underneath it?
510
00:26:22,770 --> 00:26:26,400
And then symbolically, I can
think of 'G of b' minus 'G of
511
00:26:26,400 --> 00:26:31,200
a' as being this expression
here, which I will then
512
00:26:31,200 --> 00:26:33,040
abbreviate as--
513
00:26:33,040 --> 00:26:36,090
you see, in other words, I'll
take this thing and say OK,
514
00:26:36,090 --> 00:26:41,030
this is now an abbreviation
for what we have here.
515
00:26:41,030 --> 00:26:43,020
In other words, that when you
see this thing, which we'll
516
00:26:43,020 --> 00:26:46,690
call the definite indefinite
integral, ''f of x' dx' from
517
00:26:46,690 --> 00:26:50,200
'a' to 'b', this is just an
abbreviation for what?
518
00:26:50,200 --> 00:26:55,370
'G of x', 'G of b' minus 'G of
a' where 'G' is a function
519
00:26:55,370 --> 00:26:57,520
whose derivative is 'f'.
520
00:26:57,520 --> 00:26:59,980
And with this in mind
now, I can summarize
521
00:26:59,980 --> 00:27:01,710
our lecture as follows.
522
00:27:01,710 --> 00:27:07,330
Suppose I know that 'dy dx' is
'f of x', where 'x' is defined
523
00:27:07,330 --> 00:27:10,200
on the closed interval
from 'a' to 'b'.
524
00:27:10,200 --> 00:27:14,750
Then the change in 'y' on this
interval, in other words, the
525
00:27:14,750 --> 00:27:19,080
change in 'y' as 'x' goes from
'a' to 'b' is given by the
526
00:27:19,080 --> 00:27:25,430
symbol integral 'a' to 'b' ''f
of x' dx' where this symbol is
527
00:27:25,430 --> 00:27:29,720
defined to be 'G of x' evaluated
between 'x' equals
528
00:27:29,720 --> 00:27:33,580
'a' and 'x' equals 'b', meaning
'G of b' minus 'G of
529
00:27:33,580 --> 00:27:38,700
a', where 'G' is any function
whose derivative is 'f'.
530
00:27:38,700 --> 00:27:41,810
Notice again then, in closing,
that this particular symbol
531
00:27:41,810 --> 00:27:45,530
here, as far as I'm concerned,
is artificial.
532
00:27:45,530 --> 00:27:47,430
I don't really need it.
533
00:27:47,430 --> 00:27:50,380
And whereas it might have been
unfortunate to invent the
534
00:27:50,380 --> 00:27:55,400
phrase indefinite integral, it
might be even more unfortunate
535
00:27:55,400 --> 00:27:58,720
to have two words sounding
so much alike that
536
00:27:58,720 --> 00:28:00,120
are completely different.
537
00:28:00,120 --> 00:28:04,600
In other words, notice that when
I write the integral sign
538
00:28:04,600 --> 00:28:08,040
from 'a' to 'b' ''f of x'
dx', what this thing
539
00:28:08,040 --> 00:28:09,480
denotes is a number.
540
00:28:09,480 --> 00:28:11,120
It's a change in 'y'.
541
00:28:11,120 --> 00:28:15,190
Without the two numbers here,
'a' and 'b', this denotes a
542
00:28:15,190 --> 00:28:18,660
set of functions, namely, all
functions whose derivative
543
00:28:18,660 --> 00:28:21,490
with respect to 'x'
is 'f of x'.
544
00:28:21,490 --> 00:28:24,700
Well, anyway, I hope that this
straightens out, if nothing
545
00:28:24,700 --> 00:28:27,850
more, the difference between
the words definite and
546
00:28:27,850 --> 00:28:28,910
indefinite.
547
00:28:28,910 --> 00:28:32,000
And we shall return to
the subject called--
548
00:28:32,000 --> 00:28:35,590
or turn to the subject called
integral calculus and bring
549
00:28:35,590 --> 00:28:37,260
this up in more detail
from a different
550
00:28:37,260 --> 00:28:39,640
context in a little while.
551
00:28:39,640 --> 00:28:42,860
In the meantime, we'll have
a little sojourn into the
552
00:28:42,860 --> 00:28:46,560
circular functions and revisit
trigonometry for a few days
553
00:28:46,560 --> 00:28:49,710
while we allow this idea
here to set more
554
00:28:49,710 --> 00:28:50,690
firmly in your mind.
555
00:28:50,690 --> 00:28:52,380
So until next time, goodbye.
556
00:28:52,380 --> 00:28:55,130
557
00:28:55,130 --> 00:28:58,330
Funding for the publication of
this video was provided by the
558
00:28:58,330 --> 00:29:02,380
Gabriella and Paul Rosenbaum
Foundation.
559
00:29:02,380 --> 00:29:06,560
Help OCW continue to provide
free and open access to MIT
560
00:29:06,560 --> 00:29:10,760
courses by making a donation
at ocw.mit.edu/donate.
561
00:29:10,760 --> 00:29:15,498