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PROFESSOR: Today we're going
to continue with integration.
9
00:00:24,280 --> 00:00:29,670
And we get to do the-- probably
the most important thing
10
00:00:29,670 --> 00:00:31,090
of this entire course.
11
00:00:31,090 --> 00:00:33,970
Which is appropriately named.
12
00:00:33,970 --> 00:00:50,250
It's called the fundamental
theorem of calculus.
13
00:00:50,250 --> 00:00:54,900
And we'll be abbreviating
it FTC and occasionally I'll
14
00:00:54,900 --> 00:00:58,964
put in a 1 here, because there
will be two versions of it.
15
00:00:58,964 --> 00:01:00,380
But this is the
one that you'll be
16
00:01:00,380 --> 00:01:06,420
using the most in this class.
17
00:01:06,420 --> 00:01:14,540
The fundamental theorem of
calculus says the following.
18
00:01:14,540 --> 00:01:26,875
It says that if F' =
f, so F'(x) = f(x),
19
00:01:26,875 --> 00:01:30,890
there's a capital
F and a little f,
20
00:01:30,890 --> 00:01:45,960
then the integral from a to b
of f(x) is equal to F(b) - F(a).
21
00:01:52,160 --> 00:01:52,940
That's it.
22
00:01:52,940 --> 00:01:55,600
That's the whole theorem.
23
00:01:55,600 --> 00:02:00,120
And you may recognize it.
24
00:02:00,120 --> 00:02:03,880
Before, we had the
notation that F
25
00:02:03,880 --> 00:02:10,020
was the antiderivative,
that is, capital F
26
00:02:10,020 --> 00:02:11,635
was the integral of f(x).
27
00:02:11,635 --> 00:02:12,510
We wrote it this way.
28
00:02:12,510 --> 00:02:14,860
This is this
indefinite integral.
29
00:02:14,860 --> 00:02:17,970
And now we're putting
in definite values.
30
00:02:17,970 --> 00:02:20,260
And we have a connection
between the two
31
00:02:20,260 --> 00:02:22,420
uses of the integral sign.
32
00:02:22,420 --> 00:02:24,530
But with the definite
values, we get real numbers
33
00:02:24,530 --> 00:02:26,220
out instead of a function.
34
00:02:26,220 --> 00:02:29,810
Or a function up to a constant.
35
00:02:29,810 --> 00:02:30,990
So this is it.
36
00:02:30,990 --> 00:02:32,120
This is the formula.
37
00:02:32,120 --> 00:02:35,400
And it's usually also written
with another notation.
38
00:02:35,400 --> 00:02:40,390
So I want to introduce that
notation to you as well.
39
00:02:40,390 --> 00:02:44,950
So there's a new notation here.
40
00:02:44,950 --> 00:02:47,320
Which you'll find
very convenient.
41
00:02:47,320 --> 00:02:51,010
Because we don't always
have to give a letter f
42
00:02:51,010 --> 00:02:52,710
to the functions involved.
43
00:02:52,710 --> 00:02:54,930
So it's an abbreviation.
44
00:02:54,930 --> 00:02:57,610
For right now there'll be
a lot of f's, but anyway.
45
00:02:57,610 --> 00:02:59,660
So here's the abbreviation.
46
00:02:59,660 --> 00:03:04,420
Whenever I have a difference
between a function at two
47
00:03:04,420 --> 00:03:09,430
values, I also can
write this as F(x)
48
00:03:09,430 --> 00:03:12,440
with an a down here
and a b up there.
49
00:03:12,440 --> 00:03:16,550
So that's the
notation that we use.
50
00:03:16,550 --> 00:03:19,910
And you can also, for
emphasis, and this sometimes
51
00:03:19,910 --> 00:03:23,450
turns out to be important, when
there's more than one variable
52
00:03:23,450 --> 00:03:25,750
floating around in the problem.
53
00:03:25,750 --> 00:03:28,200
To specify that
the variable is x.
54
00:03:28,200 --> 00:03:32,270
So this is the same
thing as x = a.
55
00:03:32,270 --> 00:03:34,100
And x = b.
56
00:03:34,100 --> 00:03:36,040
It indicates where
you want to plug in,
57
00:03:36,040 --> 00:03:37,450
what you want to plug in.
58
00:03:37,450 --> 00:03:41,840
And now you take the top
value minus the bottom value.
59
00:03:41,840 --> 00:03:43,380
So F(b) - F(a).
60
00:03:43,380 --> 00:03:50,290
So this is just a notation, and
in that notation, of course,
61
00:03:50,290 --> 00:03:59,420
the theorem can be written
with this set of symbols here.
62
00:03:59,420 --> 00:04:04,160
Equally well.
63
00:04:04,160 --> 00:04:06,250
So let's just give a
couple of examples.
64
00:04:06,250 --> 00:04:08,180
The first example
is the one that we
65
00:04:08,180 --> 00:04:12,260
did last time very laboriously.
66
00:04:12,260 --> 00:04:19,370
If you take the function F(x),
which happens to be x^3 / 3,
67
00:04:19,370 --> 00:04:24,090
then if you differentiate
it, you get, well,
68
00:04:24,090 --> 00:04:25,870
the the factor of 3 cancels.
69
00:04:25,870 --> 00:04:29,440
So you get x^2,
that's the derivative.
70
00:04:29,440 --> 00:04:32,760
And so by the
fundamental theorem,
71
00:04:32,760 --> 00:04:37,290
so this implies by the
fundamental theorem,
72
00:04:37,290 --> 00:04:47,140
that the integral from say, a to
b of x^3 over - sorry, x^2 dx,
73
00:04:47,140 --> 00:04:50,280
that's the derivative here.
74
00:04:50,280 --> 00:04:55,620
This is the function we're
going to use as f(x) here -
75
00:04:55,620 --> 00:05:02,530
is equal to this function
here, F(b) - F(a), that's here.
76
00:05:02,530 --> 00:05:04,220
This function here.
77
00:05:04,220 --> 00:05:13,470
So that's F(b) - F(a), and
that's equal to b^3 / 3 -
78
00:05:13,470 --> 00:05:19,480
a^3 / 3.
79
00:05:19,480 --> 00:05:23,710
Now, in this new
notation, we usually
80
00:05:23,710 --> 00:05:25,060
don't have all of these letters.
81
00:05:25,060 --> 00:05:26,310
All we write is the following.
82
00:05:26,310 --> 00:05:27,970
We write the
integral from a to b,
83
00:05:27,970 --> 00:05:29,470
and I'm going to
do the case 0 to b,
84
00:05:29,470 --> 00:05:31,803
because that was the one that
we actually did last time.
85
00:05:31,803 --> 00:05:35,940
So I'm going to set a = 0 here.
86
00:05:35,940 --> 00:05:39,590
And then, the problem we
were faced last time as this.
87
00:05:39,590 --> 00:05:41,940
And as I said we did
it very laboriously.
88
00:05:41,940 --> 00:05:47,620
But now you can see that we
can do it in ten seconds,
89
00:05:47,620 --> 00:05:48,280
let's say.
90
00:05:48,280 --> 00:05:52,300
Well, the antiderivative
of this is x^3 / 3.
91
00:05:52,300 --> 00:05:55,750
I'm going to evaluate it
at 0 and at b and subtract.
92
00:05:55,750 --> 00:06:00,540
So that's going to
be b^3 / 3 - 0^3 / 3.
93
00:06:00,540 --> 00:06:03,140
Which of course is b^3 / 3.
94
00:06:03,140 --> 00:06:06,010
And that's the end,
that's the answer.
95
00:06:06,010 --> 00:06:08,820
So this is a lot
faster than yesterday.
96
00:06:08,820 --> 00:06:10,860
I hope you'll agree.
97
00:06:10,860 --> 00:06:15,060
And we can dispense with
those elaborate computations.
98
00:06:15,060 --> 00:06:18,460
Although there's a conceptual
reason, a very important one,
99
00:06:18,460 --> 00:06:21,960
for understanding the
procedure that we went through.
100
00:06:21,960 --> 00:06:26,829
Because eventually you're
going to be using integrals
101
00:06:26,829 --> 00:06:28,370
and these quick ways
of doing things,
102
00:06:28,370 --> 00:06:32,280
to solve problems like finding
the volumes of pyramids.
103
00:06:32,280 --> 00:06:34,700
In other words, we're going
to reverse the process.
104
00:06:34,700 --> 00:06:42,100
And so we need to understand
the connection between the two.
105
00:06:42,100 --> 00:06:45,210
I'm going to give a
couple more examples.
106
00:06:45,210 --> 00:06:47,280
And then we'll go on.
107
00:06:47,280 --> 00:06:49,700
So the second
example would be one
108
00:06:49,700 --> 00:06:52,440
that would be quite difficult
to do by this Riemann sum
109
00:06:52,440 --> 00:06:55,310
technique that we
described yesterday.
110
00:06:55,310 --> 00:06:57,100
Although it is possible.
111
00:06:57,100 --> 00:06:59,860
It uses much higher
mathematics to do it.
112
00:06:59,860 --> 00:07:16,360
And that is the area under one
hump of the sine curve, sin x.
113
00:07:16,360 --> 00:07:17,920
Let me just draw
a picture of that.
114
00:07:17,920 --> 00:07:20,990
The curve goes like this, and
we're talking about this area
115
00:07:20,990 --> 00:07:21,490
here.
116
00:07:21,490 --> 00:07:24,370
It starts out at
0, it goes to pi.
117
00:07:24,370 --> 00:07:28,320
That's one hump.
118
00:07:28,320 --> 00:07:33,710
And so the answer is, it's the
integral from 0 to pi of sin
119
00:07:33,710 --> 00:07:37,849
x dx.
120
00:07:37,849 --> 00:07:39,890
And so I need to take the
antiderivative of that.
121
00:07:39,890 --> 00:07:42,970
And that's -cos x.
122
00:07:42,970 --> 00:07:46,650
That's the thing whose
derivative is sin x.
123
00:07:46,650 --> 00:07:49,820
Evaluating it at 0 and pi.
124
00:07:49,820 --> 00:07:52,070
Now, let's do this
one carefully.
125
00:07:52,070 --> 00:07:55,300
Because this is where I see
a lot of arithmetic mistakes.
126
00:07:55,300 --> 00:07:57,690
Even though this is the
easy part of the problem.
127
00:07:57,690 --> 00:08:02,100
It's hard to pay attention
and plug in the right numbers.
128
00:08:02,100 --> 00:08:04,150
And so, let's just pay
very close attention.
129
00:08:04,150 --> 00:08:05,530
I'm plugging in pi.
130
00:08:05,530 --> 00:08:08,444
That's -cos pi.
131
00:08:08,444 --> 00:08:09,360
That's the first term.
132
00:08:09,360 --> 00:08:12,240
And then I'm
subtracting the value
133
00:08:12,240 --> 00:08:19,980
at the bottom, which is -cos 0.
134
00:08:19,980 --> 00:08:22,350
There are already five
opportunities for you
135
00:08:22,350 --> 00:08:24,680
to make a transcription
error or an arithmetic
136
00:08:24,680 --> 00:08:26,740
mistake in what I just did.
137
00:08:26,740 --> 00:08:30,090
And I've seen all five of them.
138
00:08:30,090 --> 00:08:34,670
So the next one is
that this is -(-1).
139
00:08:34,670 --> 00:08:36,810
Minus negative 1, if you like.
140
00:08:36,810 --> 00:08:41,270
And then this is minus,
and here's another -1.
141
00:08:41,270 --> 00:08:44,250
So altogether we have 2.
142
00:08:44,250 --> 00:08:44,920
So that's it.
143
00:08:44,920 --> 00:08:46,810
That's the area.
144
00:08:46,810 --> 00:09:02,960
This area, which is hard
to guess, this is area 2.
145
00:09:02,960 --> 00:09:06,020
The third example
is maybe superfluous
146
00:09:06,020 --> 00:09:10,280
but I'm going to say it anyway.
147
00:09:10,280 --> 00:09:17,740
We can take the integral,
say, from 0 to 1, of x^100.
148
00:09:17,740 --> 00:09:21,260
Any power, now, is
within our power.
149
00:09:21,260 --> 00:09:24,050
So let's do it.
150
00:09:24,050 --> 00:09:32,980
So here we have the
antiderivative is x^101 / 101,
151
00:09:32,980 --> 00:09:36,550
evaluated at 0 and 1.
152
00:09:36,550 --> 00:09:42,050
And that is just 1 / 101.
153
00:09:42,050 --> 00:09:46,520
That's that.
154
00:09:46,520 --> 00:09:49,770
So that's the
fundamental theorem.
155
00:09:49,770 --> 00:09:53,670
Now this, as I say,
harnesses a lot
156
00:09:53,670 --> 00:09:58,110
of what we've already learned,
all about antiderivatives.
157
00:09:58,110 --> 00:10:05,820
Now, I want to give you an
intuitive interpretation.
158
00:10:05,820 --> 00:10:10,120
So let's try that.
159
00:10:10,120 --> 00:10:12,450
We'll talk about a proof
of the fundamental theorem
160
00:10:12,450 --> 00:10:14,170
a little bit later.
161
00:10:14,170 --> 00:10:16,210
It's not actually that hard.
162
00:10:16,210 --> 00:10:22,470
But we'll give an intuitive
reason, interpretation,
163
00:10:22,470 --> 00:10:28,040
if you like.
164
00:10:28,040 --> 00:10:37,670
Of the fundamental theorem.
165
00:10:37,670 --> 00:10:40,330
So this is going to
be one which is not
166
00:10:40,330 --> 00:10:43,990
related to area, but rather
to time and distance.
167
00:10:43,990 --> 00:10:55,600
So we'll consider x(t) is
your position at time t.
168
00:10:55,600 --> 00:11:04,020
And then x'(t), which is dx/dt,
is going to be what we know
169
00:11:04,020 --> 00:11:12,230
as your speed.
170
00:11:12,230 --> 00:11:18,150
And then what the theorem is
telling us is the following.
171
00:11:18,150 --> 00:11:25,580
It's telling us the integral
from a to b of v(t) dt -
172
00:11:25,580 --> 00:11:31,040
so, reading the relationship
- is equal to x (b) - x(a).
173
00:11:35,930 --> 00:11:40,800
And so this is some
kind of cumulative sum
174
00:11:40,800 --> 00:11:45,110
of your velocities.
175
00:11:45,110 --> 00:11:48,440
So let's interpret the
right-hand side first.
176
00:11:48,440 --> 00:11:57,290
This is the distance traveled.
177
00:11:57,290 --> 00:12:03,030
And it's also what you
would read on your odometer.
178
00:12:03,030 --> 00:12:05,550
Right, from the beginning
to the end of the trip.
179
00:12:05,550 --> 00:12:07,540
That's what you would
read on your odometer.
180
00:12:07,540 --> 00:12:19,500
Whereas this is what you would
read on your speedometer.
181
00:12:19,500 --> 00:12:23,210
So this is the interpretation.
182
00:12:23,210 --> 00:12:25,540
Now, I want to just
go one step further
183
00:12:25,540 --> 00:12:27,160
into this
interpretation, to make
184
00:12:27,160 --> 00:12:32,890
the connection with the Riemann
sums that we had yesterday.
185
00:12:32,890 --> 00:12:35,447
Because those are very
complicated to understand.
186
00:12:35,447 --> 00:12:37,280
And I want you to
understand them viscerally
187
00:12:37,280 --> 00:12:39,090
on several different levels.
188
00:12:39,090 --> 00:12:43,280
Because that's how you'll
understand integration better.
189
00:12:43,280 --> 00:12:44,960
The first thing that
I want to imagine,
190
00:12:44,960 --> 00:12:46,876
so we're going to do a
thought experiment now,
191
00:12:46,876 --> 00:12:50,090
which is that you are
extremely obsessive.
192
00:12:50,090 --> 00:12:53,410
And you're driving
your car from time a
193
00:12:53,410 --> 00:12:58,010
to time b, place Q
to place R, whatever.
194
00:12:58,010 --> 00:13:03,900
And you check your
speedometer every second.
195
00:13:03,900 --> 00:13:09,280
OK, so you've read your
speedometer in the i-th second,
196
00:13:09,280 --> 00:13:12,620
and you've read that
you're going at this speed.
197
00:13:12,620 --> 00:13:16,790
Now, how far do you
go in that second?
198
00:13:16,790 --> 00:13:19,530
Well, the answer is
you go this speed
199
00:13:19,530 --> 00:13:22,290
times the time interval,
which in this case
200
00:13:22,290 --> 00:13:24,950
we're imagining as 1 second.
201
00:13:24,950 --> 00:13:25,940
All right?
202
00:13:25,940 --> 00:13:27,980
So this is how far you went.
203
00:13:27,980 --> 00:13:29,230
But this is the time interval.
204
00:13:29,230 --> 00:13:37,070
And this is the
distance traveled
205
00:13:37,070 --> 00:13:46,210
in that-- second number
i, in the i-th second.
206
00:13:46,210 --> 00:13:48,210
The distance traveled in
the i-th second, that's
207
00:13:48,210 --> 00:13:49,640
a total distance you traveled.
208
00:13:49,640 --> 00:13:53,240
Now, what happens if you
go the whole distance?
209
00:13:53,240 --> 00:13:56,860
Well, you travel the sum
of all these distances.
210
00:13:56,860 --> 00:14:00,470
So it's some massive sum, where
n is some ridiculous number
211
00:14:00,470 --> 00:14:01,730
of seconds.
212
00:14:01,730 --> 00:14:04,250
3600 seconds or
something like that.
213
00:14:04,250 --> 00:14:05,070
Whatever it is.
214
00:14:05,070 --> 00:14:09,170
And that's going to turn out
to be very similar to what you
215
00:14:09,170 --> 00:14:11,770
would read on your odometer.
216
00:14:11,770 --> 00:14:14,140
Because during that second,
you didn't change velocity
217
00:14:14,140 --> 00:14:14,990
very much.
218
00:14:14,990 --> 00:14:17,470
So the approximation
that the speed at one
219
00:14:17,470 --> 00:14:21,360
time that you spotted it is
very similar to the speed
220
00:14:21,360 --> 00:14:22,790
during the whole second.
221
00:14:22,790 --> 00:14:24,430
It doesn't change that much.
222
00:14:24,430 --> 00:14:26,250
So this is a pretty
good approximation
223
00:14:26,250 --> 00:14:29,160
to how far you traveled.
224
00:14:29,160 --> 00:14:33,280
And so the sum is a very
realistic approximation
225
00:14:33,280 --> 00:14:34,810
to the entire integral.
226
00:14:34,810 --> 00:14:37,664
Which is denoted this way.
227
00:14:37,664 --> 00:14:39,080
Which, by the
fundamental theorem,
228
00:14:39,080 --> 00:14:43,370
is exactly how far you traveled.
229
00:14:43,370 --> 00:14:49,760
So this is x(b) - x(a) Exactly.
230
00:14:49,760 --> 00:14:55,560
The other one is approximate.
231
00:14:55,560 --> 00:15:08,950
OK, again this is
called a Riemann sum.
232
00:15:08,950 --> 00:15:17,470
All right, so that's the intro
to the fundamental theorem.
233
00:15:17,470 --> 00:15:23,900
And now what I need to do
is extend it just a bit.
234
00:15:23,900 --> 00:15:29,170
And the way I'm going to
extend it is the following.
235
00:15:29,170 --> 00:15:31,140
I'm going to do it on
this example first.
236
00:15:31,140 --> 00:15:35,530
And then we'll do
it more formally.
237
00:15:35,530 --> 00:15:39,200
So here's this example
where we went someplace.
238
00:15:39,200 --> 00:15:44,370
But now I just want to draw
you an additional picture here.
239
00:15:44,370 --> 00:15:49,360
Imagine I start here
and I go over to there
240
00:15:49,360 --> 00:15:54,650
and then I come back.
241
00:15:54,650 --> 00:15:56,100
And maybe even I
do a round trip.
242
00:15:56,100 --> 00:15:58,090
I come back to the same place.
243
00:15:58,090 --> 00:16:01,070
Well, if I come back
to the same place,
244
00:16:01,070 --> 00:16:06,120
then the position is unchanged
from the beginning to the end.
245
00:16:06,120 --> 00:16:08,140
In other words, the
difference is 0.
246
00:16:08,140 --> 00:16:12,632
And the velocity, technically
rather than the speed.
247
00:16:12,632 --> 00:16:14,840
It's the speed to the right
and the speed to the left
248
00:16:14,840 --> 00:16:16,640
maybe are the same,
but one of them
249
00:16:16,640 --> 00:16:18,330
is going in the positive
direction and one of them
250
00:16:18,330 --> 00:16:19,788
is going in the
negative direction,
251
00:16:19,788 --> 00:16:22,090
and they cancel each other.
252
00:16:22,090 --> 00:16:25,340
So if you have this
kind of situation,
253
00:16:25,340 --> 00:16:26,940
we want that to be reflected.
254
00:16:26,940 --> 00:16:28,550
We like that
interpretation and we
255
00:16:28,550 --> 00:16:32,260
want to preserve it even
when-- in the case when
256
00:16:32,260 --> 00:16:35,090
the function v is negative.
257
00:16:35,090 --> 00:16:47,280
And so I'm going to now extend
our notion of integration.
258
00:16:47,280 --> 00:17:02,320
So we'll extend integration
to the case f negative.
259
00:17:02,320 --> 00:17:04,420
Or positive.
260
00:17:04,420 --> 00:17:08,750
In other words, it
could be any sign.
261
00:17:08,750 --> 00:17:10,550
Actually, there's no change.
262
00:17:10,550 --> 00:17:12,430
The formulas are all the same.
263
00:17:12,430 --> 00:17:14,891
We just-- If this v is
going to be positive,
264
00:17:14,891 --> 00:17:16,140
we write in a positive number.
265
00:17:16,140 --> 00:17:18,640
If it's going to be negative,
we write in a negative number.
266
00:17:18,640 --> 00:17:20,270
And we just leave it alone.
267
00:17:20,270 --> 00:17:25,480
And the real-- So here's--
Let me carry out an example
268
00:17:25,480 --> 00:17:29,230
and show you how it works.
269
00:17:29,230 --> 00:17:32,750
I'll carry out the example
on this blackboard up here.
270
00:17:32,750 --> 00:17:33,670
Of the sine function.
271
00:17:33,670 --> 00:17:36,020
But we're going
to try two humps.
272
00:17:36,020 --> 00:17:38,690
We're going to try the
first hump and the one that
273
00:17:38,690 --> 00:17:39,880
goes underneath.
274
00:17:39,880 --> 00:17:40,380
There.
275
00:17:40,380 --> 00:17:43,830
So our example here is
going to be the integral
276
00:17:43,830 --> 00:17:50,910
from 0 to 2pi of sin x dx.
277
00:17:50,910 --> 00:17:55,710
And now, because the fundamental
theorem is so important, and so
278
00:17:55,710 --> 00:17:58,410
useful, and so
convenient, we just
279
00:17:58,410 --> 00:18:01,100
assume that it be true
in this case as well.
280
00:18:01,100 --> 00:18:07,170
So we insist that this is going
to be -cos x, evaluated at 0
281
00:18:07,170 --> 00:18:10,540
and 2pi, with the difference.
282
00:18:10,540 --> 00:18:12,770
Now, when we carry
out that difference,
283
00:18:12,770 --> 00:18:19,320
what we get here is
-cos 2pi - (-cos 0).
284
00:18:24,670 --> 00:18:33,670
Which is -1 - (-1), which is 0.
285
00:18:33,670 --> 00:18:39,650
And the interpretation
of this is the following.
286
00:18:39,650 --> 00:18:43,840
Here's our double hump,
here's pi and here's 2pi.
287
00:18:43,840 --> 00:18:47,590
And all that's happening is that
the geometric interpretation
288
00:18:47,590 --> 00:18:50,440
that we had before of
the area under the curve
289
00:18:50,440 --> 00:18:53,320
has to be taken with a grain
of salt. In other words,
290
00:18:53,320 --> 00:18:56,530
I lied to you before when I said
that the definite integral was
291
00:18:56,530 --> 00:18:57,610
the area under the curve.
292
00:18:57,610 --> 00:18:59,030
It's not.
293
00:18:59,030 --> 00:19:00,790
The definite
integral is the area
294
00:19:00,790 --> 00:19:03,470
under the curve when
it's above the curve,
295
00:19:03,470 --> 00:19:08,500
and it counts negatively
when it's below the curve.
296
00:19:08,500 --> 00:19:12,690
So yesterday, my geometric
interpretation was incomplete.
297
00:19:12,690 --> 00:19:19,180
And really just a plain lie.
298
00:19:19,180 --> 00:19:28,860
So the true geometric
interpretation
299
00:19:28,860 --> 00:19:43,760
of the definite integral
is plus the area
300
00:19:43,760 --> 00:19:49,890
above the axis,
above the x-axis,
301
00:19:49,890 --> 00:20:00,360
minus the area below the x-axis.
302
00:20:00,360 --> 00:20:02,580
As in the picture.
303
00:20:02,580 --> 00:20:04,230
I'm just writing
it down in words,
304
00:20:04,230 --> 00:20:08,870
but you should think
of it visually also.
305
00:20:08,870 --> 00:20:12,820
So that's the setup here.
306
00:20:12,820 --> 00:20:17,530
And now we have the complete
definition of integrals.
307
00:20:17,530 --> 00:20:19,950
And I need to list for you
a bunch of their properties
308
00:20:19,950 --> 00:20:21,740
and how we deal with integrals.
309
00:20:21,740 --> 00:20:25,270
So are there any
questions before we go on?
310
00:20:25,270 --> 00:20:25,770
Yeah.
311
00:20:25,770 --> 00:20:32,240
STUDENT: [INAUDIBLE]
312
00:20:32,240 --> 00:20:39,000
PROFESSOR: Right.
313
00:20:39,000 --> 00:20:42,733
So the question was,
wouldn't the absolute value
314
00:20:42,733 --> 00:20:45,720
of the velocity
function be involved?
315
00:20:45,720 --> 00:20:48,300
The answer is yes.
316
00:20:48,300 --> 00:20:51,730
That is, that's one
question that you could ask.
317
00:20:51,730 --> 00:20:54,430
One question you
could ask is what's
318
00:20:54,430 --> 00:20:57,360
the total distance traveled.
319
00:20:57,360 --> 00:21:00,600
And in that case,
you would keep track
320
00:21:00,600 --> 00:21:05,534
of the absolute value of
the velocity as you said,
321
00:21:05,534 --> 00:21:06,950
whether it's
positive or negative.
322
00:21:06,950 --> 00:21:13,970
And then you would get the
total length of this curve here.
323
00:21:13,970 --> 00:21:18,520
That's, however, not what the
definite integral measures.
324
00:21:18,520 --> 00:21:21,640
It measures the net
distance traveled.
325
00:21:21,640 --> 00:21:24,049
So it's another thing.
326
00:21:24,049 --> 00:21:25,340
In other words, we can do that.
327
00:21:25,340 --> 00:21:27,890
We now have the
tools to do both.
328
00:21:27,890 --> 00:21:35,180
We could also-- So if you
like, the total distance
329
00:21:35,180 --> 00:21:39,550
is equal to the
integral of this.
330
00:21:39,550 --> 00:21:40,730
From a to b.
331
00:21:40,730 --> 00:21:49,200
But the net distance is the
one without the absolute value
332
00:21:49,200 --> 00:21:54,620
signs.
333
00:21:54,620 --> 00:21:57,350
So that's correct.
334
00:21:57,350 --> 00:22:03,950
Other questions?
335
00:22:03,950 --> 00:22:04,590
All right.
336
00:22:04,590 --> 00:22:23,950
So now, let's talk about
properties of integrals.
337
00:22:23,950 --> 00:22:37,630
So the properties of integrals
that I want to mention to you
338
00:22:37,630 --> 00:22:39,030
are these.
339
00:22:39,030 --> 00:22:47,280
The first one doesn't
bear too much comment.
340
00:22:47,280 --> 00:22:53,370
If you take the cumulative
integral of a sum,
341
00:22:53,370 --> 00:22:58,820
you're just trying to get the
sum of the separate integrals
342
00:22:58,820 --> 00:23:01,867
here.
343
00:23:01,867 --> 00:23:03,200
And I won't say much about that.
344
00:23:03,200 --> 00:23:06,660
That's because sums come
out, the because the integral
345
00:23:06,660 --> 00:23:07,790
is a sum.
346
00:23:07,790 --> 00:23:15,330
Incidentally, you know
this strange symbol here,
347
00:23:15,330 --> 00:23:17,302
there's actually a reason
for it historically.
348
00:23:17,302 --> 00:23:18,760
If you go back to
old books, you'll
349
00:23:18,760 --> 00:23:21,880
see that it actually looks
a little bit more like an S.
350
00:23:21,880 --> 00:23:24,530
This capital sigma is a sum.
351
00:23:24,530 --> 00:23:27,170
S for sum, because
everybody in those days
352
00:23:27,170 --> 00:23:28,400
knew Latin and Greek.
353
00:23:28,400 --> 00:23:31,360
And this one is also
an S, but gradually it
354
00:23:31,360 --> 00:23:33,390
was such an important S
that they made a bigger.
355
00:23:33,390 --> 00:23:35,890
And then they stretched it out
and made it a little thinner,
356
00:23:35,890 --> 00:23:40,590
because it didn't fit into
one typesetting space.
357
00:23:40,590 --> 00:23:43,550
And so just for typesetting
reasons it got stretched.
358
00:23:43,550 --> 00:23:45,390
And got a little bit skinny.
359
00:23:45,390 --> 00:23:48,640
Anyway, so it's really an
S. And in fact, in French
360
00:23:48,640 --> 00:23:50,800
they call it sum.
361
00:23:50,800 --> 00:23:53,900
Even though we call it integral.
362
00:23:53,900 --> 00:23:57,690
So it's a sum.
363
00:23:57,690 --> 00:24:00,450
So it's consistent
with sums in this way.
364
00:24:00,450 --> 00:24:09,040
And similarly, similarly we can
factor constants out of sums.
365
00:24:09,040 --> 00:24:20,820
So if you have an integral like
this, the constant factors out.
366
00:24:20,820 --> 00:24:25,140
But definitely don't try to
get a function out of this.
367
00:24:25,140 --> 00:24:27,410
That won't happen.
368
00:24:27,410 --> 00:24:30,470
OK, in other words, c
has to be a constant.
369
00:24:30,470 --> 00:24:41,320
Doesn't depend on x.
370
00:24:41,320 --> 00:24:44,340
The third property.
371
00:24:44,340 --> 00:24:46,710
What do I want to call
the third property here?
372
00:24:46,710 --> 00:24:51,522
I have sort of a preliminary
property, yes, here.
373
00:24:51,522 --> 00:24:52,480
Which is the following.
374
00:24:52,480 --> 00:24:53,730
And I'll draw a picture of it.
375
00:24:53,730 --> 00:24:59,410
I suppose you have three
points along a line.
376
00:24:59,410 --> 00:25:01,300
So then I'm going to
draw a picture of that.
377
00:25:01,300 --> 00:25:03,940
And I'm going to use the
interpretation above the curve,
378
00:25:03,940 --> 00:25:05,564
even though that's
not the whole thing.
379
00:25:05,564 --> 00:25:07,990
So here's a, here's
b and here's c.
380
00:25:07,990 --> 00:25:10,490
And you can see that
the area of this piece,
381
00:25:10,490 --> 00:25:14,080
of the first two pieces
here, when added together,
382
00:25:14,080 --> 00:25:15,820
gives you the area of the whole.
383
00:25:15,820 --> 00:25:19,400
And that's the rule that
I'd like to tell you.
384
00:25:19,400 --> 00:25:23,540
So if you integrate
from a to b, and you
385
00:25:23,540 --> 00:25:28,070
add to that the
integral from b to c,
386
00:25:28,070 --> 00:25:37,344
you'll get the
integral from a to c.
387
00:25:37,344 --> 00:25:39,260
This is going to be just
a little preliminary,
388
00:25:39,260 --> 00:25:41,830
because the rule is a
little better than this.
389
00:25:41,830 --> 00:25:47,970
But I will explain
that in a minute.
390
00:25:47,970 --> 00:25:52,540
The fourth rule is
a very simple one.
391
00:25:52,540 --> 00:25:57,910
Which is that the integral
from a to a of f(x) dx
392
00:25:57,910 --> 00:26:00,624
is equal to 0.
393
00:26:00,624 --> 00:26:02,790
Now, that you can see very
obviously because there's
394
00:26:02,790 --> 00:26:04,140
no area.
395
00:26:04,140 --> 00:26:05,900
No horizontal movement there.
396
00:26:05,900 --> 00:26:08,430
The rectangle is
infinitely thin,
397
00:26:08,430 --> 00:26:10,350
and there's nothing there.
398
00:26:10,350 --> 00:26:12,020
So this is the case.
399
00:26:12,020 --> 00:26:17,860
You can also interpret
it a F(a) - F(a).
400
00:26:17,860 --> 00:26:21,860
So that's also consistent
with our interpretation.
401
00:26:21,860 --> 00:26:24,510
In terms of the fundamental
theorem of calculus.
402
00:26:24,510 --> 00:26:28,020
And it's perfectly reasonable
that this is the case.
403
00:26:28,020 --> 00:26:33,090
Now, the fifth property
is a definition.
404
00:26:33,090 --> 00:26:35,110
It's not really a property.
405
00:26:35,110 --> 00:26:38,080
But it's very important.
406
00:26:38,080 --> 00:26:46,040
The integral from a to b of f(x)
dx equal to minus the integral
407
00:26:46,040 --> 00:26:50,960
from b to a, of f( x) dx.
408
00:26:50,960 --> 00:26:57,490
Now, really, the right-hand side
here is an undefined quantity
409
00:26:57,490 --> 00:26:58,740
so far.
410
00:26:58,740 --> 00:27:02,380
We never said you
could ever do this
411
00:27:02,380 --> 00:27:05,650
where the a is less than the b.
412
00:27:05,650 --> 00:27:09,600
Because this is
working backwards here.
413
00:27:09,600 --> 00:27:12,232
But we just have a convention
that that's the definition.
414
00:27:12,232 --> 00:27:13,690
Whenever we write
down this number,
415
00:27:13,690 --> 00:27:17,120
it's the same as minus
what that number is.
416
00:27:17,120 --> 00:27:20,229
And the reason for
all of these is again
417
00:27:20,229 --> 00:27:22,520
that we want them to be
consistent with the fundamental
418
00:27:22,520 --> 00:27:23,730
theorem of calculus.
419
00:27:23,730 --> 00:27:26,250
Which is the thing that
makes all of this work.
420
00:27:26,250 --> 00:27:33,210
So if you notice the left-hand
side here is F(b) - F(a),
421
00:27:33,210 --> 00:27:36,410
capital F, the
antiderivative of little f.
422
00:27:36,410 --> 00:27:39,420
On the other hand, the
other side is minus,
423
00:27:39,420 --> 00:27:42,050
and if we just ignore that,
we say these are letters,
424
00:27:42,050 --> 00:27:44,633
if we were a machine, we didn't
know which one was bigger than
425
00:27:44,633 --> 00:27:49,980
which, we just plugged them in,
we would get here F(a) - F(b),
426
00:27:49,980 --> 00:27:50,690
over here.
427
00:27:50,690 --> 00:27:53,402
And to make these two
things equal, what we want
428
00:27:53,402 --> 00:27:54,610
is to put that minus sign in.
429
00:27:54,610 --> 00:27:59,620
Now it's consistent.
430
00:27:59,620 --> 00:28:02,610
So again, these
rules are set up so
431
00:28:02,610 --> 00:28:05,420
that everything is consistent.
432
00:28:05,420 --> 00:28:11,190
And now I want to
improve on rule 3 here.
433
00:28:11,190 --> 00:28:15,010
And point out to you - so
let me just go back to rule 3
434
00:28:15,010 --> 00:28:21,030
for a second - that now that
we can evaluate integrals
435
00:28:21,030 --> 00:28:24,920
regardless of the order, we
don't have to have a < b,
436
00:28:24,920 --> 00:28:28,070
b < c in order to make
sense out of this.
437
00:28:28,070 --> 00:28:31,740
We actually have the possibility
of considering integrals
438
00:28:31,740 --> 00:28:34,180
where the a's and the
b's and the c's are
439
00:28:34,180 --> 00:28:36,270
in any order you want.
440
00:28:36,270 --> 00:28:38,630
And in fact, with
this definition,
441
00:28:38,630 --> 00:28:43,140
with this definition 5, 3 works
no matter what the numbers are.
442
00:28:43,140 --> 00:28:44,710
So this is much more convenient.
443
00:28:44,710 --> 00:28:49,570
We don't, this is not necessary.
444
00:28:49,570 --> 00:28:51,070
Not necessary.
445
00:28:51,070 --> 00:28:57,220
It just works
using convention 5.
446
00:28:57,220 --> 00:29:04,140
OK, with 5.
447
00:29:04,140 --> 00:29:08,760
Again, before I go
on, let me emphasize:
448
00:29:08,760 --> 00:29:11,510
we really want to respect
the sign of this velocity.
449
00:29:11,510 --> 00:29:15,110
We really want the net
change in the position.
450
00:29:15,110 --> 00:29:18,180
And we don't want this
absolute value here.
451
00:29:18,180 --> 00:29:20,740
Because otherwise, all of our
formulas are going to mess up.
452
00:29:20,740 --> 00:29:22,260
We won't always
be able to check.
453
00:29:22,260 --> 00:29:24,955
Sometimes you have
letters rather than
454
00:29:24,955 --> 00:29:26,580
actual numbers here,
and you won't know
455
00:29:26,580 --> 00:29:28,160
whether a is bigger than b.
456
00:29:28,160 --> 00:29:31,370
So you'll want to know that
these formulas work and are
457
00:29:31,370 --> 00:29:36,410
consistent in all situations.
458
00:29:36,410 --> 00:29:39,550
OK, I'm going to
trade these again.
459
00:29:39,550 --> 00:29:47,250
In order to preserve the
ordering 1 through 5.
460
00:29:47,250 --> 00:29:54,200
And now I have a sixth property
that I want to talk about.
461
00:29:54,200 --> 00:30:02,010
This one is called estimation.
462
00:30:02,010 --> 00:30:05,470
And it says the following.
463
00:30:05,470 --> 00:30:18,270
If f(x) <= g(x), then the
integral from a to b of f(x) dx
464
00:30:18,270 --> 00:30:23,180
is less than or equal to the
integral from a to b of g(x)
465
00:30:23,180 --> 00:30:28,300
dx.
466
00:30:28,300 --> 00:30:36,860
Now, this one says that if I'm
going more slowly than you,
467
00:30:36,860 --> 00:30:40,840
then you go farther than I do.
468
00:30:40,840 --> 00:30:41,760
OK.
469
00:30:41,760 --> 00:30:43,710
That's all it's saying.
470
00:30:43,710 --> 00:30:47,870
For this one, you'd
better have a < b.
471
00:30:47,870 --> 00:30:49,100
You need it.
472
00:30:49,100 --> 00:30:55,360
Because we flip the signs when
we flip the order of a and b.
473
00:30:55,360 --> 00:30:59,410
So this one, it's essential
that the lower limit be smaller
474
00:30:59,410 --> 00:31:04,600
than the upper limit.
475
00:31:04,600 --> 00:31:06,930
But let me just emphasize,
because we're dealing
476
00:31:06,930 --> 00:31:08,370
with the generalities of this.
477
00:31:08,370 --> 00:31:10,350
Actually if one of
these is negative
478
00:31:10,350 --> 00:31:14,650
and the other one is
negative, then it also works.
479
00:31:14,650 --> 00:31:17,510
This one ends up being, if
f is more negative than g,
480
00:31:17,510 --> 00:31:22,380
then this added up thing is
more negative than that one.
481
00:31:22,380 --> 00:31:25,270
Again, under the assumption
that a is less than b.
482
00:31:25,270 --> 00:31:34,760
So as I wrote it it's
in full generality.
483
00:31:34,760 --> 00:31:37,490
Let's illustrate this one.
484
00:31:37,490 --> 00:31:47,030
And then we have one more
property to learn after that.
485
00:31:47,030 --> 00:31:58,970
So let me give you an
example of estimation.
486
00:31:58,970 --> 00:32:01,970
The example is the same as
one that I already gave you.
487
00:32:01,970 --> 00:32:05,080
But this time, because we
have the tool of integration,
488
00:32:05,080 --> 00:32:11,290
we can just follow our
noses and it works.
489
00:32:11,290 --> 00:32:14,400
I start with the
inequality, so I'm
490
00:32:14,400 --> 00:32:16,090
trying to illustrate
estimation, so I
491
00:32:16,090 --> 00:32:17,673
want to start with
an inequality which
492
00:32:17,673 --> 00:32:19,290
is what the hypothesis is here.
493
00:32:19,290 --> 00:32:21,480
And I'm going to
integrate the inequality
494
00:32:21,480 --> 00:32:22,900
to get this conclusion.
495
00:32:22,900 --> 00:32:25,860
And see what conclusion it is.
496
00:32:25,860 --> 00:32:30,160
The inequality that I want
to take is that e^x >= 1,
497
00:32:30,160 --> 00:32:32,800
for x >= 0.
498
00:32:32,800 --> 00:32:37,620
That's going to be
our starting place.
499
00:32:37,620 --> 00:32:39,070
And now I'm going
to integrate it.
500
00:32:39,070 --> 00:32:40,850
That is, I'm going
to use estimation
501
00:32:40,850 --> 00:32:42,420
to see what that gives.
502
00:32:42,420 --> 00:32:45,100
Well, I'm going to
integrate, say, from 0 to b.
503
00:32:45,100 --> 00:32:50,010
I can't integrate below 0
because it's only true above 0.
504
00:32:50,010 --> 00:32:54,390
This is e^x dx greater than or
equal to the integral from 0
505
00:32:54,390 --> 00:33:01,990
to b of 1 dx.
506
00:33:01,990 --> 00:33:05,980
Alright, let's work out
what each of these is.
507
00:33:05,980 --> 00:33:14,370
The first one, e^x dx, is,
the antiderivative is e^x,
508
00:33:14,370 --> 00:33:16,120
evaluated at 0 and b.
509
00:33:16,120 --> 00:33:18,930
So that's e^b - e^0.
510
00:33:18,930 --> 00:33:23,510
Which is e^b - 1.
511
00:33:23,510 --> 00:33:28,560
The other one,
you're supposed to be
512
00:33:28,560 --> 00:33:32,270
able to get by
the rectangle law.
513
00:33:32,270 --> 00:33:35,840
This is one rectangle
of base b and height 1.
514
00:33:35,840 --> 00:33:37,350
So the answer is b.
515
00:33:37,350 --> 00:33:44,100
Or you can do it by
antiderivatives, but it's b.
516
00:33:44,100 --> 00:33:49,010
That means that our inequality
says if I just combine these
517
00:33:49,010 --> 00:33:55,640
two things together,
that e^b - 1 >= b.
518
00:33:55,640 --> 00:34:02,040
And that's the same
thing as e^b >= 1 + b.
519
00:34:02,040 --> 00:34:05,010
Again, this only
works for b >= 0.
520
00:34:05,010 --> 00:34:10,520
Notice that if b were
negative, this would be a well
521
00:34:10,520 --> 00:34:13,520
defined quantity.
522
00:34:13,520 --> 00:34:18,240
But this estimation
would be false.
523
00:34:18,240 --> 00:34:22,010
We need that the b > 0 in
order for this to make sense.
524
00:34:22,010 --> 00:34:24,590
So this was used.
525
00:34:24,590 --> 00:34:28,640
And that's a good thing, because
this inequality is suspect.
526
00:34:28,640 --> 00:34:32,060
Actually, it turns out to
be true when b is negative.
527
00:34:32,060 --> 00:34:38,830
But we certainly
didn't prove it.
528
00:34:38,830 --> 00:34:42,260
I'm going to just
repeat this process.
529
00:34:42,260 --> 00:34:46,400
So let's repeat it.
530
00:34:46,400 --> 00:34:49,890
Starting from the
inequality, the conclusion,
531
00:34:49,890 --> 00:34:51,390
which is sitting right here.
532
00:34:51,390 --> 00:34:59,970
But I'll write it in a form
e^x >= 1 + x, for x >= 0.
533
00:34:59,970 --> 00:35:02,160
And now, if I
integrate this one,
534
00:35:02,160 --> 00:35:06,245
I get the integral from 0 to b,
e^x dx is greater than or equal
535
00:35:06,245 --> 00:35:12,360
to the integral from
0 to b, (1 + x) dx,
536
00:35:12,360 --> 00:35:16,020
and I remind you that we've
already calculated this one.
537
00:35:16,020 --> 00:35:19,000
This is e^b - 1.
538
00:35:19,000 --> 00:35:21,710
And the other one is
not hard to calculate.
539
00:35:21,710 --> 00:35:25,930
The antiderivative
is x + x^2 / 2.
540
00:35:25,930 --> 00:35:28,690
We're evaluating
that at 0 and b.
541
00:35:28,690 --> 00:35:34,830
So that comes out
to be b + b^2 / 2.
542
00:35:34,830 --> 00:35:43,550
And so our conclusion is that
the left side, which is e^b -
543
00:35:43,550 --> 00:35:48,300
1 >= b + b^2 / 2.
544
00:35:48,300 --> 00:35:51,930
And this is for b >= 0.
545
00:35:51,930 --> 00:36:00,510
And that's the same thing
as e^b >= 1 + b + b^2 / 2.
546
00:36:00,510 --> 00:36:04,050
This one actually is
false for b negative,
547
00:36:04,050 --> 00:36:09,280
so that's something
that you have
548
00:36:09,280 --> 00:36:15,560
to be careful with
the b positive's here.
549
00:36:15,560 --> 00:36:17,350
So you can keep on
going with this,
550
00:36:17,350 --> 00:36:20,360
and you didn't have to think.
551
00:36:20,360 --> 00:36:23,330
And you'll produce a very
interesting polynomial,
552
00:36:23,330 --> 00:36:25,340
which is a good
approximation to e^b.
553
00:36:30,720 --> 00:36:34,520
So that's it for the
basic properties.
554
00:36:34,520 --> 00:36:38,980
Now there's one tricky property
that I need to tell you about.
555
00:36:38,980 --> 00:36:47,920
It's not that tricky,
but it's a little tricky.
556
00:36:47,920 --> 00:37:07,390
And this is change of variables.
557
00:37:07,390 --> 00:37:09,600
Change of variables
in integration,
558
00:37:09,600 --> 00:37:11,220
we've actually already done.
559
00:37:11,220 --> 00:37:14,190
We called that, the last
time we talked about it,
560
00:37:14,190 --> 00:37:23,430
we called it substitution.
561
00:37:23,430 --> 00:37:26,210
And the idea here,
if you may remember,
562
00:37:26,210 --> 00:37:31,550
was that if you're faced
with an integral like this,
563
00:37:31,550 --> 00:37:38,340
you can change it to, if you put
in u = u(x) and you have a du,
564
00:37:38,340 --> 00:37:42,700
which is equal to
u'(x) du-- dx, sorry.
565
00:37:42,700 --> 00:37:45,900
Then you can change the
integral as follows.
566
00:37:45,900 --> 00:37:51,630
This is the same as
g(u(x)) u'(x) dx.
567
00:37:51,630 --> 00:37:58,920
This was the general
procedure for substitution.
568
00:37:58,920 --> 00:38:05,500
What's new today is that we're
going to put in the limits.
569
00:38:05,500 --> 00:38:10,350
If you have a limit here,
u_1, and a limit here, u_2,
570
00:38:10,350 --> 00:38:12,790
you want to know what
the relationship is
571
00:38:12,790 --> 00:38:15,690
between the limits here and
the limits when you change
572
00:38:15,690 --> 00:38:18,780
variables to the new variables.
573
00:38:18,780 --> 00:38:21,290
And it's the simplest
possible thing.
574
00:38:21,290 --> 00:38:25,760
Namely the two limits over here
are in the same relationship
575
00:38:25,760 --> 00:38:29,130
as u(x) is to this
symbol u here.
576
00:38:29,130 --> 00:38:35,830
In other words, u_1 =
u(x_1), and u_2 = u(x_2).
577
00:38:35,830 --> 00:38:39,310
That's what works.
578
00:38:39,310 --> 00:38:42,620
Now there's only
one danger here,
579
00:38:42,620 --> 00:38:48,230
there's one subtlety
which is, this only works
580
00:38:48,230 --> 00:39:02,010
if u' does not change sign.
581
00:39:02,010 --> 00:39:04,680
I've been worrying a little
bit about going backwards
582
00:39:04,680 --> 00:39:06,400
and forwards, and
I allowed myself
583
00:39:06,400 --> 00:39:08,525
to reverse and do all
kinds of stuff, right,
584
00:39:08,525 --> 00:39:09,400
with these integrals.
585
00:39:09,400 --> 00:39:11,560
So we're sort of free to do it.
586
00:39:11,560 --> 00:39:14,560
Well, this is one case where
you want to avoid it, OK?
587
00:39:14,560 --> 00:39:15,850
Just don't do it.
588
00:39:15,850 --> 00:39:17,990
It is possible, actually,
to make sense out of it,
589
00:39:17,990 --> 00:39:21,810
but it's also possible to get
yourself infinitely confused.
590
00:39:21,810 --> 00:39:24,350
So just make sure
that-- Now, it's
591
00:39:24,350 --> 00:39:27,580
OK if u' is always negative,
or always going one way,
592
00:39:27,580 --> 00:39:30,034
so OK if u' is always
positive, you're always
593
00:39:30,034 --> 00:39:31,700
going the other way,
but if you mix them
594
00:39:31,700 --> 00:39:39,960
up you'll get yourself mixed up.
595
00:39:39,960 --> 00:39:46,210
Let me give you an example.
596
00:39:46,210 --> 00:39:54,230
The example will be maybe
close to what we did last time.
597
00:39:54,230 --> 00:39:57,900
When we first did
substitution, I mean.
598
00:39:57,900 --> 00:40:02,160
So the integral from 1 to 2,
this time I'll put in definite
599
00:40:02,160 --> 00:40:09,690
limits, of x^2 plus-- sorry,
maybe I call this x^3. x^3 + 2,
600
00:40:09,690 --> 00:40:17,950
let's say, I don't know,
to the 5th power, x^2 dx.
601
00:40:17,950 --> 00:40:20,310
So this is an example
of an integral
602
00:40:20,310 --> 00:40:25,900
that we would have tried to
handle by substitution before.
603
00:40:25,900 --> 00:40:36,590
And the substitution we would
have used is u = x^3 + 2.
604
00:40:36,590 --> 00:40:38,560
And that's exactly what
we're going to do here.
605
00:40:38,560 --> 00:40:44,710
But we're just going to also
take into account the limits.
606
00:40:44,710 --> 00:40:47,730
The first step, as in any
substitution or change
607
00:40:47,730 --> 00:40:54,070
of variables, is this.
608
00:40:54,070 --> 00:40:57,056
And so we can fill
in the things that we
609
00:40:57,056 --> 00:40:58,180
would have done previously.
610
00:40:58,180 --> 00:41:01,790
Which is that this is the
integral and this is u^5.
611
00:41:01,790 --> 00:41:09,300
And then because this is
3x^2, we see that this is 3.
612
00:41:09,300 --> 00:41:13,130
Sorry, let's write
it the other way.
613
00:41:13,130 --> 00:41:17,480
1/3 du = x^2 dx.
614
00:41:17,480 --> 00:41:20,370
So that's what I'm going to
plug in for this factor here.
615
00:41:20,370 --> 00:41:26,480
So here's 1/3 du,
which replaces that.
616
00:41:26,480 --> 00:41:29,430
But now there's
the extra feature.
617
00:41:29,430 --> 00:41:31,660
The extra feature is the limits.
618
00:41:31,660 --> 00:41:35,160
So here, really in
disguise, because, and now
619
00:41:35,160 --> 00:41:38,280
this is incredibly important.
620
00:41:38,280 --> 00:41:44,070
This is one of the reasons why
we use this notation dx and du.
621
00:41:44,070 --> 00:41:47,120
We want to remind
ourselves which variable
622
00:41:47,120 --> 00:41:49,170
is involved in the integration.
623
00:41:49,170 --> 00:41:52,470
And especially if you're the
one naming the variables,
624
00:41:52,470 --> 00:41:54,760
you may get mixed
up in this respect.
625
00:41:54,760 --> 00:41:59,200
So you must know which variable
is varying between 1 and 2.
626
00:41:59,200 --> 00:42:01,510
And the answer is, it's
x is the one that's
627
00:42:01,510 --> 00:42:04,150
varying between 1 and 2.
628
00:42:04,150 --> 00:42:06,870
So in disguise, even
though I didn't write it,
629
00:42:06,870 --> 00:42:10,260
it was contained in
this little symbol here.
630
00:42:10,260 --> 00:42:11,840
This reminded us which variable.
631
00:42:11,840 --> 00:42:14,140
You'll find this amazingly
important when you
632
00:42:14,140 --> 00:42:16,480
get to multivariable calculus.
633
00:42:16,480 --> 00:42:18,810
When there are many
variables floating around.
634
00:42:18,810 --> 00:42:21,390
So this is an incredibly
important distinction to make.
635
00:42:21,390 --> 00:42:23,300
So now, over here
we have a limit.
636
00:42:23,300 --> 00:42:26,270
But of course it's supposed
to be with respect to u, now.
637
00:42:26,270 --> 00:42:29,490
So we need to calculate what
those corresponding limits are.
638
00:42:29,490 --> 00:42:33,270
And indeed it's just, I plug in
here u_1 is going to be equal
639
00:42:33,270 --> 00:42:37,430
to what I plug in for x = 1,
that's going to be 1^3 + 2,
640
00:42:37,430 --> 00:42:38,760
which is 3.
641
00:42:38,760 --> 00:42:46,390
And then u_2 is 2^3 + 2,
which is equal to 10, right?
642
00:42:46,390 --> 00:42:47,960
8 + 2 = 10.
643
00:42:47,960 --> 00:42:57,700
So this is the integral
from 3 to 10, of u^5 1/3 du.
644
00:42:57,700 --> 00:43:00,380
And now I can
finish the problem.
645
00:43:00,380 --> 00:43:06,390
This is 1/18 u^6, from 3 to 10.
646
00:43:06,390 --> 00:43:10,000
And this is where the
most common mistake occurs
647
00:43:10,000 --> 00:43:12,360
in substitutions of this type.
648
00:43:12,360 --> 00:43:14,850
Which is that if
you ignore this,
649
00:43:14,850 --> 00:43:17,020
and you plug in
these 1 and 2 here,
650
00:43:17,020 --> 00:43:20,090
you think, oh I should just
be putting it at 1 and 2.
651
00:43:20,090 --> 00:43:22,940
But actually, it
should be, the u-value
652
00:43:22,940 --> 00:43:26,400
that we're interested in,
and the lower limit is u = 3
653
00:43:26,400 --> 00:43:29,660
and u = 10 is the upper limit.
654
00:43:29,660 --> 00:43:31,405
So those are suppressed here.
655
00:43:31,405 --> 00:43:35,620
But those are the
ones that we want.
656
00:43:35,620 --> 00:43:37,360
And so, here we go.
657
00:43:37,360 --> 00:43:41,770
It's 1/18 times some ridiculous
number which I won't calculate.
658
00:43:41,770 --> 00:43:44,260
10^6 - - 3^6.
659
00:43:47,820 --> 00:43:48,630
Yes, question.
660
00:43:48,630 --> 00:44:07,250
STUDENT: [INAUDIBLE]
661
00:44:07,250 --> 00:44:10,380
PROFESSOR: So, if
you want to do things
662
00:44:10,380 --> 00:44:16,290
with where you're worrying
about the sign change,
663
00:44:16,290 --> 00:44:20,820
the right strategy is,
what you suggested works.
664
00:44:20,820 --> 00:44:23,090
And in fact I'm going to
do an example right now
665
00:44:23,090 --> 00:44:24,220
on this subject.
666
00:44:24,220 --> 00:44:30,160
But, the right strategy is
to break it up into pieces.
667
00:44:30,160 --> 00:44:34,950
Where u' has one sign
or the other, OK?
668
00:44:34,950 --> 00:44:37,430
Let me show you an example.
669
00:44:37,430 --> 00:44:40,700
Where things go wrong.
670
00:44:40,700 --> 00:44:47,900
And I'll tell you how
to handle it, roughly.
671
00:44:47,900 --> 00:44:55,790
So here's our warning.
672
00:44:55,790 --> 00:45:00,280
Suppose you're integrating
from -1 to 1, x^2 dx.
673
00:45:00,280 --> 00:45:02,810
Here's an example.
674
00:45:02,810 --> 00:45:09,654
And you have the temptation
to plug in u = x^2.
675
00:45:09,654 --> 00:45:11,570
Now, of course, we know
how to integrate this.
676
00:45:11,570 --> 00:45:16,390
But let's just pretend we
were stubborn and wanted
677
00:45:16,390 --> 00:45:19,260
to use substitution.
678
00:45:19,260 --> 00:45:26,610
Then we have du = 2x dx.
679
00:45:26,610 --> 00:45:30,170
And now if I try to
make the correspondence,
680
00:45:30,170 --> 00:45:37,130
notice that the limits
are u_1 = (-1)^2,
681
00:45:37,130 --> 00:45:38,860
that's the bottom limit.
682
00:45:38,860 --> 00:45:40,830
And u_2 is the upper limit.
683
00:45:40,830 --> 00:45:43,450
That's 1^2, that's
also equal to 1.
684
00:45:43,450 --> 00:45:45,250
Both limits are 1.
685
00:45:45,250 --> 00:45:47,750
So this is going from 1 to 1.
686
00:45:47,750 --> 00:45:53,350
And no matter what it is,
we know it's going to be 0.
687
00:45:53,350 --> 00:45:55,610
But we know this is not 0.
688
00:45:55,610 --> 00:45:58,620
This is the integral
of a positive quantity.
689
00:45:58,620 --> 00:46:03,139
And the area under a curve is
going to be a positive area.
690
00:46:03,139 --> 00:46:04,430
So this is a positive quantity.
691
00:46:04,430 --> 00:46:07,300
It can't be 0.
692
00:46:07,300 --> 00:46:12,090
If you actually plug it in,
it looks equally strange.
693
00:46:12,090 --> 00:46:16,160
You put in here this u and then,
so that would be for the u^2.
694
00:46:16,160 --> 00:46:22,110
And then to plug in for dx,
you would write dx = 1/(2x) du.
695
00:46:22,110 --> 00:46:27,760
And then you might
write that as this.
696
00:46:27,760 --> 00:46:31,500
And so what I should put in
here is this quantity here.
697
00:46:31,500 --> 00:46:33,360
Which is a perfectly
OK integral.
698
00:46:33,360 --> 00:46:37,870
And it has a value, I
mean, it's what it is.
699
00:46:37,870 --> 00:46:39,140
It's 0.
700
00:46:39,140 --> 00:46:45,270
So of course this is not true.
701
00:46:45,270 --> 00:46:51,540
And the reason is that
u was equal to x^2,
702
00:46:51,540 --> 00:46:57,340
and u'(x) was equal to 2x, which
was positive for x positive,
703
00:46:57,340 --> 00:47:00,070
and negative for x negative.
704
00:47:00,070 --> 00:47:03,470
And this was the sign change
which causes us trouble.
705
00:47:03,470 --> 00:47:08,160
If we break it off into its
two halves, then it'll be OK
706
00:47:08,160 --> 00:47:09,710
and you'll be able to use this.
707
00:47:09,710 --> 00:47:12,120
Now, there was a mistake.
708
00:47:12,120 --> 00:47:15,540
And this was essentially
what you were saying.
709
00:47:15,540 --> 00:47:19,050
That is, it's possible to see
this happening as you're doing
710
00:47:19,050 --> 00:47:21,680
it if you're very careful.
711
00:47:21,680 --> 00:47:23,610
There's a mistake
in this process,
712
00:47:23,610 --> 00:47:26,290
and the mistake is
in the transition.
713
00:47:26,290 --> 00:47:28,230
This is a mistake here.
714
00:47:28,230 --> 00:47:33,340
Maybe I haven't used
any red yet today,
715
00:47:33,340 --> 00:47:34,940
so I get to use some red here.
716
00:47:34,940 --> 00:47:36,750
Oh boy.
717
00:47:36,750 --> 00:47:38,310
This is not true, here.
718
00:47:38,310 --> 00:47:39,200
This step here.
719
00:47:39,200 --> 00:47:40,470
So why isn't it true?
720
00:47:40,470 --> 00:47:43,500
It's not true for
the standard reason.
721
00:47:43,500 --> 00:47:50,670
Which is that really, x = plus
or minus square root of u.
722
00:47:50,670 --> 00:47:53,930
And if you stick to
one side or the other,
723
00:47:53,930 --> 00:47:55,660
you'll have a coherent
formula for it.
724
00:47:55,660 --> 00:47:58,243
One of them will be the plus and
one of them will be the minus
725
00:47:58,243 --> 00:48:01,980
and it will work out when you
separate it into its pieces.
726
00:48:01,980 --> 00:48:02,880
So you could do that.
727
00:48:02,880 --> 00:48:04,100
But this is a can of worms.
728
00:48:04,100 --> 00:48:06,250
So I avoid this.
729
00:48:06,250 --> 00:48:10,112
And just do it in a place where
the inverse is well defined.
730
00:48:10,112 --> 00:48:11,570
And where the
function either moves
731
00:48:11,570 --> 00:48:13,880
steadily up or steadily down.