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HERBERT GROSS: Hi, our topic
today sounds a little bit
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00:00:36,330 --> 00:00:37,870
naughty at first glance.
12
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It's called improper
integrals.
13
00:00:40,860 --> 00:00:43,780
And rather than give a long
speech about why they're
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00:00:43,780 --> 00:00:48,015
called improper, let me try and
start our lecture a little
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00:00:48,015 --> 00:00:51,370
bit different from our usual
approach, by giving you a
16
00:00:51,370 --> 00:00:54,200
little bit of a mathematical
calculus riddle.
17
00:00:54,200 --> 00:00:56,010
Namely, let me just say this.
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00:00:56,010 --> 00:00:59,110
We'll call our lecture today
improper integrals, and we'll
19
00:00:59,110 --> 00:01:02,150
start off by saying
find the flaw.
20
00:01:02,150 --> 00:01:04,714
I would like to integrate
'dx' over 'x squared'
21
00:01:04,714 --> 00:01:06,760
from minus 1 to 1.
22
00:01:06,760 --> 00:01:09,880
Using the first fundamental
theorem of integral calculus,
23
00:01:09,880 --> 00:01:14,800
I say this is 'G of 1' minus 'G
of minus 1' where 'G prime'
24
00:01:14,800 --> 00:01:16,910
is any function whose
derivative
25
00:01:16,910 --> 00:01:18,880
is '1 over 'x squared''.
26
00:01:18,880 --> 00:01:23,840
Well, in particular, do I know a
function whose derivative is
27
00:01:23,840 --> 00:01:25,230
'1 over 'x squared''?
28
00:01:25,230 --> 00:01:31,550
Well, I guess minus
'1 over x'.
29
00:01:31,550 --> 00:01:39,070
In other words, the derivative
of 'x' to the minus 1 is minus
30
00:01:39,070 --> 00:01:40,360
'x' to the minus 2.
31
00:01:40,360 --> 00:01:42,560
With the minus in front
that makes it plus.
32
00:01:42,560 --> 00:01:43,880
'x' to the minus 2.
33
00:01:43,880 --> 00:01:45,900
That's plus '1 over
'x squared''.
34
00:01:45,900 --> 00:01:47,950
In any event, I can then
compute that 'G
35
00:01:47,950 --> 00:01:49,660
of 1' is minus 1.
36
00:01:49,660 --> 00:01:51,720
'G of minus 1' is 1.
37
00:01:51,720 --> 00:01:56,390
Therefore, 'G of 1' minus 'G of
minus 1' is minus 1 minus
38
00:01:56,390 --> 00:01:58,190
1, is minus 2.
39
00:01:58,190 --> 00:02:03,260
Therefore, the integral from
minus 1 to 1, 'dx' over 'x
40
00:02:03,260 --> 00:02:06,030
squared', is minus 2.
41
00:02:06,030 --> 00:02:09,550
And the question is,
find the flaw.
42
00:02:09,550 --> 00:02:11,090
Find the flaw.
43
00:02:11,090 --> 00:02:14,660
And since again, we're on
limited time, let me, if you
44
00:02:14,660 --> 00:02:18,420
haven't already discovered the
flaw, at least point out why
45
00:02:18,420 --> 00:02:20,550
you should be suspicious
of a flaw.
46
00:02:20,550 --> 00:02:24,950
Remember, we visualized this
as an area under a curve.
47
00:02:24,950 --> 00:02:27,510
In other words, in a way, the
base of our region is the
48
00:02:27,510 --> 00:02:30,480
closed interval from minus 1
to 1, and the height of our
49
00:02:30,480 --> 00:02:33,490
region at any point is given
by '1 over 'x squared''.
50
00:02:33,490 --> 00:02:35,870
Well, look at '1 over
'x squared''.
51
00:02:35,870 --> 00:02:38,390
Certainly, '1 over 'x squared''
can't be negative
52
00:02:38,390 --> 00:02:40,820
because for any real number
'x', its square is
53
00:02:40,820 --> 00:02:41,710
non-negative.
54
00:02:41,710 --> 00:02:44,250
And one over a non-negative
number is certainly
55
00:02:44,250 --> 00:02:45,240
non-negative.
56
00:02:45,240 --> 00:02:48,870
In particular then, since my
region never goes below the
57
00:02:48,870 --> 00:02:52,700
x-axis, I would expect that the
area represented by this
58
00:02:52,700 --> 00:02:54,200
should be positive.
59
00:02:54,200 --> 00:02:56,960
In other words, this is the sum
of positive numbers taken
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00:02:56,960 --> 00:02:57,830
in the limit.
61
00:02:57,830 --> 00:03:01,110
In other words, the fact that
the integrand is non-negative
62
00:03:01,110 --> 00:03:04,100
tells me that whenever the
integral turns out to be, it
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00:03:04,100 --> 00:03:09,470
must be at least as big as 0,
and minus 2 does not fulfill
64
00:03:09,470 --> 00:03:11,230
that criteria.
65
00:03:11,230 --> 00:03:14,600
In other words, something must
be wrong here because we get a
66
00:03:14,600 --> 00:03:18,280
negative answer in a situation
where a negative answer is
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00:03:18,280 --> 00:03:19,580
preposterous.
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00:03:19,580 --> 00:03:22,590
And the reason that I wanted
to pick this approach is I
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00:03:22,590 --> 00:03:25,400
think that the best way to
motivate that something is
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00:03:25,400 --> 00:03:29,400
wrong is precisely by winding up
with a preposterous answer
71
00:03:29,400 --> 00:03:32,250
in a situation where we know
the answer is preposterous.
72
00:03:32,250 --> 00:03:36,020
Then we know we must be on our
guard to see what went wrong.
73
00:03:36,020 --> 00:03:37,180
And what did go wrong?
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00:03:37,180 --> 00:03:39,080
Well, the key point is this.
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00:03:39,080 --> 00:03:41,590
And again, it's something that
we've said many, many times.
76
00:03:41,590 --> 00:03:44,290
But I think it takes a counter
example before it
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00:03:44,290 --> 00:03:45,560
really sinks in.
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00:03:45,560 --> 00:03:47,900
People are always quoting
the first
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00:03:47,900 --> 00:03:49,780
fundamental theorem in general.
80
00:03:49,780 --> 00:03:52,200
It's easy to remember that the
integral from 'a' to 'b', ''f
81
00:03:52,200 --> 00:03:55,530
of x' dx' is 'G of b' minus
'G of a', where 'G
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00:03:55,530 --> 00:03:56,860
prime' equals 'f'.
83
00:03:56,860 --> 00:04:00,600
But the thing that we must be
extremely careful of is that
84
00:04:00,600 --> 00:04:04,970
this definition requires that
'f' be at least piecewise
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00:04:04,970 --> 00:04:07,500
continuous on the interval
from 'a' to 'b'.
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00:04:07,500 --> 00:04:09,810
Remember when we took the limit
and put the squeeze on,
87
00:04:09,810 --> 00:04:13,150
we required that there be
continuity, so we could pass
88
00:04:13,150 --> 00:04:14,040
to the limit.
89
00:04:14,040 --> 00:04:16,730
The important point is in this
particular example that we're
90
00:04:16,730 --> 00:04:21,070
dealing with, observe that '1
over 'x squared'' is infinite.
91
00:04:21,070 --> 00:04:23,970
I write this in quotation marks
to point out you have a
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00:04:23,970 --> 00:04:27,060
1/0 form, which we usually
abbreviate as infinity.
93
00:04:27,060 --> 00:04:29,120
In other words, '1 over 'x
squared'' increases without
94
00:04:29,120 --> 00:04:31,380
bound as 'x' approaches 0.
95
00:04:31,380 --> 00:04:35,600
And notice that 0 was within the
limits of our integration
96
00:04:35,600 --> 00:04:37,260
from minus 1 to 1.
97
00:04:37,260 --> 00:04:40,490
At any rate, without further
ado, this leads to a
98
00:04:40,490 --> 00:04:41,670
fundamental definition.
99
00:04:41,670 --> 00:04:43,900
Which in terms of today's
lesson, I call
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00:04:43,900 --> 00:04:45,260
definition number 1.
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00:04:45,260 --> 00:04:49,460
And that is the integral from
'a' to 'b', 'f of x dx is
102
00:04:49,460 --> 00:04:51,280
called 'improper'.
103
00:04:51,280 --> 00:04:53,930
And there are going to be two
types of improper integrals as
104
00:04:53,930 --> 00:04:55,450
we'll find out very shortly.
105
00:04:55,450 --> 00:05:00,050
But it's called improper of the
first kind, if and only if
106
00:05:00,050 --> 00:05:04,730
'f' is infinite for at least one
'c' in the interval from
107
00:05:04,730 --> 00:05:05,700
'a' to 'b'.
108
00:05:05,700 --> 00:05:09,190
And again, let me caution you
about this and I'll summarize
109
00:05:09,190 --> 00:05:11,260
at the end of the lecture
again to remind you.
110
00:05:11,260 --> 00:05:14,340
If you look at this particular
integral, it doesn't look at
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00:05:14,340 --> 00:05:16,040
all suspicious.
112
00:05:16,040 --> 00:05:19,620
In fact, that's where the
trap came in in this
113
00:05:19,620 --> 00:05:20,790
problem over here.
114
00:05:20,790 --> 00:05:23,630
The reason that we wound up with
minus 2 is that we just
115
00:05:23,630 --> 00:05:26,430
integrated '1 over 'x squared'',
forgetting that
116
00:05:26,430 --> 00:05:30,100
there was a point at which the
integrand was infinite in the
117
00:05:30,100 --> 00:05:32,050
given interval.
118
00:05:32,050 --> 00:05:34,640
And so the question that comes
up is, how shall we handle
119
00:05:34,640 --> 00:05:37,100
this if we have an
infinity in here?
120
00:05:37,100 --> 00:05:40,120
And the answer is, let's work
around the point at which the
121
00:05:40,120 --> 00:05:43,700
integral or the integrand
is infinite.
122
00:05:43,700 --> 00:05:45,720
Let's see what that means.
123
00:05:45,720 --> 00:05:47,320
And this again, is a
continuation of the
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00:05:47,320 --> 00:05:48,950
definition, I guess.
125
00:05:48,950 --> 00:05:51,650
If 'c' is the only point in the
closed interval from 'a'
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00:05:51,650 --> 00:05:54,930
to 'b' at which 'f' is infinite,
we defined the
127
00:05:54,930 --> 00:05:57,780
integral from 'a' to 'b',
''f of x' dx' to be
128
00:05:57,780 --> 00:05:59,130
the following limit.
129
00:05:59,130 --> 00:06:01,330
It's the limit as
'h' approaches 0
130
00:06:01,330 --> 00:06:03,190
through positive values.
131
00:06:03,190 --> 00:06:06,650
The definite integral from
'a' to 'c' minus 'h'.
132
00:06:06,650 --> 00:06:09,700
In other words, you're stopping
somewhere before you
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00:06:09,700 --> 00:06:13,270
get to this bad point,
evaluating ''f of x' dx'.
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00:06:13,270 --> 00:06:16,200
And there's no harm done in here
because you see, 'f of x'
135
00:06:16,200 --> 00:06:19,540
is infinite only at 'x' equals
'c' in this problem.
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00:06:19,540 --> 00:06:23,130
Consequently, if you stop prior
to 'c', 'f of x' is at
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00:06:23,130 --> 00:06:25,720
least piecewise continuous
every place in here.
138
00:06:25,720 --> 00:06:28,860
Then you jump over the bad
point, pick up again at 'c
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00:06:28,860 --> 00:06:33,230
plus h' and say plus the
integral from 'c plus h' to
140
00:06:33,230 --> 00:06:34,910
'b', ''f of x' dx'.
141
00:06:34,910 --> 00:06:38,630
This limit as 'h' approaches
0 through positive values.
142
00:06:38,630 --> 00:06:41,250
Now again, this may sound
very abstract.
143
00:06:41,250 --> 00:06:44,050
So to give this a more realistic
meaning to you,
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00:06:44,050 --> 00:06:47,680
let's interpret this again
in terms of an area.
145
00:06:47,680 --> 00:06:49,730
What we're saying is
this pictorially.
146
00:06:49,730 --> 00:06:53,500
We have a function which is
piecewise continuous at least
147
00:06:53,500 --> 00:06:56,370
on the entire interval from
'a' to 'b', except at the
148
00:06:56,370 --> 00:06:59,370
point 'c' where we have an
infinite discontinuity, an
149
00:06:59,370 --> 00:07:00,420
infinite jump.
150
00:07:00,420 --> 00:07:04,160
What we do is, is we pick a
positive number 'h' and knock
151
00:07:04,160 --> 00:07:08,580
off the interval 'c minus h' to
'c plus h' surrounding 'c'.
152
00:07:08,580 --> 00:07:11,460
We then compute the area.
153
00:07:11,460 --> 00:07:14,450
We then compute the area
of this shaded region.
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00:07:14,450 --> 00:07:17,470
That's exactly what's inside
the brackets over here.
155
00:07:17,470 --> 00:07:19,180
You see, notice that
this sum is what?
156
00:07:19,180 --> 00:07:25,290
It's the sum of two integrals,
one of which names this area,
157
00:07:25,290 --> 00:07:28,020
and the other of which
names this area.
158
00:07:28,020 --> 00:07:32,510
And what we say is, let's put
the squeeze on and see what
159
00:07:32,510 --> 00:07:36,730
limiting value this area
approaches as we squeeze all
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00:07:36,730 --> 00:07:39,650
the space out around 'c'.
161
00:07:39,650 --> 00:07:42,960
In other words, what's really
happening over here?
162
00:07:42,960 --> 00:07:46,640
What we're really saying is, why
can't we treat an infinite
163
00:07:46,640 --> 00:07:50,140
region the same as if it were
a finite region taking the
164
00:07:50,140 --> 00:07:54,570
limit as the size of the strip
in here goes to 0?
165
00:07:54,570 --> 00:07:55,840
And here's the key point.
166
00:07:55,840 --> 00:07:59,430
You see, as 'h' approaches 'c'
from both sides, notice that
167
00:07:59,430 --> 00:08:03,450
the width of this region
approaches 0.
168
00:08:03,450 --> 00:08:06,640
By the same token, the
height of the region
169
00:08:06,640 --> 00:08:08,800
is approaching infinity.
170
00:08:08,800 --> 00:08:12,960
In essence, what we're running
into again, is the problem of
171
00:08:12,960 --> 00:08:16,610
determining the area of an
infinite region involves the
172
00:08:16,610 --> 00:08:19,150
study of infinity times 0.
173
00:08:19,150 --> 00:08:24,370
In fact, the question sort of
is, does 'h' go to 0 faster
174
00:08:24,370 --> 00:08:28,450
than 'f of 'c plus or minus
h'' goes to infinity?
175
00:08:28,450 --> 00:08:31,070
In fact, I think that's an
important enough observation,
176
00:08:31,070 --> 00:08:33,070
so I've written it down
here to make sure
177
00:08:33,070 --> 00:08:33,980
that you notice it.
178
00:08:33,980 --> 00:08:36,870
The question centers about the
question, the problem of
179
00:08:36,870 --> 00:08:40,190
whether 'h' approaches
0 faster than--
180
00:08:40,190 --> 00:08:41,380
whatever this means.
181
00:08:41,380 --> 00:08:45,710
Faster than 'f of 'c plus or
minus h'' approaches infinity.
182
00:08:45,710 --> 00:08:48,930
In our particular example,
this didn't happen.
183
00:08:48,930 --> 00:08:52,820
Now again, if you're afraid of
the symbol infinity, let's
184
00:08:52,820 --> 00:08:56,010
deal with the finite case and
take the limit later.
185
00:08:56,010 --> 00:08:59,110
In other words, instead of
looking at the integral from 0
186
00:08:59,110 --> 00:09:04,340
to 1, 'dx' over 'x squared',
let's look at the integral
187
00:09:04,340 --> 00:09:06,190
from 'h' to 1.
188
00:09:06,190 --> 00:09:09,580
Integral from 'h' to 1, 'dx'
over 'x squared', and then
189
00:09:09,580 --> 00:09:12,520
later we'll take the limit
as 'h' approaches 0.
190
00:09:12,520 --> 00:09:15,820
In particular, if this is my
region 'R', namely bounded
191
00:09:15,820 --> 00:09:19,250
above by the curve 'y' equals '1
over 'x squared'', below by
192
00:09:19,250 --> 00:09:20,290
the x-axis.
193
00:09:20,290 --> 00:09:23,590
On the left by the line 'x'
equals 'h' and on the right by
194
00:09:23,590 --> 00:09:26,910
the line 'x' equals 1, the
area of the region 'R' is
195
00:09:26,910 --> 00:09:27,490
written what?
196
00:09:27,490 --> 00:09:30,790
Integral 'h' to 1, 'dx'
over 'x squared'.
197
00:09:30,790 --> 00:09:33,880
Notice that as long as 'h' is
not 0, our integrand is
198
00:09:33,880 --> 00:09:34,900
continuous.
199
00:09:34,900 --> 00:09:37,600
Therefore, we can use the first
fundamental theorem to
200
00:09:37,600 --> 00:09:39,590
evaluate it.
201
00:09:39,590 --> 00:09:40,140
It's what?
202
00:09:40,140 --> 00:09:44,110
It's minus '1 over x' evaluated
between 'h' and 1.
203
00:09:44,110 --> 00:09:47,460
Putting in the upper limit
gives me a minus 1.
204
00:09:47,460 --> 00:09:50,580
Putting in the lower limit
gives me minus '1/h'.
205
00:09:50,580 --> 00:09:54,000
But I'm subtracting that, so
it becomes plus '1/h'.
206
00:09:54,000 --> 00:09:58,870
In other words, the area of my
region 'R' is 1/h minus 1.
207
00:09:58,870 --> 00:10:03,850
Notice then that 'A sub R' as
'h' approaches 0 from the
208
00:10:03,850 --> 00:10:07,280
right becomes the limit of '1/h'
as 'h' approaches 0 from
209
00:10:07,280 --> 00:10:09,330
the right minus 1.
210
00:10:09,330 --> 00:10:12,670
And the limit of '1/h' as 'h'
approaches 0 from the right is
211
00:10:12,670 --> 00:10:13,560
clearly infinity.
212
00:10:13,560 --> 00:10:16,510
In other words, as 'h' gets
arbitrarily close to 0 through
213
00:10:16,510 --> 00:10:20,450
positive values, '1/h' increases
without any bound.
214
00:10:20,450 --> 00:10:24,230
In other words, what this means
in plain English is that
215
00:10:24,230 --> 00:10:27,730
we can make the area of the
region 'R' as arbitrarily as
216
00:10:27,730 --> 00:10:31,110
large as we wish just by
choosing 'h' to be
217
00:10:31,110 --> 00:10:34,300
sufficiently close to 0 here.
218
00:10:34,300 --> 00:10:36,690
You see, in other words, as
'h' approaches 0, what's
219
00:10:36,690 --> 00:10:40,110
happening here is that this
curve is not approaching the
220
00:10:40,110 --> 00:10:44,160
y-axis fast enough to allow
us to get a finite area.
221
00:10:44,160 --> 00:10:47,550
And by the way, let me make
one little aside comparing
222
00:10:47,550 --> 00:10:49,740
this with our previous answer.
223
00:10:49,740 --> 00:10:52,350
You may recall, and let me just
pull this board down here
224
00:10:52,350 --> 00:10:53,730
to remind you of that.
225
00:10:53,730 --> 00:10:56,580
You may recall that when we
worked this problem before, we
226
00:10:56,580 --> 00:10:59,520
got the wrong answer,
negative 2.
227
00:10:59,520 --> 00:11:00,540
OK, now look it.
228
00:11:00,540 --> 00:11:05,370
In terms of my problem over
here, we went from minus 1 to
229
00:11:05,370 --> 00:11:08,120
1 so this is only 1/2
of the region that
230
00:11:08,120 --> 00:11:09,170
we're talking about.
231
00:11:09,170 --> 00:11:12,200
Notice that the minus
2 comes from this.
232
00:11:12,200 --> 00:11:16,390
See twice minus 1 by symmetry.
233
00:11:16,390 --> 00:11:19,030
The interesting thing is notice
here that when you take
234
00:11:19,030 --> 00:11:20,010
the limit--
235
00:11:20,010 --> 00:11:22,540
well, forgetting that there's
an infinite discontinuity
236
00:11:22,540 --> 00:11:25,590
here, when you let 'h' approach
0, you tend to say
237
00:11:25,590 --> 00:11:28,430
let's forget about the function
of 'h' in here.
238
00:11:28,430 --> 00:11:30,710
Notice here as 'h' approaches
0, this
239
00:11:30,710 --> 00:11:32,110
thing went to infinity.
240
00:11:32,110 --> 00:11:35,820
By ignoring it, that's where
you got the answer minus 2.
241
00:11:35,820 --> 00:11:38,540
In other words, by assuming
that as 'h' got small, the
242
00:11:38,540 --> 00:11:42,460
function of 'h' got small, we
threw out an infinite part of
243
00:11:42,460 --> 00:11:43,400
our answer.
244
00:11:43,400 --> 00:11:44,420
In other words, this was
the thing that we
245
00:11:44,420 --> 00:11:45,350
had to worry about.
246
00:11:45,350 --> 00:11:48,340
In this particular case, we
slurred over something that
247
00:11:48,340 --> 00:11:50,470
wasn't going to 0 fast enough.
248
00:11:50,470 --> 00:11:54,300
By the way, let me caution you
and point out that in this
249
00:11:54,300 --> 00:11:57,920
particular example it happened
that infinity
250
00:11:57,920 --> 00:12:00,370
times 0 became infinite.
251
00:12:00,370 --> 00:12:06,120
It is possible, however, that
as 'h' approaches 0, it may
252
00:12:06,120 --> 00:12:09,620
happen that 'h' approaches 0
fast enough, so that we do get
253
00:12:09,620 --> 00:12:10,560
a finite area.
254
00:12:10,560 --> 00:12:13,880
To show you this, let's just
change the problem slightly.
255
00:12:13,880 --> 00:12:17,360
Instead of taking 'y' equals
'1 over 'x squared'', let's
256
00:12:17,360 --> 00:12:20,180
take 'y' equals '1 over the
'square root of x''.
257
00:12:20,180 --> 00:12:22,700
The same problem, and by the
way, here's the subtlety of
258
00:12:22,700 --> 00:12:23,410
this again.
259
00:12:23,410 --> 00:12:26,200
If you were to sketch 'y' equals
'1 over the 'square
260
00:12:26,200 --> 00:12:32,760
root of x'' alongside of 'y'
equals '1 over 'x squared''--
261
00:12:32,760 --> 00:12:36,160
in fact, fast enough so that
even though we get an infinite
262
00:12:36,160 --> 00:12:38,620
region, it has only
a finite area.
263
00:12:38,620 --> 00:12:40,680
In particular, here's
what I'm saying now.
264
00:12:40,680 --> 00:12:44,840
Let 'R' be the region bounded
above by 'y' equals '1 over
265
00:12:44,840 --> 00:12:47,900
the 'square root of x',
below by the x-axis.
266
00:12:47,900 --> 00:12:50,360
On the left by 'x' equals
'h' and on the right
267
00:12:50,360 --> 00:12:51,850
by 'x' equals 1.
268
00:12:51,850 --> 00:12:55,450
Again, since the only place that
'1 over the 'square root
269
00:12:55,450 --> 00:12:59,140
of x'' has an infinite
discontinuity is when 'x' is
270
00:12:59,140 --> 00:13:02,430
0, and since 'h' is greater
than 0 here, we have a
271
00:13:02,430 --> 00:13:03,900
continuous integrand.
272
00:13:03,900 --> 00:13:07,490
The area of the region 'R'
therefore is given by what?
273
00:13:07,490 --> 00:13:10,170
Any function whose derivative is
'1 over the 'square root of
274
00:13:10,170 --> 00:13:12,710
x'' evaluated between
'h' and 1.
275
00:13:12,710 --> 00:13:14,600
Again, sparing you
the details.
276
00:13:14,600 --> 00:13:17,830
Observe that if you
differentiate twice 'x to the
277
00:13:17,830 --> 00:13:21,070
1/2', the 1/2 and the 2 cancel
and you get '1 over the
278
00:13:21,070 --> 00:13:22,020
'square root of x''.
279
00:13:22,020 --> 00:13:25,130
In other words, 2 square root
of 'x' is the function whose
280
00:13:25,130 --> 00:13:27,250
derivative is '1 over the
'square root of x''.
281
00:13:27,250 --> 00:13:32,120
If I now evaluate that between
1 and 'h', I get 2 minus 2
282
00:13:32,120 --> 00:13:33,530
'square root of h'.
283
00:13:33,530 --> 00:13:37,280
If I now take the limit of 'A
sub R' as 'h' approaches 0,
284
00:13:37,280 --> 00:13:40,590
notice that what happens here
is that as 'h' approaches 0,
285
00:13:40,590 --> 00:13:42,160
so does the square
root of 'h'.
286
00:13:42,160 --> 00:13:45,040
And the area turns out
to be 2 in the limit.
287
00:13:45,040 --> 00:13:47,310
In other words, the area of
the region 'R' in this
288
00:13:47,310 --> 00:13:49,980
particular case, is also
a function of 'h'.
289
00:13:49,980 --> 00:13:53,360
But the part that depends
on 'h' goes to 0
290
00:13:53,360 --> 00:13:55,480
as 'h' goes to 0.
291
00:13:55,480 --> 00:13:58,530
In fact, notice in this
particular case, and why I
292
00:13:58,530 --> 00:14:01,290
couldn't have started with this
example is that if I took
293
00:14:01,290 --> 00:14:04,170
this example and you did it the
incorrect way, in other
294
00:14:04,170 --> 00:14:08,230
words, if you failed to notice
that '1 over the 'square root
295
00:14:08,230 --> 00:14:12,260
of x'' was discontinuous at 'x'
equals 0, and you just did
296
00:14:12,260 --> 00:14:15,970
this problem mechanically, and
evaluated this between 0 and
297
00:14:15,970 --> 00:14:18,030
1, you would have got
the right answer.
298
00:14:18,030 --> 00:14:20,130
And the reason that you would
have got the right answer, is
299
00:14:20,130 --> 00:14:23,320
that in the limit process,
notice here that the part of
300
00:14:23,320 --> 00:14:26,310
the answer that depends
on 'h' does actually
301
00:14:26,310 --> 00:14:28,250
go to 0 in the limit.
302
00:14:28,250 --> 00:14:32,590
At any rate, using all of this
as motivation, we come to a
303
00:14:32,590 --> 00:14:36,000
new piece of terminology
that we should define.
304
00:14:36,000 --> 00:14:40,390
Namely, if 'f' is infinite at
'x' equals 'c', where 'c' is
305
00:14:40,390 --> 00:14:44,120
someplace in the interval from
'a' to 'b', then the improper
306
00:14:44,120 --> 00:14:49,710
integral 'a' to 'b', ''f of x'
dx' is called convergent if
307
00:14:49,710 --> 00:14:52,850
the particular limit that we
talked about previously in the
308
00:14:52,850 --> 00:14:55,550
definition of how you compute
integral from 'a' to 'b', ''f
309
00:14:55,550 --> 00:14:59,140
of x' dx', if that particular
limit exists, meaning it's a
310
00:14:59,140 --> 00:15:02,980
finite number, then we call
the integral a convergent
311
00:15:02,980 --> 00:15:04,610
improper integral.
312
00:15:04,610 --> 00:15:06,290
On the other hand,
meaning what?
313
00:15:06,290 --> 00:15:09,440
If this limit doesn't exist, if
for example, the limit is
314
00:15:09,440 --> 00:15:12,670
infinite, then we call the
integral divergent.
315
00:15:12,670 --> 00:15:14,680
In other words, in our
first example, the
316
00:15:14,680 --> 00:15:16,590
integral was divergent.
317
00:15:16,590 --> 00:15:20,130
In the second example, the
integral was convergent.
318
00:15:20,130 --> 00:15:23,720
By the way, it's rather
interesting to point out too--
319
00:15:23,720 --> 00:15:24,990
I shouldn't say it's
rather interesting.
320
00:15:24,990 --> 00:15:27,980
It's just my excuse of I don't
know of any other way of
321
00:15:27,980 --> 00:15:31,230
motivating improper integrals
of the second kind.
322
00:15:31,230 --> 00:15:33,350
But the thing that's worth
observing is this.
323
00:15:33,350 --> 00:15:37,250
324
00:15:37,250 --> 00:15:43,650
When we evaluated the area under
the curve here from 0 to
325
00:15:43,650 --> 00:15:47,380
1, we elected to pick
strips like this.
326
00:15:47,380 --> 00:15:48,550
Remember this is why we
got into that infinite
327
00:15:48,550 --> 00:15:51,580
discontinuity; we picked out
strips to look like this.
328
00:15:51,580 --> 00:15:54,290
The question that comes up is,
why couldn't we have picked
329
00:15:54,290 --> 00:15:55,790
our strips to be horizontal?
330
00:15:55,790 --> 00:15:58,390
In other words, if we had
in terms of say, inverse
331
00:15:58,390 --> 00:16:01,860
functions, inverted the role of
'x' and 'y', why couldn't
332
00:16:01,860 --> 00:16:05,940
we have computed our area by
picking strips like this and
333
00:16:05,940 --> 00:16:09,670
adding them all up from 1 to
'b', and then taking the limit
334
00:16:09,670 --> 00:16:11,360
as 'b' went to infinity?
335
00:16:11,360 --> 00:16:13,820
In other words, if I wanted to
compute the area of this
336
00:16:13,820 --> 00:16:17,960
region using horizontal strips,
I would have had what?
337
00:16:17,960 --> 00:16:20,730
I would have had the area of the
region 'R' is the integral
338
00:16:20,730 --> 00:16:22,660
from 1 to 'b'.
339
00:16:22,660 --> 00:16:26,090
And remember, if 'y' equals '1
over 'x squared'', 'x' is '1
340
00:16:26,090 --> 00:16:27,580
over the 'square root of y''.
341
00:16:27,580 --> 00:16:30,020
Therefore, the area of the
region 'R' is the interval
342
00:16:30,020 --> 00:16:34,210
from 1 to 'b', 'dy' over
the square root of 'y'.
343
00:16:34,210 --> 00:16:37,680
Notice that the square root of
'y' is certainly continuous.
344
00:16:37,680 --> 00:16:39,190
'1 over the 'square
root of y'' is
345
00:16:39,190 --> 00:16:40,610
continuous in this interval.
346
00:16:40,610 --> 00:16:44,240
At any rate, I could then have
evaluated this integrand here
347
00:16:44,240 --> 00:16:47,150
and taken the limit as 'b'
approached infinity.
348
00:16:47,150 --> 00:16:49,860
And by the way, the abbreviation
for writing the
349
00:16:49,860 --> 00:16:54,010
limit of the integral 1 to 'b'
as 'b' approaches infinity, is
350
00:16:54,010 --> 00:16:57,720
to simply write integral from
1 to infinity 'dy' over the
351
00:16:57,720 --> 00:16:58,940
square root of 'y'.
352
00:16:58,940 --> 00:17:00,370
And this is all this
thing means.
353
00:17:00,370 --> 00:17:03,130
For example, in this particular
case, notice that
354
00:17:03,130 --> 00:17:07,050
as 'b' goes to infinity, so does
the square root of 'b'.
355
00:17:07,050 --> 00:17:11,089
Therefore, 2 'square root of b'
minus 2 goes to infinity as
356
00:17:11,089 --> 00:17:12,440
'b' approaches infinity.
357
00:17:12,440 --> 00:17:16,040
And we find by a second method
that the area of
358
00:17:16,040 --> 00:17:17,460
our region is infinite.
359
00:17:17,460 --> 00:17:20,450
And again, we have another kind
of improper integral.
360
00:17:20,450 --> 00:17:23,849
But structurally, this
improper integral is
361
00:17:23,849 --> 00:17:25,880
completely different
from the improper
362
00:17:25,880 --> 00:17:27,630
integral that we had before.
363
00:17:27,630 --> 00:17:31,300
Namely, before we had an
improper integral where the
364
00:17:31,300 --> 00:17:34,900
integrand became infinite,
even though the limits of
365
00:17:34,900 --> 00:17:36,590
integration stayed finite.
366
00:17:36,590 --> 00:17:39,180
Notice what happens
in this case.
367
00:17:39,180 --> 00:17:42,160
In this particular case, notice
that the integrand
368
00:17:42,160 --> 00:17:43,240
stays finite.
369
00:17:43,240 --> 00:17:46,660
It's the limits of integration
that become infinite.
370
00:17:46,660 --> 00:17:50,450
In other words, this leads to
our next definition of a new
371
00:17:50,450 --> 00:17:52,810
type of improper integral.
372
00:17:52,810 --> 00:17:57,210
Namely, the integral say, from
'a' to infinity ''f of x' dx'.
373
00:17:57,210 --> 00:18:01,200
Whereby this we mean the
abbreviation limit as 'b'
374
00:18:01,200 --> 00:18:04,410
approaches infinity, integral
from 'a' to 'b', ''f of x'
375
00:18:04,410 --> 00:18:08,560
dx', where 'f' is continuous
in this entire range.
376
00:18:08,560 --> 00:18:11,050
In other words, for 'x' greater
than or equal to 'a'.
377
00:18:11,050 --> 00:18:15,030
That type of an integral is
called an improper integral of
378
00:18:15,030 --> 00:18:16,630
the second kind.
379
00:18:16,630 --> 00:18:19,350
In other words, in the first
kind, the integrand becomes
380
00:18:19,350 --> 00:18:22,170
infinite, but the limits of
integration are finite.
381
00:18:22,170 --> 00:18:25,970
In the second kind, the
integrand stays finite, but
382
00:18:25,970 --> 00:18:29,330
the limits of integration take
on an infinite range.
383
00:18:29,330 --> 00:18:32,120
And the important point is
that these two are very
384
00:18:32,120 --> 00:18:33,030
closely connected.
385
00:18:33,030 --> 00:18:35,520
In other words, in terms of the
example that I just showed
386
00:18:35,520 --> 00:18:38,700
you, I wanted you to see how you
can get from an integral
387
00:18:38,700 --> 00:18:41,130
of the first kind to an improper
integral of the
388
00:18:41,130 --> 00:18:44,410
second kind just by replacing
your vertical strips by
389
00:18:44,410 --> 00:18:45,720
horizontal strips.
390
00:18:45,720 --> 00:18:49,770
In fact, in terms of my first
example over here notice that
391
00:18:49,770 --> 00:18:53,240
indeed, '1 over 'x squared'' and
'1 over the 'square root
392
00:18:53,240 --> 00:18:56,960
of x'', where 'x' is at least
as big as 0, are inverses.
393
00:18:56,960 --> 00:18:59,850
By the way, the reason I say 'x'
is, in fact, bigger than
394
00:18:59,850 --> 00:19:02,400
0, otherwise I'll have
a 0 denominator here.
395
00:19:02,400 --> 00:19:05,295
The reason I point this out is
that if you don't presuppose
396
00:19:05,295 --> 00:19:09,320
that x is positive, remember,
the square root function or
397
00:19:09,320 --> 00:19:12,470
the squaring function is not
1:1, so I have to be very
398
00:19:12,470 --> 00:19:14,400
careful when I talk about
inverse functions.
399
00:19:14,400 --> 00:19:17,270
The thing I want to see however
is, the graph of 'y'
400
00:19:17,270 --> 00:19:21,410
equals '1 over 'x squared'' is
the inverse of the graph 'y'
401
00:19:21,410 --> 00:19:23,370
equals '1 over the 'square
root of x''.
402
00:19:23,370 --> 00:19:25,740
In other words, essentially, by
interchanging the role of
403
00:19:25,740 --> 00:19:29,270
the x- and y-axes, you go from
one graph to the other.
404
00:19:29,270 --> 00:19:34,750
You see, in particular, the
integral 1 to infinity, 'dx'
405
00:19:34,750 --> 00:19:36,430
over 'x squared'.
406
00:19:36,430 --> 00:19:39,280
Namely, finding the area of this
particular region is the
407
00:19:39,280 --> 00:19:43,130
inverse of finding the area
of this particular region.
408
00:19:43,130 --> 00:19:45,850
In other words, notice that
'y' equals '1 over 'x
409
00:19:45,850 --> 00:19:50,590
squared'' behaves for large
values of 'x' the same as 'y'
410
00:19:50,590 --> 00:19:53,880
equals '1 over the 'square root
of x'' behaves for small
411
00:19:53,880 --> 00:19:55,430
parts of the values of 'x'.
412
00:19:55,430 --> 00:19:59,520
Again, let me point out that
much of this will be done much
413
00:19:59,520 --> 00:20:04,090
more hard core in our learning
exercises and in the reading
414
00:20:04,090 --> 00:20:05,220
material and the like.
415
00:20:05,220 --> 00:20:08,480
And so I pass over some of this
in a fairly quick way.
416
00:20:08,480 --> 00:20:10,800
But what I want to make sure is
clear before we begin the
417
00:20:10,800 --> 00:20:13,700
exercises and the reading
material is that we have a
418
00:20:13,700 --> 00:20:16,530
good idea as to what improper
integrals mean.
419
00:20:16,530 --> 00:20:19,480
At any rate, continuing on this
way, what do I mean by
420
00:20:19,480 --> 00:20:21,170
the integral from
1 to infinity,
421
00:20:21,170 --> 00:20:22,590
'dx' over 'x squared'?
422
00:20:22,590 --> 00:20:25,290
Well, it's an improper integral
of a second type
423
00:20:25,290 --> 00:20:29,680
because the integrand is finite
between 1 and infinity.
424
00:20:29,680 --> 00:20:33,050
The limits of integration span
an infinite space over here.
425
00:20:33,050 --> 00:20:33,840
At any rate, it's what?
426
00:20:33,840 --> 00:20:36,710
The limit as 'b' approaches
infinity, integral from 1 to
427
00:20:36,710 --> 00:20:38,910
'b', 'dx' over 'x squared'.
428
00:20:38,910 --> 00:20:42,130
429
00:20:42,130 --> 00:20:44,140
At any rate, what
is the integral
430
00:20:44,140 --> 00:20:44,926
of '1 over 'x squared''.
431
00:20:44,926 --> 00:20:47,520
It's minus '1/x'.
432
00:20:47,520 --> 00:20:51,840
If I evaluate that that's minus
''1 over b' plus 1'.
433
00:20:51,840 --> 00:20:54,820
And if I now take the limit as
'b' approaches infinity that
434
00:20:54,820 --> 00:20:56,190
answer is 1.
435
00:20:56,190 --> 00:20:57,390
In other words, what this means
436
00:20:57,390 --> 00:20:59,250
pictorially is the following.
437
00:20:59,250 --> 00:21:03,570
If I look at the area of this
region 'R' for any value of
438
00:21:03,570 --> 00:21:09,080
'b' whatsoever, that area is
equal to 1 minus '1/b'.
439
00:21:09,080 --> 00:21:12,400
In other words, no matter how
big 'b' becomes, the area of
440
00:21:12,400 --> 00:21:16,240
this region 'R' can never
be as big as one unit.
441
00:21:16,240 --> 00:21:19,320
And in the limit, as 'b' goes
out as far as we wish, what
442
00:21:19,320 --> 00:21:20,850
we're really saying is what?
443
00:21:20,850 --> 00:21:24,040
That the area of this region
'R' can be made arbitrarily
444
00:21:24,040 --> 00:21:28,360
close to 1 just by choosing 'b'
to be arbitrarily large.
445
00:21:28,360 --> 00:21:30,700
And by the way, this you see
leads to an interesting
446
00:21:30,700 --> 00:21:31,710
observation.
447
00:21:31,710 --> 00:21:35,050
Namely, at this end of the
spectrum, the curve was not
448
00:21:35,050 --> 00:21:37,580
approaching the y-axis
fast enough to make
449
00:21:37,580 --> 00:21:39,250
this integral finite.
450
00:21:39,250 --> 00:21:43,640
Remember, the integral from 0
to 1, 'dx' over 'x squared'
451
00:21:43,640 --> 00:21:44,830
was infinite.
452
00:21:44,830 --> 00:21:48,570
However, for large values of
'x', the curve does approach
453
00:21:48,570 --> 00:21:51,910
the x-axis fast enough, so that
not only is the enclosed
454
00:21:51,910 --> 00:21:55,700
area finite, it can't even
be bigger than 1.
455
00:21:55,700 --> 00:21:57,750
I thought that might be an
interesting example for you to
456
00:21:57,750 --> 00:21:59,440
think about for a while.
457
00:21:59,440 --> 00:22:01,960
Another type of interesting
example worth thinking about
458
00:22:01,960 --> 00:22:04,530
for a while is this
second example.
459
00:22:04,530 --> 00:22:08,250
Let's take the region 'R' to be
bounded above by 'y' equals
460
00:22:08,250 --> 00:22:12,950
'1/x', on the left by 'x' equals
1, on the right by 'x'
461
00:22:12,950 --> 00:22:15,680
equals 'b', and below
by the x-axis.
462
00:22:15,680 --> 00:22:18,960
And let's compute the area
of the region 'R'.
463
00:22:18,960 --> 00:22:21,780
Again, the area of the region
'R' is a definite integral
464
00:22:21,780 --> 00:22:24,210
from 1 to 'b', 'dx' over 'x'.
465
00:22:24,210 --> 00:22:27,840
And that's equal to the
natural log of 'b'.
466
00:22:27,840 --> 00:22:32,390
Namely, this integral is log
absolute value of 'x'.
467
00:22:32,390 --> 00:22:34,920
Log absolute value of
'b' is just 'log b'.
468
00:22:34,920 --> 00:22:36,200
Log 1 is 0.
469
00:22:36,200 --> 00:22:39,020
So the area of the region
'R' is 'log b'.
470
00:22:39,020 --> 00:22:42,140
As 'b' increases without bound,
its log increases
471
00:22:42,140 --> 00:22:44,390
without bound, we find
that the area of the
472
00:22:44,390 --> 00:22:46,230
region 'R' is infinite.
473
00:22:46,230 --> 00:22:49,380
In fact, another way of
saying this is what?
474
00:22:49,380 --> 00:22:54,730
This is the integral 1 to
infinity 'dx' over 'x'.
475
00:22:54,730 --> 00:22:56,820
And this is, therefore, what?
476
00:22:56,820 --> 00:23:02,170
A divergent improper integral
of the second type.
477
00:23:02,170 --> 00:23:04,300
All right, so far so
good at any rate.
478
00:23:04,300 --> 00:23:08,540
Let me review our volumes of
revolution and rotation.
479
00:23:08,540 --> 00:23:11,710
Let's take the region 'R' and
rotate it about the x-axis.
480
00:23:11,710 --> 00:23:15,000
481
00:23:15,000 --> 00:23:17,820
If we rotate 'R' about the
x-axis, remember a volume of
482
00:23:17,820 --> 00:23:23,430
revolution is just pi 'y
squared' 'dx' from 1 to 'b'.
483
00:23:23,430 --> 00:23:26,530
Well, my 'y' is '1/x'.
484
00:23:26,530 --> 00:23:28,430
That's '1 over x squared'.
485
00:23:28,430 --> 00:23:32,110
The integral of '1 over x
squared' is minus '1/x'.
486
00:23:32,110 --> 00:23:35,580
Notice then as before, the
volume when I rotate 'R' about
487
00:23:35,580 --> 00:23:40,440
the x-axis is just pi times
'1 minus '1/b''.
488
00:23:40,440 --> 00:23:43,710
In other words, what this say
is as 'b' gets very, very
489
00:23:43,710 --> 00:23:45,810
large, so does 'R'.
490
00:23:45,810 --> 00:23:50,230
Yet, if I rotate 'R' about the
x-axis, no matter how big 'b'
491
00:23:50,230 --> 00:23:56,520
is the volume is just pi
times '1 minus '1/b''.
492
00:23:56,520 --> 00:24:01,090
As 'b' approaches infinity, this
goes to 0 and my volume
493
00:24:01,090 --> 00:24:02,270
is just pi.
494
00:24:02,270 --> 00:24:05,170
In other words, this area
gets infinitely large.
495
00:24:05,170 --> 00:24:08,700
Yet, the volume generated
by rotating that
496
00:24:08,700 --> 00:24:10,840
area remains finite.
497
00:24:10,840 --> 00:24:14,860
Again, another idiosyncrasy
of working with improper
498
00:24:14,860 --> 00:24:16,730
integrals and infinite
regions.
499
00:24:16,730 --> 00:24:19,820
You've got to be very, very
careful about how rapidly
500
00:24:19,820 --> 00:24:23,550
something approaches 0 or
infinity because infinity
501
00:24:23,550 --> 00:24:25,080
times 0 is indeterminate.
502
00:24:25,080 --> 00:24:28,350
In other words, here is a very
important observation, an
503
00:24:28,350 --> 00:24:33,610
infinite area can enclose or can
generate a finite volume.
504
00:24:33,610 --> 00:24:35,530
Be very, very careful
in dealing
505
00:24:35,530 --> 00:24:37,050
with improper integrals.
506
00:24:37,050 --> 00:24:41,890
And by the way, let me make
one computational aside.
507
00:24:41,890 --> 00:24:46,070
In trying to determine whether a
particular improper integral
508
00:24:46,070 --> 00:24:50,400
is convergent or not, we do not
have to be able to compute
509
00:24:50,400 --> 00:24:52,870
the limit itself.
510
00:24:52,870 --> 00:24:54,490
I mention this because
it's very important.
511
00:24:54,490 --> 00:24:57,860
Sometime you'll be given an
integral which you don't know
512
00:24:57,860 --> 00:24:59,220
how to integrate directly.
513
00:24:59,220 --> 00:25:02,010
A typical example that usually
comes up, integral ''e to the
514
00:25:02,010 --> 00:25:06,520
minus 'x squared' dx', say
between 1 and infinity.
515
00:25:06,520 --> 00:25:08,430
And you want to know whether
this converges or not.
516
00:25:08,430 --> 00:25:10,570
You don't care what the limit
is because once you know it
517
00:25:10,570 --> 00:25:13,200
converges, you can approximate
it to as great a degree of
518
00:25:13,200 --> 00:25:14,550
accuracy as you wish.
519
00:25:14,550 --> 00:25:17,140
So the question is, does this
converge or doesn't it?
520
00:25:17,140 --> 00:25:20,120
And again, the answer is, let's
see what happens on this
521
00:25:20,120 --> 00:25:21,980
particular range, for example.
522
00:25:21,980 --> 00:25:24,060
In other words, we know, for
example, that if 'x' is
523
00:25:24,060 --> 00:25:29,210
greater than or equal to 1 for
numbers in magnitude at least
524
00:25:29,210 --> 00:25:33,180
as big as 1, the square
exceeds the number.
525
00:25:33,180 --> 00:25:36,930
In particular by the way, if 'x
squared' exceeds 'x', minus
526
00:25:36,930 --> 00:25:40,270
'x squared' cannot
exceed minus 'x'.
527
00:25:40,270 --> 00:25:42,170
In other words, on this
particular interval that we're
528
00:25:42,170 --> 00:25:44,890
dealing with, say from 1 to
'b', as 'b' approaches
529
00:25:44,890 --> 00:25:47,620
infinity, 'e to the minus
'x squared'' is
530
00:25:47,620 --> 00:25:49,465
less than minus 'x'.
531
00:25:49,465 --> 00:25:52,280
532
00:25:52,280 --> 00:25:57,670
Therefore, because the
exponential function is 1:1,
533
00:25:57,670 --> 00:26:01,100
''e to the minus 'x squared'
dx' from 1 to 'b' cannot
534
00:26:01,100 --> 00:26:04,850
exceed the integral from 1 to
'b', ''e to the minus x' dx'.
535
00:26:04,850 --> 00:26:07,290
In other words, since this
exponent is less than this
536
00:26:07,290 --> 00:26:10,440
exponent, this integral
is less than this one.
537
00:26:10,440 --> 00:26:13,650
However, it turns out that we
know how to evaluate this one.
538
00:26:13,650 --> 00:26:15,630
How do we integrate 'e
to the minus x'?
539
00:26:15,630 --> 00:26:17,810
It's just minus 'e
to the minus x'.
540
00:26:17,810 --> 00:26:21,260
And if we evaluate that between
1 and 'b', we find
541
00:26:21,260 --> 00:26:25,260
that we get '1/e' minus '1
over 'e to the b'' power.
542
00:26:25,260 --> 00:26:27,040
'b' is some positive number.
543
00:26:27,040 --> 00:26:29,340
This thing here is less
than infinity.
544
00:26:29,340 --> 00:26:33,390
In fact, as 'b' approaches
infinity, this approaches 0,
545
00:26:33,390 --> 00:26:35,790
and the limiting
value is '1/e'.
546
00:26:35,790 --> 00:26:39,360
In other words, what we now know
is, is that whatever this
547
00:26:39,360 --> 00:26:43,690
improper integral is, it cannot
exceed '1/e' even
548
00:26:43,690 --> 00:26:46,600
though we don't know what
the exact answer is.
549
00:26:46,600 --> 00:26:50,110
Pictorially, you see what
happens over here is that the
550
00:26:50,110 --> 00:26:53,870
curve which I've drawn in the
accentuated chalk, this curve
551
00:26:53,870 --> 00:26:56,260
is 'y' equals 'e to the
minus 'x squared''.
552
00:26:56,260 --> 00:26:58,960
The light curve is 'y' equals
'e to the minus x'.
553
00:26:58,960 --> 00:27:02,560
Notice that these two curves
crisscross when 'x' equals 1.
554
00:27:02,560 --> 00:27:07,280
And all we're saying is that
since the white curve chops
555
00:27:07,280 --> 00:27:11,050
off a finite area and the
accentuated curve, the dark
556
00:27:11,050 --> 00:27:15,260
curve is inside the white one,
certainly the area enclosed by
557
00:27:15,260 --> 00:27:18,610
the black curve cannot exceed
that of the white curve.
558
00:27:18,610 --> 00:27:21,970
In other words, if this region
here approaches a finite value
559
00:27:21,970 --> 00:27:25,240
as 'x' goes to infinity, so
must the other region.
560
00:27:25,240 --> 00:27:27,170
And this is essentially
the theory
561
00:27:27,170 --> 00:27:29,490
behind improper integrals.
562
00:27:29,490 --> 00:27:32,700
In closing, there is one
thing I would like
563
00:27:32,700 --> 00:27:33,570
to say about this.
564
00:27:33,570 --> 00:27:37,310
It's a short summary, but
here's the whole idea.
565
00:27:37,310 --> 00:27:40,700
It may be difficult to evaluate
improper integrals of
566
00:27:40,700 --> 00:27:41,550
a second kind.
567
00:27:41,550 --> 00:27:42,830
But in the certain manner of
568
00:27:42,830 --> 00:27:45,470
speaking, they're not dangerous.
569
00:27:45,470 --> 00:27:46,660
In what manner of speaking?
570
00:27:46,660 --> 00:27:50,560
Well, when you see the limits on
the integral sign like a to
571
00:27:50,560 --> 00:27:54,490
infinity, or minus infinity to
'b', or minus infinity to
572
00:27:54,490 --> 00:27:57,135
infinity, you're
on the lookout.
573
00:27:57,135 --> 00:27:59,970
You're on the lookout
to be careful about
574
00:27:59,970 --> 00:28:01,170
something going wrong.
575
00:28:01,170 --> 00:28:03,480
There's something about the
symbol infinity, which
576
00:28:03,480 --> 00:28:07,860
hopefully strikes a certain fear
or respect in your minds
577
00:28:07,860 --> 00:28:09,870
when you work mathematically.
578
00:28:09,870 --> 00:28:13,560
See, we have a warning to
be ware in this case.
579
00:28:13,560 --> 00:28:17,120
However, the dangerous part is
that when you're given an
580
00:28:17,120 --> 00:28:21,660
improper integral of the first
type, and I don't know the
581
00:28:21,660 --> 00:28:24,210
best way of saying this other
than to say perhaps, that
582
00:28:24,210 --> 00:28:26,930
mathematics has no built-in
burglar alarms.
583
00:28:26,930 --> 00:28:29,990
What I mean by that is if you're
supposed to multiply 5
584
00:28:29,990 --> 00:28:32,930
and 3 to get the right answer to
a problem, and instead you
585
00:28:32,930 --> 00:28:36,170
add 5 and 3, you're going to
get 8 for the answer even
586
00:28:36,170 --> 00:28:37,160
though it's incorrect.
587
00:28:37,160 --> 00:28:39,160
There's not going to be a buzzer
that goes off and says,
588
00:28:39,160 --> 00:28:41,180
please do not add these
two numbers.
589
00:28:41,180 --> 00:28:44,520
In other words, if you go ahead
over here and you forget
590
00:28:44,520 --> 00:28:47,660
that 'f of x' may be infinite
and you go ahead and integrate
591
00:28:47,660 --> 00:28:49,670
this using the first fundamental
theorem, you're
592
00:28:49,670 --> 00:28:54,650
going to get an answer just like
we did in our beginning
593
00:28:54,650 --> 00:28:56,280
exercise at the beginning
of the lecture.
594
00:28:56,280 --> 00:28:58,130
It happened to be the
wrong answer.
595
00:28:58,130 --> 00:29:00,830
So in other words, the word of
warning here is whenever you
596
00:29:00,830 --> 00:29:04,310
see an integral with finite
limits of integration, check
597
00:29:04,310 --> 00:29:07,500
the integrand and make sure it
doesn't blow up, it doesn't
598
00:29:07,500 --> 00:29:08,850
become infinite.
599
00:29:08,850 --> 00:29:12,170
At any rate, with these words
we finish our block on
600
00:29:12,170 --> 00:29:13,240
integration.
601
00:29:13,240 --> 00:29:15,190
And until next time, good bye.
602
00:29:15,190 --> 00:29:17,990
603
00:29:17,990 --> 00:29:20,520
ANNOUNCER: Funding for the
publication of this video was
604
00:29:20,520 --> 00:29:25,240
provided by the Gabriella and
Paul Rosenbaum Foundation.
605
00:29:25,240 --> 00:29:29,410
Help OCW continue to provide
free and open access to MIT
606
00:29:29,410 --> 00:29:33,610
courses by making a donation
at ocw.mit.edu/donate.
607
00:29:33,610 --> 00:29:38,346