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PROFESSOR: Hi.
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The aim of our lecture today is
to present what is perhaps
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the singly most powerful
recipe for evaluating
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00:00:43,840 --> 00:00:47,160
indefinite or, for that matter,
definite integrals.
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00:00:47,160 --> 00:00:50,920
It's called integration by
parts, and therefore, today's
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00:00:50,920 --> 00:00:53,420
lecture is called 'Integration
by Parts'.
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00:00:53,420 --> 00:00:56,940
And what I want to emphasize is
essentially that this is a
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00:00:56,940 --> 00:01:00,960
technique which shows us how to
find indefinite integrals,
16
00:01:00,960 --> 00:01:03,510
but then in theory, it
changes nothing over
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00:01:03,510 --> 00:01:04,319
what we've had before.
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00:01:04,319 --> 00:01:07,530
This is essentially, as we've
mentioned, a recipe for being
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00:01:07,530 --> 00:01:12,000
able to simplify or reduce
indefinite integrals to things
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00:01:12,000 --> 00:01:12,960
that we can handle.
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00:01:12,960 --> 00:01:18,390
Well, it centers around
the rule for
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00:01:18,390 --> 00:01:19,660
differentiating a product.
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00:01:19,660 --> 00:01:22,150
And to show you what I mean by
that, let's just barge right
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00:01:22,150 --> 00:01:23,050
into this thing.
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00:01:23,050 --> 00:01:26,460
In differential notation, we
call the 'u' and 'v' our
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00:01:26,460 --> 00:01:29,260
differentiable functions
of 'x'.
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00:01:29,260 --> 00:01:32,690
The differential of 'uv' is 'u'
times the differential of
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00:01:32,690 --> 00:01:35,830
'v', plus 'v' times the
differential of 'u'.
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00:01:35,830 --> 00:01:39,190
And solving for 'udv', just
transposing, we have that
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00:01:39,190 --> 00:01:44,820
'udv' is the differential of
'u' times 'v' minus 'vdu'.
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00:01:44,820 --> 00:01:46,960
Now, what we do is
we integrate both
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00:01:46,960 --> 00:01:49,580
sides of this equality.
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00:01:49,580 --> 00:01:52,920
On the left-hand side,
we have interval udv.
34
00:01:52,920 --> 00:01:56,270
On the right-hand side, recall
that the integral of the
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00:01:56,270 --> 00:02:00,145
derivative just gives us back
the original function.
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00:02:00,145 --> 00:02:02,580
In other words, if we integrate
the differential of
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00:02:02,580 --> 00:02:05,210
'uv' up to an arbitrary
constant, we get
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00:02:05,210 --> 00:02:07,150
back 'u' times 'v'.
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00:02:07,150 --> 00:02:11,480
The integral of 'vdu' we'll just
denote by integral 'vdu'.
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00:02:11,480 --> 00:02:14,740
And observe that there's really
an arbitrary constant
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00:02:14,740 --> 00:02:17,210
coming from this term, an
arbitrary constant coming from
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00:02:17,210 --> 00:02:19,130
this term, an arbitrary
constant
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00:02:19,130 --> 00:02:20,280
coming from this term.
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00:02:20,280 --> 00:02:22,920
But since the sum and difference
of arbitrary
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00:02:22,920 --> 00:02:25,460
constants is still an arbitrary
constant, we can
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00:02:25,460 --> 00:02:28,950
neglected all the constants
here, and just lump them
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00:02:28,950 --> 00:02:31,730
together, and at the
end write plus 'c'.
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00:02:31,730 --> 00:02:35,330
Or we can leave the 'c' off,
making it implied that the
49
00:02:35,330 --> 00:02:38,330
arbitrary constant is contained
in the symbol
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00:02:38,330 --> 00:02:39,820
integral 'vdu'.
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00:02:39,820 --> 00:02:42,390
All right, now we've done
this, but a much more
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00:02:42,390 --> 00:02:45,060
important question would be--
or exclamation, for that
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00:02:45,060 --> 00:02:47,550
matter-- would be so what?
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See, who cares whether this
happens to be true or not?
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00:02:51,210 --> 00:02:52,280
The answer is we care.
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00:02:52,280 --> 00:02:54,270
That's why we're making
up a lecture on it.
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00:02:54,270 --> 00:02:56,510
And the reason that
we care is this.
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00:02:56,510 --> 00:03:01,490
That frequently, for a given 'u'
and 'v', 'udv' may have a
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00:03:01,490 --> 00:03:06,040
completely different nature
about it than does 'vdu'
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00:03:06,040 --> 00:03:09,230
insofar as being able to
find the integral.
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Now again, the best way to show
what we mean by this is
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00:03:12,490 --> 00:03:15,040
to choose a particular
example.
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Let me start, just arbitrarily
here, picking a couple of
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00:03:18,680 --> 00:03:21,930
simple functions for us to
differentiate or to integrate.
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00:03:21,930 --> 00:03:26,300
Let's pick 'u' to equal 'x'
and 'v' to equal 'sine x'.
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Now, it's simple to see from
this example that in this
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00:03:29,330 --> 00:03:33,970
case, if 'u' equals 'x' and 'v'
is 'sine x', then 'dv' is
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00:03:33,970 --> 00:03:35,356
''cosine x' dx'.
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00:03:35,356 --> 00:03:40,380
In fact, let's write that down
so we can see it here.
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00:03:40,380 --> 00:03:43,790
Notice that in this particular
case, integral 'udv' is
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00:03:43,790 --> 00:03:46,670
integral of 'x 'cosine
x' dx'--
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'udv'.
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00:03:48,260 --> 00:03:51,500
On the other hand,
what is 'vdu'?
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'v' is 'sine x'.
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00:03:53,350 --> 00:04:01,790
Then since 'u' equals
'x', 'du' is 'dx'.
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00:04:01,790 --> 00:04:06,550
Therefore, 'vdu' is just
integral ''sine x' dx'.
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00:04:06,550 --> 00:04:10,200
Now, look at these
two integrals.
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00:04:10,200 --> 00:04:13,640
This one we can integrate
just by looking at it.
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It's minus 'cosine x'.
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00:04:15,760 --> 00:04:19,649
This one, well, offhand, we
don't know a function whose
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00:04:19,649 --> 00:04:23,720
derivative with respect to
'x' is 'x 'cosine x''.
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00:04:23,720 --> 00:04:29,360
However, by this recipe, we now
have a way of showing how
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00:04:29,360 --> 00:04:36,070
to 'udv' once we know
integral 'vdu'.
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00:04:36,070 --> 00:04:39,140
In other words, in this
particular problem, letting
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00:04:39,140 --> 00:04:46,890
'u' equal 'x' and 'dv' being
''cosine x' dx', we can
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00:04:46,890 --> 00:04:48,240
proceed as follows.
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00:04:48,240 --> 00:04:51,480
Namely, by this recipe,
we get what?
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00:04:51,480 --> 00:04:57,130
Integral 'x 'cosine x' dx' is
equal to-- this is 'udv'
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00:04:57,130 --> 00:04:58,790
equals 'u' times 'v'.
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00:04:58,790 --> 00:05:02,370
Recall that 'u' was 'x', 'v' is
'sine x', so 'u' times 'v'
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00:05:02,370 --> 00:05:08,010
is 'x sine x', minus
integral 'vdu'.
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00:05:08,010 --> 00:05:12,460
But 'v' is 'sine x', 'du' is
'dx', so we wind up with this.
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00:05:12,460 --> 00:05:15,160
But now we know that the
integral of 'sine x' with
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00:05:15,160 --> 00:05:17,740
respect to 'x' is minus
the 'cosine x'.
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00:05:17,740 --> 00:05:20,640
The minus and the minus cancel
to give me a plus.
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00:05:20,640 --> 00:05:24,330
And I wind up with that the
integral of 'x 'cosine x' dx'
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00:05:24,330 --> 00:05:27,770
is 'x sine x' plus 'cosine
x' plus an
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00:05:27,770 --> 00:05:30,790
arbitrary constant, OK.
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00:05:30,790 --> 00:05:33,760
That's all integration
by parts really is.
100
00:05:33,760 --> 00:05:37,970
All we do here is we evaluate
one integral by knowing
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00:05:37,970 --> 00:05:38,970
another integral.
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00:05:38,970 --> 00:05:42,600
And by the way, what I meant
here when I said that one did
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00:05:42,600 --> 00:05:46,710
not have to change any of the
theory, what does the integral
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00:05:46,710 --> 00:05:47,720
mean here anyway?
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00:05:47,720 --> 00:05:49,600
What we mean by this is what?
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00:05:49,600 --> 00:05:53,240
Find all functions whose
derivative with respect to 'x'
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00:05:53,240 --> 00:05:55,100
is 'x cosine x'.
108
00:05:55,100 --> 00:05:58,780
Now, what integration by parts
did for us was it showed us
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00:05:58,780 --> 00:06:00,330
how to get this result.
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00:06:00,330 --> 00:06:03,620
But if we didn't know how to get
this result, observe that
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00:06:03,620 --> 00:06:06,030
we check the answer the
same way as always.
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00:06:06,030 --> 00:06:09,100
Namely, one thing that hopefully
we can do is
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00:06:09,100 --> 00:06:10,600
differentiate this thing.
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00:06:10,600 --> 00:06:12,040
And if we differentiate
it, let's see.
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00:06:12,040 --> 00:06:13,140
This is a product.
116
00:06:13,140 --> 00:06:15,490
It's the first factor times the
derivative of the second.
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00:06:15,490 --> 00:06:17,380
That's 'x cosine x'.
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00:06:17,380 --> 00:06:20,430
Plus the second factor times the
derivative of the first.
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00:06:20,430 --> 00:06:22,100
That's just 'sine x'.
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00:06:22,100 --> 00:06:24,720
The derivative of 'cosine
x' is minus 'sine x'.
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00:06:24,720 --> 00:06:26,350
The derivative of
a constant is 0.
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00:06:26,350 --> 00:06:29,440
And if we now sum up the pieces
that we have here, we
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00:06:29,440 --> 00:06:31,900
wind up with 'x cosine x'.
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00:06:31,900 --> 00:06:35,650
Exactly what we were supposed
to get, OK?
125
00:06:35,650 --> 00:06:38,190
Now, you see again, nothing
has happened in theory.
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00:06:38,190 --> 00:06:42,680
All we have now is a technique
that allows us to explicitly
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00:06:42,680 --> 00:06:45,210
solve certain problems that
we couldn't have solved
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00:06:45,210 --> 00:06:48,420
otherwise, even though we
could've stated them before.
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00:06:48,420 --> 00:06:51,570
For example, let's go back to
a problem that technically
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00:06:51,570 --> 00:06:54,150
speaking we could have presented
before we got to
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00:06:54,150 --> 00:06:56,340
this unit of our material.
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00:06:56,340 --> 00:06:59,930
Let's support that the region
'R' is the region bounded
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00:06:59,930 --> 00:07:04,240
above by 'y' equals 'cosine
x', below by the x-axis
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00:07:04,240 --> 00:07:07,490
between the y-axis at
'x' equals pi/2.
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00:07:07,490 --> 00:07:11,040
What we want to do is take this
region 'R' and revolve it
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00:07:11,040 --> 00:07:12,530
about the y-axis.
137
00:07:12,530 --> 00:07:16,100
And what we'd like to find is
what volume is generated when
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00:07:16,100 --> 00:07:19,160
the region 'R' is revolved
about the y-axis.
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00:07:19,160 --> 00:07:22,020
Now, when we talked about
volumes and cylindrical
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00:07:22,020 --> 00:07:24,420
shells, we already
had the recipe.
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00:07:24,420 --> 00:07:27,630
Recall that to rotate this
about the y-axis, just to
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00:07:27,630 --> 00:07:30,940
refresh memories here, if we
call this 'x' and we call this
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00:07:30,940 --> 00:07:33,620
'y', the element of
volume was what?
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00:07:33,620 --> 00:07:38,010
Well, you swung out a
circumference '2 pi x' times
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00:07:38,010 --> 00:07:40,460
height 'y' times
thickness 'dx'.
146
00:07:40,460 --> 00:07:43,590
In other words, that the volume
when this was rotated
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00:07:43,590 --> 00:07:50,620
about the y-axis was just 2 pi
integral 'x 'cosine x' dx' as
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00:07:50,620 --> 00:07:52,850
'x' goes from 0 to pi/2.
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00:07:52,850 --> 00:07:55,300
You see, I rigged this
particular problem so we
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00:07:55,300 --> 00:07:57,800
wouldn't have to waste the fact
that we've already found
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00:07:57,800 --> 00:08:00,040
out what the indefinite
integral of 'x
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00:08:00,040 --> 00:08:01,650
'cosine x' dx' is.
153
00:08:01,650 --> 00:08:03,870
In other words, we get down
to this step here.
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00:08:03,870 --> 00:08:07,180
Now, by the first fundamental
theorem of integral calculus,
155
00:08:07,180 --> 00:08:10,490
we know that whatever this
integral is, it's also equal
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00:08:10,490 --> 00:08:15,830
to 'G of pi/2' minus 'G of 0',
where 'G' is any function of
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00:08:15,830 --> 00:08:18,670
'x' whose derivative with
respect to 'x' is the
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00:08:18,670 --> 00:08:21,690
integrand here, which
is 'x cosine x'.
159
00:08:21,690 --> 00:08:25,040
Now, you see, this part, or up
to here, we could have got
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00:08:25,040 --> 00:08:27,610
this result without
today's lecture.
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00:08:27,610 --> 00:08:30,370
The only thing that we learned
to do today that we couldn't
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00:08:30,370 --> 00:08:34,059
do before was the idea of how
we find a function whose
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00:08:34,059 --> 00:08:35,900
derivative is 'x cosine x'.
164
00:08:35,900 --> 00:08:38,030
In fact, what we
found was what?
165
00:08:38,030 --> 00:08:42,289
A function whose derivative is
'x cosine x' is just 'x sine
166
00:08:42,289 --> 00:08:44,390
x' plus 'cosine x'.
167
00:08:44,390 --> 00:08:47,220
So by the first fundamental
theorem, the answer to our
168
00:08:47,220 --> 00:08:51,810
volume problem is that it's 2 pi
times the quantity 'x sine
169
00:08:51,810 --> 00:08:56,330
x' plus 'cosine x' evaluated
between 0 and pi/2.
170
00:08:56,330 --> 00:09:00,660
Putting in 'x' equals pi/2,
this term is pi/2.
171
00:09:00,660 --> 00:09:02,240
This term is 0.
172
00:09:02,240 --> 00:09:05,890
Putting in the lower limit when
'x' is 0, this term is 0.
173
00:09:05,890 --> 00:09:07,030
This term is 1.
174
00:09:07,030 --> 00:09:08,740
We subtract the lower limit.
175
00:09:08,740 --> 00:09:10,310
That gives us a minus 1.
176
00:09:10,310 --> 00:09:14,860
And therefore, the volume that
we get is simply 2 pi times
177
00:09:14,860 --> 00:09:16,860
pi/2 minus 1.
178
00:09:16,860 --> 00:09:19,830
Again, notice then, that the
only new thing that we had to
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00:09:19,830 --> 00:09:23,890
do here was devise a way for
finding a function whose
180
00:09:23,890 --> 00:09:27,510
derivative with respect to
'x' was 'x cosine x'.
181
00:09:27,510 --> 00:09:30,330
By the way, I think you'll
observe over here that I
182
00:09:30,330 --> 00:09:34,210
started in kind of a rigged
fashion, namely, I started
183
00:09:34,210 --> 00:09:36,040
with a specific 'u' and 'v'.
184
00:09:36,040 --> 00:09:39,290
It might be interesting to see
how would one tackle this
185
00:09:39,290 --> 00:09:43,500
problem if one had just been
given the 'x 'cosine x' dx'
186
00:09:43,500 --> 00:09:45,040
and had to work from there.
187
00:09:45,040 --> 00:09:47,960
In other words, instead of
working backwards now, let's
188
00:09:47,960 --> 00:09:50,310
present the problem as it
would come up in a real
189
00:09:50,310 --> 00:09:53,430
problem, namely, all of a
sudden, we find that we have
190
00:09:53,430 --> 00:09:55,990
to integrate 'x cosine x'.
191
00:09:55,990 --> 00:09:58,100
And the idea go something
like this.
192
00:09:58,100 --> 00:09:59,120
You see, we say lookit.
193
00:09:59,120 --> 00:10:01,790
If this 'x' weren't in
here, this would be
194
00:10:01,790 --> 00:10:02,940
pretty easy to handle.
195
00:10:02,940 --> 00:10:04,130
So why don't we do this?
196
00:10:04,130 --> 00:10:07,260
Why don't we call this piece
here 'u' and call
197
00:10:07,260 --> 00:10:10,060
this piece here 'dv'.
198
00:10:10,060 --> 00:10:14,730
Then, you see, 'du' would
just be 'dx'.
199
00:10:14,730 --> 00:10:17,200
'v' would just be 'sine x'.
200
00:10:17,200 --> 00:10:20,230
In other words, the differential
of 'sine x' is
201
00:10:20,230 --> 00:10:21,970
'cosine x dx'.
202
00:10:21,970 --> 00:10:24,640
See, if 'u' is 'x',
'du' is 'dx'.
203
00:10:24,640 --> 00:10:28,480
If 'dv' is 'cosine x dx', 'v'
is equal to 'sine x'.
204
00:10:28,480 --> 00:10:31,070
I suppose, technically speaking
here, I should put in
205
00:10:31,070 --> 00:10:35,060
plus a constant, but I'll show
later why we can disregard
206
00:10:35,060 --> 00:10:36,010
this constant.
207
00:10:36,010 --> 00:10:38,290
Roughly speaking, it just hinges
on the fact that we
208
00:10:38,290 --> 00:10:39,770
want only one solution.
209
00:10:39,770 --> 00:10:41,330
But we'll talk about
that later.
210
00:10:41,330 --> 00:10:44,380
For the time being, I'll just
leave out the fact that
211
00:10:44,380 --> 00:10:45,590
there's a whole family
of functions.
212
00:10:45,590 --> 00:10:48,170
All I want is one function
whose differential
213
00:10:48,170 --> 00:10:49,140
is 'cosine x dx'.
214
00:10:49,140 --> 00:10:54,520
At any rate, I now apply my
recipe, namely, integral 'udv'
215
00:10:54,520 --> 00:10:57,260
is equal to 'u' times 'v'--
216
00:10:57,260 --> 00:10:59,100
that's 'x sine x'--
217
00:10:59,100 --> 00:11:03,580
minus integral 'v', which
is 'sine x', times 'du',
218
00:11:03,580 --> 00:11:05,050
which is just 'dx'.
219
00:11:05,050 --> 00:11:08,740
In other words, integral 'x
'cosine x' dx' is just 'x sine
220
00:11:08,740 --> 00:11:11,210
x' minus 'sine x dx'.
221
00:11:11,210 --> 00:11:13,620
But I know how to integrate
'sine x'.
222
00:11:13,620 --> 00:11:15,500
It's just minus 'cosine x'.
223
00:11:15,500 --> 00:11:19,720
Therefore, replacing integral
'sine x dx' by minus 'cosine
224
00:11:19,720 --> 00:11:21,500
x' plus a constant.
225
00:11:21,500 --> 00:11:22,860
The minus cancels here.
226
00:11:22,860 --> 00:11:26,830
I wind up with the same result
as before, only now hopefully
227
00:11:26,830 --> 00:11:29,410
in terms of a smooth flow
not that didn't
228
00:11:29,410 --> 00:11:30,950
presuppose the answer.
229
00:11:30,950 --> 00:11:33,860
Namely, this integral
is 'x sine x' plus
230
00:11:33,860 --> 00:11:36,000
'cosine x' plus a constant.
231
00:11:36,000 --> 00:11:38,400
And by the way, I should
mention here that this
232
00:11:38,400 --> 00:11:42,060
technique does require a certain
amount of ingenuity.
233
00:11:42,060 --> 00:11:45,300
In other words, in a given
problem, where you're not told
234
00:11:45,300 --> 00:11:48,580
what to call 'u' and what to
call 'v', where you have the
235
00:11:48,580 --> 00:11:51,820
choice of picking what's going
to be 'u' and what's going to
236
00:11:51,820 --> 00:11:55,610
be 'dv', there is a certain
element of insight that one
237
00:11:55,610 --> 00:11:57,880
needs to guess wisely.
238
00:11:57,880 --> 00:12:00,470
In this particular problem,
it's kind of obvious
239
00:12:00,470 --> 00:12:01,360
which way to guess.
240
00:12:01,360 --> 00:12:04,370
But let me, just to give you the
experience, show you what
241
00:12:04,370 --> 00:12:06,270
happens when you guess
incorrectly.
242
00:12:06,270 --> 00:12:09,600
For example, given the same
problem, why couldn't I have
243
00:12:09,600 --> 00:12:14,100
said gee, I think I'll let
'cosine x' be 'u', and what's
244
00:12:14,100 --> 00:12:16,290
left over, namely, 'xdx'.
245
00:12:16,290 --> 00:12:18,790
That will be 'dv'.
246
00:12:18,790 --> 00:12:20,850
Now, notice that mathematically,
no harm is
247
00:12:20,850 --> 00:12:21,760
done this way.
248
00:12:21,760 --> 00:12:26,080
Namely, if 'u' is 'cosine x',
'du' is minus 'sine x dx'.
249
00:12:26,080 --> 00:12:30,040
If 'dv' is 'xdx', 'v'
is 1/2 'x squared'.
250
00:12:30,040 --> 00:12:34,260
And now, applying the recipe,
integral 'udv' is what?
251
00:12:34,260 --> 00:12:35,460
It's 'uv'.
252
00:12:35,460 --> 00:12:40,050
That's 1/2 'x squared' times
'cosine x' minus
253
00:12:40,050 --> 00:12:42,140
the integral 'vdu'.
254
00:12:42,140 --> 00:12:47,780
But 'vdu' is already minus 1/2
'x squared' 'sine x dx', so
255
00:12:47,780 --> 00:12:51,020
minus that is plus
1/2 integral ''x
256
00:12:51,020 --> 00:12:52,670
squared' 'sine x' dx'.
257
00:12:52,670 --> 00:12:55,920
And I wind up with the result
that the integral 'x 'cosine
258
00:12:55,920 --> 00:13:00,820
x' dx' is 1/2 'x squared'
'cosine x' plus 1/2 integral
259
00:13:00,820 --> 00:13:02,690
''x squared' 'sine x' dx'.
260
00:13:02,690 --> 00:13:05,460
And there is absolutely nothing
at all wrong with
261
00:13:05,460 --> 00:13:12,420
this, except that the chances
are if I didn't know how to
262
00:13:12,420 --> 00:13:15,640
find the function whose
derivative was 'x cosine x',
263
00:13:15,640 --> 00:13:18,330
what is the likelihood that
I know a function whose
264
00:13:18,330 --> 00:13:20,760
derivative is ''x squared'
'sine x' dx'?
265
00:13:20,760 --> 00:13:26,140
In other words, here's a case
where replacing 'udv' by 'vdu'
266
00:13:26,140 --> 00:13:29,930
gave me an equivalent integral
to evaluate which was more
267
00:13:29,930 --> 00:13:31,830
cumbersome than the first.
268
00:13:31,830 --> 00:13:35,140
In fact, somehow or other, it
would seem that by multiplying
269
00:13:35,140 --> 00:13:39,560
through here by 2 and solving
for ''x squared' 'sine x' dx',
270
00:13:39,560 --> 00:13:42,940
that it would seem that somehow
or other this would be
271
00:13:42,940 --> 00:13:45,560
a way of simplifying
the integral ''x
272
00:13:45,560 --> 00:13:47,210
squared' 'sine x' dx'.
273
00:13:47,210 --> 00:13:50,170
In other words, if you had
started with this, I think
274
00:13:50,170 --> 00:13:53,110
you'd accept the fact, well,
this still isn't as nice as
275
00:13:53,110 --> 00:13:57,430
I'd like it, but it does seem
less complicated than this.
276
00:13:57,430 --> 00:13:59,730
You see, at least the power of
'x' is only to the first power
277
00:13:59,730 --> 00:14:02,050
here, whereas here it's
the second power.
278
00:14:02,050 --> 00:14:04,790
In fact, this leads
to a technique
279
00:14:04,790 --> 00:14:07,170
called a reduction formula.
280
00:14:07,170 --> 00:14:10,270
Again, using hindsight rather
than foresight, having
281
00:14:10,270 --> 00:14:13,850
motivated the integral ''x
squared' 'sine x' dx', let's
282
00:14:13,850 --> 00:14:16,640
see how I could have solved this
particular problem if I
283
00:14:16,640 --> 00:14:20,380
didn't know what the answer was
going to be in advance.
284
00:14:20,380 --> 00:14:23,000
You see, I say to myself, you
know, this x squared here is
285
00:14:23,000 --> 00:14:23,980
bothering me.
286
00:14:23,980 --> 00:14:26,270
If that 'x squared' weren't
here, this would be pretty
287
00:14:26,270 --> 00:14:27,280
easy to handle.
288
00:14:27,280 --> 00:14:30,560
Well, the best way to knock the
'x squared' down is to let
289
00:14:30,560 --> 00:14:32,840
'u' equal 'x squared',
in which case,
290
00:14:32,840 --> 00:14:34,910
'du' will be '2xdx'.
291
00:14:34,910 --> 00:14:38,710
In other words, your second
power will now be replaced by
292
00:14:38,710 --> 00:14:39,660
a first power.
293
00:14:39,660 --> 00:14:42,810
It's not perfect, but at least
there's hope here.
294
00:14:42,810 --> 00:14:46,660
At any rate, what I'm saying is
motivated into calling 'u'
295
00:14:46,660 --> 00:14:50,670
'x squared' and therefore
letting 'dv' be the rest of
296
00:14:50,670 --> 00:14:53,870
this integrand, which is 'sine
x dx', we wind up with the
297
00:14:53,870 --> 00:14:59,190
fact that 'du' is '2xdx' and
'v' is minus 'cosine x'.
298
00:14:59,190 --> 00:15:03,105
Well, you see now, remembering
that 'u' is 'x squared', 'dv'
299
00:15:03,105 --> 00:15:07,420
is 'sine x dx', the recipe tells
us that integral 'udv'
300
00:15:07,420 --> 00:15:09,360
is 'u' times 'v'.
301
00:15:09,360 --> 00:15:16,290
That's minus 'x cosine x'
minus integral 'vdu'.
302
00:15:16,290 --> 00:15:21,480
But integral 'dvu' is integral
twice 'x 'cosine x' dx' with a
303
00:15:21,480 --> 00:15:22,550
minus sign.
304
00:15:22,550 --> 00:15:25,040
The minus sign in the recipe
makes it plus.
305
00:15:25,040 --> 00:15:28,350
And we therefore wind up with
that integral ''x squared'
306
00:15:28,350 --> 00:15:33,470
'sine x' dx' is equal to minus
''x squared' 'cosine x'' plus
307
00:15:33,470 --> 00:15:37,650
twice integral 'x
'cosine x' dx'.
308
00:15:37,650 --> 00:15:40,640
By the way, just to refresh your
memories on this, if we
309
00:15:40,640 --> 00:15:44,880
come back and compare this with
what we had over here,
310
00:15:44,880 --> 00:15:47,820
it's exactly the same thing.
311
00:15:47,820 --> 00:15:50,080
The only difference now
is that we arrived
312
00:15:50,080 --> 00:15:51,980
at this thing directly.
313
00:15:51,980 --> 00:15:56,220
Now you see the point is, having
reduced this to this, I
314
00:15:56,220 --> 00:16:00,190
can now use integration by parts
again to simplify this.
315
00:16:00,190 --> 00:16:03,150
In other words, what I would do
next is let 'u' equal 'x'
316
00:16:03,150 --> 00:16:05,490
and 'dv' be 'cosine x dx'.
317
00:16:05,490 --> 00:16:07,730
But let's face it, I've already
done this problem.
318
00:16:07,730 --> 00:16:08,950
Time is getting short.
319
00:16:08,950 --> 00:16:10,390
Why do it again?
320
00:16:10,390 --> 00:16:14,670
The point is that integral 'x
cosine x' using integration by
321
00:16:14,670 --> 00:16:18,540
parts again is going to come
out to be 'x sine x' plus
322
00:16:18,540 --> 00:16:22,060
'cosine x' plus 'c' because
that's what we got before when
323
00:16:22,060 --> 00:16:24,310
we used integration by
parts to do this.
324
00:16:24,310 --> 00:16:28,190
At any rate, what I would now
do is replace integral 'x
325
00:16:28,190 --> 00:16:32,200
cosine x' in here by this, OK?
326
00:16:32,200 --> 00:16:34,060
And if I do that, I have what?
327
00:16:34,060 --> 00:16:40,750
Integral ''x squared' 'sine
x' dx' is equal to what?
328
00:16:40,750 --> 00:16:45,470
Minus ''x squared' 'cosine x''
plus twice times this, which
329
00:16:45,470 --> 00:16:46,600
is twice this.
330
00:16:46,600 --> 00:16:47,340
That's what?
331
00:16:47,340 --> 00:16:53,050
2 'x sine x' plus 2 'cosine
x' plus a constant.
332
00:16:53,050 --> 00:16:56,310
And by the way, again notice,
if I wanted to check whether
333
00:16:56,310 --> 00:17:00,840
this was right, the one thing I
can do if somebody gives me
334
00:17:00,840 --> 00:17:02,320
this is what?
335
00:17:02,320 --> 00:17:05,690
I can differentiate it
with respect to 'x'.
336
00:17:05,690 --> 00:17:08,560
When I differentiate it with
respect to 'x', if I get ''x
337
00:17:08,560 --> 00:17:12,040
squared' 'sine x'' as my result,
this is correct.
338
00:17:12,040 --> 00:17:13,630
Otherwise, I've made
a mistake.
339
00:17:13,630 --> 00:17:16,740
But the idea is keep in mind
that these can always be
340
00:17:16,740 --> 00:17:19,250
checked because they're
inverse derivatives.
341
00:17:19,250 --> 00:17:22,670
They can always be checked by
actual differentiation.
342
00:17:22,670 --> 00:17:25,849
And again, notice the beauty
of this technique.
343
00:17:25,849 --> 00:17:29,310
If somebody said construct the
function whose derivative with
344
00:17:29,310 --> 00:17:32,730
respect to 'x' is ''x squared'
'sine x'', notice how
345
00:17:32,730 --> 00:17:37,630
difficult it would be to try to
piece this thing together.
346
00:17:37,630 --> 00:17:39,800
See, notice the difference
between being able to check
347
00:17:39,800 --> 00:17:43,060
that this is a right answer and
how one would find this.
348
00:17:43,060 --> 00:17:46,980
This is the technique of a power
of integration by parts.
349
00:17:46,980 --> 00:17:50,190
In fact, in making up exam
questions, this is exactly how
350
00:17:50,190 --> 00:17:52,000
professor sometimes do this.
351
00:17:52,000 --> 00:17:55,610
That what you start with is you
start with this, then give
352
00:17:55,610 --> 00:17:58,300
a person this problem and tell
them to differentiate this.
353
00:17:58,300 --> 00:18:00,140
And that whole mess simply
comes out to be ''x
354
00:18:00,140 --> 00:18:01,270
squared' 'sine x''.
355
00:18:01,270 --> 00:18:03,830
And that's why a lot of times
when you have a messy problem
356
00:18:03,830 --> 00:18:06,190
to differentiate and you look
at the answer in the back of
357
00:18:06,190 --> 00:18:09,060
the book, you become amazed to
see that the thing really
358
00:18:09,060 --> 00:18:09,910
simplified.
359
00:18:09,910 --> 00:18:11,330
Again, notice the idea.
360
00:18:11,330 --> 00:18:14,630
The integrand here is
relatively simple.
361
00:18:14,630 --> 00:18:18,350
What we have to differentiate to
get that relatively simple
362
00:18:18,350 --> 00:18:20,960
integrand is not quite
so simple.
363
00:18:20,960 --> 00:18:23,150
But enough about that.
364
00:18:23,150 --> 00:18:25,970
Let me just go on with
a few more examples.
365
00:18:25,970 --> 00:18:29,450
See, I have plenty of examples
in the exercises to keep you
366
00:18:29,450 --> 00:18:30,980
busy and to give you
drill on this.
367
00:18:30,980 --> 00:18:34,450
All I'm trying to do now is to
motivate places where one
368
00:18:34,450 --> 00:18:36,520
would want to use integration
by parts.
369
00:18:36,520 --> 00:18:39,700
Well, let me show you one
place it might be used.
370
00:18:39,700 --> 00:18:42,820
One place that it's always worth
trying is when you look
371
00:18:42,820 --> 00:18:45,680
at the integrand, haven't the
vaguest notion as to how to
372
00:18:45,680 --> 00:18:49,210
integrate it, but you are able
to differentiate it.
373
00:18:49,210 --> 00:18:50,600
You see, you've got
nothing to lose.
374
00:18:50,600 --> 00:18:53,370
The worst that will happen is
that the resulting integral
375
00:18:53,370 --> 00:18:57,660
vdu will be no easier to handle
than the original udv,
376
00:18:57,660 --> 00:18:59,800
but at least there's a chance
that something will work.
377
00:18:59,800 --> 00:19:01,720
But let me give you
a for instance.
378
00:19:01,720 --> 00:19:03,290
I see I've written this
rather badly.
379
00:19:03,290 --> 00:19:04,570
This is not 'Stan'.
380
00:19:04,570 --> 00:19:10,080
This is the integral of inverse
'tangent x dx', OK?
381
00:19:10,080 --> 00:19:12,210
Integral inverse
'tangent x dx'.
382
00:19:12,210 --> 00:19:15,780
What function has its derivative
equal to inverse
383
00:19:15,780 --> 00:19:16,720
'tangent x'?
384
00:19:16,720 --> 00:19:19,870
Well, you see, the thing I do
know hopefully is what the
385
00:19:19,870 --> 00:19:22,320
derivative of inverse
'tangent x' is.
386
00:19:22,320 --> 00:19:25,100
The derivative of inverse
'tangent x' with respect to
387
00:19:25,100 --> 00:19:27,430
'x' is '1 over '1 plus
x squared''.
388
00:19:27,430 --> 00:19:29,760
And by the way, if you don't
know this, there's no hope of
389
00:19:29,760 --> 00:19:32,360
even beginning the integration
by parts. in other words, we
390
00:19:32,360 --> 00:19:34,440
have to assume that we
can do something in a
391
00:19:34,440 --> 00:19:35,280
problem like this.
392
00:19:35,280 --> 00:19:38,070
At any rate, knowing this,
what do I do next?
393
00:19:38,070 --> 00:19:41,180
I say, well, since I know how
to differentiate the inverse
394
00:19:41,180 --> 00:19:44,500
tangent, that's a very natural
thing to call 'u' because I
395
00:19:44,500 --> 00:19:46,170
can certainly find 'du'.
396
00:19:46,170 --> 00:19:50,030
Moreover, the nice thing is that
calling 'dv' 'dx', if I
397
00:19:50,030 --> 00:19:52,870
can integrate anything,
it's certainly 'dx'.
398
00:19:52,870 --> 00:19:56,280
Namely, what I'm saying is if
'u' equals inverse 'tan x' and
399
00:19:56,280 --> 00:20:00,790
'dv' is 'dx', then 'du' is 'dx'
over '1 plus x squared',
400
00:20:00,790 --> 00:20:02,980
and 'v' is just equal to 'x'.
401
00:20:02,980 --> 00:20:06,320
Now, therefore, integral inverse
'tan x dx', in other
402
00:20:06,320 --> 00:20:08,870
words, integral 'udv',
is just what?
403
00:20:08,870 --> 00:20:11,150
It's still just 'u' times 'v'.
404
00:20:11,150 --> 00:20:18,200
That's 'x' times inverse
'tan x' minus 'vdu'.
405
00:20:18,200 --> 00:20:21,730
That's minus integral
'x' times 'dx'
406
00:20:21,730 --> 00:20:23,790
over '1 plus x squared'.
407
00:20:23,790 --> 00:20:25,530
Now, look what happens
in this case.
408
00:20:25,530 --> 00:20:28,580
Obviously, with time being short
and having other places
409
00:20:28,580 --> 00:20:31,520
to show you my failures, namely
in the exercises, I
410
00:20:31,520 --> 00:20:33,120
picked one that worked nicely.
411
00:20:33,120 --> 00:20:37,260
Here's the case, you see, that
the integral inverse 'tan x
412
00:20:37,260 --> 00:20:43,590
dx', by using parts, becomes
replaced by integral 'xdx'
413
00:20:43,590 --> 00:20:45,300
over '1 plus x squared'.
414
00:20:45,300 --> 00:20:47,880
This is one I can handle
from before.
415
00:20:47,880 --> 00:20:50,500
This is one where the
substitution 'u' equals '1
416
00:20:50,500 --> 00:20:53,150
plus x squared' works
very nicely for me.
417
00:20:53,150 --> 00:20:56,160
Or if you're getting glib with
these things, just observe
418
00:20:56,160 --> 00:20:59,970
that this integrand
is 1/2 natural
419
00:20:59,970 --> 00:21:01,630
log '1 plus x squared'.
420
00:21:01,630 --> 00:21:03,810
Because how would you
differentiate natural log '1
421
00:21:03,810 --> 00:21:05,010
plus x squared'?
422
00:21:05,010 --> 00:21:09,760
It would just be what? '1 over
'1 plus x squared'' times the
423
00:21:09,760 --> 00:21:11,370
derivative of this,
which is '2x'.
424
00:21:11,370 --> 00:21:14,750
The 2 and the 1/2 would cancel,
and we just have left
425
00:21:14,750 --> 00:21:17,280
'x over '1 plus x squared''.
426
00:21:17,280 --> 00:21:21,540
In other words, again, notice
that the answer here is what?
427
00:21:21,540 --> 00:21:24,710
This is the function that one
would have to differentiate
428
00:21:24,710 --> 00:21:29,730
with respect to 'x' to wind
up with inverse 'tan x'.
429
00:21:29,730 --> 00:21:31,810
I'll show you another example
in a few minutes.
430
00:21:31,810 --> 00:21:34,800
What I did just remember is that
I mentioned why we can
431
00:21:34,800 --> 00:21:36,370
drop the arbitrary constant.
432
00:21:36,370 --> 00:21:39,250
Let me just take a minute here
and show how this problem
433
00:21:39,250 --> 00:21:40,600
should have really
been worked.
434
00:21:40,600 --> 00:21:44,180
If we wanted to be precise here,
we'd say OK, 'u' equals
435
00:21:44,180 --> 00:21:45,620
inverse 'tan x'.
436
00:21:45,620 --> 00:21:48,850
Therefore, 'du' is 'dx over
'1 plus x squared''.
437
00:21:48,850 --> 00:21:50,220
That part's fine.
438
00:21:50,220 --> 00:21:52,810
Then we say, OK, 'dv' is 'dx'.
439
00:21:52,810 --> 00:21:55,850
Now, if 'dv' is 'dx',
what must 'v' be?
440
00:21:55,850 --> 00:21:59,280
Well, 'v' could be 'x', but it
must be something of the form
441
00:21:59,280 --> 00:22:01,950
'x' plus some constant,
which I'll call 'c1'.
442
00:22:01,950 --> 00:22:05,240
Could I have chosen 'v'
to be 'x plus c1'?
443
00:22:05,240 --> 00:22:06,240
Why not?
444
00:22:06,240 --> 00:22:08,490
In fact, what would have
happened if I did that?
445
00:22:08,490 --> 00:22:10,620
Notice that what would have
happened in this case is that
446
00:22:10,620 --> 00:22:15,490
'u' times 'v' now would simply
have been 'x plus c1' times
447
00:22:15,490 --> 00:22:20,060
inverse 'tan x', whereas
integral 'vdu', instead of
448
00:22:20,060 --> 00:22:23,610
being 'xdx' over '1 plus x
squared' would have been ''x
449
00:22:23,610 --> 00:22:26,840
plus c1' dx' over '1
plus x squared'.
450
00:22:26,840 --> 00:22:30,090
In other words, leaving the
constant in, I would get this.
451
00:22:30,090 --> 00:22:32,810
And now all I want you to see
here, and I've done this just
452
00:22:32,810 --> 00:22:35,860
for this particular example,
and in the exercises we'll
453
00:22:35,860 --> 00:22:37,100
work it out in general.
454
00:22:37,100 --> 00:22:38,430
The whole idea is this.
455
00:22:38,430 --> 00:22:40,880
This, of course, can be written
as two terms, one of
456
00:22:40,880 --> 00:22:44,880
which is 'x inverse tan x', and
the other of which is 'c1
457
00:22:44,880 --> 00:22:46,990
inverse tan x'.
458
00:22:46,990 --> 00:22:49,770
This also can be written
as two integrals.
459
00:22:49,770 --> 00:22:53,150
Namely, with the minus sign
kept aside here, this is
460
00:22:53,150 --> 00:22:58,390
integral of 'xdx' over '1 plus
x squared' and plus integral
461
00:22:58,390 --> 00:23:02,500
'c1 dx' over '1 plus x squared',
the minus making
462
00:23:02,500 --> 00:23:03,440
this minus.
463
00:23:03,440 --> 00:23:07,270
But we know that the integral
'dx' over '1 plus x squared'
464
00:23:07,270 --> 00:23:10,050
is inverse 'tan x'.
465
00:23:10,050 --> 00:23:12,350
In other words, if we now
separate these into
466
00:23:12,350 --> 00:23:14,030
pieces, we have what?
467
00:23:14,030 --> 00:23:17,600
'x inverse tan x' minus
integral 'xdx'
468
00:23:17,600 --> 00:23:19,660
over '1 plus x squared'.
469
00:23:19,660 --> 00:23:24,040
And notice that the 'c1' term
drops out because we have 'c1
470
00:23:24,040 --> 00:23:26,670
inverse tan x' occurring
twice.
471
00:23:26,670 --> 00:23:32,160
It's this term here,
and it's also
472
00:23:32,160 --> 00:23:33,340
this part of our integral.
473
00:23:33,340 --> 00:23:36,050
And see, with the minus sign
here, they just cancel out.
474
00:23:36,050 --> 00:23:39,020
And that's why we really don't
have to keep the arbitrary
475
00:23:39,020 --> 00:23:42,400
constant in there, but if
we wish to, we may.
476
00:23:42,400 --> 00:23:45,080
At any rate, let me summarize
with just one more problem to
477
00:23:45,080 --> 00:23:49,570
make sure that we have the
technique down fairly pat.
478
00:23:49,570 --> 00:23:53,660
Let's suppose we were given
integral 'log x dx'.
479
00:23:53,660 --> 00:23:56,890
What we certainly do know I hope
by this time is that the
480
00:23:56,890 --> 00:24:00,580
derivative of 'log x' with
respect to 'x' is '1/x'.
481
00:24:00,580 --> 00:24:04,090
Since I can differentiate 'log
x', it becomes very meaningful
482
00:24:04,090 --> 00:24:07,270
to try letting 'u'
equal 'log x'.
483
00:24:07,270 --> 00:24:09,620
If I do this, 'u' is 'log x'.
484
00:24:09,620 --> 00:24:11,520
'dv' is 'dx'.
485
00:24:11,520 --> 00:24:15,340
'du' then becomes ''1 over
x' dx', and 'v' is 'x'.
486
00:24:15,340 --> 00:24:18,240
And now I put in plus 'c' in
parentheses here to indicate
487
00:24:18,240 --> 00:24:20,080
that if you wanted to
use it, you could.
488
00:24:20,080 --> 00:24:21,900
At any rate, what happens now?
489
00:24:21,900 --> 00:24:26,150
Integral udv, integral 'log x
dx' in this case is 'u' times
490
00:24:26,150 --> 00:24:30,530
'v', which is 'x log x',
minus integral 'vdu'.
491
00:24:30,530 --> 00:24:33,830
Well, you see the 'x' times
the '1/x' is just 1.
492
00:24:33,830 --> 00:24:38,920
Therefore, 'vdu' is just
'dx', so minus that
493
00:24:38,920 --> 00:24:39,870
is just minus 'dx'.
494
00:24:39,870 --> 00:24:43,170
In other words, minus the
integral 'vdu' is just minus
495
00:24:43,170 --> 00:24:44,450
integral 'dx'.
496
00:24:44,450 --> 00:24:47,100
And now, you see I have
what? 'x log x'.
497
00:24:47,100 --> 00:24:49,020
And this is very easy
to integrate.
498
00:24:49,020 --> 00:24:51,890
It just happens to be minus
'x', then I tack on the
499
00:24:51,890 --> 00:24:53,600
arbitrary constant.
500
00:24:53,600 --> 00:24:56,830
Again, to check whether this
was the right answer, all I
501
00:24:56,830 --> 00:25:00,720
have to do is differentiate 'x
log x' minus 'x' with respect
502
00:25:00,720 --> 00:25:05,070
to 'x' and see if I get 'log
x', and I will, OK?
503
00:25:05,070 --> 00:25:07,220
Now, how would I use a problem
like this, say,
504
00:25:07,220 --> 00:25:08,740
in a practical situation?
505
00:25:08,740 --> 00:25:12,280
Well, for example, suppose I
wanted to find the area of the
506
00:25:12,280 --> 00:25:16,060
region 'R' where 'R' was the
region bounded above by the
507
00:25:16,060 --> 00:25:20,540
curve 'y' equals 'natural log
x', below by the x-axis, on
508
00:25:20,540 --> 00:25:23,710
the left by 'x' equals 1, and
on the right by 'x' equals
509
00:25:23,710 --> 00:25:26,580
'b', where 'b' can be any number
you want right now as
510
00:25:26,580 --> 00:25:28,630
long as it's bigger than
1 in this diagram.
511
00:25:28,630 --> 00:25:31,820
Notice that the definite
integral says that the area of
512
00:25:31,820 --> 00:25:34,110
the region 'R' is the definite
integral from 1
513
00:25:34,110 --> 00:25:36,250
to 'b', 'log x dx'.
514
00:25:36,250 --> 00:25:37,930
And the first fundamental
theorem of
515
00:25:37,930 --> 00:25:39,320
calculus says lookit.
516
00:25:39,320 --> 00:25:41,720
If you happen to know a function
whose derivative with
517
00:25:41,720 --> 00:25:44,100
respect to 'x' is 'log
x', you can evaluate
518
00:25:44,100 --> 00:25:45,460
this thing very quickly.
519
00:25:45,460 --> 00:25:49,140
Namely, it's just 'G of b' minus
'G of 1' where 'G prime
520
00:25:49,140 --> 00:25:50,470
of x' is 'log x'.
521
00:25:50,470 --> 00:25:54,050
Well, you see, we do know a
function now whose derivative
522
00:25:54,050 --> 00:25:57,160
with respect to 'x' is
'log x', namely, 'x
523
00:25:57,160 --> 00:25:59,050
log x' minus 'x'.
524
00:25:59,050 --> 00:26:02,550
Consequently, to solve this
problem, all I have to do is
525
00:26:02,550 --> 00:26:06,010
evaluate this between the limits
of 1 and 'b', which I
526
00:26:06,010 --> 00:26:09,780
won't bother doing
right now, OK?
527
00:26:09,780 --> 00:26:11,180
Now, here's the idea.
528
00:26:11,180 --> 00:26:14,940
What we've really learned here
is one more powerful technique
529
00:26:14,940 --> 00:26:16,870
for evaluating integrals.
530
00:26:16,870 --> 00:26:20,620
How we arrive at integrals
in no way changes in this
531
00:26:20,620 --> 00:26:21,710
particular lecture.
532
00:26:21,710 --> 00:26:23,260
All we have now is what?
533
00:26:23,260 --> 00:26:28,500
A new technique whereby we
evaluate udv by being able to
534
00:26:28,500 --> 00:26:32,080
evaluate equivalently vdu.
535
00:26:32,080 --> 00:26:36,040
At any rate, that finishes up
what we had in mind for today,
536
00:26:36,040 --> 00:26:37,760
and so until next
time, goodbye.
537
00:26:40,530 --> 00:26:43,070
NARRATOR: Funding for the
publication of this video was
538
00:26:43,070 --> 00:26:47,780
provided by the Gabriella and
Paul Rosenbaum Foundation.
539
00:26:47,780 --> 00:26:51,960
Help OCW continue to provide
free and open access to MIT
540
00:26:51,960 --> 00:26:56,160
courses by making a donation
at ocw.mit.edu/donate.