WEBVTT
1
00:00:00.240 --> 00:00:03.100 A:middle L:90%
No, this problem will use the integral test for
2
00:00:03.100 --> 00:00:05.250 A:middle L:90%
the interval test. We want for these terms to
3
00:00:05.250 --> 00:00:09.949 A:middle L:90%
eventually be strictly decreasing an absolute value to zero.
4
00:00:10.880 --> 00:00:19.300 A:middle L:90%
So for that to happen, you'd want for this
5
00:00:19.300 --> 00:00:26.289 A:middle L:90%
toe eventually, Just be increasing. So typically speaking
6
00:00:26.289 --> 00:00:28.750 A:middle L:90%
, if you know that these things air going toe
7
00:00:29.239 --> 00:00:33.090 A:middle L:90%
be going to zero, then you can guess that
8
00:00:33.090 --> 00:00:35.539 A:middle L:90%
they're going to be going to zero an absolute value
9
00:00:35.539 --> 00:00:38.340 A:middle L:90%
unless you have some trig functions involved. Because with
10
00:00:38.340 --> 00:00:42.450 A:middle L:90%
Trig functions you can imagine there being some oscillations occurring
11
00:00:43.170 --> 00:00:46.200 A:middle L:90%
. But generally speaking, it's a fairly good gas
12
00:00:46.200 --> 00:00:49.460 A:middle L:90%
. If it looks like it's going toe zero,
13
00:00:49.460 --> 00:00:51.520 A:middle L:90%
then it's probably going to be going to zero when
14
00:00:51.520 --> 00:00:53.750 A:middle L:90%
you throw the absolute value signs around it as well
15
00:00:54.369 --> 00:00:58.090 A:middle L:90%
. Like I said, the exception typically being with
16
00:00:58.549 --> 00:00:59.969 A:middle L:90%
trig functions, you have to be a little bit
17
00:00:59.969 --> 00:01:02.090 A:middle L:90%
more careful with that. Ah, but it is
18
00:01:02.090 --> 00:01:03.789 A:middle L:90%
true if you want to be really rigorous and you
19
00:01:03.789 --> 00:01:06.060 A:middle L:90%
should look at this, you could take the derivative
20
00:01:06.060 --> 00:01:08.040 A:middle L:90%
of this and show that eventually the derivative is going
21
00:01:08.040 --> 00:01:12.349 A:middle L:90%
to be positive. The derivative of this is positive
22
00:01:12.349 --> 00:01:17.959 A:middle L:90%
for and large enough then that means that eventually this
23
00:01:17.959 --> 00:01:22.900 A:middle L:90%
is just going to be decreasing to zero. Case
24
00:01:22.900 --> 00:01:26.269 A:middle L:90%
of the integral test is applicable here. So the
25
00:01:26.269 --> 00:01:30.790 A:middle L:90%
integral test says that this thing is going to be
26
00:01:30.799 --> 00:01:34.859 A:middle L:90%
is going to have the same behavior. Is this
27
00:01:34.870 --> 00:01:49.209 A:middle L:90%
integral? Okay, so this is ahh u substitution
28
00:01:49.209 --> 00:01:53.209 A:middle L:90%
problem. So the U would be Ellen of N
29
00:01:57.040 --> 00:02:01.250 A:middle L:90%
D'You would be won over and Tien So then the
30
00:02:01.250 --> 00:02:09.479 A:middle L:90%
integral and consideration turns into Alan of to infinity.
31
00:02:10.919 --> 00:02:15.330 A:middle L:90%
You to the minus one half. See you.
32
00:02:15.819 --> 00:02:20.430 A:middle L:90%
So this lower bound here, we got that from
33
00:02:20.870 --> 00:02:23.250 A:middle L:90%
looking at the lower bound over here. So over
34
00:02:23.250 --> 00:02:24.349 A:middle L:90%
here are lower bound was too. And that correspondent
35
00:02:24.349 --> 00:02:28.969 A:middle L:90%
to end. So when in is too plugin and
36
00:02:28.969 --> 00:02:30.889 A:middle L:90%
has to hear and we get that you is natural
37
00:02:30.889 --> 00:02:32.120 A:middle L:90%
log of two. So that's why when we're working
38
00:02:32.120 --> 00:02:35.389 A:middle L:90%
with d ur Lord bound his natural log of two
39
00:02:36.039 --> 00:02:38.189 A:middle L:90%
Get the upper bound here. We looked to see
40
00:02:38.189 --> 00:02:40.080 A:middle L:90%
what's happening with up around here. So hear this
41
00:02:40.080 --> 00:02:44.810 A:middle L:90%
corresponds toe in being infinity. Go on in his
42
00:02:44.810 --> 00:02:47.460 A:middle L:90%
infinity, you is natural log of infinity, which
43
00:02:47.460 --> 00:02:53.580 A:middle L:90%
is still infinity. And then we replaced the one
44
00:02:53.580 --> 00:02:57.319 A:middle L:90%
over N d n a with d u. That's
45
00:02:57.319 --> 00:03:00.419 A:middle L:90%
what this equation tells us we Khun d'Oh And then
46
00:03:00.419 --> 00:03:05.270 A:middle L:90%
we had Then we're left with Do you want over
47
00:03:05.280 --> 00:03:07.539 A:middle L:90%
square root of natural log of end, but natural
48
00:03:07.539 --> 00:03:12.379 A:middle L:90%
log of in is just you sore left with one
49
00:03:12.379 --> 00:03:15.599 A:middle L:90%
over the square root of you which is just you
50
00:03:15.710 --> 00:03:19.120 A:middle L:90%
to the minus one half. Okay, so,
51
00:03:19.129 --> 00:03:22.849 A:middle L:90%
of course it would be if you're working with D'You
52
00:03:22.849 --> 00:03:24.229 A:middle L:90%
You want everything to be written in terms of you
53
00:03:24.719 --> 00:03:27.569 A:middle L:90%
. So that's what these equations tell us to dio
54
00:03:28.020 --> 00:03:29.629 A:middle L:90%
We know what to do with the d n.
55
00:03:29.800 --> 00:03:30.650 A:middle L:90%
We know what to do with the natural lava vent
56
00:03:30.650 --> 00:03:34.310 A:middle L:90%
. We we can use these equations to express everything
57
00:03:34.310 --> 00:03:36.639 A:middle L:90%
in terms of you. And we can change the
58
00:03:36.639 --> 00:03:38.689 A:middle L:90%
bounds by, you know, looking at the bounds
59
00:03:38.689 --> 00:03:40.520 A:middle L:90%
appear and figuring out what the bounce correspond to buy
60
00:03:40.780 --> 00:03:45.659 A:middle L:90%
this top equation. And then once we have this
61
00:03:45.659 --> 00:03:46.409 A:middle L:90%
guy, this is just we could just use the
62
00:03:46.409 --> 00:03:53.280 A:middle L:90%
power rule. We get one half, you do
63
00:03:53.280 --> 00:03:58.389 A:middle L:90%
one half, and then we're evaluating this from natural
64
00:03:58.389 --> 00:04:03.694 A:middle L:90%
log of the two to infinity. And this is
65
00:04:03.694 --> 00:04:06.014 A:middle L:90%
certainly going to be something that is infinite. All
66
00:04:06.014 --> 00:04:08.865 A:middle L:90%
right. If we plug in infinity here, we're
67
00:04:08.865 --> 00:04:11.215 A:middle L:90%
going to get something that blows up. And then
68
00:04:11.215 --> 00:04:14.824 A:middle L:90%
we're subtracting what happens when we plug in natural love
69
00:04:14.835 --> 00:04:16.454 A:middle L:90%
too. And for you, So have infinity minus
70
00:04:16.454 --> 00:04:23.324 A:middle L:90%
something finite. Certainly going to be infinite. Okay
71
00:04:23.324 --> 00:04:25.774 A:middle L:90%
, So since this integral turns out to be infinite
72
00:04:26.625 --> 00:04:29.334 A:middle L:90%
, the integral test tells us that this son there's
73
00:04:29.334 --> 00:04:31.084 A:middle L:90%
also going to be infinite, which means that we
74
00:04:31.084 --> 00:04:32.524 A:middle L:90%
get divergence.