WEBVTT mathematics/calculus-ii/murray
00:00:00.000 --> 00:00:05.000
Today we are going to talk about improper integration.
00:00:05.000 --> 00:00:16.000
The idea of improper integration is the function that you are trying to integrate has either a horizontal or a vertical asymptote.
00:00:16.000 --> 00:00:20.000
For example, if you have a horizontal asymptote, that means something like this,
00:00:20.000 --> 00:00:27.000
Where the curve goes on forever and gets closer and closer to a horizontal line.
00:00:27.000 --> 00:00:38.000
What you might be trying to do is find the area under the curve but this is an area that goes on forever.
00:00:38.000 --> 00:00:48.000
The formal notation for this is to take the integral from x=a and we say the integral from a to infinity of f(x) dX.
00:00:48.000 --> 00:01:05.000
The way you really do that since we do not really know what it means to plug in infinity for a function is we cut off this integral at a certain point which we call T.
00:01:05.000 --> 00:01:10.000
Then we just find the integral from x=a to x=t.
00:01:10.000 --> 00:01:21.000
So, we are finding the area under the curve from x=a to x=t, and then we let t get bigger and bigger to go off to infinity.
00:01:21.000 --> 00:01:25.000
We take the limit as t goes to infinity.
00:01:25.000 --> 00:01:34.000
What can happen is it can turn out that that can approach a finite limit or it can diverge to infinity.
00:01:34.000 --> 00:01:44.000
The same way that the limits that you learned about in Calculus 1 can approach a finite limit or they can diverge to infinity, or negative infinity.
00:01:44.000 --> 00:01:52.000
When we do some examples here, we will be asking whether these integrals converge to a finite limit or if so what the limit is,
00:01:52.000 --> 00:01:57.000
Or whether they diverge to either negative infinity, or positive infinity, or neither.
00:01:57.000 --> 00:02:09.000
The other kind of integral we will be looking at is a vertical asymptote.
00:02:09.000 --> 00:02:14.000
That means that the function you are trying to integrate maybe goes up to infinity, in other words it approaches a vertical line.
00:02:14.000 --> 00:02:28.000
If it has a vertical asymptote at b, we might want to try to find the total area under that curve between a and b.
00:02:28.000 --> 00:02:35.000
We try to find that area but the reason the function probably goes up to infinity at b,
00:02:35.000 --> 00:02:48.000
Is because we are dividing by 0 at b and it is not legitimate to just plug in a value at b for the function.
00:02:48.000 --> 00:02:59.000
We cut off the function a little bit short of the vertical asymptote, so if there is b, and there is a,
00:02:59.000 --> 00:03:03.000
What we do is we cut off the function a little bit short.
00:03:03.000 --> 00:03:05.000
We call that value t.
00:03:05.000 --> 00:03:16.000
We cut it somewhere short of the vertical asymptote where the function is still defined and it does not blow up to infinity and we find the area between a and t.
00:03:16.000 --> 00:03:20.000
Then what we do is we gradually move t closer and closer to b,
00:03:20.000 --> 00:03:33.000
That this area approaches the area under the entire curve that we are looking at.
00:03:33.000 --> 00:03:37.000
Again, this area can turn out to be finite or infinite,
00:03:37.000 --> 00:03:46.000
So, we will talk about the integral converging to a finite number or diverging to infinity or negative infinity, or neither.
00:03:46.000 --> 00:03:50.000
Again, we handle it with limits, we take the limit as t goes to b.
00:03:50.000 --> 00:03:54.000
Since t is coming towards b from the left hand side,
00:03:54.000 --> 00:04:02.000
That is why we put a little negative sign down there to show that t is coming to b from the negative side.
00:04:02.000 --> 00:04:14.000
We can also talk about functions that approach vertical asymptotes from the right hand side.
00:04:14.000 --> 00:04:21.000
We can ask ourselves what is the area under that function between a and b where a is the vertical asymptote.
00:04:21.000 --> 00:04:30.000
Then what we would say is that the integral from a to b of f(x) dX,
00:04:30.000 --> 00:04:37.000
We would look at the integral and now we would have t be just a little bit to the right of a,
00:04:37.000 --> 00:04:45.000
And t would approach a from the right, and we would integrate from t to b of f(x) dX.
00:04:45.000 --> 00:04:54.000
Then we would take the limit as T gets closer and closer to a, but this time t is on the right-hand side of a.
00:04:54.000 --> 00:04:59.000
I am going to put a little plus sign there to show that t is approaching a from the positive side.
00:04:59.000 --> 00:05:04.000
Let us try some examples and see how it works out.
00:05:04.000 --> 00:05:10.000
Our first example is to look at the integral from 0 to infinity of 1/(x+1).
00:05:10.000 --> 00:05:25.000
The key thing here is that if you graph 1/(x+1), well, at 0 it is 1, and as x approaches infinity it goes down to 0.
00:05:25.000 --> 00:05:31.000
We are trying to find that area there and see whether that area is finite or infinite.
00:05:31.000 --> 00:05:41.000
The way we set this up is we take the integral from 0 to t, of 1/(x+1) dX.
00:05:41.000 --> 00:05:48.000
We will go ahead and do that integral and then we will take the limit as t goes to infinity.
00:05:48.000 --> 00:05:56.000
We are cutting this integral off at some value of t that will then let the t get bigger and bigger.
00:05:56.000 --> 00:06:07.000
The integral of 1/(x+1) is a pretty easy integral if you let u = x+1 and put the substitution through there, then it turns into ln(x+1).
00:06:07.000 --> 00:06:30.000
Evaluated from x=0 to x=t, and so we get ln(t+1) - ln(0+1), so ln(1) but that goes away to 0.
00:06:30.000 --> 00:06:41.000
We get ln(t+1) and we want to take the limit of that as t goes to infinity.
00:06:41.000 --> 00:06:51.000
If you remember, ln(x), y = ln(x).
00:06:51.000 --> 00:06:57.000
As you plug in bigger and bigger numbers, ln(x) goes to infinity.
00:06:57.000 --> 00:07:04.000
If we plug in bigger and bigger values for t, this limit goes to infinity.
00:07:04.000 --> 00:07:15.000
What we say about this integral is that it diverges to positive infinity.
00:07:15.000 --> 00:07:24.000
This represents the fact that there is actually infinite area under that curve.
00:07:24.000 --> 00:07:33.000
Again, the way you handle these improper integrals is you convert the infinity into a t and then you do the integration
00:07:33.000 --> 00:07:43.000
Then you take the limit as t goes to infinity and you see whether it turns out to be a finite or an infinite area.
00:07:43.000 --> 00:07:44.000
Let us try another example.
00:07:44.000 --> 00:07:51.000
Again, the same function actually, 1/(x+1).
00:07:51.000 --> 00:08:08.000
But, this time we are taking the integral from -1 to 0 so the point here is that if you plug in x=-1 to this function,
00:08:08.000 --> 00:08:20.000
It blows up because you get 0 in the denominator and because this function has a vertical asymptote at x=-1.
00:08:20.000 --> 00:08:29.000
We are trying to figure out whether the area under that function as it approaches that vertical asymptote is finite or infinite.
00:08:29.000 --> 00:08:32.000
Again, we handle this with limits.
00:08:32.000 --> 00:08:35.000
We will take a value t that is just to the right of -1,
00:08:35.000 --> 00:08:44.000
And we will integrate from t to 0, instead of from -1 to 0 because you cannot plug in -1 to that function.
00:08:44.000 --> 00:08:55.000
We will look at dX/(x+1) and then we will take the limit as t gets closer and closer to -1.
00:08:55.000 --> 00:09:03.000
I will put a little positive sign to indicate that t is approaching -1 from the positive side.
00:09:03.000 --> 00:09:15.000
Again, we will take the integral and we get ln(abs(x+1)) evaluated from x=t up to x=0.
00:09:15.000 --> 00:09:33.000
That is ln, if you plug in x=0, you get ln(1) - ln(t+1), so ln(1) goes to 0.
00:09:33.000 --> 00:09:41.000
This leaves us with -ln(t+1).
00:09:41.000 --> 00:09:49.000
Now, let us see what happens when we take values of t that are closer and closer to -1.
00:09:49.000 --> 00:10:10.000
Well, if t goes to -1, that means that t+1 is going to 0, so what we are trying to do is take the ln, I should keep my absolute values here
00:10:10.000 --> 00:10:15.000
Take the ln of something that is going to 0 from the positive side.
00:10:15.000 --> 00:10:31.000
If you remember what the graph of ln(x) looks like, when you go to 0 from the positive side, ln(x) goes down to -infinity.
00:10:31.000 --> 00:10:49.000
This is minus -infinity so we say the whole integral diverges to infinity.
00:10:49.000 --> 00:10:59.000
Which represents the fact that the area under that curve is infinite.
00:10:59.000 --> 00:11:08.000
Here we had a vertical asymptote and again the way to handle the vertical asymptote was to replace the problem value of x.
00:11:08.000 --> 00:11:12.000
The problem value of x was -1, with t.
00:11:12.000 --> 00:11:19.000
Go ahead and work through the integral and then take the limit as t approaches -1,
00:11:19.000 --> 00:11:25.000
And I should have put a little plus in here to show that it is approaching -1 from the positive side.
00:11:25.000 --> 00:11:34.000
We take that limit and we see whether it comes out to be a finite or infinite area.
00:11:34.000 --> 00:11:37.000
Our next example is a little different.
00:11:37.000 --> 00:11:39.000
We are going to look at cos(x).
00:11:39.000 --> 00:11:43.000
We are going to look at the integral from 0 to infinity of cos(x).
00:11:43.000 --> 00:11:54.000
For this one, just like the others, it is very helpful to draw a graph, so I will draw a graph of cos(x).
00:11:54.000 --> 00:12:02.000
It starts at 1 and then it goes down to 0, -1, up to 1, down to -1, and so on.
00:12:02.000 --> 00:12:08.000
So that is y=cos(x).
00:12:08.000 --> 00:12:15.000
Now what we are really doing here is we are finding the area under the curve of cos(x),
00:12:15.000 --> 00:12:23.000
Where we count area above the x axis as positive and below the x axis as negative.
00:12:23.000 --> 00:12:38.000
We are really trying to add up all of those areas where this counts positive and this counts negative, positive, negative, and so on.
00:12:38.000 --> 00:13:04.000
You can see easily just by looking at the graph here that there is going to be infinite positive area and also infinite negative area.
00:13:04.000 --> 00:13:12.000
What we are really trying to do with this integral is add up infinitely many positive things and infinitely many negative things.
00:13:12.000 --> 00:13:17.000
So, a positive infinite and a negative infinity.
00:13:17.000 --> 00:13:27.000
A common question at this point is to ask whether the positive infinity and negative infinities cancel each other.
00:13:27.000 --> 00:13:35.000
The answer is that there is no mathematically meaningful way to give rules for positive and negative infinities to cancel each other.
00:13:35.000 --> 00:14:04.000
What you do in a situation like this is you say that the integral diverges but not to infinity or negative infinity.
00:14:04.000 --> 00:14:12.000
You might be tempted to say that all of the positives and the negative infinities cancel each other and the whole integral is equal to 0.
00:14:12.000 --> 00:14:22.000
Or you might be tempted to say that, well, these two areas cancel each other and this area cancels with the next one, so all of these areas cancel
00:14:22.000 --> 00:14:28.000
Then you are just left with this area and you can find the value of that area.
00:14:28.000 --> 00:14:33.000
There are all sorts of ways that you might think you could resolve this.
00:14:33.000 --> 00:14:41.000
Because those all contradict each other, we do not sort of put the stamp of legitimacy on any of those.
00:14:41.000 --> 00:14:46.000
We just say the whole thing diverges but not to positive infinity or negative infinity.
00:14:46.000 --> 00:14:48.000
We just say it diverges.
00:14:48.000 --> 00:14:56.000
You could also do this one by actually calculating the integral from 0 to 1 of cos(x) dx.
00:14:56.000 --> 00:15:05.000
Which would give you sin(x) evaluated from 0 to 1.
00:15:05.000 --> 00:15:23.000
That in turn gives you sin(t) - sin(0), which gives you sin(t) and then you are finding your limit as t goes to infinity of sin(t).
00:15:23.000 --> 00:15:27.000
Again, we know that since sin oscillates forever, this does not exist.
00:15:27.000 --> 00:15:43.000
It is not positive infinity, it is not negative infinity, it is not 0, we just say the whole thing does not exist.
00:15:43.000 --> 00:15:50.000
It diverges.
00:15:50.000 --> 00:15:53.000
OK, let us do another example.
00:15:53.000 --> 00:15:58.000
Again this is one that if we look at the graph it is something we can resolve quickly.
00:15:58.000 --> 00:16:03.000
We are looking at the integral from negative infinity to infinity of x² dx.
00:16:03.000 --> 00:16:11.000
In this case we actually have improprieties, places where the intervals are improper at both ends.
00:16:11.000 --> 00:16:19.000
If we graph this one, y = x², it looks like that.
00:16:19.000 --> 00:16:31.000
What we are trying to do is find the area under this curve as we go to infinity in both the positive and negative directions.
00:16:31.000 --> 00:16:38.000
Now, clearly if you look at this area, the area is infinite.
00:16:38.000 --> 00:16:40.000
Both on the positive and the negative sides.
00:16:40.000 --> 00:16:44.000
Here is another common mistake that Calculus 2 students make
00:16:44.000 --> 00:16:52.000
They will look at this area over here and say wait, is that not negative area?
00:16:52.000 --> 00:17:00.000
That is a little mistake because remember, negative area is when the area is below the x axis.
00:17:00.000 --> 00:17:09.000
This area is not below the x axis, it is to the left of the y axis, but it is still above the x axis.
00:17:09.000 --> 00:17:15.000
That is not negative area.
00:17:15.000 --> 00:17:21.000
It is still positive area because it is above the x axis.
00:17:21.000 --> 00:17:44.000
If you simply look at the graph here, we are looking at positive infinite area + another batch of positive infinite area.
00:17:44.000 --> 00:17:46.000
If you put those together we have two positive infinities.
00:17:46.000 --> 00:17:54.000
Remember we learned in the last example that you cannot cancel a positive and a negative infinity, but what about 2 positive infinities?
00:17:54.000 --> 00:18:01.000
You can add those up and say that the answer will be a positive infinity.
00:18:01.000 --> 00:18:08.000
You can say that it diverges to positive infinity.
00:18:08.000 --> 00:18:15.000
You do not just say this one diverges, you say that it diverges to positive infinity.
00:18:15.000 --> 00:18:29.000
Because, if you look at these two areas we definitely have positive infinite area, positive infinite area, we put them together and we have got positive infinite area.
00:18:29.000 --> 00:18:35.000
There are a few integrals that come up very, very often when you are looking at improper integrals.
00:18:35.000 --> 00:18:44.000
The most important ones are the integral from 0 to 1 of dX/(x^p).
00:18:44.000 --> 00:18:59.000
The reason that is improper is because when x=0, you have 0 in the denominator here so that blows up that x=0, so x=0 is improper there.
00:18:59.000 --> 00:19:13.000
The other one that comes up very often is the integral from 1 to infinity of dx/(x^p), again that is improper because of the infinity there.
00:19:13.000 --> 00:19:22.000
These integrals come up so often that it is worth working them out once and then probably memorizing the answers.
00:19:22.000 --> 00:19:37.000
Especially this one, the integral of 1 to infinity of dx/x^p is worth remembering.
00:19:37.000 --> 00:19:43.000
It is worth remembering because it comes up later on in Calculus 2.
00:19:43.000 --> 00:19:48.000
When you start looking at infinite series.
00:19:48.000 --> 00:20:02.000
When you start looking at infinite series we are going to be using something called the integral test to determine whether infinite series converge or diverge.
00:20:02.000 --> 00:20:10.000
The integral test says instead of looking at a series, you look at an integral.
00:20:10.000 --> 00:20:15.000
The series that we are going to start with is 1/n^p.
00:20:15.000 --> 00:20:22.000
We will convert that into the integral of 1/x^p dx.
00:20:22.000 --> 00:20:30.000
Then we will remember what this integral was in order to determine whether the series is convergent or divergent.
00:20:30.000 --> 00:20:33.000
It is worth remembering these.
00:20:33.000 --> 00:20:36.000
Each of these integrals, you can evaluate yourself.
00:20:36.000 --> 00:20:39.000
I am not going to work through the details there.
00:20:39.000 --> 00:20:45.000
But you can evaluate yourself and it turns out that it depends on the values of p.
00:20:45.000 --> 00:20:50.000
What values of p determine whether these integrals converge or diverge.
00:20:50.000 --> 00:20:58.000
You want to remember that the first integral, if p < 1, it converges.
00:20:58.000 --> 00:21:01.000
If p = 1, it diverges.
00:21:01.000 --> 00:21:05.000
If p > 1, it also diverges.
00:21:05.000 --> 00:21:09.000
The second integral, if p < 1 it diverges.
00:21:09.000 --> 00:21:12.000
Equal to 1, it diverges, and p > 1 it converges.
00:21:12.000 --> 00:21:16.000
That seems like a lot to remember.
00:21:16.000 --> 00:21:26.000
You kind of remember, if p = 1, they both diverge, but if p < 1 or p > 1, they kind of flip flop back and forth there.
00:21:26.000 --> 00:21:37.000
We will use these on the next example to help us determine whether an interval converges or diverges.
00:21:37.000 --> 00:21:45.000
Here we have been given the integral of from 6 to 11 of 11/(x-6)x^1/3.
00:21:45.000 --> 00:21:49.000
We have been asked to determine whether it converges or diverges.
00:21:49.000 --> 00:21:56.000
The first thing to look at with this is to look at it and see if we can make a substitution to make it simpler.
00:21:56.000 --> 00:21:57.000
We can.
00:21:57.000 --> 00:22:00.000
Let us do u = x - 6.
00:22:00.000 --> 00:22:09.000
Then, whenever you make a substitution, you also have to change the differential.
00:22:09.000 --> 00:22:11.000
So, du, if u is just x - 6, that is very easy.
00:22:11.000 --> 00:22:14.000
Let us also change the limits.
00:22:14.000 --> 00:22:20.000
If x = 1, then u will be equal to 5 because that is 11 - 6.
00:22:20.000 --> 00:22:27.000
If x = 6, then u = 0.
00:22:27.000 --> 00:22:36.000
This integral converts into the integral from 0 = u to u = 5 of 11.
00:22:36.000 --> 00:22:40.000
I am just going to write the 11 on the outside.
00:22:40.000 --> 00:22:43.000
1/u^1/3 du.
00:22:43.000 --> 00:22:54.000
Now it is a little more obvious why this is an improper integral, because u = 0, you will get 0 in the denominator.
00:22:54.000 --> 00:22:58.000
That is obviously an impropriety there.
00:22:58.000 --> 00:23:00.000
Actually you might have been able to notice that from the original integral there.
00:23:00.000 --> 00:23:11.000
When x = 6, you might plug in x = 6 to the denominator you get 0 in the denominator and that is clearly improper.
00:23:11.000 --> 00:23:20.000
I am not going to worry about the 11 for the time being because it is not going to affect whether the thing converges or diverges.
00:23:20.000 --> 00:23:27.000
What we are going to do is instead of 0, I am going to change the 0 to t.
00:23:27.000 --> 00:23:40.000
We are going to integrate from t to 5, of 1 over, now u to the cube root = u^1/3 dU.
00:23:40.000 --> 00:23:48.000
Then we take the limit of that as t goes to 0 from the positive side.
00:23:48.000 --> 00:24:05.000
Now I can split this integral up from t to 1 of 1/u^1/3 du.
00:24:05.000 --> 00:24:15.000
Plus the integral from 1 to 5 of 1/u^1/3 du.
00:24:15.000 --> 00:24:20.000
Now the point about that is that this is not an improper integral.
00:24:20.000 --> 00:24:29.000
This completely proper and it will not determine whether the whole thing converges or diverges.
00:24:29.000 --> 00:24:33.000
This will give us a finite number.
00:24:33.000 --> 00:24:36.000
This is the improper part right here.
00:24:36.000 --> 00:24:42.000
What we have here is the integral of du/u^1/3.
00:24:42.000 --> 00:24:52.000
That is exactly the integral that we were looking at on the previous page.
00:24:52.000 --> 00:24:58.000
I will just write it as 0¹ for now, with a p value of 1/3.
00:24:58.000 --> 00:25:05.000
On the previous page, we learned that the integral from 0 to 1 of dx over x^p,
00:25:05.000 --> 00:25:14.000
If p < 1, converges.
00:25:14.000 --> 00:25:23.000
That tells us that this whole integral converges, or at least this part of it converges.
00:25:23.000 --> 00:25:31.000
Then, the second part with some finite number,
00:25:31.000 --> 00:25:36.000
I have not really incorporated the 11 but we can have an 11 times the whole thing.
00:25:36.000 --> 00:25:45.000
An 11 multiplied times a whole number is not going to make it more likely to go to infinity or less likely.
00:25:45.000 --> 00:25:58.000
So, the whole thing goes to it converges to a finite number.
00:25:58.000 --> 00:26:02.000
In this case, the problem only asked us whether the integral converges or diverges.
00:26:02.000 --> 00:26:06.000
At this point, we are finished.
00:26:06.000 --> 00:26:12.000
In fact, this integral is completely feasible.
00:26:12.000 --> 00:26:24.000
You could if you wanted, go back and evaluate the integral.
00:26:24.000 --> 00:26:33.000
What you should get is 3/2 × 5^3/2.
00:26:33.000 --> 00:26:40.000
That is what you get if you plug in the whole integral and then you multiply the whole thing by 11.
00:26:40.000 --> 00:26:43.000
That is what you get if you evaluate the whole integral.
00:26:43.000 --> 00:26:53.000
The important thing here is that is a finite integral, so we would say the integral converges.
00:26:53.000 --> 00:27:02.000
There is a very common sort of ambush in Calculus 2, where they will give you a sort of innocent-seeming problem.
00:27:02.000 --> 00:27:08.000
It turns out what the problem has is hidden discontinuities.
00:27:08.000 --> 00:27:10.000
Let me give you an example of that.
00:27:10.000 --> 00:27:16.000
The integral from 1 to 4 of dx/x² - 5x + 6.
00:27:16.000 --> 00:27:25.000
This is an example that is not that hard to integrate using the techniques we learned in earlier lectures.
00:27:25.000 --> 00:27:36.000
In particular, if you use partial fractions.
00:27:36.000 --> 00:27:51.000
If you use partial fractions on this integral, you get 1/x² - 5x + 6.
00:27:51.000 --> 00:27:55.000
Well, remember what we learned in partial fractions was to factor the denominator.
00:27:55.000 --> 00:28:03.000
So, 1/(x-2) × (x-3).
00:28:03.000 --> 00:28:08.000
If you do the partial fractions work, I will not spell out the details now, but we learned how to do that in a previous lecture.
00:28:08.000 --> 00:28:17.000
You get 1/(x - 3) - 1/(x - 2).
00:28:17.000 --> 00:28:23.000
So it is very tempting to do that partial fractions work and then integrate it.
00:28:23.000 --> 00:28:33.000
You integrate it and you get ln(x-3) - ln(x-2).
00:28:33.000 --> 00:28:37.000
Then you think, OK, I can plug in my bounds, x=1 and x=4.
00:28:37.000 --> 00:28:41.000
You can plug in those numbers just fine.
00:28:41.000 --> 00:28:44.000
And, you get a nice numerical answer.
00:28:44.000 --> 00:28:54.000
That is the temptation in this kind of problem.
00:28:54.000 --> 00:29:01.000
That is exactly the trap that your Calculus 2 teacher is trying to make you fall into.
00:29:01.000 --> 00:29:07.000
You work all the way through this problem and you get a nice numerical answer and you present it as your answer.
00:29:07.000 --> 00:29:12.000
The fact is that that is flat wrong.
00:29:12.000 --> 00:29:14.000
What is wrong with that?
00:29:14.000 --> 00:29:25.000
Well, what is wrong with that is that if you look at the function we had to integrate there, 1/(x-2) × (x-3).
00:29:25.000 --> 00:29:39.000
That blows up at x=2 and x=3.
00:29:39.000 --> 00:29:46.000
Because if you plug in x=2 or x=3, you get 0 in the denominator and the thing explodes.
00:29:46.000 --> 00:29:55.000
So, this integral, which looked very tame and safe and looked like a fairly easy Calculus 2 integral actually is an improper integral.
00:29:55.000 --> 00:29:59.000
Since we are talking about it in the section in improper integrals.
00:29:59.000 --> 00:30:09.000
Maybe it is obvious to look for that, but if this comes in a swarm of other integrals where you are using all kinds of other techniques,
00:30:09.000 --> 00:30:12.000
It is very easy to overlook things like this.
00:30:12.000 --> 00:30:19.000
Instead of just solving it, using the generic partial fractions idea,
00:30:19.000 --> 00:30:28.000
What you have to do is you have to split this up into pieces at the discontinuities.
00:30:28.000 --> 00:30:51.000
So, you split it up at the integral from 1 to 2 of dx over the denominator + the integral from 2 to 3 of dx over the denominator + the integral from 3 to 4 of dx over the denominator.
00:30:51.000 --> 00:31:00.000
You split it up at these two places, at 2 and 3, because those are the two discontinuities.
00:31:00.000 --> 00:31:08.000
In fact, this middle integral, is now discontinuous at both ends because 1 end is 2 and 1 end is 3.
00:31:08.000 --> 00:31:12.000
This middle integral is discontinuous in two different place.
00:31:12.000 --> 00:31:26.000
We are going to split that up from 2 to 2.5 of dx over the denominator + the integral from 2.5 to 3 of dx over the denominator.
00:31:26.000 --> 00:31:39.000
So, then we look at these 4 different improper integrals.
00:31:39.000 --> 00:31:47.000
Four improper integrals.
00:31:47.000 --> 00:31:49.000
You solve all four of them.
00:31:49.000 --> 00:32:00.000
If any one of them diverges.
00:32:00.000 --> 00:32:19.000
Or if more than one diverges, then we say the whole original integral from 1 to 4 diverges.
00:32:19.000 --> 00:32:27.000
Now, each one of these 4 integrals you could work on using partial fractions.
00:32:27.000 --> 00:32:33.000
So, using that technique that I outlined up here,
00:32:33.000 --> 00:32:52.000
Solve each one using partial fractions.
00:32:52.000 --> 00:33:07.000
It turns out that all 4 of those integrals diverge.
00:33:07.000 --> 00:33:28.000
So, we say the original integral from 1 to 4 diverges, as our answer.
00:33:28.000 --> 00:33:43.000
This is really quite dangerous because this depends on your noticing that there are these hidden discontinuities between the bounds of integration.
00:33:43.000 --> 00:33:51.000
In order not to fall into this trap, what you have to do is look at the thing being integrated and ask yourself when does that blow up.
00:33:51.000 --> 00:33:55.000
In this case that blows up at x=2 and x=3.
00:33:55.000 --> 00:33:58.000
2 and 3 would be in the bounds of integration.
00:33:58.000 --> 00:34:04.000
They are between 1 and 4 and so that is why we have this problem.
00:34:04.000 --> 00:34:19.000
If we were asked to integrate the same function from 4 to 6 of dx/x² - 5x + 6,
00:34:19.000 --> 00:34:24.000
We would have no problems here and would not have to split it up.
00:34:24.000 --> 00:34:27.000
It would not be improper.
00:34:27.000 --> 00:34:37.000
We could go back and use this regular technique of partial fractions and just plug in the bounds and we would not have to worry about limits at all.
00:34:37.000 --> 00:34:44.000
The reason there is that the place where it blows up, x=2 and x=3, is not in between 4 and 6.
00:34:44.000 --> 00:34:48.000
There would be no discontinuities in our region of interactions.
00:34:48.000 --> 00:34:50.000
That is a very dangerous kind of integral.
00:34:50.000 --> 00:34:55.000
You kind of have to watch out for when you see a denominator
00:34:55.000 --> 00:35:00.000
Ask yourself, where is the denominator 0, where does that make you function blow up.
00:35:00.000 --> 00:35:07.000
If that is inside your region of integration, then you have to split up the integral and work on each part.
00:35:07.000 --> 00:35:09.000
If any one of them diverges.
00:35:09.000 --> 00:35:14.000
Then you say the whole integral diverges.
00:35:14.000 --> 00:35:16.000
We will do some more examples later on.
00:35:16.000 --> 00:44:18.000
This has been Will Murray for educator.com.