Calculus: The Second Type of Improper Integral

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Improper Integrals
Improper Integrals
The Second Type of Improper Integral Page [1 of 2]
Okay, so to give you a sense of other types of improper behavior, with respect to integrals, I want to just give you a question. And this question is going to be really sort of peculiar in that I'm going to ask you a visual question, but we're not going to graph or look at a visual picture at all - no visual image - neither here with me, or even in the content box - nothing. Here's the question. I want you to find the area under the curve of y = between -1 and 1. Now no looking up pictures here. We're going to do this pictureless. What do we do?
Well, we set up an integral. So answer, we set up an integral from -1 to 1 of dx. And we integrate and we see , and I evaluate for -1 to 1. When I plug in 1, I see a -1. And when I plug in -1 into here, I see -1 divided by -1, which is another 1. And so I see -1 - (1). Well, -1 - (1) is -2. So the answer is -2. So there's the answer. What's the area under the curve? It's -2.
Now does that make sense? Okay, well now let's look at a picture. Now let's finally look at a picture and see. Well, now, I happen to know what that function looks like. It's sort of a hyperbola-ish looking thing that has two wings. See, since I'm squaring it, y is always positive. So we're always going to live above the x-axis. And it's very symmetric. In fact, it's an even function. So it looks like that.
And we're going to go from -1, which is right about here, and we're going to go to 1, which is right about here. And we'll look at the area, and the area is all of this stuff right here. That area's supposed to be negative. Now, what does it mean to negative area? Negative area means that you must actually be beneath the x-axis. That's the only way you can have this sort of funny negative sign appearing. But look at the picture. It's actually clear from the picture that's everything's living above. So in fact, this area's supposed to be positive. So, in fact, this answer must be wrong. So this is wrong.
Now what went wrong? We set up the right integral. We did it. There was no mistakes in this part of it. So there must have been a mistake in just going to here. That must, in fact, be not allowed. That must be not allowed. And, what was not allowed? Well, the truth is, this is really an example of another improper integral type issue. And the reason that it's improper is because when you want to use this technique that we thought about, early on in calculus, that's absolutely fine and will always work for you and is always dependable, as long as the function is sort of smooth and continuous and just goes around here and actually has no brakes.
Now if you lose that sort of continuity, if you will, and have some place where the function's not defined - for example notice, this is not defined when x = 0, because I have , which is not a number. In fact, we have an asymptote there. Then, in fact, this procedure doesn't work, as illustrated by the crazy answer that we got. So the moral here is that if you're trying to find the area under a curve, and that curve actually has a place where things are discontinuous - and in particular, in this case, where you have an asymptote, so things sort of blow up - you can't just look at the end points because something really fishy is happening here and you've got to study the fishy part too.
So let's try to study the fishy part too and see what happens. If we do that, notice what we can do. I would integrate from -1 to 0, and that would give me this piece. And I would integrate from zero to 1, and that would give me this piece. So what I'm saying is the following. If you're trying to integrate a function and it turns out there's a really bad point somewhere in there - so you have an asymptote at some point - what you really should do is not just integrate from here to here but you should break it up into two sub-functions - two sub-integrals - one from -1 to zero, and one from zero to 1. So let's try that and see what we get now. If we do that, what I would see is the following.
Oh, by the way, look at this. This picture is so symmetric, that really, if I just find the value of this, the value of this will be the same. It's just a reflection. So, in fact, all I've got to do is take the value of this and double it and I'll get the whole thing. So all I'll do is look from zero to 1.
Okay, so let's look at the integral from zero to 1 and double it. And now the real area - real area - no fooling around now - that will equal two times the integral from zero to 1 of dx. Now we already took the integral of that, so that's not a big deal. It's going to be a 2 x , which equals , and I evaluate that, from zero to 1. So when I plug in 1 into here, I just see a -2 and I subtract when I plug in zero. Whoops! When I plug in zero into here, this thing is actually going to be, what? Exploding with a negative sign in front. So this is actually approaching negative infinity, and so I see this whole thing is actually approaching plus infinity, minus 2. Well, plus infinity, minus the number 2, is infinity. And so I see that, in fact, this integral diverges. This integral diverges, which means that the actual area of this region is infinite.
So even though the things are shrinking and that spike is getting thinner and thinner and thinner, it's not getting thinner and thinner and thinner fast enough. So the reds are accruing and accruing and accruing, and it turns out that the area - the total area of the red - is infinite.
This is another example of improper behavior - of behavior where, in fact, you've got some spike - you've got some asymptote - you must be very careful to make sure that if you integrate, you integrate from this point up to the asymptote, and then from the asymptote away. You just can't integrate blindly, because you'll start to get really, really crazy answers. So this is another indication of how an integral really can be improper. So the two basic ideas of improper integrals: one is if you, instead of putting a number in one of the points, you improperly put infinity and go off to the horizon and see what happens as you go off to infinity. Another possibility for improperness is if you're actually integrating over a nice region from one number to another number, but the function spikes up up a vertical asymptote somewhere in between. Then you need to go just from here to the vertical asymptote, vertical asymptote to here, and then see if those things exist, so a lot of improper behavior with respect to integrals. I'll see you at the next lecture.