 Hello, and welcome to another screencast on improper integrals. So today we're going to be looking at whether the improper integral converges or diverges, and if it converges, let's find its exact value. All right. The integral we're going to be looking at here is from negative 2 to 1 of 1 over the square root of 1 minus x dx. All right. I graph this function in geogibre, so negative 2 starts here. So we're trying to find the area under this curve until we get to 1, and look what happens at 1. The graph shoots up and it has a vertical asymptote. So let's make a note of that. So we have a vertical asymptote at x equals 1. So 1 is a trouble spot for us, and you notice because that trouble spot is actually in our integral, we're going to want to replace that then with a variable, and then we can take a look at the limit as our variable approaches that number. Okay. So let's do the limit as how about a approaches 1, and then, so our integral goes from negative 2 to 1. So we're really going to be approaching 1 from that left side. We don't care what's happening to 1 on the right side. We just care what's happening on the left side, because we care between negative 2 and positive 1. So we're going to do the integral from negative 2 to a, 1 over the square root of 1 minus x dx. Okay. So I think all these end points and limits and stuff kind of make this integral look messy. So let's go ahead and just take a look at the indefinite integral of 1 over the square root of 1 minus x dx. Let's see what happens with this, and then we can go back and look at our end points and see what happens with our limit. Okay. So I see an inside function here. So I'm going to go ahead and do a u. And my inside function is 1 minus x. So then my du is negative dx. So I'm going to go ahead and stick that negative out front to make sure everything matches up. So I've got the negative integral of 1 over the square root of u du. Okay. We're rewriting that a little bit. So that gives me negative u to the negative 1 half du. Okay. So now I can go ahead and apply my power rule to this. So I get negative 2 u to the 1 half plus c. So then the 1 half power number is our square root. And then let me go ahead and plug my u back in as well. So I get negative 2 square root of 1 minus x plus c. Okay. This is my indefinite integral. But remember, I did have end points on this. So let's go ahead and take a look at what's happening with those end points. So I've got the limit as a approaches 1. From the left of negative 2 square root. So using the fundamental theorem of calculus, I want to go ahead and plug in my top end point first. So that's going to be a, 1 minus a, and then minus, plugging in my bottom end point. So that's going to be negative 2 square root of 1 minus a negative 2. Oh, lots of negatives floating around here. Okay. So let's take a look at then what happens with our first piece here. So the limit only applies to the first piece because only the first piece has an a in it. All right. So as a approaches 1, the square root of 1 minus a, that's going to give us a 0. Okay. So the first piece then gives us 0. And then let's look at the second piece. So negative negative. So that's going to be plus 2 square root of 1 minus a minus 2. So 1 plus 2, which is 3. Oh, look. This is a number. How about that? So that means our interval converges. Our integral converges. And its exact value is 2 square root of 3. Okay. So looking at the picture of these functions is always a good place to start. But you notice your intuition does not necessarily follow what happens with this problem. This integral looks like, I would have thought this one would diverge for sure, because it looks like this value is getting really, really big. But as it turns out, it actually converges. So that's where you've got to go through and you've got to work through the calculus and the algebra of these parts and see whether or not you get a number out of your integral or if you get something that's undefined or unbounded. All right. Thank you for watching.