 Hello and welcome to a screencast today about improper integrals. So we are going to look at whether or not the improper integral converges or diverges, and then if it converges, find its exact value. Okay, so the integral we're going to be focusing in on today is the integral from 2 to infinity of x to the 4th divided by x to the 5th minus 3 dx. Alright, so over here I drew a graph of our function in geotubra. So basically we're going to be looking at the area under this curve from 2 all the way down here to infinity. Okay, now obviously there's a horizontal asymptote there. So, you know, as it turns out this area is going to be getting smaller and smaller. So it is possible for this integral to converge. It's possible. Not likely, but possible. Okay, so the first thing we want to do whenever we have an improper integral is take a look at any endpoints that are causing problems or any points in between as well. So for this particular example, infinity is obviously causing a problem. It looks like this graph has some issue maybe between 1 and 2 as well, but our integral starts at 2, so that doesn't bother us. Okay, so I'm going to go ahead and change this then. We're going to say the limit as b approaches infinity of the integral from 2 to b of x to the 4th over x to the 5th minus 3 dx. So what we want to do then is we want to change and replace that unbounded limit with a variable then that's going to be representing a finite value. Okay, so we can deal with b's. We can't really deal with infinities. Also what we're going to do is we're just going to go ahead and look at the indefinite integral of this function and then we're going to go ahead and throw our endpoints on later. Okay, so if I look at then just the integral of x to the 4th over x to the 5th minus 3 dx. I'm looking at this one. I think I'm going to need some u substitution. So let's let u be the denominator, x to the 5th minus 3. So du is going to end up giving us 5x to the 4th dx. Well I have my x to the 4th dx in my integral. So all I have to do is get rid of this 5. So that's going to become 1 5th du is going to be x to the 4th dx. So then when I rewrite my integral that's going to give me then 1 5th, pulling that constant out front, the integral of du over u. So hopefully you guys remember then that's going to be 1 5th natural log of u plus c. So then putting our u back in that's going to give us 1 5th natural log of x to the 5th minus 3 plus c. Okay, so this is our indefinite integral. But remember we technically had a definite integral to start with. So let's take a look then what happens when we do the limit as b approaches infinity of this value, of this function, x to the 5th minus 3. And then we're going to evaluate that from 2 to b. Okay, so plugging in our endpoints it's going to be the limit as b approaches infinity, 1 5th natural log of b to the 5th minus 3 minus 1 5th natural log of 2 to the 5th minus 3. Okay, the second piece here is obviously just some number, right? It's going to be whatever 1 5th natural log of what is that 29 is going to be. So this is just some number. Okay, so that piece obviously converges. But the first piece here, I don't know about that one. Okay, that one definitely looks suspicious. So if we think about the inside of this function. So what happens to b to the 5th minus 3 as b gets really, really, really big? Well, that piece is also going to get really, really big, right? So that piece is going to go to infinity as b goes to infinity. And then what happens to your natural log function as it goes to infinity? Well, that also goes to infinity. So this piece diverges. It gets really, really, really big. So even though it doesn't necessarily look like it from our graph, this area is actually going to be divergent. So our integral is going to be divergent. Okay, so this integral diverges. So don't let that fact that that area is really, really skinny. This leads you because as it turns out, this integral does not give you a nice number when you go to integrate it. Alright, thank you for watching.