 Welcome to this quick recap of section 6.5, improper integrals. First we'll define what an improper integral is. An integral is called improper if one of two things happens. Either the interval of integration itself is unbounded because an infinity appears in one of the limits of integration, or the integrand is unbounded somewhere in the interval of integration. In the first case you can see how the infinity shows up directly in the integral sign itself. It can be at the beginning or the end of the endpoints, or it could be at both of them. In the second case we can see first how x equals zero is a problem for one over square root of x, but x equals one is a problem for one over x minus one, and that's in the middle of the range over which we're integrating. The surprising result here is that improper integrals can have finite values. It's possible that the area that they represent can come out to a finite number even though somehow it stretches over an infinitely large interval. Let's take a look at how we can calculate these. First we'll look at the case where the interval of integration itself is unbounded. To fix this we're going to replace the infinite endpoint with a variable instead. So in the first example we have an infinity in the upper level here, and so we'll replace that with a b instead which we think of as some variable, just a finite number, and then we'll let b go to infinity. In the second example we have a negative infinity, and so that turns into an a, and we take the limit as that a goes to negative infinity. We'll treat this new endpoint as if it's a constant, simply a number we can work with like in any definite integral. We'll evaluate the integral and then we'll take the limit. So in other words you should think of these as if there are parentheses around the integrals. Evaluate the integrals first, and whatever you get take the limit of that result. To understand what's going on here, let's take a look at these pictures at the bottom. The left picture represents an integral from 0 to b, where b is a finite value. Certainly that's some area and we could calculate it. Imagine b going farther and farther to the right. As it goes farther and farther we get a bunch more areas, and perhaps that list of areas has a limit. So as b goes off to infinity, we might get a limit for the area, and that's what we call the value of the integral. This is the process that we're highlighting up on the top. Let's talk about what happens when you do calculate one of these values. In this case we're only going to look at an integrand that is non-negative, that is greater than or equal to zero. That way this represents area as opposed to net area or signed area. So if an improper integral is said to converge or is convergent, that means that this limit that we calculate exists and is finite. So the limit of all of these areas comes out to a real finite number. On the other hand, if the integral does not come out to a finite value, if the limit does not exist, we say that the integral diverges or is divergent. So convergent means we have a finite area, divergent means we don't. Let's take a look at the other situation now. What if we have an unbounded integrand? So the limits of integration in this case are still really where the problem lies. We might have an unbounded integral at an endpoint, or it might happen in the middle of the interval. We'll take a look at both of those cases. On the left here the problem is at x equals zero because we can't divide by zero, and so our solution is very similar to what we did with infinite endpoints. We'll replace that zero causing trouble with a variable a, and we'll take a limit as a goes to zero from the right. In other words, as a goes to the trouble value. Why are we doing this? Let's imagine drawing out the number line from zero to one. So here we have zero, here we have one, and zero is where this integrand can't be evaluated. So instead we imagine a is somewhere very close to there, and we're going to let a go towards zero. So a will always be to the right of zero, which is why we put the plus. This is only a right hand limit. We really don't care what happens to this integral if we had a approach from the left because that's not part of our interval of integration. Let's take a look at the right here. In this case we have a problem at x equals one, which is right in the middle of the interval of integration. To fix this we're going to do an algebraic trick that turns this integral into one that has problems at the endpoints instead of in the middle. So we'll split this integral up into two parts, the integral from zero to one and the integral from one to two, and certainly those add up to give us the integral from zero to two. Now that those have problems at their endpoints when x equals one we'll replace those ones with variables and we'll take a limit. And again notice how we've split this up into two pieces, so let's look at the number line. We have zero to two, this is our interval of integration, and the problem point is right in the middle at one. For this first integral where we've replaced the upper limit with a b we imagine that b is somewhere just to the left of one and it's going to go towards the right. So that's why we have b approaching from the left. For the second integral we have a somewhere between one and two and it's going to approach one from the right. So that's why we have a goes to one from the right. We can evaluate each of those limits and figure out what the value is. This works if and only if both of the integrals written here has a finite value. If either of them is infinite or doesn't exist for some other reason then we say that this integral diverges. Now that we've seen those let's take a look at some examples of these in action.