 Another type of improper integral occurs when there is a discontinuity in the interval. So, suppose f of x is discontinuous at x equals c, where c is someplace in our interval. So first, we'll define the integral from a to c as the limit of the integral as our upper bound approaches c from below, and similarly, the integral from c to b as the limit as our lower bound approaches c from above, and provided that the two limits exist, we'll define our integral from a to b as the sum of the two integrals. So let's say we want to find this integral. We note that f of x equals 1 over square root of x is discontinuous at x equals 0, so, and so we have a type 1 improper integral. Now, since our lower limit is the point of discontinuity, we'll limit around that, and so we'll take our limit as the lower bound goes to 0 from above. So we'll find our antiderivative, we'll evaluate it, then we'll take the limit, and since the limit exists, the integral converges to this value. So let's take another integral. So when I ran into this problem the first time I was taking calculus too, I did it the following way. We find the antiderivative, we evaluate, got to remember how to add fractions, and I was very quickly able to get the wrong answer. The problem is the function has a discontinuity at x equals 1, and we can't integrate over it. So let's do this integral the right way. Since there's a discontinuity at x equals 1, we need to split up our integral around this point. And let's evaluate each of these integrals separately. So we'll take that first integral from 0 to 1, and because we have that discontinuity at 1, we'll limit away from that discontinuity. We'll find the antiderivative, we'll evaluate it, and then we'll take the limit, which diverges, which means that our integral diverges as well.