 In this video, we'll work through two examples of improper integrals that involve a point where the integrand is discontinuous. Consider the curve f of x equals 1 over x minus 2 over the interval 2 to 4, as we saw in our lesson. If we were interested in evaluating the integral of the function over that interval, we need to account for the discontinuity at x equals 2, as we see in this graph. We have a vertical asymptote at x equals 2. So we can rewrite this integral from 2 to 4 of 1 over x minus 2 dx as the limit as a approaches 2. Now, I'm going to leave a little space right here because we're going to talk a bit about where I'm approaching 2 from which direction. Now, taking a look at the graph of our function, we note that we're approaching 2 from values greater than 2 because it is the lower bound of that interval, 2 to 4. So when we're evaluating the limit, we're looking at a approaching 2 from above. So we put a little plus superscript above that 2 to indicate that we're coming towards 2 from above. So rewriting here what I just wrote on the previous slide, limit as a approaches 2 from above of the integral from a to 4 of 1 over x minus 2 dx. That's equal to the limit as a approaches 2 from above of natural log of absolute value of x minus 2 evaluated from a to 4, which is equal to the limit as a approaches 2 from above of the natural log of absolute value of 4 minus 2 minus the natural log of a minus 2. That's equal to the limit as a approaches 2 from above of the natural log of 2 minus the natural log of absolute value of a minus 2. Now, let's take a look as a approaches 2 from above, a minus 2 is going to be between 0 and 1. So as a approaches 2 from above, this value approaches 0 from above and we know that the behavior of the natural log graph is that as we're approaching 0 for the argument value, natural log of that approaches negative infinity. Since we're subtracting this off, this gives us positive infinity. So this integral diverges. Consider a second example, the integral from 0 to 2 of 1 over the cube root of x minus 1 dx. Now we notice that since this cube root of x minus 1 is in the denominator, we know there's a discontinuity at x equals 1 and this limits of integration actually include x equals 1. So in order to integrate, we're actually going to split this integral into two parts exactly at that discontinuity point. So I've got the integral from 0 to 1 of 1 over the cube root of x minus 1 plus the integral from 1 to 2 because that's my upper bound of 1 over cube root of x minus 1 dx. And then we treat each of these integrals as we did in the previous example where the discontinuity was one of the limits of integration. So let's look at this first integral first. It itself is an improper integral so we write the limit as B approaches 1 from below, the integral from 0 to B of 1 over cube root of x minus 1 dx. We integrate and find this is the limit as B approaches 1 from below of 3 halves times x minus 1 to the 2 thirds evaluated from 0 to B. This is equal to the limit as B approaches 1 from below of 3 halves times B minus 1 to the 2 thirds minus 3 halves times 0 minus 1 to the 2 thirds. Now this term as B approaches 1 from below, this term goes to 0 and this term is negative 3 halves times negative 1 to the 2 thirds. Now negative 1 to the 2 thirds is equal to 1 because we're squaring and then taking cube root. Or we can think of it as negative 1 to the 1 third which is cube root of negative 1, negative 1 squared gives me 1. So this results in an integral of 3 halves, negative 3 halves. Okay we're going to hold on to that and move to the second part of our integral. We can rewrite this as the limit as A approaches 1 from above because our discontinuity is now the lower bound from A to 2 of 1 over cube root of x minus 1 dx. Integrating as we did before, we find this is the limit as A approaches 1 from above of 3 halves times x minus 1 to the 2 thirds evaluated from A to 2. This is equal to the limit as A approaches 1 from above of 3 halves times 2 minus 1 to the 2 thirds minus 3 halves times A minus 1 to the 2 thirds. We note again as A approaches 1 from above, this term goes to 0, this term equals 3 halves. So this second integral equals 3 halves. Now since both integrals converge so does their sum and what we found is that their sum in fact gives us negative 3 halves from before plus the 3 halves we see here we get an integral of 0.