 Hello, and welcome to this screencast. In this screencast, we will evaluate an example of an improper interval that involves a function with a vertical asymptote. We will evaluate the integral from 0 to 7 of negative 1 divided by x minus 7 dx. At first glance, this doesn't look like an improper integral because both limits of integration are finite. But we notice that when x is equal to 7, the denominator of the function is 0, so the function is undefined at x equals 7. Since our function isn't continuous or even defined on the closed interval from 0 to 7, then the fundamental theorem of calculus doesn't apply. To help us visualize the region whose area that we are trying to calculate, we'll draw a graph of the function y is equal to negative 1 divided by x minus 7. We get a vertical asymptote where the function is undefined at x equals 7. When x is less than 7, our y value will be positive and it will increase without bound as x gets closer to 7 from the left side. When x is larger than 7, the y values are negative and they decrease without bound as x gets closer to 7 from the right side and we see that vertical asymptote at x equals 7. The region we're interested in is this shaded region that's under the curve between the y-axis and the vertical line x equals 7. To evaluate our improper integral, we will replace 7 with the variable b. This variable represents a value that is less than 7 so that the function will be continuous on the interval from 0 to b. This negative symbol next to the 7 shows that we're evaluating the limit as b approaches 7 from the left so our interval never includes that vertical asymptote. Our next step is to find an antiderivative for our function. We will keep the negative 1 constant multiple and then the antiderivative of 1 over x minus 7 is the natural log of the absolute value of x minus 7 and then we evaluate that from x equals 0 to x equals b. When we evaluate this antiderivative at x equals b, we get negative natural log of the absolute value of b minus 7 and then we subtract negative natural log of the absolute value of 0 minus 7. We can simplify this limit as b approaches 7 from the left of negative natural log of b minus 7 plus natural log of 7. Our next step is to consider what will happen to the quantity negative natural log of b minus the absolute value of b minus 7 as b gets closer to 7. As b approaches 7 from the left, the absolute value of b minus 7 is going to get closer and closer to 0. Then the natural log of that value of the absolute value of b minus 7 is going to get closer and closer to negative infinity because the graph of the natural log of x has a vertical asymptote as x approaches 0. If we multiply this by negative 1, then negative natural log of the absolute value of b minus 7 approaches positive infinity as b gets closer and closer to 7 from the left side. So when we take a value that's approaching infinity and add natural log of 7 to it, that result will also increase without bound. And so we get a result of infinity for our limit. To summarize the result for this screencast, even though the limits of our integration are both finite, we find that the area under the curve is actually infinite as it stretches upward without bound near that vertical asymptote. And so in this case we say that the integral diverges. Thanks for watching.