 In this video, we're going to discuss the solution to question number seven from the final exam for Math 12-20. And in this question, we're asked to set up the integral to find the area of the region under the parametric curve given as x equals 1 plus e to the 2t, and y equals e to the t, where x ranges from 1 to 2. And this question is to set up the integral type question. Although simplifications are necessary, we are not expected to actually evaluate the integral for this question. So the start of this one is to recognize that to find the area under a curve, we want to integrate the expression y dx. And this is also relevant for parametric curves, and this actually gives us a way of how we want to compute this. So the integral, now we're going to try to set up a here. We're going to integrate the, well, to get the area, we get y times dx, y is the height of the rectangle, dx is the thickness. So we're going to insert the y function as it's given. So that gives us an e to the t. Then we're going to insert dx where we take the derivative of the x there with respect to t, in which case that's going to give us 2e to the 2t dt, like so. And so by labeling, we see there's the y coordinate right here. Here is the dx right there. And so that gets us most of the way there. The last thing to do is the bounds here. So we have to identify the bounds. And in terms of the integral itself, there's not really a lot of simplification to do there. You could leave it factored in the way that you have before, or you could bring out the 2e to the t, e to the 2t, like so. I mean, you of course can add them together. The exponents e to the t times e to the 2t would be e to the 3t dt. And so again, this wouldn't be a hard anti-derivative, but we're not actually asked to set it up. I'm sorry, we are asked to set it up. We're not asked to evaluate it. So the last thing to identify are the bounds here. So as x approaches 1 and x approaches 2, what happens? So as x approaches, as x is going closer to 2, we then plug, we have to solve this equation right here. We have 2 equals 1 plus e to the 2t. It's attracting 1 from both sides. We get 1 equals e to the 2t. Taking the natural log of both sides, you're going to see that 0 equals 2t, and therefore, t would equal 0. That's going to be the upper bound of this thing. So we're going to get 0 as the upper bound. As the lower bound, we take x equal to 1, and so we have 1 equals 1 plus e to the 2t. So attracting 1, you're going to get 0 equals e to the 2t. Now, one thing I should mention is that the exponential function can never actually equal 0. So it's tempting to say that, oh, there's no solution here. But in all reality, although e to the 2t cannot equal 0, we can have that e to the 2t approaches 0. And so really, we think of this as a limit. What happens is x approaches 1. So in that situation, when can e to the 2t approach 0? So kind of amending what we said earlier. So we want 1 plus e to the 2t to approach 1. That means e to the 2t should approach 0. And that happens, of course, as 2t approaches negative infinity, the horizontal asymptote there. And so that's telling us that t is approaching negative infinity as well. And so it turns out that this integral is actually was an improper integral in disguise. Kind of like the Decepticons there, in which case the lower bound actually mean negative infinity here. So our integral would be from negative infinity to 0 of 2 times e to the 3t dt. And honestly speaking here, as this one was meant just to set it up, if you left it factored as e to the t times 2e to the 2t dt, that would be perfectly fine. You do have the bounds negative infinity to 0. So if I was looking at this one, things I'd want to grade is that definitely I want to see that you swapped from the x values to the t values. So we need the bounds to be negative infinity to be 0. You'd have the y-coordinate e to the t. You have the derivative of the x-coordinate 2e to the 2t. You also need to have the differential, the dt here. That is part of the integral. Its omission would be a curvature of a point there. And this shows you how you could set up the integral to find the area under this parametric curve.