 In this video, we provide the solution to question number six for practice exam number three for math 1220 In which case we have to set up the integral to find the area under the parametric curve x equals 1 plus e to the 2t and y equals e to the t from the region x equals 1 x equals 2 We set up this integral, but we do not have to evaluate it So some things to note here that the area under the curve in general will follow the formula a to b of y dx and that situation and so as we adapt this for the parametric form here We want to plug in y with respect to t. We're going to take the derivative of x there So doing so we're going to get for y we get e to the t Then we have to take the derivative of x with respect to t when you take the derivative of constant That's going to disappear. So you're going to get 2e to the 2t dt So far so good Now we need to evaluate it with respect to the difference now since our differential became dt You do need that for the full credit on this integral here, of course, we need to change the bounds We're not integrating from x equals 1 to 2. We have to integrate t So some things to note here. I'm gonna bring that 2 out in front You have e to the t times e to the t you can actually add the powers together And so this becomes e to the 3t dt We do need to evaluate the bounds here. So it's x equals 1 we plug in 1 for x right here So we have 1 is equal to 1 plus e to the 2t For which then you get 0 equals e to the 2t there can an exponential ever equals 0 Not precisely, but if you take the limit you can so the thing is this happens as t approaches negative infinity You can get zero so that's going to give her a lower bound right there So it turns out this is actually improper integral. We didn't see that one coming And then we also want to x equals 2 So if you plug in 2 for that one right there 2 equals 1 plus e to the 2t Subtract 1 from both sides you get 1 equals e to the 2t taking the natural log You're going to get 0 equals 2t dividing by 2 you'll get 2 equals 0 and that then gives you the upper bound like so and so then the Integral to measure the area will be 2 times integral from negative infinity to 0 of e to the 3t dt