 Hi everyone, I have for you here an example problem of finding the area between two curves. So we're asked to find the area of the region bounded by the graphs of y equals x square plus 2, y equals negative x, x equals 0, so I think that's the y axis, and then also the vertical line x equals 1. So the first thing you'll want to do is go ahead and graph these on your graph and calculator. So I have them set up under y equals on my calculator, and I just went ahead and did zoom 6 to graph it, so hopefully that's the graph you're looking at. Now remember we're only concerned with the region going from x equals 0 to x equals 1, so it might be helpful if we can zoom in a little bit on that region. A great way to do that, I don't know if you've used this all that often, is zoom box. If you hit your zoom button, you'll notice that first one says z box. So hit enter and you'll notice your little cursor there blinking at you. What you're going to do is essentially create a box, a frame around the part that you want to focus on. So you'll want to move your cursor to the upper left corner of where you will want your frame to start. So from where I am at the origin now, I'm going to go a little bit to the left, and then I'll go up a little bit, maybe go up a little bit higher than the first curve. And once you know where you want your upper left corner to start, hit enter, and then you start going to the right and down. So I'm going to go to the right a little past 1, and then I'm going to go down. Because remember it's also bounded by on the lower end that red curve, that's the line y equals negative x. Now your numbers might not be exactly like mine, that's okay. Then when you're done just hit enter, and you'll notice it draws it just focusing in on that part. So really what you're looking for is that it's part of the region in the first quadrant, a little bit there in the fourth quadrant, bounded on the upper end by the blue curve. The blue curve is y equals x squared plus 2, and at the bottom by the red curve, y equals negative x, in between x equals 0 and x equals 1. So think of the idea of the representative rectangles we talked about. Remember the idea goes back to Riemann sums, and if you could inscribe rectangles in here so that your rectangle is hitting each of the curves, in this case your rectangles would have to go vertically. That would make it a dx problem. So the way in which we're going to find the area, let's go back and write it out, our area is going to be the integral from 0 to 1, remember it's a dx problem. So we want our limits of integration to be x's, and it's going to be the curve on the top, which was x squared plus 2 minus the curve on the bottom, which is minus x dx. So we could rewrite that and simplify it. You could go ahead and do your anti-derivative, which would be evaluated from 0 to 1. You can then use your graphing calculator to do the evaluation. So let's go ahead and do that, give you some practice doing that. So let's go back, let's do our math 9 function integral. So we want the from 0 to 1, and I'm just going to type it out. So it was x squared plus 2. I could easily do y1 minus y2. It might be just as easy, and remember this turned into plus x because of minus the negative, and our answer is 2.83 repeating. If you did have to attach units, it would be square units because we're finding an area. So let me show you how to work that area between program that you have on your graphing calculator. So let's go back from your quit screen, if you bring up your program listing. Now, you probably don't have all the programs I do. That's okay. But yours should be alphabetized as mine are. So go down to area betwee. That's the one you want, area between. And when you go ahead and run it, it's going to ask you what your lower bound is, in this case it's 0. Upper bound would be 1, and there's your answer. Great way to check what you have to do by hand. That's what I really, really suggest this for you. Now one thing I do want to point out. Remember how we set up our integral was essentially y1 minus y2. It's that whole f of x minus g of x idea that was presented to you in the lesson. And if you take a look at under y equals, look at what the program's doing. Notice how under y3, it actually did y1 minus y2. So the calculator program is actually mimicking what you are doing by hand. But again, I suggest it as a really, really great way for you to check all your answers.