 Welcome back to our lecture series, Math 1210, Calculus 1 for students at Southern Utah University. As usual, I'm your professor today, Dr. Andrew Misilline. It's good to have you. We are continuing on our, in lecture 44 here, we're continuing on our discussion of area under the curve. We saw a little bit of that last time as we were trying to approximate the number pi by looking under the area under a quarter circle. We want to take the principles we saw in that lecture and generalize them to, well, to more general settings. I guess that's what the word generalize means here. The area problem is essentially the following scenario. Let's me just draw a picture real quick. We have the x-axis, we have the y-axis right here, and most importantly, we have a function. Let's say we have a function that looks something like this. This is some continuous function f. We're interested in what is the area under the curve between two values in its domain. We have some value here a, and we have some other value here b. What we want to figure out is if we take the interval from a to b, and we look at the function below, that is, we want to look at the region that's below the curve, but it's between these two boundaries. We want to figure out what's the area of this region right here. This is what we mean by this area problem. How do we go about doing that? The first idea, again, mimicking what we did in the last video, is we're going to take our domain a to b, and we're going to divide it up into n pieces. We take this domain and we subdivide it into smaller and smaller pieces here. We're going to get n subintervals in subdivisions. For the sake of simplicity, we want each of these subintervals to have the same space equal width, and that's what this number delta x is going to be. Delta x is going to be the width of each of the subintervals, and they're all going to be the same length. Since they're supposed to be all the same length, the idea is we take the length of the entire interval. The length of the entire interval would be b minus a, this value right here, b minus a. Then, in order to get equal pieces, we divide the entire length by n. Therefore, the length of a single interval, we look at one of these right here, a single interval, its length is going to be delta x. This is going to be the width of each and every rectangle. We're going to take a comment about that here. This is the width of the rectangles that will be forthcoming. Again, think about the last video, what we did here, the width of our rectangles. The next thing to consider is when we have these subdivisions here, we're going to start labeling the marks along the line here. What I mean by that is we're going to start labeling these things. We're going to have an x1, an x2, an x3. We label these things as we go along the way. I'm actually going to go dot, dot, dot. We'll have some arbitrary point in the middle. Xi is just a generic spot that's in the middle. Xi minus one would be the point that's immediately in front of that. Let me do a little bit better pinmanship there. Xi minus one. Then this continues on until we get to the very end. This last number, B, we can actually consider xn. It will be the nth mark along the way. The first one, A, we're actually going to call it x0. Going along the line, it's going to be x0, x1, x2, x3, x4 all the way up to xn, where A is just x0 and B is xn. This actually gives us a formula for the xi's, which we get right here, because the basic idea is we start off first at A, and then we take a step to the right, a step to the right, a step to the right, and each step you take is going to be the same distance delta x. If you keep track of how many steps you take, you're going to take i times delta x, add that to A. That gives you this xi that's along the journey here. For each of these intervals, you have the first interval, x0 to x1, the second interval, x1 to x2, the third interval, x2 to x3, et cetera. For each of these intervals, what we're going to do is we're going to select a representative, you might say a delegate, as some number that sits between xi minus 1 and xi. In the i-th interval, we choose some number xi, we call star. We're choosing someone in this i-th interval. What are we choosing this number for? We're choosing this number xi star because xi star as a representative of his hometown, the i-th interval, we're going to have xi star determine the height of the rectangles because the rectangle has to have a length, it has to have a height, just trying to move xi star a little bit to the side. The rectangle has to have a height. We know it's width already, the width we already determined is going to be the same distance as delta x, but how tall should the rectangle be? We want the rectangle's height to be somewhat determined by the height of the function, otherwise this won't be a very good approximation to air into the curve. The height of the rectangles is going to be determined to be this value f of xi star. This is for the i-th rectangle. As you do different intervals, xi star will change and hence the height will change as well. If we take this up, we get this point right here, this f of xi star. Then the height of the rectangle will be determined by the location, this f of xi star. We get a rectangle like this. This is our i-th rectangle. What we want to do is we want to replicate this for each and every single interval you see here. You pick some xi star, so there's an x1 star, we then have our x2 star. Again, we do this for each and every interval. We choose a delegate who will decide how tall the rectangle is going to be based upon the y-coordinate associated to that x-coordinate on the function. Then the height of the rectangle will be determined by f of xi star. The width will be determined by the delta x, which we computed earlier. It's the same width for everyone. Everyone has the same thickness that will make for an easier calculation later. Then we get all these rectangles here. We calculate the area of the first rectangle, r1, I guess I should say a1, the first area. Then we find the area of the second one, a2, then we do a3. We keep on going until we get to the very end right here. We try to approximate the area under this curve using a bunch of rectangles. We end up with this formula right here. The area of the i-th rectangle will be f of xi star. That's the height of the rectangle times delta x, which is the width of the rectangle. Areas of rectangles are lengths times widths, so that gives us a single rectangle. We might call this the area of the i-th one, ai, like so. The total area of the curve under the curve, I should say, is going to be approximated by taking the sum of all of these rectangles. The idea is you take the sum, as i goes from 1 to n, of all of these different areas. You take the area of the first rectangle plus the second rectangle plus the third rectangle. You add those all together, and you'll get the area of all the rectangles together, which should be an approximation of the area under the curve. When you add up these ai's, since the ai's as a rectangle's length times width, you'll take f of xi star times delta x, and that gives you the area of these rectangles. That should approximate the area under the curve. Let's see an example of this right here. Imagine we have the function f of x equals x cubed minus 6x, and let's say that we wanted to calculate the area under this curve. Now, we don't necessarily have to draw the picture to know what's going on here, but I'm going to draw the x-axis to give some illustration of what's going on here. Our domain is going to be from a to b, which in this case is going to be 0 to 3. These are the places we're going to mark off on the x-axis. We want to go from 0 to 3 like so. We're going to have six subdivisions. We're going to break up this line into six pieces. Let's first calculate my delta x. Delta x is going to equal 3 minus 0 all over 6, which gives us 3 6, which is 1 half, or if you prefer a decimal point 5. I'm going to actually build a table using this information right here. If we keep track of which point are we along the line, I here, you can get 0, 1, 2, 3, 4, 5, and 6. What is the xi's at this point? Well, xi, you're just going to just start at 0 because that's the a value, right? You're just going to add a half step to each and every one of them, so you get x1 is going to be 0.5, x2 is going to be 1, x3 is going to be 1.5, x4 is going to be 2, x5 is going to be 2.5, and x6 is going to be 3, which is the b-value, which is what we would expect. We get 1 half, 1, 1.5, 2, 2.5. I'm going to label these things, of course. Sorry, that was 0.5, 1, 1.5, 2, and 2.5. These are our little markers along the way. These are x0, x1, x2, x3, x4, x5, and x6. So we marked up the x-axis in the following way. Now what we have to do is we have to decide who is going to be our delegate for each of these intervals. We have the first interval, the second one, the third one, the fourth one, the fifth one, the sixth one. Who is going to be our delegate for each of these intervals here? That depends on the scheme, right? We've determined how many rectangles there's going to be, there's going to be six, but we have to decide who, how are we going to choose the delegates. In the description here, we're going to use our sampling points to be the right endpoints, the right endpoints. In this notation, we refer to this approximation of the air into the curve as r6. The r represents we're going to choose our sample points, xi star, to be the right endpoints. So for the first interval, we're going to choose the right endpoint right here, x1. For the second interval, we're going to choose the right endpoint, x2. For the third interval, we'll choose x3. For the fourth interval, we'll choose x4. For the fifth interval, we'll choose x5, and you guessed it, for the sixth one, we'll choose x6. So when it comes to the right endpoint sampling technique, rn here means you're going to choose, you're going to choose your delegate xi star to be the number xi itself, the right endpoint for each of these intervals. So what this then tells us is that the area under the curve is going to be approximately r6, which r6 is going to equal a sum where i goes from 1 to 6, because there's six rectangles here. So we have to compute f of xi times it by delta x. Normally, we would get an xi star right here, but because we chose the right endpoints, that's just going to be xi itself. And so some things to note here is the delta x, like we computed, is always a one-half. So I'm just going to stick a one-half out in front of everything. And then if we expand that sum, we have to compute f of 0, and I'm sorry, that would be x0. We have to do f of 0.5, or one-half if you prefer. Then we get f of 1 plus f of 1.5 plus f of 2 plus f of 2.5 and then finally f of 3 right there. So we have to compute each and every one of those things. Now, this is the actual part where we need the function, the function x cubed minus 6x, right? So I'm going to jot that down right here just so we know f of x equals x cubed minus 6x. So we're going to have to plug each of these numbers into the function f of x and compute each and every one of these things. So if we were to do f of x right here, so we'd look at f of 0, which would be 0. We don't actually need that calculation. We're going to take f of 1-half. That turns out to be a negative 23 over 8. Or as we're probably going to prefer for decimals here, excuse me, we get negative 2.875. We did do the next one, f of 1, which is going to be a negative 5. We're going to do f of 1.5, which gives us negative 5.625. And you get the idea. We're going to continue on in this. So as we do this calculation, we're going to have 0.5 in front. And like we said, f of 0.5 was negative 2.875. We just evaluate the function at 1-half. And then we subtract 5. That was f of 1. f of 1.5 was negative 5.625, like we said earlier. If we do f of 2, we get negative 4. If we do f of negative 2, we're going to get positive 0.625. And then lastly, we're going to get f of 3, which is equal to 9. In which case, if we add all these things together, that gives us a total. Again, if you're keeping track of fractions or decimals, I don't really care too much. You know, the 0.5 was 1-half. If you did all of these as fractions, this would add up to be negative 63 over 8. So when you times that by 2, you get, of course, negative 63 over 16. That gives you the fractional answer. Again, if you want a decimal, just write that as a decimal. No big deal. You get negative 3.9375. Now, you might be wondering, so this gives us the area under the curve. You might be wondering, how in the world do we get negative numbers here for our area? And so the idea is actually the following. Remember that the height of the rectangle was determined to be the height of the function. Now, when the function is above the x-axis, f of xi star is going to be a positive value. But when the function actually goes below the x-axis, f of xi star will actually be a negative y-coordinate. And this has the effect that gives you a negative area. Negative area right here. And, you know, that might seem kind of weird, but that's actually something we want to build into this formula because when we look at some very scientific applications and such, it might make sense to have quote-unquote negative area, which we'll talk about that a little bit. We'll talk about that in another video here. And so for our function right here, if we actually to look at its graph, its graph is going to look something like the following. Again, this isn't the best picture in the world, but it would look something like the following, like so. And so as we're calculating area, we're getting some like negative rectangles, negative rectangle, negative rectangle, negative rectangle. You know, something like this. Whoops. This is just a rough estimate, of course. And so this negative area down here is more powerful than the positive area down here. So we actually get the net area of this thing to be negative, about negative four. All right. And so we'll talk about more of this in the next video, of course, so stay tuned. I'll see you then.