 Hi there, let's do another example of a Riemann sum. This time I'm going to go back and use a function that doesn't have a formula attached to it. Just to prove to us that we don't need formulas to do these Riemann sums. So this is the function that we used in the quick review screencast. There's no formula again, but it does have a nice graph with a well positioned grid structure on this. And I would like to calculate the area that's between this graph, the x-axis, and x equals 1 and x equals 4. So that interval, we're going to be looking on the interval from 1 to 4. And I'm going to calculate the Riemann sum l sub 6. And what that means is I'm going to look at this interval from 1 to 4, split it up into 6 equal subdivisions, and then use the left hand endpoints of each subdivision to construct my rectangles. So let's walk through this step by step. So first of all, in l6 we're using n equal to 6 rectangles. Let's just draw the basis of those rectangles very simply. That's going to be this piece. Here's the second piece, the third, fourth, fifth, and sixth. So those endpoints break that interval from 1 to 4 up into 6 equal pieces. Now what are the sizes of each of those pieces? That's what we call delta x, that change in x going from one endpoint to its successive endpoint. Now the formula that you see in your book just says it's b minus a divided by n. And the way to think about that formula is all it's asking you to do is take the total length of the interval, that's what b minus a is, and then just divide it up by n, the number of rectangles we have. And so visually what's happening is I'm looking at the total length here going from 1 to 4, that's a length of 3, 4 minus 1 is 3 in this case equals 3, and I'm dividing that up into 6 pieces. And so if I have a length 3 interval and dividing it into 6 pieces, then delta x is a half. So each of these rectangles that I'm going to eventually construct here has a width of a half. Now let's go find the left-hand endpoints. There's going to be 6 left-hand endpoints, and let me just kind of put dots where they are. Here's the first one, the second, third, fourth, fifth, sixth. So in the language of the Riemann sum, this is what we call x sub 0. And sometimes we put a star up above that to denote that this is a special x value that we're picking out. This is x sub 1 star. This is x sub 2 star, x 4 star, and x 5 star. Remember, this last guy here at x equals 4 is not a left-hand endpoint. And so I'm not going to include it in any calculations. Only these 6 guys here, 1, 2, 3, 4, 5, 6, are the only left-hand endpoints I'm going to consider in this problem. I start the numbering at 0, and so it ends at 5. But there are 6 of them. Now let's actually draw the rectangles here. And this is simple to draw. I'm just going to create a base for each of these rectangles. Look at the left-hand endpoint of each of those intervals and throw it up on the curve, and then make the rectangle that high all the way across. So there is the first rectangle that I've created. The second rectangle, I would go to this left-hand endpoint, the next one, go up to its location on the curve, which is about right there. And then make the rectangle that high all the way across. The third rectangle will be the same deal. I'm going to go to this left-hand endpoint, go up to its position on the curve, and go across. And we can see how this is going to go. So I'll just go quickly through the rest of this. There's the fourth rectangle. Here's the fifth rectangle. And here is the sixth one that kind of sticks out just a little bit over the curve. Okay, so those rectangles, the areas of those rectangles, all added together is going to equal my L6 sum. So what I need to do for each of these rectangles is find the area of each of those rectangles. So the area of each one of these, say the first one, is going to be the base of that rectangle, which is delta x. That's equal to a half this time, remember. But I'm going to write delta x here just to make it look like my Riemann sum formula. And I'm going to take the height value of the left-hand endpoint. So that left-handed endpoint and then evaluate it into f to get its height. So f of x sub zero star times delta x is just notational code for the height of the rectangle times its base. It's very simple geometry. So don't let the notation throw you on this. This guy's area is going to be f of x sub one star times delta x. In other words, take this left-hand endpoint and evaluate it into f to get that height and then multiply by the base. And we just continue this process through all six of these rectangles. Now we don't have a formula to tell us how high these rectangles are, but we can look on the grid and estimate them using the scaling on the y-axis. For example, let's make a little table over here off to this blank space here on the right and just collect all these x values. So the first left-hand endpoint I need is one. Another one is one and a half, two, two and a half, three, and then three and a half. And for each of those endpoints, I'm just going to look at the graph and get a decent estimate on what I think the f value is, what the height value is. For the first couple of endpoints here and here, it appears like the height is about equal to one and a half. So I'm going to put one and a half down for each of those guys. The x equal to two endpoint, that's right here. That appears to have a height of about, oh, let's call that 2.2. We can only be so accurate here if all we've got is a graph with this level of scaling on it. x equals two and a half, that has got a height all the way up here. That's about, let's call that 3.8. x equal to three, that's here. That goes pretty high up the curve. That's almost at five. Let's call it five right on the nose. And then at three and a half, that last left-hand endpoint goes all the way up here. Let's call that 5.2. Now on the next slide, I'm just going to take all these height values, these guys, and run it through my Riemann sum formula. Sum formula laid out for us in abstraction. So what we're doing here is we have six things we're adding up. One, two, three, four, five, and six. And that corresponds to the areas of the six rectangles. Again, notice that the last term in the sum is f of x five star times delta x. There is no f of x six star in this, even though we have six rectangles. And again, that's because we start the counting here at zero. The first one is actually the zero rectangle. Now, one thing that we can do to help ourselves here is just to notice that each of these six terms we're adding up has a common factor delta x. And so we can, and that's the same delta x for every term we're using here. So I'm going to split that delta x term off and factor it out and just multiply by what's left. Now, what's left is going to be f of x zero star plus f of x one star and so forth. And what are these things that I'm adding up? Well, these are the height values of the left hand end points that I picked out. And we put those on the table on the previous slide. And so if you need to go back and look at those, just pause the video and do so. I've got these written down separately. So I'm just going to write these down very quickly. We estimated these f values off of the graph. And so all I'm going to do is just put all the numbers in that I know. Now, first of all, delta x, we do know is one half. This is going to be a half times. Now, let's write down all these f of x star values. The first one was one and a half. The second one was also one and a half. These are estimates, you know, but we do the best we can with what we've got. 2.2, 3.8, 5, and 5.2. Now, let's just all do the arithmetic and what should be done here. Now, the one half is still being multiplied. Don't forget that. And then on the inside, all these numbers add up to 19.2 if you do a quick calculation. And so a half times 19.2 is 9.6. Now, let's debrief this in reality check it to make sure that it makes sense. So this 9.6 figure is supposed to be an estimate to the area that's underneath the graph of f above the x-axis between x equals 1 and x equals 4. Let's go back and see if that's reasonable. I'm going to erase a number of things that are on this page or all the extraneous calculations. Everything but the rectangles, basically. Let me get rid of all this stuff to clean my mess up. And this goes to... Now, we said the area was about 9.6. Let's see if that's reasonable. Now, the maximum area we would get on this interval would be this rectangle that has a base equal to a 3 and a height that goes all the way up to the maximum height. So this rectangle here is definitely bigger than L6. And that rectangle has an area of about 15 because it goes a little bit higher than 5 on the y-axis and has a base of 3. So the fact that my L6 calculation came in under 15 is a really good sign. If I had forgotten, for example, to multiply by the one half here and gotten 19.2 for my L6, I could easily go back to the picture here, look at the area, and see that I made a mistake somewhere because this area cannot be any larger than roughly 15. Now, on the flip side of that, the smallest figure that we could have for area would be taking the base like so and going to the minimum value of the height, which would be about right here. So the area under the curve and above the x-axis and in this zone here is definitely bigger than three times roughly one and a half, which is four and a half. So I feel good that my L6 calculation, my Riemann sum, came out to be 9.6. So this definitely has to be between four and a half and 15, and my area figure actually is that. So that's a real simple way to just reality check the results of a Riemann sum calculation. Just ask yourself what could the area, what must the area be larger than and what must the area be smaller than? And if your L6 or whatever the Riemann sum figure is fits in between those bounds, then you're definitely on the right track. If it falls outside those bounds, then you need to go back and debug. So that's just another quick example of a Riemann sum using a function without a formula. Thanks for watching.