 Welcome back to our lecture series, Math 1210 Calculus I for Students at Southern Utah University. As usual, I am your professor today, Dr. Andrew Misseldine. This part of our lecture series starts what we'll call lecture 45, which is definitely a step in the right direction. That is to say that we are continuing our discussion in 5.2 about the definite integral for which we are going to officially define at this moment. In the last several portions of this lecture series, we've talked a lot about it, and so what we can imagine with the problem we have is the following. We have some function f, which is given to us and we have some nice continuous function f right here. Then consider underneath the function our x-axis right here, for which we have some specific numbers a and b, for which we're interested in what does the area under the function f between a and b look like? How do we calculate such a thing like that? Well, as we've seen in these previous sections, is that we can approximate the area under the curve by taking rectangles, right? If we have this interval a to b, let's say we want to divide this interval up into n, we have 1, 2, 3, keep on going all the way up. We have these even pieces right here, which we're going to call the first one x1, x2, x3, continue on till we get somewhere in the middle xi-1, xi, all the way up to b. Where b here represents the last number in this sequence, so we'll call that xn, and actually a represents the beginning there, which we could call x0. And so each of these pieces is supposed to be evenly spaced, so the distance between x2 to x3, for example, we're calling this delta x, which if we want these to be evenly spaced, what we would do is we would take the total amount that we have, so b-a is the length of the entire, and we divide that by n. This gives us the amount that each little segment should be sliced into b-a over n. So each of these little slices is gonna be exactly delta x in thickness. We got this one right here, this one right here, this one right here, these are all delta x thickness. All right, and so then in each interval, we select a representative. So we pick some number in the first interval, some number in the second interval, some number in the third interval, some number in the fourth interval, some number in the next one, some number in the eighth interval right here, and this number we are gonna refer to as xi star. It's the delegate, it's the delegate for the eighth interval. And so using this number will take the height of the function at that value. So you have a point right here on the curve, which has a height of f of xi star, f of xi star right there. That's a little bit messy. Let me try that again, xi star. And we do this for each of these xi stars. We take the height of the function at that location. And let's not worry about the stratagem we use to decide xi star at this moment. We can come back and revisit that issue in just a second. But we select the height of the function based upon these different choices of xi star for each of the different intervals. So you get something like this. And so then once we have these xi stars selected, we can then compute xi star and we're gonna create rectangles whose thickness is delta x and then whose height is given by the f of xi star. So for each of these, we would get a, so for the first rectangle, it's something like this. For the next one, we get this one. For the next one, we get this one. This one, this one, this one, this one, this one. And then one more, we get something like this. And so we have all these rectangles, these green rectangles here, for which we can calculate the areas of these rectangles respectively. So if you take an individual rectangle, like this one right here, the area of the ith rectangle would become f of xi star times delta x. That's the area of a single rectangle. When we add these all back together, the area under the curve would be approximately the sum. Oh boy, what happened to that sigma there? Started off so good. It would be the sum where i ranges from one to n of all of these different areas, ai's. So you have all of these areas right there, which of course we could be more specific about. We can write f of xi star times delta x, like so. And so this right here, this summation notation see right here is what we have referred into the past or what we call now a Riemann sum. A Riemann sum. Approximates the area under the curve. But it's only an approximation, right? The fact that it's not exact, that is we only have approximately equal to a, comes to the fact that we have these rectangles, which there's a little bit of overlap that happens there. There's some places that are too low, some places that are too high, too high, right? And so we often see that there's some error, right? We're overestimating that parts, we're underestimating that parts. And so we're not gonna have a precise calculation of the area under the curve. But things we can do to improve this is if we can choose xi star more efficiently, this potentially could improve the, I wanna say efficiently that is, can we choose it so that it is calculating the area more accurately? And then the other option is we could start adding more and more and more rectangles. The more rectangles we have, the fit under this curve here. So what's the best number of rectangles to use? Well, if we keep on choosing better and better and better, that is by choosing more and more and more, we could actually consider the limit as n goes to infinity here. We could take an infinite number of rectangles and then we have our Riemann sum right here. If we take the limit of the Riemann sum, this is what we call a definite integral. We take the integral of f of x as x ranges from a to b. And so that's how you wanna read this expression right here. We have our integrand right here. We have our differential dx. You have the integral symbol, but you also have these numbers a and b right here. These are often referred to as the limits, oops, the limits of the integral. And they're gonna tell us the bounds of the domain. So as x ranges from a to b right here. And so other notation we should mention out here is that for simplicity's sake, we're gonna choose xi star to be the number xi, which xi is just a plus i delta x. You start at a and you take i steps to the right by a length of delta x. X i star, this of course is just the choice in the right point method for selecting the heights of the Riemann sums. Cause the idea here is that if a function is integrable, it's integrable means that this limit exists. The limit of a Riemann sum, I mean, as it's a limit, its existence is somewhat suspect. And so if the limit of Riemann sum exists, we say our function's integrable. And when it's integrable, it doesn't matter how you choose xi star. And so we're gonna choose it just to be xi, which is a plus i delta x. And so we've seen this definition before. And the definite integral gives us the area under the curve. So equality's obtained that the integral is the area. Now in previous examples, we were able to use geometric formulas to help us compute these. And so I wanna show you how one can compute the area in the curve for something that doesn't have a cute little formula to it. So like for example, if we were to consider this function right here, y equals x cubed minus six x, crudely speaking, our graph would look something like this. And honestly, I'm not placing the location of the x-axis very correctly. Let me try that again. It might look something more like maybe this right here. Again, this is just a rough estimate. This is not like me looking at a graphing calculator. Just by virtue of the formula, you get something like this, right? It goes up, it's up, and then, you know, like this sort of the standard cubic function. And so we wanna find the area as x goes from zero to three, all right? And so we wanna kind of see what's going on here with this area, and this is like a cubic function. This isn't a circle, this isn't a rectangle. We're gonna have to try something else and this Riemann sum approach is what we're gonna have to do. And so to calculate this, there's a few things we wanna probably evolve. So first, let's look at delta x. Delta x, by the formula we had before, is b minus a over n. Now, in this setting, the value a and the value b will be given to you as the limits of the integral. So you get zero and three right there. So we take three minus zero. The number n is gonna be left as a variable. n is the variable of the limit. And so we have to keep the n around for a little bit, at least until we compute the limit. So delta x is gonna turn out to be three over n. The next thing to compute is our xi star. And again, for the sake of simplicity in these calculations, there's gonna be a huge advantage to using xi as our xi star, which remember, it looks like a plus i times delta x. And so plug in those values, a was zero. That was very generous of this example to provide a equals zero. And then we're gonna get i times three over n right here. Simplifying that we get xi equals just three i over n. And so the significance of this is the integral from zero to three of this function x cubed minus six x dx. This right here is gonna equal the limit as n goes to infinity. We take this sum, it's the Riemann sum, where i goes from one to n here. We're gonna then take f of xi, so f of three i over n, and then we times that by delta x, which is itself three over n. And of course, it probably goes without saying here, but when we talk about the function f here, we're talking about x cubed minus six x. And so we are going to plug this three i over n into this expression right here for these values of x. So let me slide it down a little bit. And so upon doing that, we're gonna still get the limit. Make sure you keep your notation with you the long of the way. So to keep on writing the limit, keep on writing the sum, we're gonna need those and we'll take care of them. All right, so when we plug the three i over n in, we're gonna get three i over n cubed minus six times three i over n. This is our f of xi right here, and then we have this three over n for the delta x right there. Apply the power of the three right here and multiply by the six. Doing so, we still get the limit as n goes to infinity. I'm going to factor out also this three over n, because this three over n, in terms of the sigma, three over n is a constant. Three doesn't depend as, three doesn't change as i changes, and neither does the n. So we can actually factor out those constants there. And this always happens with delta x. It doesn't depend on i whatsoever. So you can bring delta x outside of the sum, in which case then inside the sum, we're gonna get 27 i cubed over n cubed, and then minus 18 i over n, as i ranges from one to n. And so in this sigma here, in this sum, we want to break this, and so we have the two parts, one involving the i cubed and one involving just the i. So breaking up into two sums and pulling out the constants, we would get something like the following. Take the limit as n goes to infinity. We get this three over n from the delta x. And then when we break up the sum, we can bring the 27 over n cubed out of the first sum, because those, again, are constants with respect to the sum, which it's the sum's variable as i. You're gonna get the sum of i cubed. And then we're gonna subtract from this 18 over n times the sum of i as i goes from one to n, right here. And so at this moment, we have to recall the formulas we had for sums of powers of i, the so-called sigma power rule. And so some things to remember is when you take the sum of the number i as i goes from one to n, this will always look like n times n plus one. Over two. And when you do this for i cubed, again, i goes from one to n, you end up with the exact same formula, n times n plus one over two, but everything is squared in that setting. So we're gonna plug those in for the sigma here as for these power rules. And so let's see what we have again. We got our limit as n goes to infinity. We get three over n. And then plugging these in for the formulas there, you get 27 over n cubed. We're gonna get n times n plus one over two squared minus 18 over n times n times n plus one over two. So we plug those in there. And now I'm gonna try to factor this thing. Again, for the sake of the limit calculation, let's try to factor everything. So some things to notice is that there's an n times n plus one over two that shows up at both of them. There's some n's in the denominators. There's like a factor of nine between the 27 and the 18. So if we factor all of this stuff out, identifying this least common denominator, we'll take the limit as n approaches infinity. Like I said, between the coefficients 27 and 18, we can take out a nine. So we get three times nine. The three was already out there. We were bringing out an n, we're bringing out an n plus one. And that should be everything from the numerators. We had an n that's already sitting out. We're gonna bring out another n and we're gonna bring out a two. That was common to all of these things. So what was left behind in the aftermath here? What was left behind? Well, nine take away from 27 would leave three. We took an n and n plus one. So we're left with an n and n plus one. And then we also took away a two from the denominator. So there's still two down there. And then we're left behind with an n squared right there. Just a quick note here, we can simplify this fraction, the n, and then cancel right there. For the second fraction, we took away the nine from the 18, so at least behind the two. We took away the n, the n plus one. And then we also took away the two n in the denominator. So we just left with a negative two right there. That's all. We sort of decimated that second fraction there. So this is good news, right? I guess I should mention we could cancel the n right here as well now that I mentioned it. Now I think of it. And so next, rewriting our thing one more time. These things do take a while to go through. I'll tell you that. Take the limit as n goes to infinity. We get three times nine, which is 27. This will sit above n plus one over two n. But we're trying to simplify the expression right here. What is left behind? We have this three over, sorry, three n plus one that sits above a two n, and we're subtracting from that two. So I would want to combine those together, together if I could, to a single fraction. That's one option we could pursue. But another thing to remember is we're trying to take the limit as n goes to infinity. When you look at a rational expression like this, you could break it up into two terms, right? And you could also redistribute this through if you wanted to. You could leave it factored or distributed, however you prefer. If we did, if we did then. And because we, I mean, we could go through this effort to try to explain what's going on. We could go through this effort to try to combine these two fractions together, which there's two horrible, but we're just trying to calculate the limit as n goes to infinity. So if we allow n to go to infinity, what's happening here? Well, as n goes to infinity, if you look at the first part, the 27, you have this 27 over n plus one over two n. That right there is a balanced rational function. You'll notice that you have a power of n on top and a power of n on the bottom. So this thing that's gonna converge towards the coefficients ratio, that is the leading coefficient of bottom, that gives you this 27 over two, right? And then for the next part, again, you have this balance rational function as n goes to infinity, you have n over n here. This will likewise converge towards the ratio of coefficients, three over two. And then the negative two, as it's a constant, it'll stay negative two as we go through this. So we can simplify this thing dramatically as we've done right here. And so we do have to add the fractions finally. We can write this two over two right there, but we don't have to do, we don't have to add all of the variable ends together to get this thing to work. So we get 27 over two. And so we're gonna get three over two, minus four over two, in which case we end up with negative one-half right there. And so then putting that together, we end up with negative 27 over four. And this would calculate the integral there, the area under the curve. And this is calculated area after all. It might seem a little bit weird that the area under the curve in this situation turned out to be negative. But after all, when we look at these, when we look at these integrals, the one thing that we wanna remember is that it is a net area because the height of our rectangles could be positive, they could be negative. And so if we go here, let's slide it back up here. If we go back to our original picture, the thing is some portion, some portion of, and some portion could be below. And so when it came to this picture, it turned out more of it was below than it was above. And so we end up with this negative value. This is quite common here. And so looking at this again, wow, I mean, there's a lot. There's a lot going on on this exercise right here. Computing the definite integral by the definition is a massive task. It involves taking limits. It involves simplifying these Riemann sums using those sigma power rules. It really is a capstone type problem for all these type of critical things that we've done here in calculus so far in this course. I wanna do some more examples of that, but check out the next couple of videos to see some more examples of computing definite integrals by the definition. I will see you.