Everyone, and welcome back. Up to this point, we've spent a lot of time figuring out how we can approximate the area underneath the curve of a function by filling it with rectangles and adding the area of each rectangle. Now, in this video, we're going to learn how we can find the exact area underneath the curve. So we're no longer approximating. Now the results we get are going to be exact using this process of filling it with a bunch of rectangles. And this might sound like it's impossible to do, but it turns out there is a process that allows us to do this with what we've learned. So pay close attention, because I'm going to walk you through these steps and show you how to solve these types of problems in this course. Let's get right into things. The main thing we need to think about is how exactly we could use rectangles to find the exact area. Well, for approximating with rectangles, we're in essence just filling this up and calculating the area of each rectangle, base times height. Suppose I were to try to do an approximation with 2 rectangles. It would look something like this, where we'd have one rectangle right here and another rectangle right there, using right endpoints. Notice, if I were to try and find this area, we would get a severe over-approximation because we have all this extra area above the curve that's sticking out, and this would give us an overapproximation. But now let's say I wanted to fill this with more rectangles, so rather than using 2, we use 4. So four rectangles would look something like this. Again, I'll use these right endpoints to approximate. We'd have one rectangle here, one there, one right about there, and one right here. Notice, when doing this, we still get an over-approximation. Because if you look, we have all this extra area that's still above the curve, but this is a better approximation than the area we had before. As we draw more rectangles as n gets bigger, the width or the base of these rectangles, I should say, gets smaller, and we get a more accurate approximation. We need to increase the number of rectangles to get the most accurate approximation possible. Let's say that we had 100 or 1000 or even 1000000 rectangles. This would give us a very accurate approximation. We would have very little error, very little extra area popping up above the curve. So if we could somehow find a way to get this n value, this number of rectangles to approach infinity, that would allow us to get a truly accurate approximation or the exact amount. And in this case, we actually can do that because we can't necessarily just say, oh, we have an infinite number of rectangles, but what we can do is apply a lim n → ∞ to this equation that we've already been using, and say, what happens if this limit approaches infinity, the number of rectangles? Well, in that case, we're going to get the most accurate approximation we can, meaning that that's going to be the exact area underneath the curve. So let's actually see how we can use this equation to solve the type of problem where we want to find the exact area. In this example down here, it says find the exact area underneath the function f ( x ) = 1/2 x from x = 0 to x = 6 , that's our interval, using the limit definition. Now we have this argument out in front right here. This argument says we need to add a certain number of rectangles, and this time we're going to an infinite number of rectangles. What I need to find is delta x to go in here. We know how to calculate delta x; it's this equation that we've been using, b - a / n . So delta x is going to be b = 6 minus a = 0 , all divided by the number of rectangles, which in this case, we have an infinite number of rectangles. But I can't just put infinity in the bottom of the fraction. It's not really a number, so I have to leave this as just n. So what we end up getting is that delta x is 6-0 n = 6 n . Now, here we have delta x calculated, so I can go ahead and put 6 n into this summation. But now we need to find f of x k. f of x k is the height because remember this is the base, that's the height, but how do I find the height of each of these rectangles? What you first need to do is find what this x of k inside the function is, and x of k is going to be a, the start of your interval, plus k times delta x. This might look kind of confusing, but all this is saying is that if we're using, say, right endpoints to approximate, well, in this case, we would just keep counting up for each rectangle we have. And this k would be like, well, we have 1 delta x or 2 delta x or 3 delta x. We've already seen that before, so that's all this is saying. We're just using variables to describe that. For x of k, what I need to do is take a, which is the start of our interval 0, and then add this to k times delta x, which we figured out was 6 n . So what we're going to end up having for x of k is that this is all equal to 6 k n . Now that we have delta x and x of k calculated, what I can now do is plug in this function here f of x k, which is going to be f of this value 6 k n . So now we have everything plugged into our equation, and at this point, I just need to do some algebra to get this figured out. Now, this argument right here that you see, going to want to keep this consistent every time you are doing a step here. But because I want to save some space here, I'm just going to say, remember this argument is out in front for each step that we do because I'm mostly just going to focus on simplifying this part right here. So we have 6 n , that's our delta x. And to find f of 6 k n , well, that means I need to take this argument right here and plug it into our function. Our function is 1/2 x, so that means we're going to have 1/2 of this value 6 k n . Now what I can do is take this 6 right here and divide it by this 2, that's going to give us 3. So we're going to have 3 n , and keep in mind this is all going to be what this area is equal to, so we're going to have 3 n and that's going to be times 6 k n . Now 3 times 6, that's going to give us 18. So we're going to have 18 k, and n times n is n squared. So this is what everything simplifies to when we go ahead and plug this into the function, and we reduce this down as far as we can get. Now keep in mind this argument we're matching is still out in front here. And what I can actually do at this point is I'm trying to get this into a certain form with my summation because this summation stays in the argument. And if I can somehow get just the k that we have right here to be by itself, well, that's going to allow me to get a form of a summation that I can simplify with the n's that I have right here. So what I'm going to do is take this 18 n 2 and I'm going to bring it outside of the summation. So I'm going to have 18 n n 2 on the outside, meaning all we're going to have left over is this k on the inside. And this is perfect right here because remember this summation that we have there. We know this is a familiar form; it's n × ( n
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- 0. Fundamental Concepts of Algebra3h 29m
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23. Intro to Derivatives & Area Under the Curve
Area Under a Curve
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