 Welcome back everyone to our lecture series Math 1210, Calculus 1 for students at Southern Utah University. As usual, I'm your professor today, Dr. Andrew Missildine. As you can see from the title slide that's on the screen right now, we're in section 46. We're getting really near the end of our series here, and we're going to talk about the funnel mill theorem of calculus. We're going to break this up over actually two different lectures, look for lecture 47 as well. And when we talk about the funnel mill theorem of calculus, it often has a natural division into two parts. So in lecture 46, we're going to talk about part one of the funnel mill theorem of calculus, or we might call it FTC for short. And then part two will show up in lecture 47. So take a look for those links forthcoming. Now the funnel mill theorem of calculus is very appropriately named because it actually establishes a connection between the two branches of calculus. What's called differential calculus, which focuses on derivatives and integral calculus, which focuses on integrals, which we've been talking about in this chapter, chapter five here. Differential calculus arose from the tangent problem essentially, whereas integral calculus arose from a seemingly unrelated problem, this so-called area problem that we've been working with in chapter five so far. The funnel mill theorem of calculus gives the precise inverse relationship between derivatives and integrals. Essentially, we can equate integrals with antiderivatives in a manner of speaking, and again, that'll be more clear with FTC part two. Well, I mean, it comes up in FTC part one as well. In particular, the funnel theorem enables us to compute areas and integrals very easily without having to compute them as limits of Riemann sums as we were doing beforehand. And I also want to give a little bit of a historical connection right here. People often contribute the invention of calculus to Sir Isaac Newton and Godric Leibniz, who were contemporary mathematicians a couple of centuries ago. And that itself is a little bit of a myth that we really can't say that Newton invented the calculus, because people were doing calculus for hundreds and thousands of years prior to the advent of Newton himself. What credit deserves to go to Newton is that we had these different notions of calculus that existed, these very prototypes to what we nowadays call calculus, which could be described using modern languages like the integrals and derivatives and such. But what Newton really did is that he established this so-called fundamental theorem of calculus. He was able to connect the ideas of integrals with derivatives using the notion of a limit. So really, calculus as we know it today could be attributed to Newton, but he certainly wasn't the first one to come up with it. And really was Newton's contribution was this fundamental theorem of calculus for which he can receive the credit for such things. Now before we jump straight into the fundamental theorem, let me kind of give you an idea of where we're going, right? Definite integrals remember measure the area under a curve. And so if we have a function and we want to calculate the area under it, we can actually define a new function using integrals. It has these so-called area functions. So imagine we have a function f of x and it's continuous on some domain a to b. Well, with this we can define a new integral function or a new area function. In the following way, let's integrate the function f of t dt as we go from a up to x. And so you'll notice here that our lower bound in terms of the integral is this number a. It will be the lower bound of the domain of the function there. And then the upper bound we define to be x. That is, this is a variable that's to be inserted later on. And so this g of x, we're going to calculate the area from a, the left farce point up to x, which x can vary based upon location. And so we want to find the area under the function f. So this right here, g of x is calculating the area under f from a to x, sort of like from c to shining c type of thing right there. So let's look at a specific example. We take g of x to equal the integral function 0 to x of the function 2x dx. Now you'll notice that the upper limit here is a variable. It's typically going to be called x. And then inside the integrand, we have this function 2t. You'll have to see a different variable. We're not going to see an x right there. We're going to see a different variable. And this variable typically we'll call it t. It's also serving as a dummy variable. Kind of like when we look at sigma notation, you take i equals one to n and we had like i squared or something like that. The i there is just this dummy variable. Just to keep track of where you are in the sequence, t is doing the same thing here. It's keeping track of where you are in the area calculation. So if you look at the graph of the function f of, if you take f of t to equal 2t and you see we can graph the function over here. f of t equals 2t. This is just a line that passes through the origin right there. It has a slope of two. And therefore you can see it here in yellow in that graph that's displayed. If we look for the area that can be found under this curve, under this line so to speak, as you can kind of see illustrated, what would the area from 0 to 2 look like if we took the area under the curve? The area under this curve is going to be a triangle, right? In fact, it's going to be a right triangle. And so if we wanted to calculate the area of this thing, we could try to use formulas for areas of triangles. Now I first want to mention that if we were to look at the area from 0 to 0, so if you look at g of 0, what does that mean? This would mean you would go take the area from 0 to 0 of your function 2t. And by properties of integration that we've talked about before, if you integrate from a number to itself, you'll always end up with 0. And geometrically we're saying, well, what's the area under the dot that I just drew there? Well, there's nothing. That's just a single point. There's no area to that. You get area of 0. Well, how about g of 1? g of 1 says we're going to go from 0 to 1 right here. We want to capture the area of this triangle right here. Well, again, this is a right triangle. And so the area we can get, we can find the area here by taking 1 half base times height. All right. The base of this triangle would be 1. The height of this triangle would be its y-coordinate. So you have this 0.2 comma f of 2. Notice f of 2 is itself 2, 1 times 2 there. And so then you're, so you buy the area formula, you get 1 half, 1 times 2, which gives us an area of 1. So g of 1 would equal 1. All right. Let's do a couple of more examples. If we do g of 2 this time, g of 2, which is actually what you see illustrated to the right here, we would want to be finding the area from 0 to 2 and thus taking this right triangle. The base of that would be 2. The height of that is going to be 2 times 2, which is equal to 4. You double in it because that's what our function does here, y equals 2t there. And so we end up with 1 half, 2 times 4, in which case that ends up just giving us a 4. All right. Another example, take g of 3. We're just trying to do a couple of examples here to keep a, to find if we can identify a pattern of what's going on here. If you want to find the area from 0 to 3, that goes off my picture a little bit, but we could do it. If you went from 0 to 3, we're looking at the area of this triangle right here. The base is going to just be a 3. And then the height will be 2 times 3, which is equal to 6. And so if we put those together, 1 half times 3 times 6, that's equal to 9. Right. And that's finds the area of that triangle right there. Well, you can see that's as far as my picture goes. What if we kept on going? What's the pattern that seems to happen? If we go some distance out along the x-axis, let's call that value x, right? We go some distance far out, right? That's going to give us the base. And so, and the distance there is going to be x minus 0, right? So it'll just be an x, 1 half x. But then as we go up here, this distance, we have to look at f of x. f of x here, which is equal to 2 times x. We end up with this 2 times x that's right there. The 1 half and the 2 are going to cancel each other out. And you can see that this thing always equals x squared. That whatever we choose for x, the area under this curve, which is a triangle, the area formula could actually be x squared. So we could actually simplify this. Instead of involving the integral whatsoever, we could call this an x squared. It still captures the area under this line, which gives us a triangle, but we can remove the calculus and just look at the function x squared. g of x is just x squared. One thing I can't help but notice, and I have to point out to you, is that if we take this function g prime of x, if we take the derivative of this function, we end up with a 2x. Although the variable is different, 2t versus 2x, but notice that this 2x is exactly the integrand of the example. It turns out this is not a mere coincidence. In fact, this connection that the derivative of our function g gives you the function we were integrating in the first place, this is no coincidence. And this is actually the idea behind the fundamental theorem of calculus that these integrals and derivatives are connected to each other. Stay tuned. We'll actually make this connection more explicit in the next video. See you then.