 In the previous video, we saw how to compute average value of a function. It's fairly simple. It's just the value one over b minus a times the integral of f from a to b. That's all there is to it. And so as long as the integral is easy to compute, average value is fairly simple to compute as well. But why is one interested in average value? Like if you're looking for what's the average homework score, you can see that you're trying to find what's the middle of all the homework you've done and that has consequences, perhaps on a final grade in a calculus class or something like that. And the average value of function is trying to do the same thing. It's trying to find what's the center of the function? What's the center of the data? Now, this idea of average value we talked about is related to the mean value theorem we saw in calculus one. The mean value theorem from calculus one tells us that if a function is differentiable, then there exists a value of x, such that the derivative is equal to the average rate of change of the function. That is, there's a moment where the instantaneous rate of change matches the average rate of change. A similar question actually applies in the following context. Is there a value of x for which the function obtains its average value? And for a continuous function, the answer is going to be yes. And so what we want to do is restate the mean value theorem, not for derivatives, but for integrals. If f is the continuous function on the interval a to b, then there exists some number c that sits between a and b, such that the average value, which we defined earlier, is actually equal to f of c. So the function obtains its average value at some location. Now, if you take this equation right here and you times both sides by b minus a, you've got to do it on the left-hand side as well. Then you get the integral, like you see right here. The integral is equal to f of c times b minus a. Now, this second equation has a very interesting geometric interpretation, so I want to give some explanation about that. So if you look at the left-hand side, the left-hand side right here is the area under the curve. So if your function f is given right here, you see that the area under the curve is this region colored in right there. The right-hand side, though, is factored. It's f of c times b minus a. Because we have a product to two numbers that's equal to an area, this makes me think that this is likewise an area. And as it's a product of two numbers, I think it's the area of a rectangle. A rectangle whose width is b minus a and whose height is f of c. So if we find the guaranteed point f of c right here, that doesn't look like a c. Try that again. If we take the guaranteed point f of c, is the point on the graph which has the same value as the average value of the function. In this situation, if we take a height that and we take the thickness, so here's a and here's b, if we take this rectangle right here whose width is b minus a, whose height is f of c, what we're saying is the mean value theorem for integrals tells us is that the area under a curve is actually equal to the area of a rectangle. And the height of the rectangle is the average value. So basically how we wanna take this geometrically here is the average value is saying if we re, to redistribute the area under the curve in a uniform distribution, it will form a rectangle the height of the rectangle is that average value. So why does such a point exist? Well, like we saw in calculus one, what we can do to kind of prove this argument is we can introduce a integral function, a function f of x defined as integrate from a to x, the function f of t dt. So when we talk about the fundamental theorem of calculus, we can define functions using integrals where the upper limit is itself a variable. All right, then the average rate of change of the function capital F on the interval a to b is gonna be given by the following capital, the delta capital F over delta x right here. Well, this will look like f of b minus f of a, and these are capital F's, not lowercase f. This will sit above b minus a, that's just the average rate formula there. Now as capital F is itself an integral, f of b is the integral from a to b of f of t dt. And f of a is the integral from a to a of f of t dt. Like so, this all sits above b minus a. Now notice that second one right there, the integral from a to a, that's gonna be a zero. If you integrate from a to a, that's always gonna be a zero. And so the numerator is just gonna be the integral. And if we rewrite this thing, you're gonna see that we get one over b minus a times the integral from a to b f of t dt. And now if this feels like deja vu, it's not quite that. This is, we've seen this before, this is the average value of F. So what we see here is that the average rate of change of capital F, the anti-derivative of little f, is the average value of little f. So we see that relationship right there. Now what we also know by the fundamental theorem of calculus, so by the fundamental theorem of calculus, FTC, we do know that capital F prime of x, this is gonna equal, well, if we write out the details, d dx of the integral from a to x f of t dt. What we know from the fundamental calculus is this is gonna be little f of x, like so. And so if we apply the mean value theorem for derivatives to this capital F, right? So we apply the mean value theorem from calculus one to capital F. What this tells us is that there exists some number C such that the average rate of change, delta F, delta F over delta x is equal to the derivative of capital F at this number C, right? But the left-hand side is just the average value, which then proves the argument we're trying to look for, that there will be some location where the function obtains its average value. Now, to see that, let's take the example we saw in the previous video, take the function F of x equals one plus x squared, we consider the interval negative one to two, and as we saw previously, the average value here was exactly two. Where does this function F obtain this value two? Well, just solve the equation one plus x squared equals two, I replace the x with a c here. But if you take one plus x, c squared equals two, subtract one from both sides, you're gonna get c squared equals one, taking the square root, which notice there are two square roots, the positive and negative one, we get that c equals plus or minus one. And so if we look at F of one or F of negative one, those will both equal two. Now, notice that there were two values, and this mean value theorem guarantees there's at least one. There could be numerous, right? There could be 12 dozen, there could be 5,280 numbers that obtain the average value. But we know there's at least one. But I do wanna draw your attention to the fact that if we look at the interval, negative one to two, the mean value guarantees that there will be at least one number in this interval between negative one and two that obtained the average value. And there does turn out to be two. Negative one's on the boundary and one is in the middle somewhere. So we do get multiple values that do that. And if we look at these numbers geometrically, we see those points right here. When x equals one, the y coordinate is two. When x equals one, negative one, or when x equals one, this is x equals one, we see the y coordinates equal to two. And if we look at the area below the curve, we look at the area below the curve right there, we see, we calculated earlier that this area was equal to six. And so, if you notice the length of the interval right here is equal to three, the average value is two. If we look at this rectangle right here, this rectangle likewise has a volume of six, not volume, an area of six, excuse me. So this rectangle has the same area as the area under this curve from negative one to two. And this is a rectangle whose width is the length of the interval and whose height will pass through these guaranteed points, these average values of the function. So we can always turn the area under a curve into a rectangle using the average value. And that actually brings us to the end of our discussion of average value. It's a really short section. It's cute and adorable, but that's it. If you have any questions, like always, please, please, please, you can post your questions in the comments of this video below. Now be aware that if you're watching this video through a learning management system like Canvas, you might have to actually open up the YouTube page. Of course, if you're on YouTube already, you can see the comments. Post your comments below. I would, your questions or comments, I'd be glad to address them, answer any questions you have with any of these videos. If you like what you're seeing, please click the like button, subscribe to see new videos in the future, and I will see you next time. Keep on calculating everyone, bye.