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Calculating average value of function over interval

Here we find the average value of x^2+1 on the interval between 0 and 3.
Video transcript
Let's say that we have the function, F of X is equal to X squared plus one. And what we want to do is we wanna figure out the average value of our function our function F on the interval, on the closed interval between zero and, let's say between zero and three. And I encourage you to pause this video, and especially if you see in the other videos on introducing the idea of the average value of a function, figure out what this is. What is the average value of our function F over this interval? So I'm assuming you've had a go at it, let's just visualize what's going on and then we can, we can actually find the average. So, that's my Y axis. This is my X axis. Now over the interval between zero and three, so lets say this is zero, this is one, two, three. It's a close interval. When, when X is zero, F of zero is going to be one. So, we're going to be, we're going to be right over here. F of one is two, so it's gonna be, so this is one, two, three. Actually, let me make my scale a little bit smaller on that. I have to go all the way up to nine, up to ten. So, this is gonna be ten. This is gonna be five and then one, two, three. Actually let me, it's hardest part. It's making this even. So let's see. This gonna be in the middle. Pretty good. And then let's see, in the middle and then we have that. Oh good enough. All right. So we're gonna be, we're gonna be there. We're gonna be there. I have obviously different scales for X and Y axis. Two squared plus one is five. Three squared plus one is ten. Three squared plus one is ten. So it's going to look something like this. This is what our function is going to look like. So that is the graph of Y is equal to F of X. And we care about the average value of the interval, close interval, between zero and three, between zero and three. So one way to think about it, you can apply the formula. But it's very important to think about what does that formula actually mean? And once again, you shouldn't memorize this formula because it kind of actually falls out of what it actually means. So, the average of our function is going to be, it's going to be equal to the definite integral over this interval. So, essentially, the area under this curve. So, it's going to be the definite integral from zero to three of F of X, which is X squared plus one DX. And we're gonna take this area, we're gonna take this area right over here and we're gonna divide it by the width of our interval to essentially come up with the average height or the average value of our function. So we're gonna divide it by B minus A, or three minus zero, which is just going to be three. And so now we just have to evaluate this. So this is going to be equal to one third times, see, the antiderivative of X squared is X to the third, or three. Antiderivative of one is X. And we're going to evaluate it from zero to three. And so this is going to be equal to one third times, when we evaluate it at three, let me use another color here. When we evaluate it at three, it's going to be three to the third divided by three. Well that's just going to be 27 divided by three is nine plus three. And then when we evaluate at zero minus zero minus zero. So it's just a minus. Minus when you evaluate zero, it's just going to be zero. And so, we are left with, I want to make the brackets that same color. This is going to be one third times 12. One-third times 12 which is equal to four, which is equal to four. So this is the average value of our function. The average value of our function over this interval, over this interval is equal, the average value of our function to four. And notice, our function actually hits that value at some point in the interval. At some point in the interval, something lower than two but bigger than one, we could maybe call that C. It looks like our function hits that value. And this is actually, this comes, this is, this is actually a, a generally true thing. This is a mean value theorem for integrals that, and we'll go into more depth there. But you can see that this kind of does look like it's average value. That if you imagine the box, if you multiply this height, this average value times this width, you would have this area right over here. And this area right over here, is the same, this area that I'm highlighting in yellow right over here; is the same as the area under the curve. Cuz we have the average height times the width is the same thing as the area under the curve. So anyway, hopefully you've found that interesting.