And so now we just have to evaluate this. So, we're going to divide it by B minus A, or three minus zero, which is just going to be three. With the average height, or the average value of our function. We're going to take this area right over here and we're going to divide it by the width of our interval to essentially come up
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The definite intergral from zero to three of F of X which is X squared plus one DX. So, essentially the area under this curve. So the average of our function is going to be equal to the definite integral over this interval. So, one way to think about it, you could apply the formula, but it's very important to think about what does that formula actually mean? Once again, you shouldn't memorize this formula because it actually kind of falls out out of And we care about the average value on the closed intervalīetween zero and three.
So, that's the graph of Y is equal to F of X. This is what our function is going to look like. So it's going to look something like this. I have obviously different scales for X and Y axis. So see this is going to be in the middle. This is the hardest part is making this even. Actually, let me make my scale a little bit smaller on that. When X is zero F of zero is going to be one. Now over the interval between zero and three, so let's say this is the zero, this is one, two, three. Let's just visualize what's going on and then we can actually find the average. Value of our function F over this interval? So, I'm assuming you've had a go at it. I encourage you to pause this video especially if you've seen the other videos on introducing the idea of an average value of a function and figure out what this is.
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 want to figure out the average value of our function F on the interval, on the closed interval between zero and let's say between zero and three.
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