What is the average value of y x 2

I think i'm struggling actually taking the integral. I think I've seen this very same problem before Sign up to join this community. The best answers are voted up and rise to the top. Stack Overflow for Teams — Collaborate and share knowledge with a private group.

Create a free Team What is Teams? Learn more. Asked 7 years, 6 months ago. Active 7 years, 6 months ago. Viewed 8k times. Add a comment. Active Oldest Votes. I'm just missing some big concept here I think. It makes it easier to calculate, especially on the part of the AP test which forbids the use of a calculator. John Joy John Joy 7, 1 1 gold badge 21 21 silver badges 30 30 bronze badges.

Sign up or log in Sign up using Google. F of one is two. So it's one, two, three. Actually, let me make my scale a little bit smaller on that. I have to go all the way up to So this is going to be This is going to be five. And then one, two, three. This is the hardest part is making this even. So see this is going to be in the middle. Pretty good, and then let's see in the middle.

Then we have that. Good enough. All right. So, we're going to be there. We're going to be there. I have obviously different scales for X and Y axis. Two squared plus one is five. Three squared plus one is So it's going to look something like this. This is what our function is going to look like. So, that's the graph of Y is equal to F of X. And we care about the average value on the closed interval between zero and three.

Between zero and three. 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 what it actually means. So the average of our function is 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 intergral from zero to three of F of X which is X squared plus one DX. Then we're going to take this area. 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 with the average height, or the average value of our function. So, we're going to 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 -- Let's see the antiderivative of X squared is X to the third over three. Antiderivative of one is X, and we're going to evaluate it from zero to three.

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. That's nine plus three and then when we evaluated zero, minus zero minus zero.