If you're seeing this message, it means we're having trouble loading external resources on our website.

Εάν είστε πίσω από ένα web φίλτρο, παρακαλούμε να βεβαιωθείτε ότι οι τομείς *. kastatic.org και *. kasandbox.org δεν είναι αποκλεισμένοι.

Κύριο περιεχόμενο

Calculating average value of function over interval

Here we find the average value of x^2+1 on the interval between 0 and 3.

Θέλετε να συμμετάσχετε σε μια συζήτηση;

Δεν υπάρχουν αναρτήσεις ακόμα.
Μπορείς να διαβάσεις στα Αγγλικά; Κάνε κλικ εδώ για να δείτε περισσότερες συζητήσεις που συμβαίνουν στην αγγλική ιστοσελίδα της Khan Academy.

Απομαγνητοφώνηση βίντεο

- 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. 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. 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 actually find the average. So that's my Y axis. This is my X axis. Now over the interval between zero and three, so let's say this is the zero, this is one, two, three. It's a closed interval. When X is zero F of zero is going to be one. So, we're going to be right over here. 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 10. So this is going to be 10. 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 10. Three squared plus one is 10. 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. So, it's just minus. When you evaluated zero it's just going to be zero. And so, we are left with -- I'm going 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 is equal to four. Notice, our function actually hits that value at some point in the interval. At some point in the interval, something lower than two but greater than one. We can maybe call that C. It looks like our function hits that value. This is actually a generally true thing. This is a mean value theorem for integrals and we'll go into more depth there. But you can see this kind of does look like it's average value. That if you imagine the box, if you multiplied 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 because we have the average height times the width is the same thing as the area under the curve. So, anyway hopefully you found that interesting.