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## Calculus 2

### Unit 4: Lesson 1

Average value of a function

# Calculating average value of function over interval

AP.CALC:
CHA‑4 (EU)
,
CHA‑4.B (LO)
,
CHA‑4.B.1 (EK)
Here we find the average value of x^2+1 on the interval between 0 and 3.

## Want to join the conversation?

• is the mean value (c) always in the middle of the bounds (b and a)?
• No. The mean value theorem states that there exists some point "c" that the tangent to the arc is parallel to the secant through the endpoints. This does not imply that it is always in the middle of [a, b]. If the graph has really strange things going on (for instance shoots wayyy up and then mellows out) it would be at a different location.
• When I hear the average value of a function over closed interval, the first thing that come to my mind is to plug the start and the endpoint of that interval into the function then sum the two values and divide it by 2.

Following up the values which was given on the video :
`(1 + 10) / 2= 5.5`
Why didn't I come up with the same value? and how that idea differs from what is said on the video ?
• If the function was linear, your idea would work.
Consider the graphs of x²+1 and the linear function 3x+1.
Both are equal to 1 at x=0 and both are equal to 10 at x=3.
If you use the same approach on 3x+1 as is done in the video, you will get 11/2.
So why does your idea not work with x²+1?
It is because between 0 and 3, x²+1 doesn't "grow" as fast as 3x+1, so in that interval, the output is less than or equal to 3x+1 for all x in the interval [0,3].
Here is a graph of the two functions. The shaded area represents the area for which 3x+1 is greater than x²+1.
I hope it helps you to visually see the reason why your idea does not work with non linear functions.
https://www.desmos.com/calculator/drjxoub87g
• At 3;14 Sir Sal took anti derivative of 1 as x .Please explain how he did that ?
• That is a definition of integration: the integration of a constant "c" is equal to "cx".
• Where Did you get 1/3 From? Can anyone explain Please?
• Each interval, or each space for each rectangle was 1 unit wide. This gives you the one on top. The bottom, as Sal says at , is B-A. B=3 (The top number on the S dx), and A=0 (the bottom one). 3-0 is 3, and the three goes under the one. Hence, 1/3.
• shouldn't f(x) be marked as f'(x)
• What about if the graph goes above and below the x-axis, for example the integral of x^3-x from -1 to 1 do we find the area of the whole (thus the positive part cancel out with the negative part) hence the average height will be zero, or we consider the positive parts for each section and then find the average height?
(1 vote)
• The average height in your scenario will be 0, because the integral of x^3 - x over [-1, 1] is 0. Conceptually, you can look at the graph and realize that for any height, there is a corresponding height of the same magnitude but different signs, thus cancelling out each other in the average of the heights. Hope that I helped.
• suppose if we get the valu of function like f(-1)=17,f(0)=5,f(1)=1,f(2)=5 how can we plot in graph
(1 vote)
• You have a bunch of points: (-1, 17), (0,5), (1,1), (2,5)
Plot them just like you would in algebra?

Sorry, not sure if I'm fully understanding your question.
• is there a video finding the average volume of trig functions?
(1 vote)
• I don't think there is a video on that in particular, but you would do it the same way as was demonstrated in this video (I assume you mean the average height, i.e. average value). Take the definite integral of your function from a to b which gives you the area under the curve, and then divide the area by (a-b). If you're unsure how to take the definite integral of a trig function, you might want to look at the "Integrals" playlist.

However, you should note that for y-values less than zero, the area between the curve and the x-axis is counted as negative. For this reason the definite integral of sin(x) or cos(x) will always be between 2 and -2 and the average value of a longer interval will go towards zero. Which makes sense, as both sin(x) and cos(x) just goes up and down along the x-axis.
• Wait does this mean the average value of a function is not just the average value of the y or the f(x) as I thought and was taught but the average of the relationship between the x values and the corresponding y or f(x) values? because if you look at the graph, just visually of just the line o the y values it's clear the average should be bigger than 4, especially since the values range from 1 to 10, that means you have everything from 4, 5, 6,..all the way up to 10 being greater than 4..but the way he illustrates it in the video it seems like the avergae they are talking about is the average area..in other words f(x) times the x value, which I would argue is different from the average value of the y or f(x) values by itself. This needs to be clarified. Because then the average would be greater than 4 as I stated, I'm just not sure how you would calculate that..since there are infinitely many values..
(1 vote)
• (average y value)(B-A) =
integral of x^2+1 with bounds A to B.

Then divide by B-A to get

1/(B-A) integral x^2 + 1 from bounds A to B where B = 3 and A = 0.

So you get the formula 1/(B-A) integral f(x) with bounds from A to B by comparing area of the rectangle
(B-A)(Average y-value) with the area under f(x)
• Ok, so I input 0 through 4 into the fubction and divided by 4 to get an average of 4.5. Am I right in saying that the smaller the intervals become the closer it gets to 4? Like summing 0.01 through to 4 in 0.01 intervals.
(1 vote)
• Not really. If you input 0 through 4 into the function, multiplying every outcome by whatever interval you're testing with, say 0.01, add them all together and then divide all of it by 4 you'll close in to ~25.33.

You'll close in to 4 the same way if you input values on the interval from 0 to 3, just like in the video.

By the way, this video just skips over the idea of calculating the area under a curve. If you have trouble understanding clearly what's going on there review some of the earlier videos like this one for example - https://www.khanacademy.org/math/ap-calculus-ab/ab-accumulation-riemann-sums/modal/v/riemann-sums-and-integrals

There's really nothing to memorize about the average value of a function when you know what taking an integral of a function over an interval really means.
(1 vote)

## 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 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.