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### Course: Integral Calculus>Unit 3

Lesson 1: Average value of a function

# Calculating average value of function over interval

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)
• No, f'(x) is the derivative of f(x), not a function f(x) that is the derivative of another function.
(1 vote)
• 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)