 Let's start now the final section of chapter six six point five about the average value of a function This is gonna be our final application of integration in this chapter. We'll actually return to Applications of integration again in chapter eight But before we do that we're gonna develop some more techniques of integration Because as we've noticed in many of these examples we've done although setting up the integral was a huge challenge The actual computation integral was a fairly it was fairly simple, right? And that's not a principle. It's gonna be true for forever, but that's a discussion for another day What I want to talk about this idea is average values to give you some understanding what that is Suppose we have some data set and it's a very simple data set So we have numbers v1 v2 v3 up to some final value v in like these could be like homework averages you have in like a hypothetical calculus 2 class, right? If you want to find your average Homework score. Well the idea is you just take v1 plus v2 Plus v3 all the way up to vn so you add together all the values and then you divide by the total number of Data points you have there how many homework assignments were there in a slightly more compact form This looks like 1 over n times the sum as I goes from 1 to n of the vi's And so this this is how one would find the average Homework score or the mean homework score right there. It's a very fairly simple idea So could but that this works only if you have a finite sum could we turn this into an infinite sum of some kind? Such as an integral and the add the idea is going to be the following Let's take our x-axis right here And we'll have some function That sits above it. Maybe you look something like this and We have two specific values in mind We have this x equals a and x equals b and so looking at the region below the function here We're interested in what would the average value? the average value of F on the interval a to b What would that look like well trying to mimic the the average formula we talked about here for this Finite list of data points Let's try the technique of accumulation if we subdivide the domain into smaller pieces So we have some x1 x2 Xi minus 1 xi you get the basic idea here look at just a single interval right here Let us consider This as a let's let's continue the subdivision of the interval right here Let's let's consider this as like a few isolated finite points right here. So if we choose xi Star we have this f of xi star right here and we choose just represented as for each interval Then we have this we have a bunch of values right here We have this f of x1 star plus f of x2 star and we proceed all the way up to This f of xn star we have all these data points, right? And so just assume that instead of a continuous function. We just have this isolated sequence of points Right, and if we take the average of those things well, we would take that sum and divide it by n That would give us the average and like we saw above This would look like 1 over n times the sum of f of xi star as i ranges from 1 to n And so this would give us the value we could then take then our average value Which is often denoted f sub a v e The average value then we can take the limit as n goes to infinity of 1 over n times the sum As i goes from 1 to n of f of xi star And so that would give us the average value of the function taking the limit of these Approximate solutions, right because as n gets bigger and bigger bigger you take more data points and better accurately and more accurately Calculates the average value thus taking the limit gives the true value, but unlike other applications of the integral of integrals that we've seen in the past And the present form this limit it is an infinite sum, but it's not a Riemann sum And therefore this doesn't immediately translate into a It doesn't immediately translate into an integral because a Riemann sum we need a delta x which we don't have here Now it turns out it's not too hard to fix that because all we have to do is multiply this thing by b minus a over b minus a That is multiply by the length of the of the interval over itself The reason we do that is because if you take the b minus a on top And couple it with the 1 minus n that's already present those two powers combined to make A captain delta x right here And so if you if you kind of put that around you get your delta x and then you're going to keep this b minus a sticking around And so with this adjustment We can rewrite this as 1 over b minus a I took the b minus a over here out because it's constant with respect to n Then we get the limit As n goes to infinity Of the sum Where i ranges from 1 to n f of x i star times delta x Now the limit of this sum, which is now a Riemann sum will be the integral We get 1 over b minus a times the integral from a to b Of f of x dx And so we get a very nice simple formula for this idea of average value The average value of a function will be the area under the curve Divided by the length of the interval So we get 1 over b minus a times the integral of f of x here And this is summarized in this average value formula that we see right here. The average value Is going to be the length of interval That is the area under the function divided by the length of the interval Let's see an example of this. It doesn't get much more complicated than what you've seen already here so So Consider the function f of x equals 1 plus x squared Let's find the average value of f on the integral negative 1 to 2 In terms of a formula the average value f of the average value of f This is going to look this will look like 1 over 2 minus a negative 1 Times the integral from negative 1 to 2 Of f of x, which is 1 plus x squared Dx that's all there is to it I'm simplifying the interval there 2 minus a negative 1 is a is 3 so the length of the interval is 3 Find an anti-derivative for 1 plus x squared. We get x plus x cubed over 3 Plug it in negative 1 and 2 When you plug in 2 you're going to get 2 plus 8 thirds Then we plug in negative 1 you're going to get a negative 1 minus a third And so of course negative 1 Minus a negative 1 is actually a plus 1 there so you get 2 plus 1 which is a 3 And then we're going to get Like I'll likewise mean distribute the negative sign on to the negative 1 third You're going to get a positive 1 third plus 8 thirds is 9 thirds Which of course is the same thing as just a 3 And so you get 3 plus 3 which is 6 6 over 3 and we see that the average value is going to be a 2 And so in terms of computing average value That's what we get and so in essence what we've now computed is that the middle y coordinate for this function would be the y coordinate y equals 2 this is the middle of the function