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I have found the equation for the average of a function to be $$ \bar{f} = \frac{1}{b-a}\int_a^b f(x) \mathrm{d}x $$ however, this equation doesn't make complete sense to me. I interperet it as being the same as $$ \lim_{n \to \infty} \sum_{i=1}^n f\left(a + i\frac{b-a}{n}\right) $$ This equation would blow up to infinity, since there is an infinite quantity of numbers larger than 0 that are being summed.

The average of a set of numbers is found by adding them up and then dividing the sum by the quantity of numbers added. So, for $n$ numbers, individually represented by $x_i$, their average would be $$ \frac{1}{n}\sum_{i=1}^{n} x_i $$ The way that I understand the first equation, the $\frac{1}{b-a}$ would only remove the $\mathrm{d}x$ from the integral. Each $\mathrm{d}x$ represents a ininitely small interval; if you were to add up the infinite number of these $\mathrm{d}x$'s you would end up with a total value of the full interval $b - a$; therefore, in the original equation, to remove the summed $\mathrm{d}x$'s we simply divide out $b - a$; however, this does not account for the sum of $f(x)$ over the interval $b-a$. When the integral equation is calculated I understand its result as simply the sum of $f(x)$ over the interval. How is it not this? (I know that it can't be, as an infinite sum of numbers larger than 0 would always go to infinity, but I don't understand why it is not.)

If I were to construct an equation to yield the average of a function I would write the following $$ \bar{f} = \lim_{n \to \infty} \frac{1}{n} \sum_{i=1}^{n} f\left(a + i\frac{b-a}{n}\right) $$ How would this get turned into the first equation?

I understand the problem in terms of another definition that states that the average of a function would yield the same area under the curve as that of the original function over the same interval $$ \int_a^b\bar{f}\mathrm{d}x = \int_a^bf(x)\mathrm{d}x $$ but I just cant understand the intution behind the original equation.

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    $\begingroup$ Note that $\frac{1}{b-a}\int_a^b f(x) \mathrm{d}x =\lim_{n \to \infty} \frac{1}{n} \sum_{i=1}^{n} f\left(a + i\frac{b-a}{n}\right)$, and not $\lim_{n \to \infty} \sum_{i=1}^{n} f\left(a + i\frac{b-a}{n}\right)$ as you claim. $\endgroup$
    – Martin R
    Sep 12, 2022 at 5:09
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    $\begingroup$ "an infinite sum of numbers larger than 0 would always go to infinity" Not necessarily. Eg, $$\sum_{i=1}^{\infty}\frac1{2^i}$$ isn't infinite. $\endgroup$
    – PM 2Ring
    Sep 12, 2022 at 5:11
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    $\begingroup$ @MartinR ah okay this is where my confusion is stemming from, then. I currently can't see how those two are equal, and I also don't understand why it is not equal to the second sum that I mentioned. $\endgroup$
    – Kalcifer
    Sep 12, 2022 at 5:14
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    $\begingroup$ @PM2Ring Noted! Although, I am unsure of what the more proper way of stating what I intended would be. $\endgroup$
    – Kalcifer
    Sep 12, 2022 at 5:14
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    $\begingroup$ The Riemann sums for the integral are $\frac{b-a}{n} \sum_{i=1}^{n} f\left(a + i\frac{b-a}{n}\right)$, since $\frac{b-a}{n} $ is the distance between the values where the function is evaluated. Now divide that by $b-a$ ... $\endgroup$
    – Martin R
    Sep 12, 2022 at 5:16

3 Answers 3

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The “right Riemann sum” of $f$ over $[a, b]$ with the partition $a = x_0 < x_1 < \cdots < x_n = b$ is $$ \sum_{k=1}^n f(x_n) (x_n - x_{n-1}) \, . $$ For equidistant partition points $x_k = a + k (b-a)/n$ this is $$ \frac{b-a}{n}\sum_{k=1}^n f(a+ k\frac{b-a}{n})\, . $$ If $f$ is Riemann-integrable, then any sequence of Riemann sums whose maximum width converges to zero, converges to the Riemann integral: $$ \int_a^b f(x) \, dx = \lim_{n \to \infty}\frac{b-a}{n}\sum_{k=1}^n f(a+ k\frac{b-a}{n}) \, , $$ so that the “average” is $$ \tilde f = \frac{1}{b-a }\int_a^b f(x) \, dx = \lim_{n \to \infty}\frac{1}{n}\sum_{k=1}^n f(a+ k\frac{b-a}{n}) \, . $$ So the average is (as one might expect) the limit of the arithmetic mean of the function values at $n$ equidistant points if $n$ tends to infinity.

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Does it perhaps make more sense like this?
$$ \bar{f} = \frac{\Big(\int_a^b f(x) \mathrm{d}x \Big)}{b-a} $$ On top we have an integral representing the area-under-the-curve between points a-and-b, and in the denominator we have the length from a-to-b. This is essentially dividing the total area by it's total width to get an average height.

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  • $\begingroup$ Personally, I can't get the formula yet. However, your explanation makes geometrical sense only maybe for square or rectangle shapes! Maybe if one considers the Physics applications such as speed, distance and acceleration together with the meaning of the area under the curve of such functions, the concept can be more clear. $\endgroup$
    – NoChance
    Sep 12, 2022 at 19:49
  • $\begingroup$ @NoChance Does that mean your issue in the main question lies more in understanding an integral as "the area under the curve" in-general more so than how "this formula defines the average" in-particular? $\endgroup$
    – AmateurDotCounter
    Sep 12, 2022 at 20:05
  • $\begingroup$ I get the concept of integration and limited integral meaning well. The concept of the average and its relationship to limited integral is a bit tricky for me in the mathematical sense. However, it is a bit more mild from physical perspective as in here: math24.net/rectilinear-motion.html $\endgroup$
    – NoChance
    Sep 12, 2022 at 20:16
  • $\begingroup$ @NoChance "...your explanation makes geometrical sense only maybe for square or rectangle shapes!" Maybe this is the issue. All I am trying to say is that $\int_a^b f(x) dx$ defines a shape with an area (even if it is hard to compute exactly 'how big' that area is for now), and that $b-a$ defines a length. So, the definition of $\bar{f}$ is just a constant function with height value just right to satisfy that the area of $\bar{f}$*(b-a) is equal to the area defined by $\int_a^b f(x) dx$. That concept of $A_1=A_2$ is all I'm trying to get across. $\endgroup$
    – AmateurDotCounter
    Sep 13, 2022 at 16:50
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    $\begingroup$ @NoChance Reread my sentence about "the area of those individual strips" again carefully. The f(x) value at each point represents a height, but each strip also has an infinitesimal width! That is how each strip has an individual area that relates back to the total area! $\endgroup$
    – AmateurDotCounter
    Sep 15, 2022 at 2:15
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Your last limit (which is correct) is nothing but the limit of the Riemann sums for the integral representing the average (that is, including $b-a$).

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    $\begingroup$ okay, but my original question remains unanswered. How do we account for the infinite sum in the first integral? why does the answer not blow up to infinity in some way? when we remove the infinitesimal widths that create the slices of area, how does this not simply result in an infinite sum? $\endgroup$
    – Kalcifer
    Sep 12, 2022 at 5:00
  • $\begingroup$ If $f$ is a nice function (say, continuous on $[a,b]$), its integral is a finite number. You are adding infinitely many infinitesimal quantities, namely $f(x)dx$, getting a finite result in the end. I am not sure if you understand this idea. This happens because the size of the addends and their amount balance each other. For example, $1/2+1/2=1$, $1/3+1/3+1/3=1$, etc. You can increase the number of addends, reducing accordingly their size, and you get $1$ in the limit, that is $\int_0^1 dx=1$. If your infinitesimals are too small, the sum is zero, $\int_0^1dx^2=0$. $\endgroup$
    – GReyes
    Sep 12, 2022 at 5:08

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