Are there better methods for finding the average value of a function?

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The discussion focuses on the limitations of the average value formula for functions, particularly when values are unevenly distributed. The standard formula, (1/(b-a)) ∫a to b f(x) dx, can yield misleading averages if a function has extreme values, as small numbers can disproportionately affect the outcome. Alternative statistical measures, such as variance, skewness, and kurtosis, provide deeper insights into data distribution and can be more informative than the arithmetic mean. The conversation suggests that understanding these concepts is better suited for a statistics course, which would cover various methods for analyzing averages. Overall, the average value of a function is most reliable when the data is evenly distributed.
Square1
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Hey so I am in a second semester calculus class and we are learning about the average value of the function. We are presented the following formula:

(1/(b-a)) ∫a to b f(x) dx for a ≤ x ≤ b

This is basically doing (x1+...xn)/n, except for infinitely many numbers over the interval
[a -b]

Naturally we need to learn from some starting point, but in terms of averages of functions, this is where we end in the course.

Now for the question. Aren't there better formulas for functions that are not evenly distributed?

Say you were using this formula to get the average value of some function where most its values are some big big numbers, but the remaining minority of numbers are very tiny numbers, so much so that average value using the above method is brought to a value that is very misleading. Misleading as in you would never have guessed for example that an average value of 5 was produced by a function where let's say most of its values are in the thousands...and this happens because you have a small collection of very small numbers.

If I'm not making sense here just let me know haha. I'll try to explain it another way.

So are there other techniques of taking averages? Would these be in a statistics class?
 
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The next quantity you can define to characterize a function or data set would be something like "the average deviation from the average". The way you define it is as:

\frac{1}{b-a}\int_{a}^b f(x) (x - c) dx

where c is the average value of f(x), defined by your integral.

This expression is called the variance of the function f(x), and it's used very frequently in statistics.

You can go further and look at "average deviation from the average deviation" (skewness), and even higher order deviations (kurtosis). These things are called moments. If you know all the moments of a function (mean, variance, etc) you will be able to derive the function itself from these values.
 
I think you are right, a discussion of the best way to measure "the average value" in a particular situation belongs in a statistics class rather than pure math. the median and the mode are two other simple ones. Google for more information about them.

You are right that the arithmetic mean (or the integral version of it) doesn't necessarily agree with "common sense". For example, over 99.9% of humans have more than the average number of arrms and legs. There are so few surviving humans with more than two arms or legs that you can ignore them out of a total of 7 billion, but there are certainly people with fewer than two as the result of accidents, birth defects, etc. So the average number will be a bit less than 2.
 
I see, thanks! So I should enroll in statistics courses to learn about this sort of analysis? Topics in particular you can think of?

So the results of "the average of a function" stand alone really only when the interval in question is evenly distributed?
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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