Question

What is the average value of the function `1/x` between x=2/3 and x=8/3?

Answer 1

The average value of a continuous function `f(x)` over an interval `[a,b]` is `\frac{1}{b-a}\int_{a}^{b}f(x)\ dx`.

For this example, the average value is therefore `\frac{1}{8/3-2/3}\int_{2/3}^{8/3}1/x\ dx=\frac{1}{2}ln|x| |_{2/3}^{8/3}=\frac{1}{2}(ln(8/3)-ln(2/3))`

`=\frac{1}{2}ln(\frac{8/3}{2/3})=\frac{1}{2}ln(4)=\ln(4^{1/2})=\ln(2)\approx 0.693.`

One interpretation of this fact is geometric. The average height of `f(x)=1/x` over the interval `[2/3,8/3]` is close to 0.7. In other words, the area of region under the graph of `f(x)` from `x=2/3` to `x=8/3` is the same as the area of the rectangle based on the same interval of width 2 and height `ln(2)\approx 0.7`.

Note that this is smaller than the average of the maximum and minimum values of `f(x)=1/x` on this interval, `\frac{3/2+3/8}{2}=12/16+3/16=15/16=0.9375`. The reason is that the graph of `f(x)` is concave up, so it "spends more time" in regions where its value is lower than 0.9375 than it does in regions where its value is higher than 0.9375.

Another way to interpret average value is statistically. If you enter =1/(2*RAND()+2/3) into a spreadsheet cell, and then copy and paste this over, for instance, `n=100` cells total, the average (mean) of these 100 numbers should be close to `ln(2)\approx 0.693`.

I did it myself a few times. Here is a list of the averages (means) I got when I did this process: 0.677, 0.631, 0.699, 0.687, 0.650, 0.665, 0.694, 0.735.

Of course, you could also consider "averaging the averages" to get even closer to 0.693 (on average...i.e., with less variability). The average (mean) of the 8 numbers above is 0.680.

Side note: =RAND() generates random numbers between 0 and 1. By multiplying it by 2 and then adding 2/3, a random number between 2/3 and 8/3 is generated. By taking the reciprocal of this number, we obtain a random output of `f(x)=1/x`.