Well, one of the mistakes:
> If the side lengths have to be integers, then it's just (1 + 8 + 27 + 125)/4 = 40.25 inches
If I'm not mistaken, 4 is an integer, and so should not have cruelly snubbed when you sent out invitations to your summation of the cubes of the integers from 1 to 5. Sure, 4 is not a cool integer like the others in 1 to 5...he's got a reputation as a square, but that doesn't mean he doesn't have feelings.
But seriously, you write rather well on Quora & I'm sure the crowd would want to know what happens to the volume were you to choose a normal distribution instead of a uniform, so I hope you respond.
Let X be a random variable and g be any function.
If X is continuous, then the expectation of g(X) is defined as,
E[g(X)] = \int_−∞^∞ g(x)f(x) dx,
where f is the probability density function of X.
---
For us, with mean 3 and sigma 2/3, f = 3/2 e^(-9/8 (x-3)^2)/sqrt(2 π)
We need to integrate numerically; using Wolfram [1] we get an estimate of 30.8099.
If we ask for the expected value over the entire distribution we get 31.
[0] http://imai.princeton.edu/teaching/files/Expectation.pdf
[1] http://www.wolframalpha.com/input/?i=integrate+3%2F2+e%5E%28...
[2] http://www.wolframalpha.com/input/?i=expected+value+of+x%5E3...
In a bit more detail: take the integral of x^3 (the volume function) from 1 to 5 and you get 5^4/4 - 1^4/4 = 624/4 = 156. Then we divide by the length of the line segment (4) to get 39.
If the side lengths have to be integers, then it's just (1 + 8 + 27 + 64 + 125)/5 = 45 inches.
Edit: corrected division mistake, thanks to dxbydt
Edit 2: and corrected a second mistake thanks to tzs.