posted by Anon .
The average area of all squares with sides between a inches and b inches (b>a) is ____ in^2.
∫[a,b] x^2 dx
= 1/3 x^3 [a,b]
= 1/3 (b^3 - a^3)
Consider only integer values. Then that will give you
a^2 + (a+1)^2 + ... + (b-1)^2 + b^2
Now, we know that
∑ k^2 = n(n+1)(2n+1)/6
So, we want sum from a-b = sum from 1-b - sum from 1-a, or
b(b+1)(2b+1)/6 - a(a+1)(2a+1)/6
= 1/6 (2b^3+3b^2+b - 2a^3-3a^2-a)
= 1/3 (b^3-a^3) plus some lower-order junk
So, our integral exact value appears to agree with the integer approximation.