Step-by-step explanation:
Area, A = Length,x × Breadth,y
A=xy
When x and y are independent, E(xy) = E(x)E(y)
As x and y have the same distribution, U[L-A,L+A], they have the same mean.
We could argue by symmetry that E(x) = L and E(y) + L, also.
We can also reason this from the fact that, if X ~ U[L-A, L+A], f(x) = 1/(2A) from L-A to L+A
Therefore
![E(x)= \int\limits {f(x)} \, dx \\\\=\int\limits^{(L+A)}_{(L-A)} {\frac{1}{2A}x } \, dx](https://tex.z-dn.net/?f=E%28x%29%3D%20%5Cint%5Climits%20%7Bf%28x%29%7D%20%5C%2C%20dx%20%5C%5C%5C%5C%3D%5Cint%5Climits%5E%7B%28L%2BA%29%7D_%7B%28L-A%29%7D%20%7B%5Cfrac%7B1%7D%7B2A%7Dx%20%7D%20%5C%2C%20dx)
![=\int\limits^{(L+A)}_{(L-A)} {\frac{1}{2A}x } \, dx \\\\=[\frac{1}{4A}x^2 ]\limits^{(L+A)}_{(L-A)}](https://tex.z-dn.net/?f=%3D%5Cint%5Climits%5E%7B%28L%2BA%29%7D_%7B%28L-A%29%7D%20%7B%5Cfrac%7B1%7D%7B2A%7Dx%20%7D%20%5C%2C%20dx%20%5C%5C%5C%5C%3D%5B%5Cfrac%7B1%7D%7B4A%7Dx%5E2%20%5D%5Climits%5E%7B%28L%2BA%29%7D_%7B%28L-A%29%7D)
=1/(4A)(L+A)2 - 1/(4A)(L-A)2
= 1/(4A)(L2 + 2AL + A2) - 1/(4A)(L2 - 2AL + A2)
=1/(4A)(2AL+2AL)
= 1/(4A)(4AL)
= L
Thus, E(xy) = E(x)E(y) = L×L = L²