Expected value with two uniformly distributed random variables A surveyor wishes to lay out a square region with each side having length L. However, because of a measurement error, he instead lays out a rectangle in which the north–south sides both have length X and the east–west sides both have length Y. Suppose that X and Y are independent and that each is uniformly distributed on the interval [L - A, L + A] (where 0 < A < L). What is the expected area of the resulting rectangle?
I tried using $E(x,y) = E(x)E(y) + Cov(x,y) = E(x)E(y)$ (since they are independent)
I ended up with $4A^2L^2$, but the answer should be just $L^2$

So you want to find $E( X\cdot Y)$, so we need to integrate $\int_{L-A}^{L+A} \int_{L-A}^{L+A} X \cdot Y \cdot P(X \cap Y) dY dX$
and since X and Y are independent $P(X \cap Y) = P(X) \cdot P(Y)$ and the integrals are separable (note in this case $P(X)=\frac{1}{b-a}=\frac{1}{(L+A)-(L-A)}=\frac{1}{2A}$ which also doesn't depend on X or Y). Once you find they are separable you notice each is the just the definition of expected value ( $E(x)=\int X\cdot P(X) dX$ ).
