This may be a dumb question, but I thought that to get the expected value, we were supposed to calculate
$$A = \int_0^1 \int_x^1 y 2(x+y) dydx$$
which gave me a different answer to
$$B = \int_0^1 \int_0^y y 2(x+y) dxdy$$
As far as I know the second expression is finding the marginal density $f_y$ first, then straightforwardly finding the expectation for that.
I keep finding that A = 5/12 while B = 3/4, but I am confident that A should also work, too.
So, it would really help if you could give me the following information.
1), Is there something wrong with my calculation?
2), If not, what DID I find with the expression A?
I get $\frac{3}{4}$ for A.
We are integrating $2xy+2y^2$ with respect to $y$. We get antiderivative $xy^2+\frac{2}{3}y^3$. Plugging in the bounds we get $x+\frac{2}{3}-\frac{5}{3}x^3$.
Integrating with respect to $x$ gives $\frac{x^2}{2}+\frac{2}{3}x-\frac{5}{12}x^4$.
Insert the bounds. We get $\frac{1}{2}+\frac{2}{3}-\frac{5}{12}$. Simplify.
For B we also get $\frac{3}{4}$. The calculation is shorter.
Remark: One can view the second way as finding the density function of $Y$ first. I prefer to think of the expectation as a double integral, evaluated in two ways as an iterated ntegral.