Discussion
\(E(X)\) and \(E(Y)\) are, respectively,
Answer: B
Both means are \(3/5\): \(\frac{3}{5}\) and \(\frac{3}{5}\).
Integrate over the unit square, splitting each integrand into products of one-variable integrals:
\[ \begin{aligned} E(X) &= \tfrac{6}{5}\int\!\!\int x(x + y^{2}) \\ &= \tfrac{6}{5}\left(\tfrac{1}{3} + \tfrac{1}{2}\cdot\tfrac{1}{3}\right) = \tfrac{6}{5}\cdot\tfrac{1}{2} = \tfrac{3}{5} \end{aligned} \]
\[ \begin{aligned} E(Y) &= \tfrac{6}{5}\int\!\!\int y(x + y^{2}) \\ &= \tfrac{6}{5}\left(\tfrac{1}{2}\cdot\tfrac{1}{2} + \tfrac{1}{4}\right) = \tfrac{6}{5}\cdot\tfrac{1}{2} = \tfrac{3}{5} \end{aligned} \]
The density integrates to \(\frac{6}{5}(\frac{1}{2} + \frac{1}{3}) = 1\), as it must. Both variables lie in [0, 1], so a mean of \(6/5\) is impossible, which rules out two options at once; \(2/5\) for \(E(X)\) is an arithmetic slip.
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