The following problem is from the book "Probability and Statistics" which is part of the Schaum's outline series. It can be found on page 104 and is problem number 3.50.
Problem:
Let $X$ and $Y$ be random variables having joint density function: $$ f(x,y) = \begin{cases} \left( \frac{3}{5} \right) x(x+y) & \text{for } 0 \leq x \leq 1, \, 0 \leq y \leq 2 \\ 0, & \text{otherwise } \end{cases} $$ Find $E(X)$.
Answer:
\begin{align*} E(X) &= \int_0^{ 1 } \int_0^2 \left( \frac{1}{2} \right) x^2 (x+y) \, \, dy \, dx \\ E(X) &= \int_0^{ 1 } \int_0^2 \frac{x^3}{2} + \frac{x^2 y}{2} \, \, dy \, dx \\ E(X) &= \int_0^{ 1 } \frac{x^3y}{2} + \frac{x^2 y^2}{4} \Bigg|_{y = 0}^{y = 2} \, dx \\ E(X) &= \int_0^{ 1 } \frac{2x^3}{2} + \frac{4x^2}{4} \, dx = \int_0^{ 1 } x^3 + x^2 \, dx \\ E(X) &= \frac{x^4}{4} + \frac{x^3}{3} \Bigg|_{0}^{1} = \frac{1}{4} + \frac{1}{3} \\ E(X) &= \frac{7}{12} \end{align*} However, the book's answer is $\frac{7}{10}$. Where did I go wrong?
Below is my updated answer. I would like somebody to confirm that my answer is now correct. \begin{align*} E(X) &= \int_0^{ 1 } \int_0^2 \left( \frac{3}{5} \right) x^2 (x+y) \, \, dy \, dx \\ E(X) &= \int_0^{ 1 } \int_0^2 \frac{3x^3}{5} + \frac{3x^2 y}{5} \, \, dy \, dx \\ E(X) &= \int_0^{ 1 } \frac{3x^3y}{5} + \frac{3x^2 y^2}{10} \Bigg|_{y = 0}^{y = 2} \, dx \\ E(X) &= \int_0^{ 1 } \frac{3x^3(2)}{5} + \frac{3x^2(4)}{10} \, dx \\ E(X) &= \int_0^{ 1 } \frac{6x^3}{5} + \frac{12x^2}{10} \, dx \\ E(X) &= \frac{6x^4}{20} + \frac{12x^3}{30} \, \Bigg|_{0}^{1} = \frac{3x^4}{10} + \frac{6x^3}{15} \, \Bigg|_{0}^{1} \\ E(X) &= \frac{3}{10} + \frac{6}{15} = \frac{3(3) + 6(2)}{30} \\ E(X) &= \frac{7}{10} \end{align*}
For one thing, the given joint density doesn't even integrate to $1$: $$\int_{x=0}^1 \int_{y=0}^2 \frac{x(x+y)}{2} \, dy \, dx = \frac{5}{6} < 1.$$ What is interesting is that if we multiply $7/12$ by $6/5$, we get $7/10$ as claimed by the book. Double-check that you have specified the correct density.