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*}
