Naive question about continuous random variable: solution verification The probability density function of the random variable $X$ is given by
$$f(x) =
\begin{cases} 
          a + bx^{2} & 0 < x < 1 \\
          0 & \text{otherwise} 
\end{cases}$$
If $\textbf{E}(X) = 3/5$, determine the values of $a$ and $b$
MY SOLUTION
Since $f_{X}$ is a probability density function, we must have
\begin{align*}
\int_{0}^{1}(a + bx^{2})\mathrm{d}x = a + \frac{b}{3} = 1
\end{align*}
On the other hand, according to the definition of expected value, we  get
\begin{align*}
\int_{0}^{1}(ax+ bx^{3})\mathrm{d}x = \frac{a}{2} + \frac{b}{4} = \frac{3}{5}
\end{align*}
Solving the linear system of equations obtained, it results that $(a,b) = (3/5,6/5)$.

Everything's correct; solution is short, but totally understandable!
