Naive question about continuous random variable: solution verification

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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 $(a,b) = (3/5,6/5)$.

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