Probability density function with two parameters

For the continuous probability density function to be a valid one, the following condition must hold true:

$\int_{-\infty}^{\infty} f_{X}(x) \ dx=1$ $\int_{-\infty}^{-1} f_{X}(x) \ dx +\int_{-1}^{3} f_{X}(x) \ dx + \int_{3}^{\infty} f_{X}(x) \ dx=1$ $\implies 0 + \int_{-1}^{3} f_{X}(x) \ dx + 0 = 1$ $\implies 4\alpha \beta + 4\alpha = 1$ $\implies 4\beta + 4 = \frac{1}{\alpha} \dots (1)$

Now, the expected value of the distribution is defined as:

$E(X)=\int_{-\infty}^{\infty} xf_{X}(x) \ dx$ $E(X)= \frac{5}{3}=\int_{-\infty}^{-1} xf_{X}(x) \ dx +\int_{-1}^{3} xf_{X}(x) \ dx + \int_{3}^{\infty} xf_{X}(x) \ dx$ $E(X)= \frac{5}{3}=0+\int_{-1}^{3} xf_{X}(x) \ dx + 0$ $\implies 28\alpha\beta+12\alpha = 5$ $\implies \frac{28\beta + 12}{5} = \frac{1}{\alpha} \dots (2)$

Equating (1) and (2) by eliminating $\alpha$ and solving for $\beta$ yields $\boxed{\beta = 1}$