Find the value of two constants in a probability density function I have this exercise:
The continuous random variable X has p.d.f. f(x) where:
$f(x) = \begin{cases} \frac{x}{3} - \frac{2}{3}, & \mbox{if } 2 \le x \le 3 \\ \alpha, & \mbox{if }3 \le x \le 5 \\ 2-\beta x & \mbox{if } 5 \le x \le 6 \\ 0 & \mbox{otherwise} \end{cases}$
find $\alpha$ and $\beta$.
How can I find $\alpha$ and $\beta$ ?
I tried to do that with the following equation:
$\int_2^3(\frac{x}{3}- \frac{2}{3})dx + \int_3^5 \alpha \, dx + \int_5^6 (2-\beta x)dx = 1 \\ \frac{1}{3} \int_2^3x \, dx - \frac{2}{3} \int_2^3 dx + \alpha \int_3^5 dx + 2\int_5^6 dx - \beta \int_5^6 x \, dx = 1 \\ \frac{1}{3}[\frac{x^2}{2}]_2^3 - \frac{2}{3}[x]_2^3 + \alpha[x]_3^5 + 2[x]_5^6- \beta[\frac{x^2}{2}]_5^6 = 1 \\ \frac{5}{6} - \frac{2}{3} + 2\alpha + 2 - \frac{11}{2} \beta = 1$
arrived here I have tried also to maintain $\alpha$ and $\beta$ to the left side and trying to do something in addition, and I have multiplied by 6 both sides to obtain a integer number:
$2\alpha - \frac{11}{2} \beta = -\frac{7}{6} \\ 12 \alpha - \frac{11}{2} \cdot 6 \beta = -7 \\ 12 \alpha - 33 \beta = -7$
arrived here I really don't know how to continue.
Please, can you help me? Thanks!

With the hint of Med you get two equations. The argument is that  you have $\texttt{overlapping intervals}$. And the value of the function must be equal for equal $x$ values.
1. The value of the function must be equal for the upper limit of the first interval and the lower limit of the second interval (common point):
$$\frac33-\frac23=\alpha$$
2. The value of the function must be equal for the upper limit of the second interval and the lower limit of the third interval (common point):
$$\alpha=2-5\beta$$
With these two equations you can evaluate the values of $\alpha$ and $\beta$.
Finally you can check if these values satisfy your equation $12 \alpha - 33 \beta = -7$
