TITLE: Find $\alpha$ and $\beta$ so that $f_X(x)$ can be a density function.
QUESTION [0 upvotes]: \begin{equation*} f_X(x) = \begin{cases} \frac{4x^2}{5}               
& \text{ , if } 0 < x \leq 1\\
\alpha(5-2x)         & \text{ , if } 1 \leq x < 2\\ \beta x^2            & \text{ , if } 2 \leq x < 3\\ 0         
& \text{ , else } \end{cases} \end{equation*}
I want to find $\alpha$ and $\beta$ in a way that $f_X(x)$ can be a density function.
I tried integrating it knowing that: $\int_{-\infty}^\infty f_X(x) dx = 1$, and dividing it into multiple integrals and I was able to get an equation with $\alpha$ and $\beta$. But I would need another one. I know I should use somehow that $P(a \leq x \leq b) = \int_a^b f_X(x) dx$ but I don't know how am I supposed to do that. Note that if $x = 1$ then it has two corresponding conditions in $f_X(x)$.

REPLY [1 votes]: Considering that at $x=1$, we have two corresponding conditions, you could impose continuity, i.e., $4/5 \times 1^2 = \alpha \times  3$, hence $\alpha = 4/15.$ You would then use your other condition to find $\beta.$
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