I am reading the course by Finan, in Example 22.5, on p.213 one can see the Solution:
Determine the constant $c$ so that the following function is a pdf. $$f(t) = \begin{cases} \frac{15}{64} + \frac{x}{64}, & -2 \leq x \leq 0, \\ \frac{3}{8} + cx, & 0 < x \leq 3, \\ 0, \text{otherwise.} \end{cases}$$ Calculate $P(-1 \leq X \leq 1)$ and find $F(x)$.
Solution
Observe that $f$ is discontinuous at the points $-2$ and $0$, and is potentially also discontinuous at the point 3. We first find the value of $c$ that makes $f$ a pdf. \begin{equation*} \begin{split} 1 &= \int_{-2}^0 \left( \frac{15}{64} + \frac{x}{64} \right) dx + \int_{0}^3 \left( \frac{3}{8} + cx \right) dx = \\ &= \left( \frac{15}{64}x + \frac{x^2}{128} \right) \bigg\rvert^0_{-2} + \left( \frac{3}{8}x + \frac{cx^2}{8} \right) \bigg\rvert^3_{0} = \\ \newline &= \frac{30}{64} - \frac{2}{64} + \frac{9}{8} + \frac{9c}{2} = \frac{100}{64} + \frac{9c}{2}. \end{split} \end{equation*}
Solving for $c$ we find $c = -\frac{1}{8}.$
The probability $P(-1 \leq X \leq 1)$ is calculated as follows. $$P(-1 \leq X \leq 1) = \int_{-1}^0 \left( \frac{15}{64} + \frac{x}{64} \right) dx + \int_{0}^1 \left( \frac{3}{8} + \frac{x}{8} \right) dx = \frac{69}{128}.$$
For $-2 \leq x \leq 0$ we have $$ F(x) = \int_{-2}^x \left( \frac{15}{64} + \frac{t}{64} \right) dt = \frac{x^2}{128} + \frac{15}{64}x + \frac{7}{16} $$ and for $0 < x \leq 3$ $$ F(x) = \int_{-2}^0 \left( \frac{15}{64} + \frac{x}{64}\right)dx + \int_0^x \left(\frac{3}{8} - \frac{t}{8}\right) dt = \frac{7}{16} + \frac{3}{8}x - \frac{1}{16}x^2. $$
Question. Why in the last equation for the $0 < x \leq 3$ were used two integrals? And why is the first integral definite?