I'm trying to work out the cdf given this probability density function:
$$
f(x)=\begin{cases}
\frac{1}{4} & {0<x<2}\\
\frac{1}{2}x+c & {2<x<3}\\
0, & \text{elsewhere}
\end{cases}
$$
While this being the given solution:
$$
F(x)=\int^x_{-\infty}f(t)dt=\begin{cases}
\int^x_0 \frac{1}{4}dt=\frac{x}{4}\\
\int^x_2(\frac{1}{2}t-\frac{3}{4})dt+k=\frac{x^2-3x}{4}+\frac{1}{2}+k\\
\end{cases}
$$
Under the two levels, F(2) must be the same (reason for introducing k).
$\rightarrow F(2)=\frac{1}{2}=k$
$$
\therefore
F(x)=\begin{cases}
0, & {x<0}\\
\frac{x}{4} & {0 \leqslant x \leqslant 2}\\
\frac{x^2-3x}{4}+1 & {2 \leqslant x \leqslant 3}\\
1, & {x>2}
\end{cases}
$$
I know how to work out cdfs but could someone explain to me by demonstration how $\frac{1}{2}x+c$ in the pdf changes to $\frac{x^2-3x}{4}+1$ in the cdf and also why the inequality signs denoting the ranges of x also change respectively.
Recall that a probability density function must integrate to $1$. So we have \begin{align} 1 &= \int_{\mathbb R}f(x)\ \mathsf dx\\ &= \int_0^2 \frac14\ \mathsf dx + \int_2^3\left(\frac12 x+c\right)\ \mathsf dx\\ &= \frac12 + \frac14(3^2-2^2)+c\\ &= \frac74 + c, \end{align} so that $c = -\frac34$. To find the distribution function $F(x) = \mathbb P(X\leqslant x)$, we integrate the density; for $0\leqslant x\leqslant 2$ we have $$ F(x) = \int_0^x f(y)\ \mathsf dy = \int_0^x \frac14\ \mathsf dy = \frac x4. $$ For $2\leqslant x\leqslant 3$ we have \begin{align} F(x) &= \int_0^x f(y)\ \mathsf dy\\ &= \int_0^2 f(y)\ \mathsf dy + \int_2^x f(x)\ \mathsf dy\\ &= \frac 12 + \int_2^x \left( \frac 12 y-\frac34\right)\ \mathsf dy\\ &= \frac 12 + \frac14(x-2)(x-1)\\ &= \frac{2+(x-1)(x-2)}4\\ &= \frac{x^2-3x}4 + 1. \end{align} Note that $F(0)=0$ and $F(3)=1$, as we should expect.