Practice: the definite integral of a floor function.

$\begin{aligned} f(x) & = \left \lfloor \color{#3D99F6}{x} - 2 \left \lfloor \frac {\color{#3D99F6}{x}}2 \right \rfloor \right \rfloor & \small \color{#3D99F6}{\text{Note that }x = \lfloor x \rfloor + \{x\}} \\ & = \left \lfloor \color{#3D99F6}{\lfloor x \rfloor + \{x\}} - 2 \left \lfloor \frac {\color{#3D99F6}{\lfloor x \rfloor + \{x\}}}2 \right \rfloor \right \rfloor \end{aligned}$

$\begin{aligned} \implies f(x) & = \begin{cases} \left \lfloor \lfloor x \rfloor + \{x\} - \lfloor x \rfloor \right \rfloor & = 0 & \text{if } \lfloor x \rfloor \text{ is even} \\ \left \lfloor \lfloor x \rfloor + \{x\} - (\lfloor x \rfloor - 1) \right \rfloor & = 1 & \text{if } \lfloor x \rfloor \text{ is odd} \end{cases} \end{aligned}$

$\begin{aligned} \implies \int_{-13}^{26} f(x) \ dx & = \underbrace{1+0+1+0+...+1+0}_{-13 \text{ to }26 \ = \ 40 \text{ terms}} = \boxed{20} \end{aligned}$