What is the area bounded by the graph of $ \lfloor x \rfloor + \lfloor y \rfloor =2 $ with the $x$ and $y$-axis? Question: If $x \ge 0$ and $y \ge 0$, then the area bounded by the graph of $ \lfloor  x \rfloor + \lfloor y \rfloor =2 $ with the $x$ and $y$-axis is?
Answer provided: $3$ units$^2$.
My doubt: How do we graph this relation in order to find the area?

Note that
$$
\left\lfloor x \right\rfloor  = a\quad  \Rightarrow \quad a \leqslant x < a + 1
$$
and therefore
$$
\begin{gathered}
  \left\{ \begin{gathered}
  \left\lfloor x \right\rfloor  + \left\lfloor y \right\rfloor  = 2 \hfill \\
  0 \leqslant \left\lfloor x \right\rfloor ,\left\lfloor y \right\rfloor  \hfill \\ 
\end{gathered}  \right.\quad  \Rightarrow \quad \left\{ \begin{gathered}
  0 \leqslant a \leqslant x < a + 1 \hfill \\
  0 \leqslant 2 - a \leqslant y < 2 - a + 1 \hfill \\ 
\end{gathered}  \right. \hfill \\
   \Rightarrow \quad \left\{ \begin{gathered}
  0 \leqslant \text{integer}\;a \leqslant 2 \hfill \\
  a \leqslant x < a + 1 \hfill \\
  2 - a \leqslant y < 2 - a + 1 \hfill \\ 
\end{gathered}  \right. \hfill \\ 
\end{gathered} 
$$
Therefore the answer $3$ refers to the area of the region 
defined by the condition ${\left\lfloor x \right\rfloor  + \left\lfloor y \right\rfloor  = 2}$
in itself.
$$
\begin{gathered}
  R = \left\{ {(x,y):\;\left\lfloor x \right\rfloor  + \left\lfloor y \right\rfloor  = 2} \right\}\quad  \Rightarrow \quad \int\limits_{\left\lfloor x \right\rfloor  + \left\lfloor y \right\rfloor  = 2} {dxdy}  =  \hfill \\
   = \sum\limits_{0\, \leqslant \,a\, \leqslant 2} {\int_{x = a}^{a + 1} {\int_{y = 2 - a}^{2 - a + 1} {dxdy} } }  = \sum\limits_{0\, \leqslant \,a\, \leqslant 2} 1  = 3 \hfill \\ 
\end{gathered} 
$$
The area bounded by the given condition and the $x$ and $y$ axes will be instead
$$
\begin{gathered}
  S = \left\{ {(x,y):\;\left\{ \begin{gathered}
  \;0 \leqslant \text{integer}\;a \leqslant 2 \hfill \\
  0 \leqslant x < a + 1 \hfill \\
  0 \leqslant y < 2 - a + 1 \hfill \\ 
\end{gathered}  \right.\quad } \right\}\quad  \Rightarrow \quad \int\limits_S {dxdy}  =  \hfill \\
   = \sum\limits_{0\, \leqslant \,a\, \leqslant 2} {\int_{y = 0}^{2 - a + 1} {dy} }  = \sum\limits_{0\, \leqslant \,a\, \leqslant 2} {\left( {2 - a + 1} \right)}  =  \hfill \\
   = 3 + 2 + 1 = 6 \hfill \\ 
\end{gathered} 
$$
