Question

$1 \leq x \leq 4$

Area enclosed by $[|x|]+[|y|]=2$

Answer 1

4

Answer 2

The problem asks for the area enclosed by the equation $[|x|]+[|y|]=2$ within the range $1 \leq x \leq 4$. Here, $[.]$ denotes the greatest integer function.

The condition $1 \leq x \leq 4$ implies that $x$ is positive, so $|x|=x$. Thus, the equation becomes $[x]+[|y|]=2$ for $1 \leq x \leq 4$.

We need to consider the possible values of $[x]$ for $x$ in the interval $[1, 4]$. The interval $[1, 4]$ can be split into sub-intervals based on the value of $[x]$:

1. $1 \leq x < 2$: In this range, $[x] = 1$. The equation becomes $1 + [|y|] = 2$, which simplifies to $[|y|] = 1$. For $[|y|] = 1$, we must have $1 \leq |y| < 2$. This inequality splits into two parts: $1 \leq y < 2$ or $-2 < y \leq -1$. So, for $1 \leq x < 2$, the points $(x,y)$ satisfying the equation are in the regions:

   • $1 \leq x < 2$ and $1 \leq y < 2$. This is a square with vertices (1,1), (2,1), (2,2), (1,2). Its area is $(2-1) \times (2-1) = 1$.
   • $1 \leq x < 2$ and $-2 < y \leq -1$. This is a square with vertices (1,-1), (2,-1), (2,-2), (1,-2). Its area is $(2-1) \times (-1 - (-2)) = 1$.

2. $2 \leq x < 3$: In this range, $[x] = 2$. The equation becomes $2 + [|y|] = 2$, which simplifies to $[|y|] = 0$. For $[|y|] = 0$, we must have $0 \leq |y| < 1$. This inequality splits into two parts: $0 \leq y < 1$ or $-1 < y < 0$. So, for $2 \leq x < 3$, the points $(x,y)$ satisfying the equation are in the regions:

   • $2 \leq x < 3$ and $0 \leq y < 1$. This is a rectangle with vertices (2,0), (3,0), (3,1), (2,1). Its area is $(3-2) \times (1-0) = 1$.
   • $2 \leq x < 3$ and $-1 < y < 0$. This is a rectangle with vertices (2,-1), (3,-1), (3,0), (2,0). Its area is $(3-2) \times (0 - (-1)) = 1$.

3. $3 \leq x < 4$: In this range, $[x] = 3$. The equation becomes $3 + [|y|] = 2$, which simplifies to $[|y|] = -1$. Since $|y| \geq 0$, $[|y|]$ cannot be negative. Thus, there are no solutions in this range.

4. $x = 4$: In this case, $[x] = 4$. The equation becomes $4 + [|y|] = 2$, which simplifies to $[|y|] = -2$. Again, $[|y|]$ cannot be negative. Thus, there are no solutions for $x=4$.

The set of points $(x,y)$ satisfying the given conditions consists of four disjoint rectangular regions:

• Region 1: $[1, 2) \times [1, 2)$ with area 1.
• Region 2: $[1, 2) \times (-2, -1]$ with area 1.
• Region 3: $[2, 3) \times [0, 1)$ with area 1.
• Region 4: $[2, 3) \times (-1, 0)$ with area 1.

The "area enclosed by" refers to the total area of these regions. Summing the areas of these four disjoint regions: Total Area = Area(Region 1) + Area(Region 2) + Area(Region 3) + Area(Region 4) Total Area = $1 + 1 + 1 + 1 = 4$.

Visual Representation: The regions are visualized as four unit squares/rectangles in the xy-plane.

• Two squares for $1 \leq x < 2$: one in the first quadrant ($1 \leq y < 2$) and one in the fourth quadrant ($-2 < y \leq -1$).
• Two rectangles for $2 \leq x < 3$: one in the first quadrant ($0 \leq y < 1$) and one in the fourth quadrant ($-1 < y < 0$).

% Visualize the regions on a 2D plane.
% The x-axis is from 1 to 4.
% For 1 <= x < 2, y is in [1, 2) or (-2, -1].
% For 2 <= x < 3, y is in [0, 1) or (-1, 0).

\begin{tikzpicture}
    % Axes and labels
    \draw[->] (0,0) -- (4.5,0) node[right] {$x$};
    \draw[->] (0,-2.5) -- (0,2.5) node[above] {$y$};
    \node[below] at (1,0) {1};
    \node[below] at (2,0) {2};
    \node[below] at (3,0) {3};
    \node[below] at (4,0) {4};
    \node[left] at (0,1) {1};
    \node[left] at (0,2) {2};
    \node[left] at (0,-1) {-1};
    \node[left] at (0,-2) {-2};

    % Region 1: 1 <= x < 2, 1 <= y < 2
    \fill[blue!20, opacity=0.5] (1,1) rectangle (2,2);
    \node at (1.5, 1.5) {R1};

    % Region 2: 1 <= x < 2, -2 < y <= -1
    \fill[red!20, opacity=0.5] (1,-2) rectangle (2,-1);
    \node at (1.5, -1.5) {R2};

    % Region 3: 2 <= x < 3, 0 <= y < 1
    \fill[green!20, opacity=0.5] (2,0) rectangle (3,1);
    \node at (2.5, 0.5) {R3};

    % Region 4: 2 <= x < 3, -1 < y < 0
    \fill[orange!20, opacity=0.5] (2,-1) rectangle (3,0);
    \node at (2.5, -0.5) {R4};

    % Indicate the x-range
    \draw[thick, <->] (1, -2.3) -- (4, -2.3) node[midway, below] {$1 \leq x \leq 4$};
\end{tikzpicture}

The final answer is $\boxed{4}$.