Area Under the Curve
Area enclosed by curves with GIF
Grade 12

Question:

<p>Find the area enclosed by the curve \([|x|] + [|y|] = 3\) where [.] denotes the greatest integer function.</p>

Step-by-Step Solution

Key Concept: The equation [|x|] + [|y|] = 3 defines regions where the sum of floor values of absolute coordinates equals 3. You must identify all integer pairs (a,b) where a + b = 3 and a,b ≥ 0, then determine the rectangular regions each pair represents in the xy-plane.
<p><strong>Step 1:</strong> Identify all valid pairs (a,b) where [|x|] = a, [|y|] = b, and a + b = 3 with a,b ≥ 0.</p><p>Valid pairs: (0,3), (1,2), (2,1), (3,0)</p><p><strong>Step 2:</strong> For each pair (a,b), determine the region:</p><ul><li>[|x|] = 0 means 0 ≤ |x| < 1, so -1 < x < 1 (width = 2)</li><li>[|x|] = 1 means 1 ≤ |x| < 2, so x ∈ [-2,-1) ∪ [1,2) (total width = 2)</li><li>[|x|] = 2 means 2 ≤ |x| < 3, so x ∈ [-3,-2) ∪ [2,3) (total width = 2)</li><li>[|x|] = 3 means 3 ≤ |x| < 4, so x ∈ [-4,-3) ∪ [3,4) (total width = 2)</li></ul><p>Similarly for [|y|] = n: width = 2 for each value.</p><p><strong>Step 3:</strong> Calculate area for each case:</p><ul><li>(0,3): width 2 × height 2 = 4 (region: -1 < x < 1, 3 ≤ y < 4)</li><li>(1,2): width 2 × height 2 = 4 (regions: (±[1,2)) × [2,3))</li><li>(2,1): width 2 × height 2 = 4 (regions: (±[2,3)) × [1,2))</li><li>(3,0): width 2 × height 2 = 4 (regions: (±[3,4)) × (0 ≤ y < 1))</li></ul><p><strong>Step 4:</strong> By symmetry in all four quadrants, total area = 4 × (4 + 4 + 4 + 4) = 4 × 16 = <strong>64</strong></p><p>∴ Answer: <strong>64 square units</strong></p>
Correct Answer: 0

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