<p>The area of the region \(A = \{(x, y)\mid y \geq x^2 - 5x + 4,\ x + y \geq 1,\ y \leq 0\}\) is</p>
Step-by-Step Solution
Key Concept: Identify the three boundary curves, find their intersection points, and integrate to find the area of the region bounded by the parabola, the line, and the x-axis constraint.
<p><strong>Step 1:</strong> Identify the boundary curves:</p><ul><li>Parabola: y = x² - 5x + 4 = (x-1)(x-4)</li><li>Line: x + y = 1, so y = 1 - x</li><li>Horizontal constraint: y ≤ 0</li></ul><p><strong>Step 2:</strong> Find intersection points of parabola with x-axis (y = 0):</p><p>x² - 5x + 4 = 0 → (x-1)(x-4) = 0 → x = 1, 4</p><p><strong>Step 3:</strong> Find where line y = 1 - x meets the parabola:</p><p>1 - x = x² - 5x + 4</p><p>x² - 4x + 3 = 0 → (x-1)(x-3) = 0 → x = 1, 3</p><p>At x = 3: y = 1 - 3 = -2 (intersection point at (3, -2))</p><p><strong>Step 4:</strong> Determine the active boundaries:</p><ul><li>For x ∈ [1, 3]: Region between parabola (below) and line (above), with y ≤ 0</li><li>For x ∈ [3, 4]: Region between parabola (below) and y = 0 (above)</li></ul><p><strong>Step 5:</strong> Calculate the area:</p><p>Area = ∫₁³ [(1-x) - (x² - 5x + 4)] dx + ∫₃⁴ [0 - (x² - 5x + 4)] dx</p><p>= ∫₁³ (-x² + 4x - 3) dx + ∫₃⁴ (-x² + 5x - 4) dx</p><p>= [-x³/3 + 2x² - 3x]₁³ + [-x³/3 + (5x²/2) - 4x]₃⁴</p><p>= [(-9 + 18 - 9) - (-1/3 + 2 - 3)] + [(-64/3 + 40 - 16) - (-9 + 45/2 - 12)]</p><p>= [0 - (-4/3)] + [(-64/3 + 24) - (45/2 - 21)]</p><p>= 4/3 + [8/3 - 3/2] = 4/3 + 16/6 - 9/6 = 8/6 + 7/6 = <strong>5/2</strong></p><p>∴ Answer: A</p>
Correct Answer: A