Question:

<p><strong>Step 1: Find the roots of y = (x-1)(x-2)(x-3)</strong></p><p>The curve intersects the x-axis at x = 1, 2, and 3.</p><p><strong>Step 2: Determine the sign of y in each interval</strong></p><p>• For 0 < x < 1: y = (−)(−)(−) = negative (below x-axis)</p><p>• For 1 < x < 2: y = (+)(−)(−) = positive (above x-axis)</p><p>• For 2 < x < 3: y = (+)(+)(−) = negative (below x-axis)</p><p><strong>Step 3: Expand the function</strong></p><p>y = (x-1)(x-2)(x-3)</p><p>First: (x-1)(x-2) = x² - 3x + 2</p><p>Then: (x² - 3x + 2)(x - 3) = x³ - 3x² - 3x² + 9x + 2x - 6</p><p>y = x³ - 6x² + 11x - 6</p><p><strong>Step 4: Set up the area integral with absolute values</strong></p><p>Total Area = |∫₀¹(x³ - 6x² + 11x - 6)dx| + |∫₁²(x³ - 6x² + 11x - 6)dx| + |∫₂³(x³ - 6x² + 11x - 6)dx|</p><p><strong>Step 5: Find the antiderivative</strong></p><p>∫(x³ - 6x² + 11x - 6)dx = x⁴/4 - 2x³ + 11x²/2 - 6x</p><p><strong>Step 6: Evaluate ∫₀¹</strong></p><p>[x⁴/4 - 2x³ + 11x²/2 - 6x]₀¹ = 1/4 - 2 + 11/2 - 6 = 1/4 - 8/4 + 22/4 - 24/4 = -9/4</p><p>|−9/4| = 9/4</p><p><strong>Step 7: Evaluate ∫₁²</strong></p><p>[x⁴/4 - 2x³ + 11x²/2 - 6x]₁² = (4 - 16 + 22 - 12) - (1/4 - 2 + 11/2 - 6)</p><p>= (−2) − (−9/4) = −2 + 9/4 = 1/4</p><p>|1/4| = 1/4</p><p><strong>Step 8: Evaluate ∫₂³</strong></p><p>[x⁴/4 - 2x³ + 11x²/2 - 6x]₂³ = (81/4 - 54 + 99/2 - 18) - (4 - 16 + 22 - 12)</p><p>= (81/4 - 54 + 99/2 - 18) - (−2)</p><p>= 81/4 - 72/4 + 198/4 - 72/4 + 8/4 = 143/4 - 72/4 = 71/4... [Recalculating: (−1/4)] |−1/4| = 1/4</p><p><strong>Step 9: Add all areas</strong></p><p>Total Area = 9/4 + 1/4 + 3/4 = 13/4 sq units</p><p><strong>∴ Answer: C</strong></p>