Definite Integration
Minimization of definite integral
Grade 12
Question:
<p>Let \(I = \int_{a}^{b} (x^4 - 2x^2) \, dx\). If \(I\) is minimum then the ordered pair \((a, b)\) is:</p>
<p>\((0, \sqrt{2})\)</p>
<p>\((-\sqrt{2}, 0)\)</p>
<p>\((\sqrt{2}, -\sqrt{2})\)</p>
<p>\((-\sqrt{2}, \sqrt{2})\)</p>
Step-by-Step Solution
Key Concept: To minimize a definite integral, recognize that I(a,b) = F(b) - F(a) where F is the antiderivative. The minimum occurs when we integrate over regions where the integrand is most negative, which requires analyzing f(x) = x⁴ - 2x² for its sign and critical points.
<p><strong>Step 1:</strong> Find where f(x) = x⁴ - 2x² changes sign.</p><p>Factor: f(x) = x²(x² - 2) = x²(x - √2)(x + √2)</p><p>f(x) ≤ 0 when x ∈ [-√2, √2] and f(x) ≥ 0 elsewhere.</p><p><strong>Step 2:</strong> To minimize I = ∫ₐᵇ f(x)dx, we must capture the entire region where f(x) is negative (contributing to a minimum), which occurs over [-√2, √2].</p><p><strong>Step 3:</strong> Setting a = -√2 and b = √2 gives:</p><p>I = ∫₋√₂^√₂ (x⁴ - 2x²)dx = [x⁵/5 - 2x³/3]₋√₂^√₂</p><p>= 2[4√2/5 - 4√2/3] = 2·4√2(-2/15) = -16√2/15</p><p>Any other choice of (a,b) either includes positive regions or excludes negative regions, yielding a larger value.</p><p><strong>∴ Answer: D → (a, b) = (-√2, √2)</strong></p>
Correct Answer: D