<p>Given \( I = \int_a^b (x^4 - 2x^2)\,dx \) and \( f(x) = x^2(x^2 - 2) \). Find the ordered pair \((a, b)\) such that \(I\) is minimum.</p>
<p>\( (0, \sqrt{2}) \)</p>
<p>\( (-\sqrt{2}, 0) \)</p>
<p>\( (0, -\sqrt{2}) \)</p>
<p>\( (-\sqrt{2}, \sqrt{2}) \)</p>
Step-by-Step Solution
Key Concept: The integral of f(x) is minimized when we integrate over an interval where f(x) is most negative. Since f(x) = x²(x²-2), it's negative when x²-2 < 0, i.e., when |x| < √2. The minimum occurs at the symmetric interval [-√2, √2].
<p><strong>Step 1:</strong> Analyze f(x) = x²(x² - 2). Find where f(x) is negative:</p><p>f(x) < 0 when x²(x² - 2) < 0</p><p>Since x² ≥ 0 always, we need x² - 2 < 0</p><p>This gives x² < 2, so |x| < √2, meaning x ∈ (-√2, √2)</p><p><strong>Step 2:</strong> For an interval [a,b] where f(x) < 0 throughout, ∫f(x)dx is negative. The integral is minimum (most negative) when we maximize the area under the curve below the x-axis.</p><p><strong>Step 3:</strong> By symmetry of f(x) = x²(x² - 2) (even function), and the fact that f(x) is negative on (-√2, √2), the integral ∫_{-√2}^{√2} f(x)dx captures all the negative area.</p><p><strong>Step 4:</strong> Calculate: ∫_{-√2}^{√2} (x⁴ - 2x²)dx = [x⁵/5 - 2x³/3]_{-√2}^{√2}</p><p>= 2[((√2)⁵/5 - 2(√2)³/3)] = 2[(4√2/5 - 4√2/3)] = 2√2(4/5 - 4/3) = 2√2(-8/15) < 0</p><p>Any other interval [a,b] containing [-√2, √2] would include positive contributions from outside this region, making I less negative (larger).</p><p>∴ Answer: (a,b) = (-√2, √2)</p>
Correct Answer: D