<p>If \(I_1 = \displaystyle\int_0^1 e^{-x}\cos^2 x\, dx\), \(I_2 = \displaystyle\int_0^1 e^{-x^2}\cos^2 x\, dx\) and \(I_3 = \displaystyle\int_0^1 e^{-x^3}\, dx\) then</p>
<p>\(I_2 > I_3 > I_1\)</p>
<p>\(I_2 > I_1 > I_3\)</p>
<p>\(I_3 > I_2 > I_1\)</p>
<p>\(I_3 > I_1 > I_2\)</p>
Step-by-Step Solution
Key Concept: Compare integrals by analyzing the behavior of exponential decay functions over [0,1]. Since e^(-x) > e^(-x²) > e^(-x³) for all x ∈ (0,1), and cos²x ∈ [0,1], we can establish the ordering through careful analysis of which function decays fastest.
<p><strong>Step 1:</strong> Analyze exponential decay on [0,1]. For x ∈ (0,1): x > x² > x³, so -x < -x² < -x³, which means e^(-x) > e^(-x²) > e^(-x³).</p><p><strong>Step 2:</strong> Since cos²x ≤ 1 for all x, we have I₁ = ∫₀¹ e^(-x)cos²x dx ≤ ∫₀¹ e^(-x) dx (with strict inequality since cos²x < 1 on most of [0,1]).</p><p><strong>Step 3:</strong> For I₂: ∫₀¹ e^(-x²)cos²x dx ≤ ∫₀¹ e^(-x²) dx, and since e^(-x) > e^(-x²) on (0,1), we get I₁ > I₂.</p><p><strong>Step 4:</strong> For I₃: ∫₀¹ e^(-x³) dx has the slowest decay. Comparing I₂ with I₃: e^(-x²) > e^(-x³) on (0,1), and even with the cos²x factor in I₂, since e^(-x²) decays much faster than e^(-x³), we have I₂ > I₃ (the exponential difference dominates).</p><p><strong>Step 5:</strong> Therefore: <strong>I₁ > I₂ > I₃</strong></p><p>∴ Answer: C</p>
Correct Answer: C