Definite Integration
Comparison of integrals
Grade 12
Question:
<p>For \( x \in (0,0) \), let \( I_3 = \int_0^1 e^{-x^3} dx \), \( I_2 = \int_0^1 e^{-x^2} \cos^2 x \, dx \), \( I_1 = \int_0^1 e^{-x} \cos^2 x \, dx \). Then which of the following is true?</p><p>(1) \(I_3 > I_2 > I_1\) (2) \(I_1 > I_2 > I_3\) (3) \(I_3 > I_2 > I_1\) (4) \(I_2 > I_1 > I_3\)</p>
<p>\(I_3 > I_2 > I_1\)</p>
<p>\(I_1 > I_2 > I_3\)</p>
<p>\(I_3 > I_2 > I_1\)</p>
<p>\(I_2 > I_1 > I_3\)</p>
Step-by-Step Solution
Key Concept: Compare integrals by analyzing the behavior of exponential decay rates: e^(-x) decays fastest, e^(-x²) decays moderately, and e^(-x³) decays slowest on (0,1). Since 0 < x < 1, we have x³ < x² < x, making e^(-x³) > e^(-x²) > e^(-x), which dominates the comparison despite the cos²x factor.
<p><strong>Step 1: Analyze exponential behavior on (0,1)</strong></p><p>For x ∈ (0,1): x³ < x² < x, therefore -x³ > -x² > -x</p><p>This gives us: e^(-x³) > e^(-x²) > e^(-x) for all x ∈ (0,1)</p><p><strong>Step 2: Compare I₃ and I₂</strong></p><p>Since e^(-x³) > e^(-x²) and 0 < cos²x ≤ 1:</p><p>∫₀¹ e^(-x³) dx > ∫₀¹ e^(-x²) cos²x dx</p><p>Therefore I₃ > I₂</p><p><strong>Step 3: Compare I₂ and I₁</strong></p><p>Since e^(-x²) > e^(-x) on (0,1) and cos²x appears in both:</p><p>∫₀¹ e^(-x²) cos²x dx > ∫₀¹ e^(-x) cos²x dx</p><p>Therefore I₂ > I₁</p><p><strong>Step 4: Establish ordering</strong></p><p>Combining results: I₃ > I₂ > I₁</p><p>∴ Answer: (3) I₃ > I₂ > I₁</p>
Correct Answer: C