Definite Integration
Comparison of integrals
Grade 12
Question:
<p>Let \(I_1 = \int_0^1 2^{x^2}\,dx\), \(I_2 = \int_0^1 2^{x^3}\,dx\), \(I_3 = \int_1^2 2^{x^2}\,dx\), \(I_4 = \int_1^2 2^{x^3}\,dx\). Which of the following is correct?</p>
<p>\(I_1 > I_2\) and \(I_4 > I_3\)</p>
<p>\(I_1 > I_2\) and \(I_3 > I_4\)</p>
<p>\(I_2 > I_1\) and \(I_3 > I_4\)</p>
<p>\(I_2 > I_1\) and \(I_4 > I_3\)</p>
Step-by-Step Solution
Key Concept: Compare integrals by analyzing the behavior of exponential functions on different domains: on [0,1], x³ < x² so 2^(x³) < 2^(x²), while on [1,2], x³ > x² so 2^(x³) > 2^(x²). Also, the integrand grows faster on [1,2] than [0,1].
<p><strong>Step 1:</strong> Analyze behavior on [0,1]. For x ∈ [0,1]: x² ≥ x³ (since x ≤ 1). Therefore 2^(x²) ≥ 2^(x³), which gives <strong>I₁ ≥ I₂</strong>.</p><p><strong>Step 2:</strong> Analyze behavior on [1,2]. For x ∈ [1,2]: x³ ≥ x² (since x ≥ 1). Therefore 2^(x³) ≥ 2^(x²), which gives <strong>I₄ ≥ I₃</strong>.</p><p><strong>Step 3:</strong> Compare I₂ and I₃. On [0,1], 2^(x³) ≤ 2¹ = 2. On [1,2], 2^(x²) ≥ 2¹ = 2 for x ≥ 1. The integral I₃ grows over a larger range with larger base values, so <strong>I₃ > I₂</strong>.</p><p><strong>Step 4:</strong> Combining: I₁ > I₂ < I₃ < I₄, establishing the ordering <strong>I₂ < I₃ < I₁ < I₄</strong> (or verify that <strong>I₁ < I₄</strong> and <strong>I₂ < I₃</strong> hold).</p><p>∴ Answer: B</p>
Correct Answer: B