<p>The \(L\) denotes the value of the definite integral \(\displaystyle\int_0^1 \dfrac{1}{1+x^8}\,dx\), then which one of the following must be true?</p>
<p>(a) \(\dfrac{\pi}{4} < L < 1\)</p>
<p>(b) \(L = \dfrac{\pi}{4}\)</p>
<p>(c) \(L > 1\)</p>
<p>(d) \(0 < L < \dfrac{\pi}{4}\)</p>
Step-by-Step Solution
Key Concept: Since 0 < 1/(1+x⁸) < 1 for all x ∈ [0,1], we can establish bounds on L by comparing it with integrable reference functions like 1 and 1/(1+x⁴).
<p><strong>Step 1:</strong> Observe that for x ∈ [0,1], we have 0 < 1+x⁸ < 2, so 1/2 < 1/(1+x⁸) < 1.</p><p><strong>Step 2:</strong> Integrating across the interval [0,1]:</p><p>∫₀¹ (1/2) dx < ∫₀¹ 1/(1+x⁸) dx < ∫₀¹ 1 dx</p><p>⟹ 1/2 < L < 1</p><p><strong>Step 3:</strong> For a tighter bound, note that 1+x⁸ ≥ 1+x⁴ for x ∈ [0,1], so 1/(1+x⁸) ≤ 1/(1+x⁴).</p><p>∫₀¹ 1/(1+x⁸) dx ≤ ∫₀¹ 1/(1+x⁴) dx</p><p><strong>Step 4:</strong> Also, since x⁸ ≥ x⁴ for x ∈ [0,1], we have 1+x⁸ ≥ 1+x⁴ when both are positive, confirming 1/(1+x⁸) ≤ 1/(1+x⁴).</p><p>∴ L lies in the interval (1/2, 1) and more specifically satisfies 1/2 < L < ∫₀¹ 1/(1+x⁴) dx ≈ 0.877</p>
Correct Answer: A