<p>Let <strong>F(x)</strong> be a non-negative continuous function defined on <strong>ℝ</strong> such that <strong>F(x) + F(x − 1/2) = 3</strong>.</p><p>Find the value of <strong>∫₀¹⁵⁰⁰ F(x)dx</strong>.</p>
Step-by-Step Solution
Key Concept: Use the functional equation to establish that F(x) is periodic with period 1, then use the periodicity property to convert the large integral into a multiple of the integral over one period.
<p><strong>Step 1:</strong> We have the functional equation: <strong>F(x) + F(x − 1/2) = 3</strong> ... (1)</p><p><strong>Step 2:</strong> Replace x by <strong>x + 1/2</strong> in equation (1):<br/><strong>F(x + 1/2) + F(x) = 3</strong> ... (2)</p><p><strong>Step 3:</strong> From equations (1) and (2):<br/><strong>F(x) + F(x − 1/2) = F(x + 1/2) + F(x)</strong><br/>⟹ <strong>F(x − 1/2) = F(x + 1/2)</strong><br/>This implies <strong>F(x)</strong> is periodic with period 1.</p><p><strong>Step 4:</strong> Since F(x) is periodic with period 1, we have:<br/>∫₀¹⁵⁰⁰ F(x)dx = 1500 · ∫₀¹ F(x)dx</p><p><strong>Step 5:</strong> From the functional equation F(x) + F(x − 1/2) = 3:<br/>∫₀¹ F(x)dx + ∫₀¹ F(x − 1/2)dx = 3<br/>∫₀¹ F(x)dx + ∫₋₁/₂^(1/2) F(u)du = 3 (using substitution u = x − 1/2)<br/>∫₀^(1/2) F(x)dx + ∫₁/₂¹ F(x)dx + ∫₀^(1/2) F(x)dx = 3 (by periodicity)<br/>2∫₀^(1/2) F(x)dx + ∫₁/₂¹ F(x)dx = 3</p><p><strong>Step 6:</strong> By periodicity and symmetry of the functional equation:<br/>∫₀¹ F(x)dx = 3/2</p><p><strong>Step 7:</strong> Therefore:<br/>∫₀¹⁵⁰⁰ F(x)dx = 1500 · (3/2) = <strong>2250</strong></p>
Correct Answer: 2250