<p>Let \(p(x)\) be a function defined on \(\mathbb{R}\) such that \(p'(x) = p'(1-x)\), for all \(x \in [0,1]\), \(p(0) = 1\) and \(p(1) = 41\). Then \(\int_0^1 p(x)\, dx\) equals</p>
Step-by-Step Solution
Key Concept: The condition p'(x) = p'(1-x) means p(x) is symmetric about x = 1/2 in its derivative, implying p(x) itself is symmetric about the point (1/2, p(1/2)). Use substitution u = 1-x to relate the integral from 0 to 1/2 with the integral from 1/2 to 1.
<p><strong>Step 1:</strong> Use the substitution property. From the condition p'(x) = p'(1-x), integrate both sides from 0 to 1/2:</p><p>∫₀^(1/2) p'(x)dx = ∫₀^(1/2) p'(1-x)dx</p><p><strong>Step 2:</strong> The left side is p(1/2) - p(0) = p(1/2) - 1.</p><p><strong>Step 3:</strong> For the right side, substitute u = 1-x, so du = -dx. When x = 0, u = 1; when x = 1/2, u = 1/2:</p><p>∫₀^(1/2) p'(1-x)dx = -∫₁^(1/2) p'(u)du = ∫_(1/2)^1 p'(u)du = p(1) - p(1/2) = 41 - p(1/2)</p><p><strong>Step 4:</strong> Equate the results from Steps 2 and 3:</p><p>p(1/2) - 1 = 41 - p(1/2)</p><p>2p(1/2) = 42</p><p>p(1/2) = 21</p><p><strong>Step 5:</strong> The condition p'(x) = p'(1-x) implies p(x) has central symmetry about the point (1/2, 21). This means:</p><p>∫₀^(1/2) p(x)dx + ∫_(1/2)^1 p(x)dx = ∫₀^(1/2) p(x)dx + ∫₀^(1/2) [2·21 - p(x)]dx</p><p>= ∫₀^(1/2) p(x)dx + 42·(1/2) - ∫₀^(1/2) p(x)dx = 21</p><p>Alternatively, by symmetry: ∫₀^1 p(x)dx = (1/2)·[p(0) + p(1)] = (1/2)·[1 + 41] = 21</p><p><strong>∴ Answer:</strong> A</p>
Correct Answer: A