Definite Integration
Properties of definite integrals
Grade 12
Question:
<p>Let \(f:[0,5] \to \mathbb{R}\) be such that \(f''(x) = f''(5-x)\), \(\forall\, x \in [0,5]\), \(f'(0) = 1\) and \(f'(5) = 7\), then the value of \(\displaystyle\int_1^4 f'(x)\, dx\) is:</p>
<p>(a) 4</p>
<p>(b) 6</p>
<p>(c) 8</p>
<p>(d) 12</p>
Step-by-Step Solution
Key Concept: The symmetry condition f''(x) = f''(5-x) means f''(x) is symmetric about x = 2.5, which implies f'(x) is antisymmetric about x = 2.5, allowing us to use properties of symmetric functions over symmetric intervals.
<p><strong>Step 1:</strong> From f''(x) = f''(5-x), integrate both sides:</p><p>f'(x) = f'(5-x) + C₁</p><p>At x = 0: f'(0) = f'(5) + C₁ → 1 = 7 + C₁ → C₁ = -6</p><p>Therefore: <strong>f'(x) + f'(5-x) = 7</strong></p><p><strong>Step 2:</strong> Use substitution to find the integral. Let I = ∫₁⁴ f'(x)dx</p><p>Substitute u = 5-x, so du = -dx:</p><p>I = ∫₁⁴ f'(x)dx = ∫₄¹ f'(5-u)(-du) = ∫₁⁴ f'(5-u)du</p><p><strong>Step 3:</strong> Add the original integral to this transformed integral:</p><p>2I = ∫₁⁴ [f'(x) + f'(5-x)]dx = ∫₁⁴ 7 dx = 7[x]₁⁴ = 7(3) = 21</p><p><strong>Step 4:</strong> Solve for I:</p><p>I = 21/2</p><p>∴ Answer: D</p>
Correct Answer: D