Definite Integration
Fundamental Theorem of Calculus
Grade 12
Question:
<p>Let <span>\(a, b\)</span> and <span>\(c\)</span> be non-zero real numbers such that <span>\(\int_0^3 (3ax^2 + 2bx + c) dx = \int_1^3 (3ax^2 + 2bx + c) dx\)</span>, then</p>
<p>(a) <span>\(a + b + c = 3\)</span></p>
<p>(b) <span>\(a + b + c = 1\)</span></p>
<p>(c) <span>\(a + b + c = 0\)</span></p>
<p>(d) <span>\(a + b + c = 2\)</span></p>
Step-by-Step Solution
Key Concept: The given condition states that the integral from 0 to 3 equals the integral from 1 to 3. Using the additivity property of definite integrals, ∫₀³ = ∫₀¹ + ∫₁³, we can deduce that ∫₀¹ must equal zero, which gives us a constraint on a, b, and c.
<p><strong>Step 1:</strong> Use the additivity property of integrals. We have:</p><p>∫₀³ (3ax² + 2bx + c) dx = ∫₀¹ (3ax² + 2bx + c) dx + ∫₁³ (3ax² + 2bx + c) dx</p><p><strong>Step 2:</strong> Given that ∫₀³ = ∫₁³, we can substitute:</p><p>∫₀¹ (3ax² + 2bx + c) dx + ∫₁³ (3ax² + 2bx + c) dx = ∫₁³ (3ax² + 2bx + c) dx</p><p><strong>Step 3:</strong> Subtract ∫₁³ from both sides:</p><p>∫₀¹ (3ax² + 2bx + c) dx = 0</p><p><strong>Step 4:</strong> Compute the antiderivative: F(x) = ax³ + bx² + cx</p><p><strong>Step 5:</strong> Evaluate from 0 to 1:</p><p>[ax³ + bx² + cx]₀¹ = a(1)³ + b(1)² + c(1) - 0 = a + b + c = 0</p><p><strong>∴ Answer:</strong> c</p>
Correct Answer: c