Definite Integration
Differentiation under integral sign / properties
Grade 12
Question:
<p>Suppose \(g'(x) < 0\) \(\forall\, x \geq 0\) and \(\displaystyle\int_0^x tg'(t)\,dt\) \(\forall\, x \geq 0\). Which of the following statement(s) are <b>correct</b>?</p>
<p>(a) \(f\) is not increasing</p>
<p>(b) \(f\) is continuous \(\forall\, x > 0\)</p>
<p>(c) \(f(x) = xg(x) - \displaystyle\int_0^x g(t)\,dt\)</p>
<p>(d) \(f'(x)\) exists \(\forall\, x > 0\)</p>
Step-by-Step Solution
Key Concept: Use the Leibniz rule for differentiating integrals with variable limits: d/dx[∫f(t)dt from a(x) to b(x)] = f(b(x))·b'(x) - f(a(x))·a'(x). This directly connects g'(x) to properties of the integrand.
<p><strong>Step 1:</strong> Apply Leibniz rule to g'(x) = d/dx[∫₀ˣ f(t)dt]</p><p>g'(x) = f(x)·(d/dx)[x] - f(0)·(d/dx)[0] = f(x)·1 - 0 = f(x)</p><p><strong>Step 2:</strong> This shows g'(x) = f(x), meaning g is an antiderivative of f on [0,x]</p><p><strong>Step 3:</strong> Verify properties:</p><p>• g(0) = ∫₀⁰ f(t)dt = 0 ✓</p><p>• g'(x) = f(x) for all x in domain ✓</p><p>• g is continuous (antiderivatives are continuous) ✓</p><p>• The fundamental theorem of calculus is satisfied ✓</p><p>∴ Answer: A, B, C, D (all properties hold for the integral function)</p>
Correct Answer: A,B,C,D