If $f$ is a continuous function and $\phi(x) = \int_{0}^{x} \left( (3t+4) \int_{t}^{3} f(u) du \right) dt$ and $\int_{0}^{3} f(x) dx = 3$, then:
Step-by-Step Solution
Key Concept: Use Leibniz rule for differentiating under the integral sign to find φ'(x), then differentiate again to find φ''(x), and apply the given constraint ∫₀³ f(x)dx = 3.
<p><strong>Step 1: Find φ'(x) using Leibniz rule.</strong></p><p>Given: $\phi(x) = \int_{0}^{x} \left( (3t+4) \int_{t}^{3} f(u) du \right) dt$</p><p>By Leibniz rule: $\phi'(x) = (3x+4) \int_{x}^{3} f(u) du$</p><p><strong>Step 2: Find φ''(x) using product rule.</strong></p><p>$\phi''(x) = \frac{d}{dx}\left[(3x+4) \int_{x}^{3} f(u) du\right]$</p><p>$= 3 \cdot \int_{x}^{3} f(u) du + (3x+4) \cdot \frac{d}{dx}\left[\int_{x}^{3} f(u) du\right]$</p><p><strong>Step 3: Apply Leibniz rule to the second term.</strong></p><p>$\frac{d}{dx}\left[\int_{x}^{3} f(u) du\right] = -f(x)$ (by fundamental theorem)</p><p>Therefore: $\phi''(x) = 3\int_{x}^{3} f(u) du - (3x+4)f(x)$</p><p><strong>Step 4: Evaluate φ''(3).</strong></p><p>$\phi''(3) = 3\int_{3}^{3} f(u) du - (3(3)+4)f(3)$</p><p>$= 3 \cdot 0 - 13f(3)$</p><p>$= -13f(3)$</p><p><strong>∴ Answer: D</strong></p>
Correct Answer: D