Definite Integration
Application of definite integral
Grade 12
Question:
<p>Let \(f:(0,\infty)\to R\) and \(F(x)=\displaystyle\int_0^x t\,f(t)\,dt\). If \(F(x^2)=x^4+x^5\), then \(\displaystyle\sum_{r=1}^{12} f(r^2)\) is equal to \(k\). Find \(\dfrac{k}{73}\).</p>
Step-by-Step Solution
Key Concept: Use Leibniz rule on F(x²) = x⁴ + x⁵ to find F'(x²), then extract f by comparing with the definition F'(x) = xf(x). The sum ∑f(r²) telescopes when expressed through F'(r²).
<p><strong>Step 1: Differentiate F(x²) using chain rule</strong></p><p>Given F(x²) = x⁴ + x⁵, differentiate both sides with respect to x:</p><p>F'(x²)·2x = 4x³ + 5x⁴</p><p>Therefore: F'(x²) = (4x³ + 5x⁴)/(2x) = 2x² + (5x³)/2</p><p><strong>Step 2: Extract f(x) using F'(x) = xf(x)</strong></p><p>Since F(x) = ∫₀ˣ tf(t)dt, by Leibniz rule: F'(x) = xf(x)</p><p>Thus: F'(x²) = x²f(x²)</p><p>So: x²f(x²) = 2x² + (5x³)/2</p><p>Therefore: f(x²) = 2 + (5x)/2</p><p><strong>Step 3: Find f(r²) for integer r</strong></p><p>Substituting x = r:</p><p>f(r²) = 2 + (5r)/2</p><p><strong>Step 4: Calculate the telescoping sum</strong></p><p>∑ᵣ₌₁¹² f(r²) = ∑ᵣ₌₁¹² [2 + (5r)/2]</p><p>= 12(2) + (5/2)∑ᵣ₌₁¹² r</p><p>= 24 + (5/2)·[12·13/2]</p><p>= 24 + (5/2)·78</p><p>= 24 + 195 = 219</p><p>So k = 219</p><p><strong>Step 5: Find k/73</strong></p><p>k/73 = 219/73 = 3</p><p>∴ Answer: <strong>3</strong></p>
Correct Answer: 3