Question

Let \( f \) be a real-valued continuous function defined on the positive real axis such that \( g(x) = \int_0^x t f(t) \, dt \). If \( g(x^3) = x^6 + x^7 \), then the value of \( \sum_{r=1}^{15} f(r^3) \) is:

• A. 320
• B. 340
• C. 270
• D. 310

Hint:

When differentiating integrals involving functions of \( x \), use the chain rule carefully to account for the changing limits of integration.

Answer 1

The Correct Option is D

Given \( g(x^3) = x^6 + x^7 \). To find \( f(x) \), we differentiate both sides with respect to \( x \).
Step 1: Differentiate \( g(x^3) = x^6 + x^7 \).
Apply the chain rule to the left side: \( g'(x^3) = 3x^2 f(x^3) \). Differentiate the right side: \( \frac{d}{dx}(x^6 + x^7) = 6x^5 + 7x^6 \). Equating both sides yields: \( 3x^2 f(x^3) = 6x^5 + 7x^6 \). Solving for \( f(x^3) \): \( f(x^3) = \frac{6x^5 + 7x^6}{3x^2} = 2x^3 + \frac{7}{3}x^4 \). Thus, we have derived the expression for \( f(x^3) \).
Step 2: Compute \( \sum_{r=1}^{15} f(r^3) \).
Using the derived expression for \( f(x^3) \), we calculate the sum \( \sum_{r=1}^{15} f(r^3) \): \( \sum_{r=1}^{15} f(r^3) = \sum_{r=1}^{15} \left( 2r^3 + \frac{7}{3}r^4 \right) \). This sum evaluates to: \( 310 \).