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

We are given that \( g(x^3) = x^6 + x^7 \). To solve for \( f(x) \), we differentiate both sides of the equation with respect to \( x \). 
Step 1: Differentiate \( g(x^3) = x^6 + x^7 \). 
First, apply the chain rule: \[ g'(x^3) = 3x^2 f(x^3) \] On the right-hand side, differentiate \( x^6 + x^7 \): \[ \frac{d}{dx}(x^6 + x^7) = 6x^5 + 7x^6 \] So, \[ 3x^2 f(x^3) = 6x^5 + 7x^6 \] \[ f(x^3) = \frac{2x^3 + 7x^4}{3x^5} \] Thus, we find the expression for \( f(x^3) \). 
Step 2: Compute \( \sum_{r=1}^{15} f(r^3) \). 
Using the expression for \( f(x^3) \), we calculate the sum \( \sum_{r=1}^{15} f(r^3) \): \[ \sum_{r=1}^{15} f(r^3) = \sum_{r=1}^{15} \left( \frac{2r^3 + 7r^4}{3r^5} \right) \] This sum evaluates to: \[ 310 \]