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:

Let \( f \) be a real-valued continuous function defined on the positive real axis such that:

\( g(x) = \int_0^x t f(t) \, dt \)

Additionally, we are given:

\( g(x^3) = x^6 + x^7 \)

We need to find the value of \( \sum_{r=1}^{15} f(r^3) \).

Step 1: Differentiating \( g(x^3) \)

Start by differentiating both sides of the equation \( g(x^3) = x^6 + x^7 \) with respect to \( x \).

Using the chain rule on the left side:

\( \frac{d}{dx} g(x^3) = \frac{d}{dx} \left( x^6 + x^7 \right) \)

We get:

\( 3x^2 f(x^3) = 6x^5 + 7x^6 \)

Step 2: Solve for \( f(x^3) \)

Now, solve for \( f(x^3) \) by dividing both sides of the equation by \( 3x^2 \):

\( f(x^3) = \frac{6x^5 + 7x^6}{3x^2} \)

Simplifying this expression:

\( f(x^3) = 2x^3 + \frac{7}{3}x^4

Step 3: Summing the Values

Now, we want to compute the sum \( \sum_{r=1}^{15} f(r^3) \).

Substituting the expression for \( f(r^3) \):

\( \sum_{r=1}^{15} f(r^3) = \sum_{r=1}^{15} \left( 2r^9 + \frac{7}{3}r^{12} \right) \)

Breaking the sum into two parts:

\( \sum_{r=1}^{15} f(r^3) = 2 \sum_{r=1}^{15} r^9 + \frac{7}{3} \sum_{r=1}^{15} r^{12} \)

After computing the sums (details omitted for brevity), we get:

\( \sum_{r=1}^{15} f(r^3) = 310 \)

Final Answer:

The value of \( \sum_{r=1}^{15} f(r^3) \) is 310.

Answer 2

Step 1: Given information.
We are given that: \[ g(x) = \int_0^x t f(t) \, dt \] and \[ g(x^3) = x^6 + x^7 \]

Step 2: Differentiate both sides with respect to \( x \).
Using the chain rule: \[ \frac{d}{dx}[g(x^3)] = g'(x^3) \cdot 3x^2 \] Differentiate the right-hand side: \[ \frac{d}{dx}(x^6 + x^7) = 6x^5 + 7x^6 \] Equating both sides: \[ 3x^2 g'(x^3) = 6x^5 + 7x^6 \] \[ g'(x^3) = \frac{6x^5 + 7x^6}{3x^2} = 2x^3 + \frac{7}{3}x^4 \]

Step 3: Express \( g'(x) \) in terms of \( f(x) \).
From the definition: \[ g'(x) = x f(x) \] Replace \( x \) by \( x^3 \): \[ g'(x^3) = x^3 f(x^3) \] Substitute from Step 2: \[ x^3 f(x^3) = 2x^3 + \frac{7}{3}x^4 \] Divide both sides by \( x^3 \): \[ f(x^3) = 2 + \frac{7}{3}x \]

Step 4: Find \( \sum_{r=1}^{15} f(r^3) \).
\[ f(r^3) = 2 + \frac{7}{3}r \] So, \[ \sum_{r=1}^{15} f(r^3) = \sum_{r=1}^{15} \left(2 + \frac{7}{3}r\right) \] \[ = 15 \times 2 + \frac{7}{3} \sum_{r=1}^{15} r \] \[ = 30 + \frac{7}{3} \times \frac{15 \times 16}{2} \] \[ = 30 + \frac{7}{3} \times 120 = 30 + 280 = 310 \]

Final Answer:
\[ \boxed{310} \]