Question

Let $f : (0, \infty) \to \mathbb{R}$ and $F(x) = \int_{0}^{x} tf(t) \, dt$. If $F(x^2) = x^4 + x^5$, then \[\sum_{r=1}^{12} f(r^2)\]is equal to:

Answer 1

Correct Answer: 219

To determine \( f(t) \), we first differentiate the given expression \( F(x^2) = x^4 + x^5 \). Let \( u = x^2 \), so \( F(u) \). The derivative is:

\[ \frac{d}{dx}[F(x^2)] = \frac{dF}{du} \cdot \frac{du}{dx} = 2x \cdot F'(x^2) \]

Differentiating \( x^4 + x^5 \) with respect to \( x \) yields:

\[ \frac{d}{dx}[x^4 + x^5] = 4x^3 + 5x^4 \]

Equating these derivatives:

\[ 2x \cdot F'(x^2) = 4x^3 + 5x^4 \]

Solving for \( F'(x^2) \):

\( F'(x^2) = \frac{4x^3 + 5x^4}{2x} \)

\[ F'(x^2) = 2x^2 + \frac{5}{2}x^3 \]

By definition, \( F'(x^2) = x^2 \cdot f(x^2) \). Setting this equal to the derived expression:

\[ x^2 \cdot f(x^2) = 2x^2 + \frac{5}{2}x^3 \]

Solving for \( f(x^2) \):

\[ f(x^2) = 2 + \frac{5}{2}x \]

To calculate \(\sum_{r=1}^{12} f(r^2)\), we use the relationship \( f(r^2) = 2 + \frac{5}{2}r \):

\(\sum_{r=1}^{12} f(r^2) = \sum_{r=1}^{12} \left(2 + \frac{5}{2}r\right)\)

We sum the constant term:

\[ \sum_{r=1}^{12} 2 = 2 \times 12 = 24 \]

And the variable term:

\[ \sum_{r=1}^{12} \frac{5}{2}r = \frac{5}{2} \sum_{r=1}^{12} r \]

The sum of the first 12 positive integers is:

\(\sum_{r=1}^{12} r = \frac{12 \cdot 13}{2} = 78\)

Calculating the sum of the variable term:

\[ \frac{5}{2} \times 78 = 5 \times 39 = 195 \]

The total sum is the sum of the constant and variable parts:

\(\sum f(r^2) = 24 + 195 = 219\)

The computed value of 219 falls within the specified range [219, 219].