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

We start by differentiating the given expression \( F(x^2) = x^4 + x^5 \) to find \( f(t) \). Consider \( F(u) \) where \( u = x^2 \), leading to:

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

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

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

Equating both derivatives gives:

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

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

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

Now, by definition,

\[ F'(x^2) = (x^2) \cdot f(x^2) \]

Setting them equal:

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

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

To find \(\sum_{r=1}^{12} f(r^2)\), notice:

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

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

Sum the constant term:

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

Sum the variable term:

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

The known sum of the first 12 positive integers is:

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

Calculating the weighted sum:

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

Adding both parts gives:

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

The computed value is within the given range [219, 219]. Therefore, the answer is 219.

Answer 2

Step 1. Fundamental Theorem of Calculus and the Given Information:

We are given that $F(x) = \int_0^x t \cdot f(t) dt$. The Fundamental Theorem of Calculus states that the derivative of a definite integral with respect to its upper limit is the integrand evaluated at that upper limit. Therefore, we have:

$F'(x) = x \cdot f(x)$

We're also given that $F(x^2) = x^4 + x^5$. Let's substitute $t = x^2$:

$F(t) = t^2 + t^{5/2}$

Step 2. Finding f(t):

Now we differentiate $F(t)$ to find $F'(t)$:

$F'(t) = \frac{d}{dt}(t^2 + t^{5/2}) = 2t + \frac{5}{2}t^{3/2}$

Since $F'(t) = t \cdot f(t)$, we can solve for $f(t)$:

$t \cdot f(t) = 2t + \frac{5}{2}t^{3/2}$

$f(t) = \frac{2t + \frac{5}{2}t^{3/2}}{t} = 2 + \frac{5}{2}t^{1/2}$

Step 3. Evaluating the Summation:

We want to find $\sum_{r=1}^{12} f(r^2)$. Substituting our expression for $f(t)$:

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

Step 4. Summation Properties and Simplification:

We can split the summation into two separate sums:

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

The first term is simply $2$ added 12 times: $2 \times 12 = 24$. The second term is the sum of the integers from 1 to 12, which can be calculated using the formula for the sum of an arithmetic series: $\sum_{r=1}^n r = \frac{n(n+1)}{2}$.

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

Step 5. Final Calculation:

Substituting back into our equation:

$\sum_{r=1}^{12} f(r^2) = 24 + \frac{5}{2} (78) = 24 + 5(39) = 24 + 195 = 219$

Therefore, the final answer is:

$\sum_{r=1}^{12} f(r^2) = 219$

Final Answer: The final answer is $\boxed{219}$