Let `f:(0,oo)to R` be a continuous function such that `F(x)=int_(0)^(x^(2)) tf(t)dt. "If "F(x^(2))=x^(4)+x^(5),"then "sum_(r=1)^(12) f(r^(2))=`

Options

A. 216
B. 219
C. 222
D. 225

Explanation

To solve the problem step by step, we will first analyze the given information and then derive the required result.

### Step 1: Understand the given function
We are given a function \( f: (0, \infty) \to \mathbb{R} \) defined by the integral:
\[ F(x) = \int_0^{x^2} t f(t) \, dt \]
It is also given that:
\[ F(x^2) = x^4 + x^5 \]

### Step 2: Differentiate both sides
To find \( f(x) \), we differentiate both sides of the equation \( F(x^2) = x^4 + x^5 \) with respect to \( x \).

Using the chain rule on the left side:
\[ \frac{d}{dx} F(x^2) = F'(x^2) \cdot \frac{d}{dx}(x^2) = 2x F'(x^2) \]

For the right side:
\[ \frac{d}{dx}(x^4 + x^5) = 4x^3 + 5x^4 \]

Setting both derivatives equal gives:
\[ 2x F'(x^2) = 4x^3 + 5x^4 \]

### Step 3: Solve for \( F'(x^2) \)
Dividing both sides by \( 2x \) (for \( x > 0 \)):
\[ F'(x^2) = 2x^2 + \frac{5}{2} x^3 \]

### Step 4: Relate \( F'(x) \) to \( f(x) \)
From the Fundamental Theorem of Calculus, we know that:
\[ F'(x) = x f(x) \]
Thus, substituting \( x^2 \) into this gives:
\[ F'(x^2) = x^2 f(x^2) \]

### Step 5: Set the equations equal
Now we can set the two expressions for \( F'(x^2) \) equal:
\[ x^2 f(x^2) = 2x^2 + \frac{5}{2} x^3 \]

### Step 6: Solve for \( f(x^2) \)
Dividing both sides by \( x^2 \) (for \( x > 0 \)):
\[ f(x^2) = 2 + \frac{5}{2} x \]

### Step 7: Find \( \sum_{r=1}^{12} f(r^2) \)
Now we need to calculate:
\[ \sum_{r=1}^{12} f(r^2) = \sum_{r=1}^{12} \left( 2 + \frac{5}{2} r \right) \]

This can be split into two separate sums:
\[ \sum_{r=1}^{12} f(r^2) = \sum_{r=1}^{12} 2 + \sum_{r=1}^{12} \frac{5}{2} r \]

### Step 8: Calculate the sums
1. The first sum:
\[ \sum_{r=1}^{12} 2 = 2 \times 12 = 24 \]

2. The second sum:
\[ \sum_{r=1}^{12} r = \frac{12 \times (12 + 1)}{2} = \frac{12 \times 13}{2} = 78 \]
Thus,
\[ \sum_{r=1}^{12} \frac{5}{2} r = \frac{5}{2} \times 78 = 195 \]

### Step 9: Combine the results
Now, we combine the results from both sums:
\[ \sum_{r=1}^{12} f(r^2) = 24 + 195 = 219 \]

### Final Answer
Thus, the final answer is:
\[ \sum_{r=1}^{12} f(r^2) = 219 \]