The greater value of `F(x)=int_(1)^(x) |t|dt` on the interval `[-1//2,1//2]`, is

Options

A. `(3)/(8)`
B. `(1)/(2)`
C. `-(3)/(8)`
D. `-(1)/(2)`

Answer

To solve the problem of finding the greater value of the function \( F(x) = \int_{1}^{x} |t| \, dt \) on the interval \([-1/2, 1/2]\), we will follow these steps:

### Step 1: Analyze the Integral
The function \( |t| \) is defined as:
- \( |t| = t \) when \( t \geq 0 \)
- \( |t| = -t \) when \( t < 0 \)

Since we are integrating from 1 to \( x \), we need to consider the value of \( x \) in relation to 1.

### Step 2: Determine the Cases for Integration
1. **Case 1:** If \( x \geq 1 \), then \( |t| = t \) for all \( t \) in the interval \([1, x]\).
\[ F(x) = \int_{1}^{x} t \, dt \]

2. **Case 2:** If \( x < 1 \), then we need to consider the point where \( t \) becomes negative. Since \( x \) is in the interval \([-1/2, 1/2]\), we will need to evaluate \( F(x) \) for \( x < 0 \) and \( x \geq 0 \).

### Step 3: Calculate the Integral for \( x \geq 1 \)
For \( x \geq 1 \):
\[ F(x) = \int_{1}^{x} t \, dt = \left[ \frac{t^2}{2} \right]_{1}^{x} = \frac{x^2}{2} - \frac{1^2}{2} = \frac{x^2}{2} - \frac{1}{2} \]

### Step 4: Calculate the Integral for \( x < 1 \)
For \( x < 1 \) and \( x \geq 0 \):
\[ F(x) = \int_{1}^{x} |t| \, dt = \int_{1}^{0} t \, dt + \int_{0}^{x} -t \, dt \]
This integral can be split as:
\[ F(x) = -\int_{0}^{1} t \, dt + \int_{0}^{x} -t \, dt = -\left[ \frac{t^2}{2} \right]_{0}^{1} + \left[ -\frac{t^2}{2} \right]_{0}^{x} \]
Calculating these gives:
\[ F(x) = -\frac{1^2}{2} + \left(-\frac{x^2}{2}\right) = -\frac{1}{2} - \frac{x^2}{2} \]

### Step 5: Evaluate \( F(x) \) at the Endpoints and Critical Points
Now we evaluate \( F(x) \) at the endpoints of the interval \([-1/2, 1/2]\):
1. **At \( x = -1/2 \)**:
\[ F(-1/2) = -\frac{1}{2} - \frac{(-1/2)^2}{2} = -\frac{1}{2} - \frac{1/8} = -\frac{4}{8} - \frac{1}{8} = -\frac{5}{8} \]

2. **At \( x = 0 \)**:
\[ F(0) = -\frac{1}{2} - \frac{0^2}{2} = -\frac{1}{2} \]

3. **At \( x = 1/2 \)**:
\[ F(1/2) = -\frac{1}{2} - \frac{(1/2)^2}{2} = -\frac{1}{2} - \frac{1/8} = -\frac{4}{8} - \frac{1}{8} = -\frac{5}{8} \]

### Step 6: Compare Values
Now we compare the values:
- \( F(-1/2) = -\frac{5}{8} \)
- \( F(0) = -\frac{1}{2} = -\frac{4}{8} \)
- \( F(1/2) = -\frac{5}{8} \)

The maximum value among these is:
\[ -\frac{4}{8} > -\frac{5}{8} \]

### Conclusion
The greater value of \( F(x) \) on the interval \([-1/2, 1/2]\) is:
\[ \boxed{-\frac{1}{2}} \]