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

Options

A. `(3)/(8)`
B. `(1)/(4)`
C. `-(3)/(8)`
D. `-(1)/(2)`

Answer

To find the greater value of the integral \( I = \int_{-\frac{1}{2}}^{x} |t| \, dt \) on the interval \( [-\frac{1}{2}, \frac{1}{2}] \), we will evaluate the integral step by step.

### Step 1: Understand the modulus function
The modulus function \( |t| \) behaves differently depending on the sign of \( t \):
- For \( t < 0 \), \( |t| = -t \)
- For \( t \geq 0 \), \( |t| = t \)

### Step 2: Split the integral based on the limits
Since we are integrating from \( -\frac{1}{2} \) to \( x \), we need to consider two cases based on the value of \( x \):
1. When \( x < 0 \)
2. When \( x \geq 0 \)

### Step 3: Case 1: \( x < 0 \)
In this case, we will integrate from \( -\frac{1}{2} \) to \( x \):
\[ I = \int_{-\frac{1}{2}}^{x} -t \, dt \]
Calculating this integral:
\[ I = -\left[ \frac{t^2}{2} \right]_{-\frac{1}{2}}^{x} = -\left( \frac{x^2}{2} - \frac{(-\frac{1}{2})^2}{2} \right) = -\left( \frac{x^2}{2} - \frac{1}{8} \right) \]
This simplifies to:
\[ I = \frac{1}{8} - \frac{x^2}{2} \]

### Step 4: Case 2: \( x \geq 0 \)
In this case, we will integrate from \( -\frac{1}{2} \) to \( 0 \) and then from \( 0 \) to \( x \):
\[ I = \int_{-\frac{1}{2}}^{0} -t \, dt + \int_{0}^{x} t \, dt \]
Calculating the first integral:
\[ \int_{-\frac{1}{2}}^{0} -t \, dt = -\left[ \frac{t^2}{2} \right]_{-\frac{1}{2}}^{0} = -\left( 0 - \frac{1}{8} \right) = \frac{1}{8} \]
Calculating the second integral:
\[ \int_{0}^{x} t \, dt = \left[ \frac{t^2}{2} \right]_{0}^{x} = \frac{x^2}{2} \]
Thus, combining both parts:
\[ I = \frac{1}{8} + \frac{x^2}{2} \]

### Step 5: Determine the maximum value of \( I \)
Now we have two expressions for \( I \):
- For \( x < 0 \): \( I = \frac{1}{8} - \frac{x^2}{2} \)
- For \( x \geq 0 \): \( I = \frac{1}{8} + \frac{x^2}{2} \)

We need to evaluate these at the endpoints of the interval \( [-\frac{1}{2}, \frac{1}{2}] \):
1. At \( x = -\frac{1}{2} \):
\[ I = \frac{1}{8} - \frac{(-\frac{1}{2})^2}{2} = \frac{1}{8} - \frac{1/8}{2} = \frac{1}{8} - \frac{1}{16} = \frac{2}{16} - \frac{1}{16} = \frac{1}{16} \]
2. At \( x = 0 \):
\[ I = \frac{1}{8} \]
3. At \( x = \frac{1}{2} \):
\[ I = \frac{1}{8} + \frac{(\frac{1}{2})^2}{2} = \frac{1}{8} + \frac{1/4}{2} = \frac{1}{8} + \frac{1}{8} = \frac{2}{8} = \frac{1}{4} \]

### Conclusion
The greater value of \( I \) on the interval \( [-\frac{1}{2}, \frac{1}{2}] \) is:
\[ \boxed{\frac{1}{4}} \]