If `f(x)=int_(-1)^(x)|t|dt`, then for any `xge0,f(x)` equals

Options

A. `1/2(1-x^(2))`
B. `1/2x^(2)`
C. `1/2(1+x^(2))`
D. none of these

Explanation

To solve the problem, we need to evaluate the integral \( f(x) = \int_{-1}^{x} |t| \, dt \) for \( x \geq 0 \).

### Step-by-Step Solution:

1. **Understanding the Absolute Value**:
The function \( |t| \) can be expressed piecewise:
- For \( t < 0 \), \( |t| = -t \)
- For \( t \geq 0 \), \( |t| = t \)

2. **Setting Up the Integral**:
Since we are interested in the interval from \(-1\) to \(x\) and \(x \geq 0\), we can split the integral at \(0\):
\[ f(x) = \int_{-1}^{0} |t| \, dt + \int_{0}^{x} |t| \, dt \]

3. **Calculating the First Integral**:
For the first integral, from \(-1\) to \(0\):
\[ \int_{-1}^{0} |t| \, dt = \int_{-1}^{0} -t \, dt \]
Evaluating this:
\[ = -\left[ \frac{t^2}{2} \right]_{-1}^{0} = -\left( 0 - \frac{(-1)^2}{2} \right) = -\left( 0 - \frac{1}{2} \right) = \frac{1}{2} \]

4. **Calculating the Second Integral**:
For the second integral, from \(0\) to \(x\):
\[ \int_{0}^{x} |t| \, dt = \int_{0}^{x} t \, dt \]
Evaluating this:
\[ = \left[ \frac{t^2}{2} \right]_{0}^{x} = \frac{x^2}{2} - 0 = \frac{x^2}{2} \]

5. **Combining the Results**:
Now we combine both integrals:
\[ f(x) = \frac{1}{2} + \frac{x^2}{2} \]
This simplifies to:
\[ f(x) = \frac{1 + x^2}{2} \]

### Final Answer:
Thus, for any \( x \geq 0 \):
\[ f(x) = \frac{1 + x^2}{2} \]