Let `f(x) = int_(-2)^(x)|t + 1|dt`, then

Options

A. f(x) is continuous in [-1, 1]
B. f(x) is differentiable in [-1, 1]
C. f'(x) is continuous in [-1, 1]
D. f'(x) is differentiable in [-1, 1]

Answer

To solve the problem, we need to evaluate the function defined by the integral and analyze its continuity and differentiability.

### Step 1: Define the function
We start with the function:
\[ f(x) = \int_{-2}^{x} |t + 1| \, dt \]

### Step 2: Analyze the absolute value
The expression \( |t + 1| \) changes its behavior at \( t = -1 \). Therefore, we need to split the integral into two parts based on this point:
- For \( t < -1 \), \( |t + 1| = -(t + 1) = -t - 1 \)
- For \( t \geq -1 \), \( |t + 1| = t + 1 \)

### Step 3: Set up the integral
We can split the integral at \( t = -1 \):
\[ f(x) = \int_{-2}^{-1} |t + 1| \, dt + \int_{-1}^{x} |t + 1| \, dt \]
This becomes:
\[ f(x) = \int_{-2}^{-1} (-t - 1) \, dt + \int_{-1}^{x} (t + 1) \, dt \]

### Step 4: Evaluate the first integral
Calculating the first integral:
\[ \int_{-2}^{-1} (-t - 1) \, dt = \int_{-2}^{-1} (-t) \, dt - \int_{-2}^{-1} 1 \, dt \]
Calculating each part:
1. \(\int_{-2}^{-1} -t \, dt = \left[-\frac{t^2}{2}\right]_{-2}^{-1} = -\frac{(-1)^2}{2} + \frac{(-2)^2}{2} = -\frac{1}{2} + 2 = \frac{3}{2}\)
2. \(\int_{-2}^{-1} 1 \, dt = [-t]_{-2}^{-1} = -(-1) - (-2) = 1\)

Thus, the first integral evaluates to:
\[ \int_{-2}^{-1} (-t - 1) \, dt = \frac{3}{2} - 1 = \frac{1}{2} \]

### Step 5: Evaluate the second integral
Now we evaluate the second integral:
\[ \int_{-1}^{x} (t + 1) \, dt = \left[\frac{t^2}{2} + t\right]_{-1}^{x} = \left(\frac{x^2}{2} + x\right) - \left(\frac{(-1)^2}{2} - 1\right) = \frac{x^2}{2} + x - \left(\frac{1}{2} - 1\right) = \frac{x^2}{2} + x + \frac{1}{2} \]

### Step 6: Combine results
Combining both parts, we have:
\[ f(x) = \frac{1}{2} + \left(\frac{x^2}{2} + x + \frac{1}{2}\right) = \frac{x^2}{2} + x + 1 \]

### Step 7: Check continuity and differentiability
The function \( f(x) = \frac{x^2}{2} + x + 1 \) is a polynomial function, which is continuous and differentiable everywhere.

### Step 8: Find the derivative
Now, we differentiate \( f(x) \):
\[ f'(x) = \frac{d}{dx}\left(\frac{x^2}{2} + x + 1\right) = x + 1 \]

### Step 9: Check continuity and differentiability of the derivative
The derivative \( f'(x) = x + 1 \) is also a polynomial, hence it is continuous and differentiable everywhere.

### Conclusion
Thus, we conclude that \( f(x) \) is continuous and differentiable, and \( f'(x) \) is also continuous and differentiable.

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