If `f(x)= |2^(x)-1| + |x-1|`, then `int_(-2)^(2) f(x) dx`=

Options

A. `5-(9)/(4) log 2`
B. `5+ (9)/(4 log 2)`
C. `-5-(9)/(4) log 2`
D. none

Answer

To solve the integral \( \int_{-2}^{2} f(x) \, dx \) where \( f(x) = |2^x - 1| + |x - 1| \), we need to analyze the function \( f(x) \) over the interval from \(-2\) to \(2\).

### Step 1: Identify the points where the expression inside the absolute values changes sign.
1. **For \( |2^x - 1| \)**:
- \( 2^x - 1 = 0 \) when \( x = 0 \).
- \( 2^x < 1 \) for \( x < 0 \) and \( 2^x > 1 \) for \( x > 0 \).

2. **For \( |x - 1| \)**:
- \( x - 1 = 0 \) when \( x = 1 \).
- \( x < 1 \) for \( x < 1 \) and \( x > 1 \) for \( x > 1 \).

### Step 2: Break the integral into intervals based on the points identified.
The critical points are \( x = 0 \) and \( x = 1 \). Thus, we will break the integral into three parts:
- From \(-2\) to \(0\)
- From \(0\) to \(1\)
- From \(1\) to \(2\)

### Step 3: Evaluate \( f(x) \) in each interval.

1. **For \( x \in [-2, 0] \)**:
- \( 2^x < 1 \) and \( x < 1 \)
- Therefore, \( f(x) = -(2^x - 1) - (x - 1) = -2^x + 1 - x + 1 = -2^x - x + 2 \).

2. **For \( x \in [0, 1] \)**:
- \( 2^x \geq 1 \) and \( x < 1 \)
- Therefore, \( f(x) = (2^x - 1) - (x - 1) = 2^x - 1 - x + 1 = 2^x - x \).

3. **For \( x \in [1, 2] \)**:
- \( 2^x \geq 1 \) and \( x \geq 1 \)
- Therefore, \( f(x) = (2^x - 1) + (x - 1) = 2^x - 1 + x - 1 = 2^x + x - 2 \).

### Step 4: Set up the integral.
Now we can write the integral as:
\[ \int_{-2}^{2} f(x) \, dx = \int_{-2}^{0} (-2^x - x + 2) \, dx + \int_{0}^{1} (2^x - x) \, dx + \int_{1}^{2} (2^x + x - 2) \, dx \]

### Step 5: Calculate each integral.

1. **Calculate \( \int_{-2}^{0} (-2^x - x + 2) \, dx \)**:
\[ = \left[-\frac{2^x}{\ln 2} - \frac{x^2}{2} + 2x \right]_{-2}^{0} \]
Evaluating at the limits:
\[ = \left[-\frac{1}{\ln 2} - 0 + 0\right] - \left[-\frac{2^{-2}}{\ln 2} - 2 + 4\right] = -\frac{1}{\ln 2} + \frac{1/4}{\ln 2} + 2 = -\frac{3}{4\ln 2} + 2 \]

2. **Calculate \( \int_{0}^{1} (2^x - x) \, dx \)**:
\[ = \left[\frac{2^x}{\ln 2} - \frac{x^2}{2}\right]_{0}^{1} = \left[\frac{2}{\ln 2} - \frac{1}{2}\right] - \left[\frac{1}{\ln 2} - 0\right] = \frac{2 - 1}{\ln 2} - \frac{1}{2} = \frac{1}{\ln 2} - \frac{1}{2} \]

3. **Calculate \( \int_{1}^{2} (2^x + x - 2) \, dx \)**:
\[ = \left[\frac{2^x}{\ln 2} + \frac{x^2}{2} - 2x\right]_{1}^{2} = \left[\frac{4}{\ln 2} + 2 - 4\right] - \left[\frac{2}{\ln 2} + \frac{1}{2} - 2\right] = \left[\frac{4 - 4\ln 2}{\ln 2}\right] - \left[\frac{2 - 2\ln 2}{\ln 2}\right] = \frac{2}{\ln 2} - \frac{1}{2} \]

### Step 6: Combine the results.
Now we sum the three integrals:
\[ \int_{-2}^{2} f(x) \, dx = \left(-\frac{3}{4\ln 2} + 2\right) + \left(\frac{1}{\ln 2} - \frac{1}{2}\right) + \left(\frac{2}{\ln 2} - \frac{1}{2}\right) \]
Combine like terms:
\[ = \left(-\frac{3}{4\ln 2} + \frac{1 + 2}{\ln 2}\right) + \left(2 - 1\right) = \left(-\frac{3}{4\ln 2} + \frac{3}{\ln 2}\right) + 1 = \left(\frac{12 - 3}{4\ln 2}\right) + 1 = \frac{9}{4\ln 2} + 1 \]

### Final Answer:
Thus, the value of the integral \( \int_{-2}^{2} f(x) \, dx = 5 + \frac{9}{4\ln 2} \).

---