`int_(-1)^(1) {|x-1| +|x|} dx` is equal to

Options

A. 1
B. 2
C. 3
D. none

Answer

To solve the integral \( \int_{-1}^{1} \left( |x-1| + |x| \right) dx \), we will break it down into manageable parts based on the properties of absolute values.

### Step 1: Identify the intervals for absolute values
The expression \( |x-1| \) changes at \( x = 1 \) and \( |x| \) changes at \( x = 0 \). Therefore, we will break the integral into three intervals: \( [-1, 0] \), \( [0, 1] \).

### Step 2: Rewrite the integral based on the intervals
We can express the integral as:
\[ \int_{-1}^{1} \left( |x-1| + |x| \right) dx = \int_{-1}^{0} \left( |x-1| + |x| \right) dx + \int_{0}^{1} \left( |x-1| + |x| \right) dx \]

### Step 3: Evaluate the integral on the interval \([-1, 0]\)
For \( x \in [-1, 0] \):
- \( |x-1| = 1 - x \) (since \( x-1 < 0 \))
- \( |x| = -x \) (since \( x < 0 \))

Thus, the integral becomes:
\[ \int_{-1}^{0} \left( (1-x) + (-x) \right) dx = \int_{-1}^{0} (1 - 2x) dx \]

Now, calculate this integral:
\[ \int (1 - 2x) dx = x - x^2 \]
Evaluating from \(-1\) to \(0\):
\[ \left[ x - x^2 \right]_{-1}^{0} = \left( 0 - 0^2 \right) - \left( -1 - (-1)^2 \right) = 0 - (-1 - 1) = 0 + 2 = 2 \]

### Step 4: Evaluate the integral on the interval \([0, 1]\)
For \( x \in [0, 1] \):
- \( |x-1| = 1 - x \) (since \( x-1 < 0 \))
- \( |x| = x \) (since \( x \geq 0 \))

Thus, the integral becomes:
\[ \int_{0}^{1} \left( (1-x) + x \right) dx = \int_{0}^{1} 1 dx \]

Calculating this integral:
\[ \int 1 dx = x \]
Evaluating from \(0\) to \(1\):
\[ \left[ x \right]_{0}^{1} = 1 - 0 = 1 \]

### Step 5: Combine the results
Now, we combine the results from both intervals:
\[ \int_{-1}^{1} \left( |x-1| + |x| \right) dx = 2 + 1 = 3 \]

### Final Answer
Thus, the value of the integral is:
\[ \int_{-1}^{1} \left( |x-1| + |x| \right) dx = 3 \]