`int_(0)^(4) |x-1| dx`=

Options

A. 1
B. 2
C. 3
D. None of these

Answer

To solve the integral \( \int_{0}^{4} |x-1| \, dx \), we need to break it down based on the definition of the absolute value function.

### Step 1: Identify the points where the expression inside the absolute value changes sign.
The expression \( |x-1| \) changes sign at \( x = 1 \). Therefore, we will split the integral at this point.

### Step 2: Split the integral into two parts.
We can write the integral as:
\[ \int_{0}^{4} |x-1| \, dx = \int_{0}^{1} |x-1| \, dx + \int_{1}^{4} |x-1| \, dx \]

### Step 3: Determine the expression for \( |x-1| \) in each interval.
- For \( 0 \leq x < 1 \), \( x - 1 < 0 \) so \( |x-1| = -(x-1) = 1-x \).
- For \( 1 \leq x \leq 4 \), \( x - 1 \geq 0 \) so \( |x-1| = x-1 \).

### Step 4: Rewrite the integral with the new expressions.
Now we can rewrite the integral:
\[ \int_{0}^{4} |x-1| \, dx = \int_{0}^{1} (1-x) \, dx + \int_{1}^{4} (x-1) \, dx \]

### Step 5: Calculate the first integral \( \int_{0}^{1} (1-x) \, dx \).
\[ \int_{0}^{1} (1-x) \, dx = \left[ x - \frac{x^2}{2} \right]_{0}^{1} = \left( 1 - \frac{1^2}{2} \right) - \left( 0 - 0 \right) = 1 - \frac{1}{2} = \frac{1}{2} \]

### Step 6: Calculate the second integral \( \int_{1}^{4} (x-1) \, dx \).
\[ \int_{1}^{4} (x-1) \, dx = \left[ \frac{x^2}{2} - x \right]_{1}^{4} = \left( \frac{4^2}{2} - 4 \right) - \left( \frac{1^2}{2} - 1 \right) = \left( \frac{16}{2} - 4 \right) - \left( \frac{1}{2} - 1 \right) \]
Calculating this gives:
\[ = (8 - 4) - \left( \frac{1}{2} - 1 \right) = 4 - \left( \frac{1}{2} - 1 \right) = 4 - \left( -\frac{1}{2} \right) = 4 + \frac{1}{2} = 4.5 = \frac{9}{2} \]

### Step 7: Combine the results from both integrals.
Now we add the results of both integrals:
\[ \int_{0}^{4} |x-1| \, dx = \frac{1}{2} + \frac{9}{2} = \frac{10}{2} = 5 \]

### Final Answer:
Thus, the value of the integral \( \int_{0}^{4} |x-1| \, dx \) is \( 5 \).

---