The value of integral `int_(a)^(b)(|x|)/(x)dx`, `a lt b` is :

Options

A. `b -a` if `a gt 0`
B. `a-b` if `b lt 0`
C. `b +a` if `a lt 0 lt b`
D. `|b | - |a|`

Answer

To solve the integral \( \int_a^b \frac{|x|}{x} \, dx \) where \( a < b \), we need to consider different cases based on the values of \( a \) and \( b \).

### Step 1: Identify Cases
We will analyze three cases based on the signs of \( a \) and \( b \):
1. Case 1: \( a < 0 \) and \( b < 0 \)
2. Case 2: \( a < 0 \) and \( b > 0 \)
3. Case 3: \( a > 0 \) and \( b > 0 \)

### Step 2: Case 1: \( a < 0 \) and \( b < 0 \)
In this case, \( |x| = -x \) for \( x < 0 \). Therefore, the integral becomes:
\[ \int_a^b \frac{|x|}{x} \, dx = \int_a^b \frac{-x}{x} \, dx = \int_a^b -1 \, dx \]
The integral of \(-1\) is:
\[ -1 \cdot (b - a) = -(b - a) = a - b \]

### Step 3: Case 2: \( a < 0 \) and \( b > 0 \)
Here, we split the integral at 0:
\[ \int_a^b \frac{|x|}{x} \, dx = \int_a^0 \frac{|x|}{x} \, dx + \int_0^b \frac{|x|}{x} \, dx \]
For \( x < 0 \), \( |x| = -x \):
\[ \int_a^0 \frac{-x}{x} \, dx = \int_a^0 -1 \, dx = -1 \cdot (0 - a) = -(-a) = a \]
For \( x > 0 \), \( |x| = x \):
\[ \int_0^b \frac{x}{x} \, dx = \int_0^b 1 \, dx = 1 \cdot (b - 0) = b \]
Combining both parts:
\[ \int_a^b \frac{|x|}{x} \, dx = a + b \]

### Step 4: Case 3: \( a > 0 \) and \( b > 0 \)
In this case, \( |x| = x \):
\[ \int_a^b \frac{|x|}{x} \, dx = \int_a^b \frac{x}{x} \, dx = \int_a^b 1 \, dx = 1 \cdot (b - a) = b - a \]

### Conclusion
From the three cases, we have:
1. If \( a < 0 \) and \( b < 0 \): \( a - b \)
2. If \( a < 0 < b \): \( a + b \)
3. If \( a > 0 \) and \( b > 0 \): \( b - a \)

Thus, the value of the integral \( \int_a^b \frac{|x|}{x} \, dx \) depends on the signs of \( a \) and \( b \).