The value of the integral `overset(e )underset(1//e)int |logx|dx`, is

Options

A. `1-1//e`
B. `3(l-1//e)`
C. `e^(-1)-1`
D. None of these

Answer

To solve the integral \( \int_{1/e}^{e} |\log x| \, dx \), we will follow these steps:

### Step 1: Analyze the function \( |\log x| \)

The function \( \log x \) is negative when \( x < 1 \) and positive when \( x > 1 \). Since our limits of integration are from \( \frac{1}{e} \) to \( e \), we note that:
- At \( x = \frac{1}{e} \), \( \log \frac{1}{e} = -1 \) (negative)
- At \( x = 1 \), \( \log 1 = 0 \) (zero)
- At \( x = e \), \( \log e = 1 \) (positive)

Thus, we can break the integral into two parts:
1. From \( \frac{1}{e} \) to \( 1 \), where \( |\log x| = -\log x \)
2. From \( 1 \) to \( e \), where \( |\log x| = \log x \)

### Step 2: Set up the integral

We can express the integral as:
\[ \int_{1/e}^{e} |\log x| \, dx = \int_{1/e}^{1} -\log x \, dx + \int_{1}^{e} \log x \, dx \]

### Step 3: Calculate the first integral \( \int_{1/e}^{1} -\log x \, dx \)

Using integration by parts, let:
- \( u = \log x \) → \( du = \frac{1}{x} \, dx \)
- \( dv = dx \) → \( v = x \)

Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \):
\[ \int -\log x \, dx = -\left( x \log x - \int x \cdot \frac{1}{x} \, dx \right) = -\left( x \log x - x \right) = -x \log x + x \]

Now, we evaluate from \( \frac{1}{e} \) to \( 1 \):
\[ \left[-x \log x + x\right]_{1/e}^{1} = \left[-1 \cdot \log 1 + 1\right] - \left[-\frac{1}{e} \cdot \log \frac{1}{e} + \frac{1}{e}\right] \]
\[ = [0 + 1] - \left[-\frac{1}{e} \cdot (-1) + \frac{1}{e}\right] = 1 - \left[\frac{1}{e} + \frac{1}{e}\right] = 1 - \frac{2}{e} \]

### Step 4: Calculate the second integral \( \int_{1}^{e} \log x \, dx \)

Using the same integration by parts as before:
\[ \left[x \log x - x\right]_{1}^{e} = \left[e \cdot \log e - e\right] - \left[1 \cdot \log 1 - 1\right] \]
\[ = [e \cdot 1 - e] - [0 - 1] = [e - e] - [-1] = 0 + 1 = 1 \]

### Step 5: Combine the results

Now, we combine both parts:
\[ \int_{1/e}^{e} |\log x| \, dx = \left(1 - \frac{2}{e}\right) + 1 = 2 - \frac{2}{e} \]

### Final Answer

Thus, the value of the integral is:
\[ \int_{1/e}^{e} |\log x| \, dx = 2 - \frac{2}{e} \]