`int_(1//2)^(2) |log_(10)x|dx` equals to

Options

A. `log10(8//e)`
B. `(1)/(2)log10(8//e)`
C. `log10(2//e)`
D. none of these

Answer

To solve the integral \( I = \int_{\frac{1}{2}}^{2} |\log_{10} x| \, dx \), we will follow these steps:

### Step 1: Analyze the function \( |\log_{10} x| \)

The logarithmic function \( \log_{10} x \) changes sign at \( x = 1 \):
- For \( x < 1 \), \( \log_{10} x < 0 \) (negative).
- For \( x = 1 \), \( \log_{10} x = 0 \).
- For \( x > 1 \), \( \log_{10} x > 0 \) (positive).

Thus, we can express the absolute value function as:
\[ |\log_{10} x| = \begin{cases} -\log_{10} x & \text{if } x \in \left[\frac{1}{2}, 1\right) \\ \log_{10} x & \text{if } x \in (1, 2] \end{cases} \]

### Step 2: Split the integral

We can split the integral at \( x = 1 \):
\[ I = \int_{\frac{1}{2}}^{1} -\log_{10} x \, dx + \int_{1}^{2} \log_{10} x \, dx \]

### Step 3: Calculate each integral

#### Integral from \( \frac{1}{2} \) to \( 1 \)
\[ I_1 = \int_{\frac{1}{2}}^{1} -\log_{10} x \, dx = -\left[ x \log_{10} x - x \right]_{\frac{1}{2}}^{1} \]

Calculating the limits:
- At \( x = 1 \):
\[ 1 \log_{10} 1 - 1 = 0 - 1 = -1 \]
- At \( x = \frac{1}{2} \):
\[ \frac{1}{2} \log_{10} \frac{1}{2} - \frac{1}{2} = \frac{1}{2} \cdot (-\log_{10} 2) - \frac{1}{2} = -\frac{1}{2} \log_{10} 2 - \frac{1}{2} \]

Thus,
\[ I_1 = -\left(-1 + \left(-\frac{1}{2} \log_{10} 2 - \frac{1}{2}\right)\right) = 1 - \left(-\frac{1}{2} \log_{10} 2 - \frac{1}{2}\right) \]
\[ = 1 + \frac{1}{2} \log_{10} 2 + \frac{1}{2} = \frac{3}{2} + \frac{1}{2} \log_{10} 2 \]

#### Integral from \( 1 \) to \( 2 \)
\[ I_2 = \int_{1}^{2} \log_{10} x \, dx = \left[ x \log_{10} x - x \right]_{1}^{2} \]

Calculating the limits:
- At \( x = 2 \):
\[ 2 \log_{10} 2 - 2 \]
- At \( x = 1 \):
\[ 1 \log_{10} 1 - 1 = -1 \]

Thus,
\[ I_2 = \left(2 \log_{10} 2 - 2\right) - (-1) = 2 \log_{10} 2 - 2 + 1 = 2 \log_{10} 2 - 1 \]

### Step 4: Combine the results

Now, combining both parts:
\[ I = I_1 + I_2 = \left(\frac{3}{2} + \frac{1}{2} \log_{10} 2\right) + \left(2 \log_{10} 2 - 1\right) \]
\[ = \frac{3}{2} - 1 + \frac{1}{2} \log_{10} 2 + 2 \log_{10} 2 = \frac{1}{2} + \frac{5}{2} \log_{10} 2 \]

### Final Answer
Thus, the value of the integral is:
\[ I = \frac{1}{2} + \frac{5}{2} \log_{10} 2 \]