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

Options

A. `log_(10) (8//e)`
B. `(1)/(2) log_(10) (8//e)`
C. `log_(10) (2//e)`
D. None of these

Answer

To solve the integral \( I = \int_{\frac{1}{2}}^{2} |\log_{10} x| \, dx \), we need to consider the behavior of the function \( \log_{10} x \) over the interval \([\frac{1}{2}, 2]\).

### Step 1: Analyze the Function
The logarithmic function \( \log_{10} x \) is negative for \( x \) in the interval \(\left(0, 1\right)\) and positive for \( x \) in the interval \(\left(1, \infty\right)\). Therefore, we can break the integral into two parts:
- From \(\frac{1}{2}\) to \(1\), \( \log_{10} x < 0 \) so \( |\log_{10} x| = -\log_{10} x \).
- From \(1\) to \(2\), \( \log_{10} x \geq 0 \) so \( |\log_{10} x| = \log_{10} x \).

Thus, we can rewrite the integral as:
\[ I = \int_{\frac{1}{2}}^{1} -\log_{10} x \, dx + \int_{1}^{2} \log_{10} x \, dx \]

### Step 2: Calculate Each Integral
Now we will calculate each integral separately.

#### Integral from \(\frac{1}{2}\) to \(1\):
\[ I_1 = \int_{\frac{1}{2}}^{1} -\log_{10} x \, dx \]
Using integration by parts, let:
- \( u = -\log_{10} x \) ⇒ \( du = -\frac{1}{x \ln 10} \, dx \)
- \( dv = dx \) ⇒ \( v = x \)

Using integration by parts formula \( \int u \, dv = uv - \int v \, du \):
\[ I_1 = \left[-x \log_{10} x\right]_{\frac{1}{2}}^{1} - \int_{\frac{1}{2}}^{1} x \left(-\frac{1}{x \ln 10}\right) \, dx \]
\[ = \left[-1 \cdot \log_{10} 1 + \frac{1}{2} \log_{10} \frac{1}{2}\right] - \left[-\frac{1}{\ln 10} \int_{\frac{1}{2}}^{1} dx\right] \]
\[ = \left[0 + \frac{1}{2} \log_{10} \frac{1}{2}\right] + \frac{1}{\ln 10} \left[1 - \frac{1}{2}\right] \]
\[ = \frac{1}{2} \log_{10} \frac{1}{2} + \frac{1}{2 \ln 10} \]

#### Integral from \(1\) to \(2\):
\[ I_2 = \int_{1}^{2} \log_{10} x \, dx \]
Using integration by parts again:
- \( u = \log_{10} x \) ⇒ \( du = \frac{1}{x \ln 10} \, dx \)
- \( dv = dx \) ⇒ \( v = x \)

\[ I_2 = \left[x \log_{10} x\right]_{1}^{2} - \int_{1}^{2} x \left(\frac{1}{x \ln 10}\right) \, dx \]
\[ = \left[2 \log_{10} 2 - 1 \cdot \log_{10} 1\right] - \frac{1}{\ln 10} \left[2 - 1\right] \]
\[ = 2 \log_{10} 2 - 0 - \frac{1}{\ln 10} \]
\[ = 2 \log_{10} 2 - \frac{1}{\ln 10} \]

### Step 3: Combine the Results
Now we combine \( I_1 \) and \( I_2 \):
\[ I = I_1 + I_2 \]
\[ = \left(\frac{1}{2} \log_{10} \frac{1}{2} + \frac{1}{2 \ln 10}\right) + \left(2 \log_{10} 2 - \frac{1}{\ln 10}\right) \]
\[ = \frac{1}{2} \log_{10} \frac{1}{2} + 2 \log_{10} 2 - \frac{1}{2 \ln 10} \]

### Step 4: Simplify Further
Using the property of logarithms, \( \log_{10} \frac{1}{2} = -\log_{10} 2 \):
\[ = \frac{1}{2} (-\log_{10} 2) + 2 \log_{10} 2 - \frac{1}{2 \ln 10} \]
\[ = \left(-\frac{1}{2} + 2\right) \log_{10} 2 - \frac{1}{2 \ln 10} \]
\[ = \frac{3}{2} \log_{10} 2 - \frac{1}{2 \ln 10} \]

### Final Answer
Thus, the final result for the integral is:
\[ I = \frac{3}{2} \log_{10} 2 - \frac{1}{2 \ln 10} \]