`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 \( \int_{\frac{1}{2}}^{2} |\log_{10} x| \, dx \), we will break it down into manageable steps.

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

The logarithm function \( \log_{10} x \) is negative for \( x \in (0, 1) \) and positive for \( x > 1 \). Therefore, we can break the integral into two parts based on the behavior of the logarithm:

\[ \int_{\frac{1}{2}}^{2} |\log_{10} x| \, dx = \int_{\frac{1}{2}}^{1} -\log_{10} x \, dx + \int_{1}^{2} \log_{10} x \, dx \]

### Step 2: Compute the first integral \( \int_{\frac{1}{2}}^{1} -\log_{10} x \, dx \)

To compute this integral, we use integration by parts. Let:
- \( u = -\log_{10} x \) (thus \( du = -\frac{1}{x \ln(10)} \, dx \))
- \( dv = dx \) (thus \( v = x \))

Using integration by parts:

\[ \int u \, dv = uv - \int v \, du \]

We have:

\[ \int -\log_{10} x \, dx = -x \log_{10} x + \int x \cdot \frac{1}{x \ln(10)} \, dx = -x \log_{10} x + \frac{x}{\ln(10)} \]

Now, we evaluate this from \( \frac{1}{2} \) to \( 1 \):

\[ \left[-x \log_{10} x + \frac{x}{\ln(10)}\right]_{\frac{1}{2}}^{1} \]

Calculating at the limits:

At \( x = 1 \):
\[ -1 \cdot \log_{10} 1 + \frac{1}{\ln(10)} = 0 + \frac{1}{\ln(10)} = \frac{1}{\ln(10)} \]

At \( x = \frac{1}{2} \):
\[ -\frac{1}{2} \log_{10} \frac{1}{2} + \frac{\frac{1}{2}}{\ln(10)} = -\frac{1}{2} \left(-\log_{10} 2\right) + \frac{1}{2 \ln(10)} = \frac{1}{2} \log_{10} 2 + \frac{1}{2 \ln(10)} \]

Thus, the first integral becomes:

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

Simplifying this gives:

\[ \frac{1}{\ln(10)} - \frac{1}{2 \ln(10)} - \frac{1}{2} \log_{10} 2 = \frac{1}{2 \ln(10)} - \frac{1}{2} \log_{10} 2 \]

### Step 3: Compute the second integral \( \int_{1}^{2} \log_{10} x \, dx \)

Using integration by parts again:

Let:
- \( u = \log_{10} x \) (thus \( du = \frac{1}{x \ln(10)} \, dx \))
- \( dv = dx \) (thus \( v = x \))

Then:

\[ \int \log_{10} x \, dx = x \log_{10} x - \int x \cdot \frac{1}{x \ln(10)} \, dx = x \log_{10} x - \frac{x}{\ln(10)} \]

Now evaluate from \( 1 \) to \( 2 \):

\[ \left[x \log_{10} x - \frac{x}{\ln(10)}\right]_{1}^{2} \]

Calculating at the limits:

At \( x = 2 \):
\[ 2 \log_{10} 2 - \frac{2}{\ln(10)} \]

At \( x = 1 \):
\[ 1 \cdot \log_{10} 1 - \frac{1}{\ln(10)} = 0 - \frac{1}{\ln(10)} = -\frac{1}{\ln(10)} \]

Thus, the second integral becomes:

\[ \left(2 \log_{10} 2 - \frac{2}{\ln(10)}\right) - \left(-\frac{1}{\ln(10)}\right) = 2 \log_{10} 2 - \frac{2}{\ln(10)} + \frac{1}{\ln(10)} = 2 \log_{10} 2 - \frac{1}{\ln(10)} \]

### Step 4: Combine both integrals

Now we combine the results from Step 2 and Step 3:

\[ \int_{\frac{1}{2}}^{2} |\log_{10} x| \, dx = \left(\frac{1}{2 \ln(10)} - \frac{1}{2} \log_{10} 2\right) + \left(2 \log_{10} 2 - \frac{1}{\ln(10)}\right) \]

Simplifying this expression:

Combine the terms:

\[ = \frac{1}{2 \ln(10)} - \frac{1}{2} \log_{10} 2 + 2 \log_{10} 2 - \frac{1}{\ln(10)} \]

Combine like terms:

\[ = \frac{1}{2 \ln(10)} - \frac{2}{2 \ln(10)} + \frac{3}{2} \log_{10} 2 = -\frac{1}{2 \ln(10)} + \frac{3}{2} \log_{10} 2 \]

### Final Answer

Thus, the final answer is:

\[ \int_{\frac{1}{2}}^{2} |\log_{10} x| \, dx = -\frac{1}{2 \ln(10)} + \frac{3}{2} \log_{10} 2 \]