To solve the integral \( \int_{1}^{3} |(2-x) \log_e x| \, dx \), we first need to analyze the expression inside the absolute value.
### Step 1: Determine the intervals for the absolute value
The expression \( 2 - x \) changes sign at \( x = 2 \):
- For \( x \in [1, 2] \), \( 2 - x \geq 0 \) so \( |2 - x| = 2 - x \).
- For \( x \in [2, 3] \), \( 2 - x < 0 \) so \( |2 - x| = -(2 - x) = x - 2 \).
### Step 2: Split the integral into two parts
We can write the integral as:
\[
\int_{1}^{3} |(2-x) \log_e x| \, dx = \int_{1}^{2} (2-x) \log_e x \, dx + \int_{2}^{3} (x-2) \log_e x \, dx
\]
### Step 3: Evaluate the first integral \( \int_{1}^{2} (2-x) \log_e x \, dx \)
Using integration by parts, let:
- \( u = \log_e x \) so \( du = \frac{1}{x} \, dx \)
- \( dv = (2-x) \, dx \) so \( v = 2x - \frac{x^2}{2} \)
Now, applying integration by parts:
\[
\int (2-x) \log_e x \, dx = (2x - \frac{x^2}{2}) \log_e x \bigg|_{1}^{2} - \int (2x - \frac{x^2}{2}) \cdot \frac{1}{x} \, dx
\]
Calculating the boundary term:
At \( x = 2 \):
\[
(2(2) - \frac{2^2}{2}) \log_e 2 = (4 - 2) \log_e 2 = 2 \log_e 2
\]
At \( x = 1 \):
\[
(2(1) - \frac{1^2}{2}) \log_e 1 = (2 - \frac{1}{2}) \cdot 0 = 0
\]
Thus, the boundary term contributes \( 2 \log_e 2 - 0 = 2 \log_e 2 \).
Now we need to evaluate the integral:
\[
\int (2 - x) \, dx = 2x - \frac{x^2}{2} \bigg|_{1}^{2} = (4 - 2) - (2 - \frac{1}{2}) = 2 - 1.5 = 0.5
\]
So we have:
\[
\int_{1}^{2} (2-x) \log_e x \, dx = 2 \log_e 2 - (0.5) = 2 \log_e 2 - 0.5
\]
### Step 4: Evaluate the second integral \( \int_{2}^{3} (x-2) \log_e x \, dx \)
Using integration by parts again:
- Let \( u = \log_e x \) so \( du = \frac{1}{x} \, dx \)
- Let \( dv = (x-2) \, dx \) so \( v = \frac{x^2}{2} - 2x \)
Applying integration by parts:
\[
\int (x-2) \log_e x \, dx = \left( \frac{x^2}{2} - 2x \right) \log_e x \bigg|_{2}^{3} - \int \left( \frac{x^2}{2} - 2x \right) \cdot \frac{1}{x} \, dx
\]
Calculating the boundary term:
At \( x = 3 \):
\[
\left( \frac{3^2}{2} - 2(3) \right) \log_e 3 = \left( \frac{9}{2} - 6 \right) \log_e 3 = \left( \frac{9}{2} - \frac{12}{2} \right) \log_e 3 = -\frac{3}{2} \log_e 3
\]
At \( x = 2 \):
\[
\left( \frac{2^2}{2} - 2(2) \right) \log_e 2 = (2 - 4) \log_e 2 = -2 \log_e 2
\]
Thus, the boundary term contributes:
\[
-\frac{3}{2} \log_e 3 + 2 \log_e 2
\]
Now we evaluate the integral:
\[
\int (x-2) \, dx = \frac{x^2}{2} - 2x \bigg|_{2}^{3} = \left( \frac{9}{2} - 6 \right) - \left( 2 - 4 \right) = -\frac{3}{2} - (-2) = -\frac{3}{2} + 2 = \frac{1}{2}
\]
So we have:
\[
\int_{2}^{3} (x-2) \log_e x \, dx = \left(-\frac{3}{2} \log_e 3 + 2 \log_e 2\right) - \frac{1}{2}
\]
### Step 5: Combine results
Now we combine both integrals:
\[
\int_{1}^{3} |(2-x) \log_e x| \, dx = \left(2 \log_e 2 - 0.5\right) + \left(-\frac{3}{2} \log_e 3 + 2 \log_e 2 - \frac{1}{2}\right)
\]
Combining like terms:
\[
= 4 \log_e 2 - \frac{3}{2} \log_e 3 - 1
\]
### Final Result
Thus, the value of the integral \( \int_{1}^{3} |(2-x) \log_e x| \, dx \) is:
\[
4 \log_e 2 - \frac{3}{2} \log_e 3 - 1
\]
