Evaluate: `int_0^(3//2) |x sin pi x| dx`.

Answer

To evaluate the integral \( I = \int_0^{\frac{3}{2}} |x \sin(\pi x)| \, dx \), we need to analyze the expression inside the absolute value to determine where it is positive and where it is negative.

### Step 1: Determine the intervals of positivity and negativity
1. The function \( x \) is positive for \( x \in [0, \frac{3}{2}] \).
2. The function \( \sin(\pi x) \) oscillates between -1 and 1. We need to find where \( \sin(\pi x) = 0 \) within the interval \( [0, \frac{3}{2}] \):
- \( \sin(\pi x) = 0 \) at \( x = 0, 1, 2 \).
- In the interval \( [0, \frac{3}{2}] \), \( \sin(\pi x) \) is positive for \( x \in [0, 1] \) and negative for \( x \in (1, \frac{3}{2}] \).

### Step 2: Split the integral
Given the analysis, we can split the integral at \( x = 1 \):
\[ I = \int_0^1 x \sin(\pi x) \, dx - \int_1^{\frac{3}{2}} x \sin(\pi x) \, dx \]

### Step 3: Evaluate the first integral \( \int_0^1 x \sin(\pi x) \, dx \)
We will use integration by parts:
Let \( u = x \) and \( dv = \sin(\pi x) \, dx \).
Then, \( du = dx \) and \( v = -\frac{1}{\pi} \cos(\pi x) \).

Using integration by parts:
\[ \int u \, dv = uv - \int v \, du \]
\[ \int_0^1 x \sin(\pi x) \, dx = \left[ -\frac{x}{\pi} \cos(\pi x) \right]_0^1 + \frac{1}{\pi} \int_0^1 \cos(\pi x) \, dx \]

Calculating the boundary term:
\[ \left[ -\frac{x}{\pi} \cos(\pi x) \right]_0^1 = -\frac{1}{\pi} \cos(\pi) - 0 = \frac{1}{\pi} \]

Now, we compute the remaining integral:
\[ \int_0^1 \cos(\pi x) \, dx = \left[ \frac{1}{\pi} \sin(\pi x) \right]_0^1 = \frac{1}{\pi}(\sin(\pi) - \sin(0)) = 0 \]

Thus,
\[ \int_0^1 x \sin(\pi x) \, dx = \frac{1}{\pi} + 0 = \frac{1}{\pi} \]

### Step 4: Evaluate the second integral \( \int_1^{\frac{3}{2}} x \sin(\pi x) \, dx \)
Again using integration by parts:
Let \( u = x \) and \( dv = \sin(\pi x) \, dx \).
Then, \( du = dx \) and \( v = -\frac{1}{\pi} \cos(\pi x) \).

Using integration by parts:
\[ \int_1^{\frac{3}{2}} x \sin(\pi x) \, dx = \left[ -\frac{x}{\pi} \cos(\pi x) \right]_1^{\frac{3}{2}} + \frac{1}{\pi} \int_1^{\frac{3}{2}} \cos(\pi x) \, dx \]

Calculating the boundary term:
\[ \left[ -\frac{x}{\pi} \cos(\pi x) \right]_1^{\frac{3}{2}} = -\frac{\frac{3}{2}}{\pi} \cos\left(\frac{3\pi}{2}\right) + \frac{1}{\pi} \cos(\pi) = -\frac{\frac{3}{2}}{\pi}(0) + \frac{1}{\pi}(-1) = -\frac{1}{\pi} \]

Now, we compute the remaining integral:
\[ \int_1^{\frac{3}{2}} \cos(\pi x) \, dx = \left[ \frac{1}{\pi} \sin(\pi x) \right]_1^{\frac{3}{2}} = \frac{1}{\pi}(\sin(\frac{3\pi}{2}) - \sin(\pi)) = \frac{1}{\pi}(-1 - 0) = -\frac{1}{\pi} \]

Thus,
\[ \int_1^{\frac{3}{2}} x \sin(\pi x) \, dx = -\frac{1}{\pi} - \frac{1}{\pi} = -\frac{2}{\pi} \]

### Step 5: Combine the results
Now we can combine our results:
\[ I = \frac{1}{\pi} - \left(-\frac{2}{\pi}\right) = \frac{1}{\pi} + \frac{2}{\pi} = \frac{3}{\pi} \]

### Final Answer
Thus, the value of the integral is:
\[ \boxed{\frac{3}{\pi}} \]