1. Key Concepts and Formulas

• Absolute Value Function: $|f(x)|$ is equal to $f(x)$ when $f(x) \ge 0$ and $-f(x)$ when $f(x) < 0$. To evaluate integrals involving absolute values, we must determine the intervals where the expression inside the absolute value is positive or negative.
• Integration by Parts: The formula for integration by parts is $\int u \, dv = uv - \int v \, du$. This is used to integrate products of functions, particularly when one function becomes simpler upon differentiation and the other is easily integrable.
• Definite Integral Properties: The definite integral $\int_a^b f(x) \, dx$ represents the signed area under the curve $y = f(x)$ from $x=a$ to $x=b$. When dealing with absolute values, we split the integral into sub-intervals where the integrand's sign is constant.

2. Step-by-Step Solution

Step 1: Analyze the integrand and the interval of integration. We need to evaluate $\int\limits_{-1}^{\frac{3}{2}} \left(| \pi^2 x \sin(\pi x) \right|) dx$. The integrand is $|\pi^2 x \sin(\pi x)|$. Since $\pi^2$ is a positive constant, the sign of $x \sin(\pi x)$ determines the sign of the integrand. We need to find the intervals where $x \sin(\pi x) \ge 0$ and $x \sin(\pi x) < 0$ within $[-1, \frac{3}{2}]$.

Step 2: Determine the sign of $x \sin(\pi x)$ in the interval $[-1, \frac{3}{2}]$. The zeros of $\sin(\pi x)$ occur when $\pi x = n\pi$, which means $x = n$ for integer $n$. In our interval, these zeros are $x = -1, 0, 1$. The function $x$ changes sign at $x=0$.

Let's analyze the sign of $x \sin(\pi x)$ in the sub-intervals:

• Interval $[-1, 0]$:

  • For $x \in (-1, 0)$, $x$ is negative.
  • For $x \in (-1, 0)$, $\pi x \in (-\pi, 0)$. In this range, $\sin(\pi x)$ is negative.
  • Therefore, for $x \in (-1, 0)$, $x \sin(\pi x) = (\text{negative}) \times (\text{negative}) = \text{positive}$.
  • At $x=-1$ and $x=0$, $x \sin(\pi x) = 0$.
  • So, on $[-1, 0]$, $x \sin(\pi x) \ge 0$.

• Interval $[0, 1]$:

  • For $x \in (0, 1)$, $x$ is positive.
  • For $x \in (0, 1)$, $\pi x \in (0, \pi)$. In this range, $\sin(\pi x)$ is positive.
  • Therefore, for $x \in (0, 1)$, $x \sin(\pi x) = (\text{positive}) \times (\text{positive}) = \text{positive}$.
  • At $x=0$ and $x=1$, $x \sin(\pi x) = 0$.
  • So, on $[0, 1]$, $x \sin(\pi x) \ge 0$.

• Interval $[1, \frac{3}{2}]$:

  • For $x \in (1, \frac{3}{2})$, $x$ is positive.
  • For $x \in (1, \frac{3}{2})$, $\pi x \in (\pi, \frac{3\pi}{2})$. In this range, $\sin(\pi x)$ is negative.
  • Therefore, for $x \in (1, \frac{3}{2})$, $x \sin(\pi x) = (\text{positive}) \times (\text{negative}) = \text{negative}$.
  • At $x=1$, $x \sin(\pi x) = 0$.
  • So, on $[1, \frac{3}{2}]$, $x \sin(\pi x) \le 0$.

Step 3: Split the integral based on the sign analysis. From Step 2, we have:

• $|x \sin(\pi x)| = x \sin(\pi x)$ for $x \in [-1, 1]$.
• $|x \sin(\pi x)| = -x \sin(\pi x)$ for $x \in [1, \frac{3}{2}]$.

So, the integral becomes: $\int_{-1}^{\frac{3}{2}} \left(| \pi^2 x \sin(\pi x) \right|) dx = \pi^2 \left( \int_{-1}^{1} x \sin(\pi x) \, dx + \int_{1}^{\frac{3}{2}} (-x \sin(\pi x)) \, dx \right)$ $= \pi^2 \left( \int_{-1}^{1} x \sin(\pi x) \, dx - \int_{1}^{\frac{3}{2}} x \sin(\pi x) \, dx \right)$

Step 4: Evaluate the indefinite integral $\int x \sin(\pi x) \, dx$ using integration by parts. Let $u = x$ and $dv = \sin(\pi x) \, dx$. Then $du = dx$ and $v = \int \sin(\pi x) \, dx = -\frac{1}{\pi} \cos(\pi x)$.

Using the integration by parts formula: $\int x \sin(\pi x) \, dx = x \left(-\frac{1}{\pi} \cos(\pi x)\right) - \int \left(-\frac{1}{\pi} \cos(\pi x)\right) \, dx$ $= -\frac{x}{\pi} \cos(\pi x) + \frac{1}{\pi} \int \cos(\pi x) \, dx$ $= -\frac{x}{\pi} \cos(\pi x) + \frac{1}{\pi} \left(\frac{1}{\pi} \sin(\pi x)\right) + C$ $= -\frac{x}{\pi} \cos(\pi x) + \frac{1}{\pi^2} \sin(\pi x) + C$

Step 5: Evaluate the definite integrals.

First integral: $\int_{-1}^{1} x \sin(\pi x) \, dx$ Let $F(x) = -\frac{x}{\pi} \cos(\pi x) + \frac{1}{\pi^2} \sin(\pi x)$. $\int_{-1}^{1} x \sin(\pi x) \, dx = F(1) - F(-1)$ $F(1) = -\frac{1}{\pi} \cos(\pi) + \frac{1}{\pi^2} \sin(\pi) = -\frac{1}{\pi}(-1) + \frac{1}{\pi^2}(0) = \frac{1}{\pi}$ $F(-1) = -\frac{-1}{\pi} \cos(-\pi) + \frac{1}{\pi^2} \sin(-\pi) = \frac{1}{\pi}(-1) + \frac{1}{\pi^2}(0) = -\frac{1}{\pi}$ $\int_{-1}^{1} x \sin(\pi x) \, dx = \frac{1}{\pi} - \left(-\frac{1}{\pi}\right) = \frac{2}{\pi}$

Second integral: $\int_{1}^{\frac{3}{2}} x \sin(\pi x) \, dx$ $\int_{1}^{\frac{3}{2}} x \sin(\pi x) \, dx = F\left(\frac{3}{2}\right) - F(1)$ $F\left(\frac{3}{2}\right) = -\frac{\frac{3}{2}}{\pi} \cos\left(\frac{3\pi}{2}\right) + \frac{1}{\pi^2} \sin\left(\frac{3\pi}{2}\right) = -\frac{3}{2\pi}(0) + \frac{1}{\pi^2}(-1) = -\frac{1}{\pi^2}$ We already found $F(1) = \frac{1}{\pi}$. $\int_{1}^{\frac{3}{2}} x \sin(\pi x) \, dx = -\frac{1}{\pi^2} - \frac{1}{\pi}$

Step 6: Substitute the definite integral values back into the main expression. $\int_{-1}^{\frac{3}{2}} \left(| \pi^2 x \sin(\pi x) \right|) dx = \pi^2 \left( \int_{-1}^{1} x \sin(\pi x) \, dx - \int_{1}^{\frac{3}{2}} x \sin(\pi x) \, dx \right)$ $= \pi^2 \left( \frac{2}{\pi} - \left(-\frac{1}{\pi^2} - \frac{1}{\pi}\right) \right)$ $= \pi^2 \left( \frac{2}{\pi} + \frac{1}{\pi^2} + \frac{1}{\pi} \right)$ $= \pi^2 \left( \frac{3}{\pi} + \frac{1}{\pi^2} \right)$ $= \pi^2 \left(\frac{3\pi + 1}{\pi^2}\right)$ $= 3\pi + 1$

3. Common Mistakes & Tips

• Sign Errors with Absolute Value: Carefully determine the sign of the expression inside the absolute value over each sub-interval. A common mistake is to assume the sign is constant or to incorrectly identify the intervals.
• Integration by Parts Application: Ensure the correct choice of $u$ and $dv$. Differentiating $x$ simplifies it to a constant, which is generally a good strategy. Also, be meticulous with the signs and constants during integration by parts.
• Evaluating Trigonometric Functions: Accurately recall or calculate the values of trigonometric functions at the limits of integration, especially for angles like $\pi, -\pi, \frac{3\pi}{2}$.

4. Summary

The problem involved integrating a function with an absolute value. We first analyzed the sign of $x \sin(\pi x)$ over the given interval $[-1, \frac{3}{2}]$. We found that $x \sin(\pi x)$ is non-negative on $[-1, 1]$ and non-positive on $[1, \frac{3}{2}]$. This allowed us to split the integral into two parts, removing the absolute value by either keeping the original expression or negating it. We then used integration by parts to find the indefinite integral of $x \sin(\pi x)$. Finally, we evaluated the definite integrals over the respective sub-intervals and combined the results, multiplying by $\pi^2$ at the end, to arrive at the final answer.

5. Final Answer

The final answer is $\boxed{1 + 3\pi}$. This corresponds to option (C).