The integral $\int_{-1}^{\frac{3}{2}} \left| \pi^2 x \sin(\pi x) \right| \, dx$ is equal to:

Step 1: Analyze the integrand and absolute value

The integrand is $f(x) = \pi^2 x \sin(\pi x)$. We need to evaluate the integral of its absolute value. The absolute value function changes the sign of negative values to positive. Therefore, we need to determine the intervals where $f(x)$ is positive and where it is negative within the integration limits of $x = -1$ to $x = \frac{3}{2}$.

Step 2: Determine the sign of the integrand

We analyze the sign of $x \sin(\pi x)$ in different intervals. For $x \in [-1, 0]$, both $x$ and $\sin(\pi x)$ are non-positive, so their product is non-negative. For $x \in [0, 1]$, both $x$ and $\sin(\pi x)$ are non-negative, so their product is non-negative. For $x \in [1, \frac{3}{2}]$, $x$ is positive but $\sin(\pi x)$ is non-positive (since $\pi x$ is in $[\pi, \frac{3\pi}{2}]$), so their product is non-positive. Thus, $f(x) \ge 0$ for $x \in [-1, 1]$ and $f(x) \le 0$ for $x \in [1, \frac{3}{2}]$.

Step 3: Split the integral

Based on the sign analysis, we split the integral into two parts. For $x \in [-1, 1]$, $|f(x)| = f(x)$, and for $x \in [1, \frac{3}{2}]$, $|f(x)| = -f(x)$. We can factor out $\pi^2$ from the integrals.

Step 4: Apply integration by parts

We use integration by parts for $\int x \sin(\pi x) \, dx$. Let $u = x$ and $dv = \sin(\pi x) \, dx$. Then $du = dx$ and $v = -\frac{1}{\pi} \cos(\pi x)$. Applying the formula $\int u \, dv = uv - \int v \, du$, we get the indefinite integral.

Step 5: Evaluate the definite integrals

Now we substitute the limits of integration into the antiderivative. For the first integral from $-1$ to $1$: $\left( -\frac{1}{\pi} \cos(\pi) + \frac{1}{\pi^2} \sin(\pi) \right) - \left( -\frac{-1}{\pi} \cos(-\pi) + \frac{1}{\pi^2} \sin(-\pi) \right) = \left( \frac{1}{\pi} + 0 \right) - \left( \frac{1}{\pi} + 0 \right) = 0$. For the second integral from $1$ to $\frac{3}{2}$: $\left( -\frac{3/2}{\pi} \cos(\frac{3\pi}{2}) + \frac{1}{\pi^2} \sin(\frac{3\pi}{2}) \right) - \left( -\frac{1}{\pi} \cos(\pi) + \frac{1}{\pi^2} \sin(\pi) \right) = \left( 0 - \frac{1}{\pi^2} \right) - \left( \frac{1}{\pi} + 0 \right) = -\frac{1}{\pi^2} - \frac{1}{\pi}$. Combining these with the negative sign for the second integral and multiplying by $\pi^2$, we get $I = \pi^2 \left[ 0 - \left( -\frac{1}{\pi^2} - \frac{1}{\pi} \right) \right] = \pi^2 \left[ \frac{1}{\pi^2} + \frac{1}{\pi} \right] = 1 + \pi$. Wait, let's recheck the first integral. $\left[ -\frac{x}{\pi} \cos(\pi x) + \frac{1}{\pi^2} \sin(\pi x) \right] \Big|_{-1}^{1} = \left( -\frac{1}{\pi} \cos(\pi) + \frac{1}{\pi^2} \sin(\pi) \right) - \left( -\frac{-1}{\pi} \cos(-\pi) + \frac{1}{\pi^2} \sin(-\pi) \right) = \left( -\frac{1}{\pi}(-1) + 0 \right) - \left( \frac{1}{\pi}(-1) + 0 \right) = \frac{1}{\pi} - \left( -\frac{1}{\pi} \right) = \frac{2}{\pi}$. So the first part is $\pi^2 \left( \frac{2}{\pi} \right) = 2\pi$. The second part is $-\pi^2 \left[ \left( 0 - \frac{1}{\pi^2} \right) - \left( \frac{1}{\pi} + 0 \right) \right] = -\pi^2 \left[ -\frac{1}{\pi^2} - \frac{1}{\pi} \right] = 1 + \pi$. Adding them: $2\pi + 1 + \pi = 3\pi + 1$. Let's re-evaluate the first integral carefully. $I_1 = \int_{-1}^{1} \pi^2 x \sin(\pi x) \, dx$. Since $x \sin(\pi x)$ is an even function ($(-x)\sin(-\pi x) = (-x)(-\sin(\pi x)) = x\sin(\pi x)$), $I_1 = 2 \pi^2 \int_{0}^{1} x \sin(\pi x) \, dx$. $2 \pi^2 \left[ -\frac{x}{\pi} \cos(\pi x) + \frac{1}{\pi^2} \sin(\pi x) \right] \Big|_{0}^{1} = 2 \pi^2 \left[ \left( -\frac{1}{\pi} \cos(\pi) + \frac{1}{\pi^2} \sin(\pi) \right) - \left( 0 + 0 \right) \right] = 2 \pi^2 \left[ -\frac{1}{\pi}(-1) + 0 \right] = 2 \pi^2 \left[ \frac{1}{\pi} \right] = 2\pi$. This is correct. Now for the second integral: $I_2 = \int_{1}^{\frac{3}{2}} -\pi^2 x \sin(\pi x) \, dx = -\pi^2 \left[ -\frac{x}{\pi} \cos(\pi x) + \frac{1}{\pi^2} \sin(\pi x) \right] \Big|_{1}^{\frac{3}{2}}$. $I_2 = -\pi^2 \left[ \left( -\frac{3/2}{\pi} \cos(\frac{3\pi}{2}) + \frac{1}{\pi^2} \sin(\frac{3\pi}{2}) \right) - \left( -\frac{1}{\pi} \cos(\pi) + \frac{1}{\pi^2} \sin(\pi) \right) \right]$. $I_2 = -\pi^2 \left[ \left( 0 + \frac{1}{\pi^2}(-1) \right) - \left( -\frac{1}{\pi}(-1) + 0 \right) \right]$. $I_2 = -\pi^2 \left[ -\frac{1}{\pi^2} - \frac{1}{\pi} \right] = 1 + \pi$. Total integral $I = I_1 + I_2 = 2\pi + (1 + \pi) = 3\pi + 1$.

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