1. We need to evaluate $I=\int_{-1}^{3/2} \left|\pi^2 x\sin(\pi x)\right|\,dx.$

Since $\pi^2>0$, this becomes $I=\pi^2\int_{-1}^{3/2} |x\sin(\pi x)|\,dx.$

2. Determine the sign of $x\sin(\pi x)$ on the interval $[-1,3/2]$.

Critical points are where either $x=0$ or $\sin(\pi x)=0$, i.e. at integer values. In the interval, these are $x=-1,\;0,\;1.$ So split into:

• $[-1,0]$
• $[0,1]$
• $[1,3/2]$

Now check signs:

• For $x\in(-1,0)$: $x<0$ and $\sin(\pi x)<0$, so $x\sin(\pi x)>0$.
• For $x\in(0,1)$: $x>0$ and $\sin(\pi x)>0$, so $x\sin(\pi x)>0$.
• For $x\in(1,3/2)$: $x>0$ and $\sin(\pi x)<0$, so $x\sin(\pi x)<0$.

Hence, $I=\pi^2\left(\int_{-1}^{0}x\sin(\pi x)\,dx+\int_0^1 x\sin(\pi x)\,dx-\int_1^{3/2}x\sin(\pi x)\,dx\right).$

3. First find the antiderivative of $x\sin(\pi x)$.

Using integration by parts: $\int x\sin(\pi x)\,dx.$ Let $u=x,\quad dv=\sin(\pi x)dx.$ Then $du=dx,\quad v=-\frac{\cos(\pi x)}{\pi}.$ So, $\int x\sin(\pi x)\,dx=-\frac{x\cos(\pi x)}{\pi}+\frac{1}{\pi}\int \cos(\pi x)\,dx$ $=-\frac{x\cos(\pi x)}{\pi}+\frac{\sin(\pi x)}{\pi^2}+C.$

Thus, $F(x)=-\frac{x\cos(\pi x)}{\pi}+\frac{\sin(\pi x)}{\pi^2}.$

4. Evaluate each part.

(i) On $[-1,0]$

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

(ii) On $[0,1]$

$\int_{0}^{1} x\sin(\pi x)\,dx=F(1)-F(0).$ Now, $F(1)=-\frac{1\cdot\cos\pi}{\pi}+\frac{\sin\pi}{\pi^2}=\frac{1}{\pi},$ so $\int_{0}^{1} x\sin(\pi x)\,dx=\frac{1}{\pi}.$

(iii) On $[1,3/2]$

$\int_{1}^{3/2} x\sin(\pi x)\,dx=F\left(\frac32\right)-F(1).$ Now, $F\left(\frac32\right)=-\frac{\frac32\cos\frac{3\pi}{2}}{\pi}+\frac{\sin\frac{3\pi}{2}}{\pi^2}=0-\frac{1}{\pi^2}=-\frac{1}{\pi^2},$ $F(1)=\frac{1}{\pi}.$ Hence, $\int_{1}^{3/2} x\sin(\pi x)\,dx=-\frac{1}{\pi^2}-\frac{1}{\pi}.$ Therefore, $-\int_{1}^{3/2} x\sin(\pi x)\,dx=\frac{1}{\pi^2}+\frac{1}{\pi}.$

5. Add all pieces: $I=\pi^2\left(\frac{1}{\pi}+\frac{1}{\pi}+\frac{1}{\pi^2}+\frac{1}{\pi}\right)$ $=\pi^2\left(\frac{3}{\pi}+\frac{1}{\pi^2}\right)$ $=3\pi+1.$

So, $I=1+3\pi.$

6. Compare with options:

• A: $2+3\pi$
• B: $4+\pi$
• C: $1+3\pi$ ✓
• D: $3+2\pi$

Therefore, the correct option is C.