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

Write the integrand in a simpler form:
$$\left|\pi^{2}x\sin(\pi x)\right| = \pi^{2}\,|x\sin(\pi x)|$$
Hence

$$I=\int_{-1}^{3/2}\left|\pi^{2}x\sin(\pi x)\right|dx=\pi^{2}\int_{-1}^{3/2}|x\sin(\pi x)|dx$$

First locate the points where $$x\sin(\pi x)=0$$ inside $$[-1,\,3/2]$$.
Zeros occur when $$x=0$$ or $$\sin(\pi x)=0\;(\Rightarrow x=\text{integer})$$.
Thus the critical points are $$x=-1,\,0,\,1$$.

Check the sign of $$f(x)=x\sin(\pi x)$$ on each sub-interval:

• $$(-1,0):$$ choose $$x=-\tfrac12$$ ⇒ $$f(x)=(-\tfrac12)\sin(-\tfrac{\pi}{2})=( -\tfrac12)(-1)\gt0$$
• $$(0,1):$$ choose $$x=\tfrac12$$ ⇒ $$f(x)=(\tfrac12)\sin(\tfrac{\pi}{2})=(\tfrac12)(1)\gt0$$
• $$(1,\,3/2):$$ choose $$x=\tfrac32$$ ⇒ $$f(x)=(\tfrac32)\sin(\tfrac{3\pi}{2})=(\tfrac32)(-1)\lt0$$

Thus
Case 1: $$x\in[-1,1]\; \Rightarrow\; |f(x)|=f(x)$$
Case 2: $$x\in[1,3/2]\; \Rightarrow\; |f(x)|=-f(x)$$

Split the integral accordingly:

$$I=\pi^{2}\left[\int_{-1}^{1}x\sin(\pi x)\,dx-\int_{1}^{3/2}x\sin(\pi x)\,dx\right]$$

Find an antiderivative of $$x\sin(\pi x)$$ using integration by parts.
Let $$u=x,\;dv=\sin(\pi x)dx$$.
Then $$du=dx,\;v=-\dfrac{\cos(\pi x)}{\pi}$$.

$$\int x\sin(\pi x)dx=-\dfrac{x\cos(\pi x)}{\pi}+\dfrac{\sin(\pi x)}{\pi^{2}}$$ $$-(1)$$

Integral over $$[-1,1]$$
Using $$(1):$$
$$F(x)=-\dfrac{x\cos(\pi x)}{\pi}+\dfrac{\sin(\pi x)}{\pi^{2}}$$

$$\begin{aligned} \int_{-1}^{1}x\sin(\pi x)dx &=F(1)-F(-1)\\ &=\left[-\dfrac{1\cdot(-1)}{\pi}+0\right]-\left[-\dfrac{(-1)\cdot(-1)}{\pi}+0\right]\\ &=\dfrac{1}{\pi}-\left(-\dfrac{1}{\pi}\right)=\dfrac{2}{\pi} \end{aligned}$$

Integral over $$[1,3/2]$$

$$\begin{aligned} \int_{1}^{3/2}x\sin(\pi x)dx &=F\!\left(\tfrac32\right)-F(1)\\ &=\left[0-\dfrac{1}{\pi^{2}}\right]-\dfrac{1}{\pi}\\ &=-\left(\dfrac{1}{\pi}+\dfrac{1}{\pi^{2}}\right) \end{aligned}$$

Now compute $$I$$:

$$\begin{aligned} I&=\pi^{2}\left[\dfrac{2}{\pi}-\Bigl(-\dfrac{1}{\pi}-\dfrac{1}{\pi^{2}}\Bigr)\right]\\[6pt] &=\pi^{2}\left[\dfrac{2}{\pi}+\dfrac{1}{\pi}+\dfrac{1}{\pi^{2}}\right]\\[6pt] &=\pi^{2}\left[\dfrac{3}{\pi}+\dfrac{1}{\pi^{2}}\right]\\[6pt] &=3\pi+1 \end{aligned}$$

Therefore, $$\displaystyle\int_{-1}^{3/2}\left|\pi^{2}x\sin(\pi x)\right|dx = 1 + 3\pi$$.

Option C is correct.