Integrating $|x \sin(\pi x)|$ over $-1 \leq x \leq 3/2$ $$\int_{-1}^{3/2} |x \sin(\pi x)| \, dx.$$
The second part of the sum I can do.
The problem is to set the limits before spliting.
I mean before spliting the integral into two integrals, how to set the limits?

You can make your life easier if you make the substitution
$$
\pi x=t
$$
so the integral becomes
$$
\frac{1}{\pi^2}\int_{-\pi}^{3\pi/2}|t\sin t|\,dt
$$
and take into account that between $-\pi$ and $0$ the sine is negative, as well as between $\pi$ and $3\pi/2$, so you can write it as
$$
\int_{-\pi}^0 t\sin t\,dt
+
\int_{0}^\pi t\sin t\,dt
-
\int_{\pi}^{3\pi/2} t\sin t\,dt
=
\int_{-\pi}^\pi t\sin t\,dt
-
\int_{\pi}^{3\pi/2} t\sin t\,dt
$$
dividing the result by $\pi^2$.
The antiderivative
$$
\int t\sin t\,dt
$$
can be computed by parts.
