1. Interpret the integrand carefully

We need to evaluate

$$\pi^2\int_0^2 \left(\sin \frac{\pi x}{2}\right)(x-[x])^{[x]}\,dx.$$

Here $[x]$ is the greatest integer function.

Since the interval is $[0,2]$, we split according to the value of $[x]$:

• For $0\le x<1$, $[x]=0$
• For $1\le x<2$, $[x]=1$

(The single point $x=2$ does not affect the integral.)

So,

$$(x-[x])^{[x]}= \begin{cases} (x-0)^0=1, & 0\le x<1,\\[4pt] (x-1)^1=x-1, & 1\le x<2. \end{cases}$$

Thus

$$I=\pi^2\int_0^2 \left(\sin \frac{\pi x}{2}\right)(x-[x])^{[x]}dx =\pi^2\left[\int_0^1 \sin\frac{\pi x}{2}\,dx+\int_1^2 (x-1)\sin\frac{\pi x}{2}\,dx\right].$$

---

2. First integral

Let

$$I_1=\int_0^1 \sin\frac{\pi x}{2}\,dx.$$

Using

$$\int \sin(ax)\,dx=-\frac{\cos(ax)}{a},$$

with $a=\frac{\pi}{2}$,

$$I_1=\left[-\frac{2}{\pi}\cos\frac{\pi x}{2}\right]_0^1 =-\frac{2}{\pi}\left(\cos\frac{\pi}{2}-\cos 0\right) =-\frac{2}{\pi}(0-1)=\frac{2}{\pi}.$$

---

3. Second integral

Let

$$I_2=\int_1^2 (x-1)\sin\frac{\pi x}{2}\,dx.$$

Put

$$t=x-1 \quad \Rightarrow \quad x=t+1,$$

so when $x=1$, $t=0$, and when $x=2$, $t=1$. Then

$$I_2=\int_0^1 t\sin\frac{\pi(t+1)}{2}\,dt.$$

Now,

$$\sin\left(\frac{\pi t}{2}+\frac{\pi}{2}\right)=\cos\frac{\pi t}{2},$$

so

$$I_2=\int_0^1 t\cos\frac{\pi t}{2}\,dt.$$

Use integration by parts:

• $u=t \Rightarrow du=dt$
• $dv=\cos\frac{\pi t}{2}\,dt \Rightarrow v=\frac{2}{\pi}\sin\frac{\pi t}{2}$

Hence

$$I_2=\left[t\cdot \frac{2}{\pi}\sin\frac{\pi t}{2}\right]_0^1-\int_0^1 \frac{2}{\pi}\sin\frac{\pi t}{2}\,dt.$$

Evaluate the boundary term:

$$\left[t\cdot \frac{2}{\pi}\sin\frac{\pi t}{2}\right]_0^1 =\frac{2}{\pi}\sin\frac{\pi}{2}=\frac{2}{\pi}.$$

Now,

$$\int_0^1 \sin\frac{\pi t}{2}\,dt=\frac{2}{\pi}$$

(from the same calculation as $I_1$), so

$$I_2=\frac{2}{\pi}-\frac{2}{\pi}\cdot \frac{2}{\pi} =\frac{2}{\pi}-\frac{4}{\pi^2}.$$

---

4. Add the two parts

Therefore,

$$I=\pi^2\left(I_1+I_2\right) =\pi^2\left(\frac{2}{\pi}+\frac{2}{\pi}-\frac{4}{\pi^2}\right) =\pi^2\left(\frac{4}{\pi}-\frac{4}{\pi^2}\right).$$

Simplify:

$$I=4\pi-4=4(\pi-1).$$

---

5. Compare with options

The value is

$$\boxed{4(\pi-1)}.$$

So the correct option is B.

---

6. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

So they agree.