If $$x$$ is the greatest integer $$\leq x$$, then $$\pi^2 \int_0^2 \sin\frac{\pi x}{2} x - x^{[x]} dx$$ is equal to:

The greatest integer function $$[x]$$ changes its value at integer points. We split the integral into two intervals: $$[0, 1)$$ and $$[1, 2)$$.

• For $$0 \le x < 1$$: $$[x] = 0$$.

  The integrand becomes $$0 \cdot x \sin\left(\frac{\pi x}{2}\right) = 0$$.

• For $$1 \le x < 2$$: $$[x] = 1$$.

  The integrand becomes $$1 \cdot x \sin\left(\frac{\pi x}{2}\right) = x \sin\left(\frac{\pi x}{2}\right)$$.

So, the integral simplifies to:

$$I = \pi^2 \left( \int_{0}^{1} 0 \, dx + \int_{1}^{2} x \sin\left(\frac{\pi x}{2}\right) dx \right) = \pi^2 \int_{1}^{2} x \sin\left(\frac{\pi x}{2}\right) dx$$

We use the formula $$\int u \, dv = uv - \int v \, du$$.

Let $$u = x \implies du = dx$$.

Let $$dv = \sin\left(\frac{\pi x}{2}\right) dx \implies v = -\frac{2}{\pi} \cos\left(\frac{\pi x}{2}\right)$$.

Applying the formula:

$$\int x \sin\left(\frac{\pi x}{2}\right) dx = \left[ -x \cdot \frac{2}{\pi} \cos\left(\frac{\pi x}{2}\right) \right] + \int \frac{2}{\pi} \cos\left(\frac{\pi x}{2}\right) dx$$

$$= -\frac{2x}{\pi} \cos\left(\frac{\pi x}{2}\right) + \frac{4}{\pi^2} \sin\left(\frac{\pi x}{2}\right)$$

Now, evaluate this from $$1$$ to $$2$$:

$$\int_{1}^{2} x \sin\left(\frac{\pi x}{2}\right) dx = \left[ -\frac{2x}{\pi} \cos\left(\frac{\pi x}{2}\right) + \frac{4}{\pi^2} \sin\left(\frac{\pi x}{2}\right) \right]_{1}^{2}$$

$$= \left( -\frac{4}{\pi} \cos(\pi) + \frac{4}{\pi^2} \sin(\pi) \right) - \left( -\frac{2}{\pi} \cos\left(\frac{\pi}{2}\right) + \frac{4}{\pi^2} \sin\left(\frac{\pi}{2}\right) \right)$$

$$= \left( -\frac{4}{\pi}(-1) + 0 \right) - \left( 0 + \frac{4}{\pi^2}(1) \right)$$

$$= \frac{4}{\pi} - \frac{4}{\pi^2}$$

Multiply the result by the $$\pi^2$$ coefficient outside the integral:

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

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

The value of the integral is $$4(\pi - 1)$$.

The correct option is (B).