Question

If \( [x] \) is the greatest integer \( \le x \), then \( \pi^2 \int_{0}^{2} \left( \sin \frac{\pi x}{2} \right) (x - [x])^{[x]} dx \) is equal to :

• A. \( 2(\pi + 1) \)
• B. \( 2(\pi - 1) \)
• C. \( 4(\pi + 1) \)
• D. \( 4(\pi - 1) \)

Hint:

Always split integrals involving Greatest Integer Function \( [x] \) or Fractional Part Function \( \{x\} \) at integer boundary points.
Remember \( (x-[x])^0 = 1 \) for all \( x \) except integers where it might be undefined, but for integration, we care about the interval interior.

Answer 1

The Correct Option is D

To solve the given integral, we need to evaluate \( \pi^2 \int_{0}^{2} \left( \sin \frac{\pi x}{2} \right) (x - [x])^{[x]} \, dx \). We will break this integral into segments based on the greatest integer function, also known as the floor function \([x]\).

1. The function \([x]\) takes specific integer values over specific intervals. Specifically:
   • \([x] = 0\) for \(0 \leq x < 1\) 
   • \([x] = 1\) for \(1 \leq x < 2\)
2. Hence, we split the integral into two parts: \[ \pi^2 \int_{0}^{2} \left( \sin \frac{\pi x}{2} \right) (x - [x])^{[x]} \, dx = \pi^2 \int_{0}^{1} \left( \sin \frac{\pi x}{2} \right) x^0 \, dx + \pi^2 \int_{1}^{2} \left( \sin \frac{\pi x}{2} \right) (x-1)^1 \, dx \]
3. Calculate the first integral: \[ \pi^2 \int_{0}^{1} \left( \sin \frac{\pi x}{2} \right) \cdot 1 \, dx = \pi^2 \int_{0}^{1} \sin \frac{\pi x}{2} \, dx \] This integral can be solved using substitution. Let \(u = \frac{\pi x}{2}\), then \(du = \frac{\pi}{2} dx\) or \(dx = \frac{2}{\pi} du\). Change limits of integration from \(x = 0\) to \(x = 1\), which correspond to \(u = 0\) to \(u = \frac{\pi}{2}\). \[ = \pi^2 \cdot \frac{2}{\pi} \int_{0}^{\frac{\pi}{2}} \sin u \, du = 2\pi \left[-\cos u \right]_{0}^{\frac{\pi}{2}} = 2\pi [0 - (-1)] = 2\pi \]
4. Calculate the second integral: \[ \pi^2 \int_{1}^{2} \left( \sin \frac{\pi x}{2} \right) (x-1) \, dx \] Apply integration by parts where \(u = x-1\) and \(dv = \sin \frac{\pi x}{2} \, dx\). Then \(du = dx\) and using substitution, the integral of \(dv\) is computed as follows:
   • Substitute: \(v = -\frac{2}{\pi}\cos\frac{\pi x}{2}\)
   • By parts: \(\int u \, dv = uv - \int v \, du\)
   • Thus: \[ = \pi^2 \left[ -(x-1)\frac{2}{\pi}\cos \frac{\pi x}{2} \bigg|_{1}^{2} + \frac{2}{\pi} \int_{1}^{2} \cos \frac{\pi x}{2} \, dx \right] \] Evaluating: \[-\pi^2\left[\frac{2}{\pi}(x-1)\cos \frac{\pi x}{2}\right]\bigg|_{1}^{2} = 0\] Remaining integral: \[ + \pi^2 \cdot \frac{2}{\pi} \cdot \frac{2}{\pi} \int_{1}^{2} \cos \frac{\pi x}{2} \, dx = \frac{4\pi}{\pi^2}[\sin u]\bigg|_{\frac{\pi}{2}}^{\pi} = \frac{4}{\pi} [0 - 1] = -\frac{4}{\pi} \]
5. The answer is thus \(4(\pi - 1)\).