Question 16

1. For $x \in [0, 2\pi]$, $x \ge 0$, so $|x \sin x| = x |\sin x|$.

2. Determine the sign of $\sin x$:

   • For $x \in [0, \pi]$: $\sin x \ge 0 \implies |\sin x| = \sin x$.
   • For $x \in [\pi, 2\pi]$: $\sin x \le 0 \implies |\sin x| = -\sin x$.

3. Split the integral:

   $$I = \int_0^\pi x \sin x\,dx - \int_\pi^{2\pi} x \sin x\,dx.$$

4. Indefinite integration by parts: $\int x \sin x\,dx = -x \cos x + \sin x$.

5. Evaluate each definite integral:

   • First interval $[0, \pi]$:
     $$\left[ -x \cos x + \sin x \right]_0^\pi = (-\pi(-1) + 0) - (0) = \pi.$$
   • Second interval $[\pi, 2\pi]$:
     $$\left[ -x \cos x + \sin x \right]_\pi^{2\pi} = (-2\pi(1) + 0) - (-\pi(-1) + 0) = -2\pi - \pi = -3\pi.$$

6. Combine the results:

   $$I = \pi - (-3\pi) = \pi + 3\pi = 4\pi.$$

Since $I = k\pi$, we have $k = 4$.