Question 57

We evaluate $I = \int_0^\pi x^2 \cos x \, dx$.

1. First Integration by Parts:
   Let $u = x^2 \implies du = 2x \, dx$, and $dv = \cos x \, dx \implies v = \sin x$.

   $$I = \left[ x^2 \sin x \right]_0^\pi - 2 \int_0^\pi x \sin x \, dx.$$

   Evaluating the boundary term:

   $$\left[ x^2 \sin x \right]_0^\pi = \pi^2 \sin(\pi) - 0 = 0.$$

   Thus:

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

2. Second Integration by Parts:
   For $\int_0^\pi x \sin x \, dx$, let $u = x \implies du = dx$, and $dv = \sin x \, dx \implies v = -\cos x$.

   $$\int_0^\pi x \sin x \, dx = \left[ -x \cos x \right]_0^\pi - \int_0^\pi (-\cos x) \, dx$$
   $$= \left[ -\pi \cos(\pi) - 0 \right] + \left[ \sin x \right]_0^\pi$$
   $$= -\pi(-1) + (\sin(\pi) - \sin(0)) = \pi + 0 = \pi.$$

3. Combine Results:

   $$I = -2(\pi) = -2\pi.$$

This matches option A.