If $\displaystyle f(x) = \int_0^x t \sin t\, dt$, then $f''(\pi) = ?$
- $-\pi$
- $\pi$
- $0$
- $1$
Show answer and solution
Correct answer: A. $-\pi$
By FTC, $f'(x) = x\sin x$.
**Differentiating:** $f''(x) = \sin x + x\cos x$.
At $x = \pi$: $f''(\pi) = \sin\pi + \pi\cos\pi = 0 + \pi(-1) = -\pi$.