设 $f\in C^2[0,\pi]$, 且 $f(\pi)=2$, $\dps{\int_0^\pi [f(x)+f''(x)]\sin x\rd x=5}$. 求 $f(0)$.
解答: 由 $$\beex \bea 5&=\int_0^\pi [f(x)+f''(x)]\sin x\rd x\\ &=\int_0^\pi f(x)\sin x\rd x +\int_0^\pi \sin x\rd f'(x)\\ &=\int_0^\pi f(x)\sin x\rd x -\int_0^\pi \cos x\rd f'(x)\\ &=\int_0^\pi f(x)\sin x\rd x-\sez{ -f(\pi)-f(0)-\int_0^\pi (-\sin x)f(x)\rd x }\\ &=2+f(0) \eea \eeex$$ 知 $f(0)=3$.
