设 $f\in C^2[0,1]$, $$\bex f(0)=-1,\quad f'(1)=3,\quad \int_0^1 xf''(x)\rd x=1. \eex$$ 试求 $f(1)$.
解答: $$\beex \bea 1&=\int_0^1 x\rd f'(x)\\ &=xf'(x)|_0^1-\int_0^1 f'(x)\rd x\\ &=f'(1)-[f(1)-f(0)]\\ &=f'(1)+f(0)-f(1)\\ &=3-1-f(1) \eea \eeex$$ 知 $f(1)=1$.
