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