问题及解答

设 $f(x)\in C^2$, 且满足

\[\int_0^{\pi}(f(x)+f''(x))\sin xdx=3,\]

求 $f(0)$.

\[
\begin{split}
\int_0^{\pi}f''(x)\sin xdx&=\int_0^{\pi}\sin xdf'(x)\\
&=f'(x)\sin x\biggr|_{0}^{\pi}-\int_0^{\pi}f'(x)d\sin x\\
&=0-\int_0^{\pi}f'(x)\cos xdx\\
&=-\int_0^{\pi}\cos x df(x)\\
&=-\cos x f(x)\biggr|_{0}^{\pi}+\int_0^{\pi}f(x)d\cos x\\
&=-[\cos\pi\cdot f(\pi)-\cos 0\cdot f(0)]-\int_0^{\pi}f(x)\sin xdx\\
&=f(0)+f(\pi)-\int_0^{\pi}f(x)\sin xdx\\
\end{split}
\]

所以

\[
3=\int_0^{\pi}(f(x)+f''(x))\sin xdx=f(0)+f(\pi)=f(0)+1,
\]

故 $f(0)=2$.