已知f(π)=1f(x)具有二阶连续导数，且∫上限π，下限0（f(x)+f''(x))sinxdx=3求f(0)

Question

已知f(π)=1 f(x)具有二阶连续导数，且∫上限π，下限0 （f(x)+f''(x))sinxdx=3 求f(0)

Answer 1

∫[f(x)+f''(x)]sinxdx
=∫f(x)sinxdx+∫f''(x)sinxdx
=-∫f(x)dcosx+∫sinxdf'(x)
=-f(x)cosx+∫f'(x)cosxdx+f'(x)sinx-∫f'(x)cosxdx
=-f(x)cosx+f(x)sinx
∫[0,π] (f(x)+f''(x))sinxdx
=-f(x)cosx+f(x)sinx|[0,π]
=f(π)
=1