设f(x)在(0,1)上具有二阶连续导数,若f(π)=2,∫ (0到π)[f(x)+f"(x)]sinxdx=5,求f(0)

∫(0→π)f''(x)sinxdx=∫(0→π)sinxd(f'(x))
=sinxf'(x)|(0→π)-∫(0→π)f'(x)cosxdx
=-∫(0→π)cosxd(f(x))
=-cosxf(x)|(0→π)-∫(0→π)f(x)sinxdx
所以左边=-cosxf(x)|(0→π)=f(π)+f(0)
所以f(0)=5-f(π)=3