设f"(x)连续,f(π)=2,且有∫0-π [f(x)+f"(x)]sinxdx=5,求f(0)

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设f"(x)连续,f(π)=2,且有∫0-π [f(x)+f"(x)]sinxdx=5,求f(0)
f(pa)=2 定积分范围在0到pa

∫[0,π] f(x) sinx dx = ∫[0,π] f(x) d(-cosx),将sinx积分
= -f(x) cosx_[0,π] + ∫[0,π] f'(x) cosx dx,分部积分法
= f(0) cos(0) - f(π) cos(π) + ∫[0,π] f'(x) d(sinx),将cosx积分
= f(0) - (2)(-1) + f'(x) sinx_[0,π] - ∫[0,π] f''(x) sinx dx,分部积分法
∫[0,π] f(x) sinx dx + ∫[0,π] f''(x) sinx dx = f(0) + 2
∫[0,π] [f(x) + f''(x)] sinx dx = f(0) + 2
5 = f(0) + 2
f(0) = 3

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