设在[0,1]上连续,在(0,1)内可导且∫0到1f(x)dx=∫0到1xf(x)dx=0,证明:存在ξ∈(0,1)使得f(ξ)=0
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设在[0,1]上连续,在(0,1)内可导且∫0到1f(x)dx=∫0到1xf(x)dx=0,证明:存在ξ∈(0,1)使得f(ξ)=0
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证:设g(x) = ∫(0到x) (1-x) f(x) dx∫0到1f(x)dx=∫0到1xf(x)dx=0 ,∫(0到1) (1-x)f(x)dx =0 即 g(1) =0又g(0) =0g(x) 在[0,1]上连续,在(0,1)内可导,满足罗尔定理条件存在ξ∈(0,1)使得 g'(ξ) =0即 g'(ξ...
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