设f(x),g(x)在[a,b]上连续,在(a,b)内可导证明在(a,b)内存在一点ξ,使得f(a)*g(b)-g(a)*f(b)=(b-a)(f(a)*g'(ξ)-g(a)*f'(ξ))
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![设f(x),g(x)在[a,b]上连续,在(a,b)内可导证明在(a,b)内存在一点ξ,使得f(a)*g(b)-g(a)*f(b)=(b-a)(f(a)*g'(ξ)-g(a)*f'(ξ))](/uploads/image/z/7165075-67-5.jpg?t=%E8%AE%BEf%28x%29%2Cg%28x%29%E5%9C%A8%5Ba%2Cb%5D%E4%B8%8A%E8%BF%9E%E7%BB%AD%2C%E5%9C%A8%28a%2Cb%29%E5%86%85%E5%8F%AF%E5%AF%BC%E8%AF%81%E6%98%8E%E5%9C%A8%28a%2Cb%29%E5%86%85%E5%AD%98%E5%9C%A8%E4%B8%80%E7%82%B9%CE%BE%2C%E4%BD%BF%E5%BE%97f%28a%29%2Ag%28b%29-g%28a%29%2Af%28b%29%3D%28b-a%29%28f%28a%29%2Ag%27%28%CE%BE%29-g%28a%29%2Af%27%28%CE%BE%29%29)
设f(x),g(x)在[a,b]上连续,在(a,b)内可导证明在(a,b)内存在一点ξ,使得f(a)*g(b)-g(a)*f(b)=(b-a)(f(a)*g'(ξ)-g(a)*f'(ξ))
设f(x),g(x)在[a,b]上连续,在(a,b)内可导
证明在(a,b)内存在一点ξ,使得f(a)*g(b)-g(a)*f(b)=(b-a)(f(a)*g'(ξ)-g(a)*f'(ξ))
设f(x),g(x)在[a,b]上连续,在(a,b)内可导证明在(a,b)内存在一点ξ,使得f(a)*g(b)-g(a)*f(b)=(b-a)(f(a)*g'(ξ)-g(a)*f'(ξ))
构造辅助函数F(x)=(x-a) [f(a)*g(b)-g(a)*f(b)]/(b-a)-f(a)g(x)+g(a)f(x)
因为有F(a)=0,F(b)=0,所以存在F‘(ξ)=0,ξ∈(a,b)
[f(a)*g(b)-g(a)*f(b)]/(b-a)-f(a)*g'(ξ)+g(a)*f'(ξ)=0
即f(a)*g(b)-g(a)*f(b)=(b-a)(f(a)*g'(ξ)-g(a)*f'(ξ))
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