设f(x)在[a,b]连续,在(a,b)可导,f'(x)≤0,F(x)=[∫(a→x)f(t)dt]/(x-a),证明在(a,b)有F'(x)≤0

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设f(x)在[a,b]连续,在(a,b)可导,f'(x)≤0,F(x)=[∫(a→x)f(t)dt]/(x-a),证明在(a,b)有F'(x)≤0
设f(x)在[a,b]连续,在(a,b)可导,f'(x)≤0,F(x)=[∫(a→x)f(t)dt]/(x-a),证明在(a,b)有F'(x)≤0

设f(x)在[a,b]连续,在(a,b)可导,f'(x)≤0,F(x)=[∫(a→x)f(t)dt]/(x-a),证明在(a,b)有F'(x)≤0
设H(x)为f(x)的一个原函数
则∫(a->x)f(t)dt=H(x)-H(a)
[∫(a->x)f(t)dt]’=H’(x)=f(x)
欲证
F’(x)≤0 ⟺
{[∫(a->x)f(t)dt]/(x-a)]’ ≤0⟺
H’(x)(x-a)-∫(a->x)f(t)dt≤0⟺
H’(x)(x-a) ≤ H(x)-H(a) ⟺
H’(x)≤[ H(x)-H(a)]/(x-a) ⟺
H’(x) ≤ H’(w) (w∈[a,x]) ⟺
即f(x) ≤f(w) (w≤x)
此有f’(x) ≤0知函数单调递减易知

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