若f(x)在[a,b]上连续,在(a,b)内可导,且f′(x)<0,令F(x)=1x?a∫xaf(t)dt,a<x<b,证明

2024-11-06 04:48:32
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回答1:

证明:∵F(x)=

1
x?a
f(t)dt,
F(x)=
f(x)(x?a)?
f(t)dt
(x?a)2

∵f′(x)<0,
∴f(x)在[a,b]上单调减少,
f(t)dt>f(x)(x?a),?x∈(a,b)

∴F′(x)<0,a<x<b,
∴F(x)在(a,b)上单调减少.