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F(x) = ∫(0->x) (x-3t) f(t) dt
(1)
let
u=-t
du =-dt
F(-x)
=∫(0->-x) (-x-3t) f(t) dt
=∫(0->x) (u-3t) f(-u) (-du)
=-∫(0->x) (u-3t) f(u) du
=-F(x)
=> F 奇函数
(2)
F(x)
=∫(0->x) (x-3t) f(t) dt
=x∫(0->x) f(t) dt -∫(0->x) 3tf(t) dt
F'(x)
=∫(0->x) f(t) dt + xf(x) -3xf(x)
=∫(0->x) f(t) dt - 2xf(x)
f(x) 连续
=> ∃ξ∈(0,x),st ∫(0->x) f(t) dt = xf(ξ)
F'(x)
=∫(0->x) f(t) dt - 2xf(x)
=xf(ξ) -2xf(x)
=x(f(ξ) - 2f(x) ) ( f(x) is increasting )
<0
=>
F(x) 是递减 [0, + ∞)
(1)
let
u=-t
du =-dt
F(-x)
=∫(0->-x) (-x-3t) f(t) dt
=∫(0->x) (u-3t) f(-u) (-du)
=-∫(0->x) (u-3t) f(u) du
=-F(x)
=> F 奇函数
(2)
F(x)
=∫(0->x) (x-3t) f(t) dt
=x∫(0->x) f(t) dt -∫(0->x) 3tf(t) dt
F'(x)
=∫(0->x) f(t) dt + xf(x) -3xf(x)
=∫(0->x) f(t) dt - 2xf(x)
f(x) 连续
=> ∃ξ∈(0,x),st ∫(0->x) f(t) dt = xf(ξ)
F'(x)
=∫(0->x) f(t) dt - 2xf(x)
=xf(ξ) -2xf(x)
=x(f(ξ) - 2f(x) ) ( f(x) is increasting )
<0
=>
F(x) 是递减 [0, + ∞)
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