Question

设fx在ab上连续,且单调递增,证明∮(a,b)xf(x)dx

Answer 1

记：g(x)=s[a,x]tf(t)dt-[(a+x)/2]s[a,x]f(t)dt,a<=t<=x,g'(x)=xf(x)-(1/2)[s[a,x]f(t)dt+f(x)(a+x)]=(1/2)[f(x)(x-a)-s[a,x]f(t)dt]=(1/2)s[a,x][f(x)-f(t)]dt>=0,（其中f(x)单增）可得g(x)在x>=a上单调不减，于是g(x)>=g(a)=0，取x=b则原命题得证。

Answer 2

简单计算一下即可，答案如图所示

Answer 3

设g(x)
=
∫

f(t)dt,
则g'(x)
=
f(x),
g"(x)
=
f'(x).
g(x)在[a,b]二阶连续可导,
且g(a)
=
0,
g'(a)
=
f(a)
=
0.
由带Lagrange余项的Taylor展开,
存在c∈(a,b)使
g(b)
=
g(a)+g'(a)(b-a)+g"(c)(b-a)²/2
=
f'(c)(b-a)²/2.
即有|
∫

f(t)dt|
=
|g(b)|
=
|f'(c)|·(b-a)²/2
≤
max{|f'(x)|}·(b-a)²/2.

Answer 4

记：g(x)=s[a,x]tf(t)dt-[(a+x)/2]s[a,x]f(t)dt,a<=t<=x,g'(x)=xf(x)-(1/2)[s[a,x]f(t)dt+f(x)(a+x)]=(1/2)[f(x)(x-a)-s[a,x]f(t)dt]=(1/2)s[a,x][f(x)-f(t)]dt>=0,（其中f(x)单增）可得g(x)在x>=a上单调不减，于是g(x)>=g(a)=0，取x=b则原命题得证。