利用二重积分证明积分不等式
利用二重积分证明积分不等式
例6.设f (x ), g (x ) 在[a , b ]上连续且单调增加,求证: (b -a )
⎰
b
a
g (x ) f (x ) dx ≥⎰f (x ) dx ⎰g (x ) dx
a
a
b
b
b
b b
分析:右端出现了两个积分,若将两个积分的积分变量换成不同符号则可化为二重积分:
⎰
b
a
f (x ) dx ⎰g (x ) dx =⎰f (y ) dy ⎰g (x ) dx =⎰
a
a
a
b
a
⎰
b
a
f (y ) g (x ) dxdy =⎰
b b a a
b
a
⎰
b
a
f (x ) g (y ) dxdy
而左边亦可化为二重积分:(b -a ) 这样就化为二重积分的比较了。 证:令 I =(b -a ) 则 I =
⎰
b
a
g (x ) f (x ) dx =⎰
b
b
⎰
f (x ) g (x ) dxdy =
⎰⎰
a
b b
a
f (y ) g (y ) dxdy
⎰
b
a
g (x ) f (x ) dx -⎰f (x ) dx ⎰g (x ) dx
a
a
b a
⎰⎰
a
b b
a
f (x ) g (x ) dxdy -⎰⎰
b
a
f (x ) g (y ) dxdy =⎰
b
a
⎰
b
a
f (x )[g (x ) -g (y )]dxdy
同样可得 I ==
⎰⎰
a
b b
a
f (y )[g (y ) -g (x )]dxdy
b a
两式相加得 2I =故 I =(b -a ) 结论得证。
⎰⎰[f (x ) -f (y )][g (x ) -g (y )]dxdy ≥0
a
b
b
a
a
b
⎰
b
a
g (x ) f (x ) dx -⎰f (x ) dx ⎰g (x ) dx ≥0
例1设函数f (x )为[0,1]上的单调减少且大于0的连续函数,
xf (x )dx ⎰f (x )dx ⎰≤求证: ⎰xf (x )dx ⎰f (x )dx
2
2
1
1
1
1
证明:令I =⎰xf (x )dx ⎰f 2(x )dx -⎰xf 2(x )dx ⎰f (x )dx
1111
=⎰xf (x )dx ⎰f
11
2
(y )dy -⎰0xf (x )dx ⎰0f (y )dy
2
11
=⎰
同理I=⎰2I=⎰
1100
1100
11
00
⎰xf (y )f (x )(f (y )-f (x ))dxdy
⎰
yf (x )f (y )(f (x )-f (y ))dxdy 两边相加整理得
⎰f (y )f (x )(x -y )[f (y )-f (x )]dxdy ,
f (x )>0且在[0,1]上单调减少, ∴(x -y )[f (y )-f (x )]≥0
∴I ≥0命题得证。
总结:当题设条件中告知被积函数减少或增加时,并没有指明是否可导,且积分区间相同时,将命题化为差式利用变量的对称式化为二重积分来进行证明。