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$\bf命题：$设连续函数$f,g：

[0,1]→[0,1]$，且$f(x)$单调递增，则

\int_0^1 {f\left( {g\left( x \right)} \right)dx} \le \int_0^1 {f\left( x

\right)dx} + \int_0^1 {g\left( x \right)dx} $$

证明：由积分中值定理知，存在$\xi \in \left[

{0,1} \right]$，使得

\[\int_0^1 {\left[ {f\left( {g\left( x \right)} \right) -

g\left( x \right)} \right]dx} = f\left( {g\left( \xi \right)} \right) -

g\left( \xi \right) = f\left( u \right) - u\]

其中$u = g\left( \xi \right)

\in \left[ {0,1} \right]$，而由$f\left( x \right)$的取值范围与单调性知

\[\int_0^1 {f\left(

x \right)dx} \ge \int_u^1 {f\left( x \right)dx} \ge f\left( u \right)\left( {1

- u} \right) \ge f\left( u \right) - u\]

从而命题成立