$\bf命题：$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7686,布布扣,bubuko.com
7686
原文：http://www.cnblogs.com/ly758241/p/3712685.html
