📝 题目
例 20 设 $f\left( x\right) \geq 0$ 在 $\left\lbrack {0,1}\right\rbrack$ 上连续, $f\left( 1\right) = 0$ . 求证: 存在 $\xi \in$ (0,1),使得
$$ f\left( \xi \right) = {\int }_{0}^{\xi }f\left( x\right) \mathrm{d}x. \tag{3.33} $$
💡 答案与解析
证(1)如果 $f\left( x\right) \equiv 0$ ,则 (3.33) 式显然成立.
(2)如果 $f\left( x\right) ≢ 0$ ,则
$$ M = \mathop{\max }\limits_{{x \in \left\lbrack {0,1}\right\rbrack }}f\left( x\right) > 0,\;\text{ 且 }\;{\int }_{0}^{1}f\left( x\right) \mathrm{d}x > 0. $$
令 $F\left( x\right) = f\left( x\right) - {\int }_{0}^{x}f\left( t\right) \mathrm{d}t$ ,则
$$ F\left( 1\right) = f\left( 1\right) - {\int }_{0}^{1}f\left( t\right) \mathrm{d}t < 0. \tag{3.34} $$
设 ${x}_{M} \in \lbrack 0,1)$ ,使得 $f\left( {x}_{M}\right) = M$ ,则
$$ F\left( {x}_{M}\right) = M - {\int }_{0}^{{x}_{M}}f\left( t\right) \mathrm{d}t\left\{ \begin{array}{ll} = M > 0, & {x}_{M} = 0, \\ \geq \left( {1 - {x}_{M}}\right) M > 0, & {x}_{M} > 0. \end{array}\right. $$
(3.35)
联合 (3.34) 与 (3.35) 式,根据连续函数的中间值定理可知存在 $\xi \in$ $\left( {{x}_{M},1}\right)$ ,使得 ${F}^{\prime }\left( \xi \right) = 0$ ,即 (3.33) 式成立.