📝 题目
例 21 设 $f\left( x\right)$ 在 $\left\lbrack {a,b}\right\rbrack$ 上连续、不恒等于常数,且
$$ f\left( a\right) = \mathop{\min }\limits_{{a \leq t \leq b}}f\left( t\right) = f\left( b\right) . $$
求证: $\exists \xi \in \left( {a,b}\right)$ ,使得
$$ {\int }_{a}^{\xi }f\left( x\right) \mathrm{d}x = \left( {\xi - a}\right) f\left( \xi \right) . $$
💡 答案与解析
证 对 $\forall t \in \left( {a,b}\right)$ ,令
$$ F\left( t\right) = \left( {t - a}\right) f\left( t\right) - {\int }_{a}^{t}f\left( x\right) \mathrm{d}x. $$
为了证明本题,只要证明 $\exists \xi \in \left( {a,b}\right)$ ,使得 $F\left( \xi \right) = 0$ . 因为 $f\left( x\right)$ 在 $\left\lbrack {a,b}\right\rbrack$ 上连续、不恒等于常数,且 $f\left( a\right) = \mathop{\min }\limits_{{a \leq t \leq b}}f\left( t\right) = f\left( b\right)$ ,所以 $\exists {t}_{0} \in$ (a, b),使得
$$ f\left( {t}_{0}\right) = \mathop{\max }\limits_{{a \leq t \leq b}}f\left( t\right) . $$
于是, 我们有
$$ F\left( {t}_{0}\right) = \left( {{t}_{0} - a}\right) f\left( {t}_{0}\right) - {\int }_{a}^{{t}_{0}}f\left( x\right) \mathrm{d}x $$
$$ > \left( {{t}_{0} - a}\right) f\left( {t}_{0}\right) - {\int }_{a}^{{t}_{0}}f\left( {t}_{0}\right) \mathrm{d}x = 0; $$
$$ F\left( b\right) = \left( {b - a}\right) f\left( b\right) - {\int }_{a}^{b}f\left( x\right) \mathrm{d}x $$
$$ < \left( {b - a}\right) f\left( b\right) - {\int }_{a}^{b}f\left( b\right) \mathrm{d}x = 0, $$
从而 $F\left( b\right) < 0 < F\left( {t}_{0}\right)$ . 根据连续函数的中间值定理,有
$$ \exists \xi \in \left( {{t}_{0},b}\right) \subset \left( {a,b}\right) \;\text{ 使得 }\;F\left( \xi \right) = 0, $$
即得 $\displaystyle{\int }_{a}^{\xi }f\left( x\right) \mathrm{d}x = \left( {\xi - a}\right) f\left( \xi \right)$ .