📝 题目
例 16 设 $f\left( x\right) \geq 0$ 在 $\left\lbrack {0,1}\right\rbrack$ 上连续,且单调下降, $0 < \alpha < \beta < 1$ . 求证:
$$ {\int }_{0}^{\alpha }f\left( x\right) \mathrm{d}x \geq \frac{\alpha }{\beta }{\int }_{\alpha }^{\beta }f\left( x\right) \mathrm{d}x. \tag{3.26} $$
💡 答案与解析
解 由积分中值定理,存在 $\xi \in \left\lbrack {0,\alpha }\right\rbrack$ ,使得
$$ \frac{1}{\alpha }{\int }_{0}^{\alpha }f\left( x\right) \mathrm{d}x = f\left( \xi \right) , $$
存在 $\eta \in \left\lbrack {\alpha ,\beta }\right\rbrack$ ,使得
$$ \frac{1}{\beta - \alpha }{\int }_{\alpha }^{\beta }f\left( x\right) \mathrm{d}x = f\left( \eta \right) . $$
注意到 $f\left( x\right)$ 单调下降,因此 $\xi \leq \eta \Rightarrow f\left( \xi \right) \geq f\left( \eta \right)$ ,即
$$ \frac{1}{\alpha }{\int }_{0}^{\alpha }f\left( x\right) \mathrm{d}x \geq \frac{1}{\beta - \alpha }{\int }_{\alpha }^{\beta }f\left( x\right) \mathrm{d}x. $$
又因为 $\alpha > 0$ ,所以
$$ {\int }_{0}^{a}f\left( x\right) \mathrm{d}x \geq \frac{\alpha }{\beta - \alpha }{\int }_{a}^{\beta }f\left( x\right) \mathrm{d}x \geq \frac{\alpha }{\beta }{\int }_{a}^{\beta }f\left( x\right) \mathrm{d}x. $$