📝 题目
例 2 (1) 假设定积分定义中采用等分的方法,并且 ${\xi }_{k}$ 取中点, 试写出 $f\left( x\right)$ 的黎曼和.
(2)又设 $f\left( x\right)$ 为凹函数,求证: $\displaystyle{\int }_{a}^{b}f\left( x\right) \mathrm{d}x \geq \left( {b - a}\right) f\left( \frac{a + b}{2}\right)$ .
💡 答案与解析
解 (1) 将 $\left\lbrack {a,b}\right\rbrack n$ 等分,每一个小区间的长度为 $\frac{b - a}{n}$ ,第 $k$ 个小区间的中点坐标为 $a + \frac{\left( {{2k} - 1}\right) \left( {b - a}\right) }{2n}\left( {k = 1,2,\cdots ,n}\right)$ . 故
$$ {\int }_{a}^{b}f\left( x\right) \mathrm{d}x = \mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{b - a}{n}\mathop{\sum }\limits_{{k = 1}}^{n}f\left( {a + \frac{\left( {{2k} - 1}\right) \left( {b - a}\right) }{2n}}\right) . \tag{2.1} $$
(2)因为 $f\left( x\right)$ 为凹函数,所以
$$ f\left\lbrack {\frac{1}{n}\mathop{\sum }\limits_{{k = 1}}^{n}\left( {a + \frac{\left( {{2k} - 1}\right) \left( {b - a}\right) }{2n}}\right) }\right\rbrack $$
$$ \leq \frac{1}{n}\mathop{\sum }\limits_{{k = 1}}^{n}f\left( {a + \frac{\left( {{2k} - 1}\right) \left( {b - a}\right) }{2n}}\right) . $$
于是根据极限不等式,由 (2.1) 式推出
$$ {\int }_{a}^{b}f\left( x\right) \mathrm{d}x \geq \mathop{\lim }\limits_{{n \rightarrow \infty }}\left( {b - a}\right) f\left\lbrack {\frac{1}{n}\mathop{\sum }\limits_{{k = 1}}^{n}\left( {a + \frac{\left( {{2k} - 1}\right) \left( {b - a}\right) }{2n}}\right) }\right\rbrack $$
$$ = \left( {b - a}\right) f\left( {a + \frac{b - a}{2}}\right) = \left( {b - a}\right) f\left( \frac{a + b}{2}\right) . $$