方企勤 第三章 一元函数积分学 第8题

证 由分部积分公式, 我们有

$$ {\int }_{a}^{b}f\left( x\right) \mathrm{d}x = {\int }_{a}^{b}f\left( x\right) \mathrm{d}\left( {x - a}\right) $$

$$ = {\left. f\left( x\right) \left( x - a\right) \right| }_{a}^{b} - {\int }_{a}^{b}\left( {x - a}\right) {f}^{\prime }\left( x\right) \mathrm{d}x, \tag{3.6} $$

$$ {\int }_{a}^{b}f\left( x\right) \mathrm{d}x = {\int }_{a}^{b}f\left( x\right) \mathrm{d}\left( {x - b}\right) $$

$$ = {\left. f\left( x\right) \left( x - b\right) \right| }_{a}^{b} - {\int }_{a}^{b}\left( {x - b}\right) {f}^{\prime }\left( x\right) \mathrm{d}x, \tag{3.7} $$

(3.6) 与 (3.7) 式相加除以 2, 得

$$ {\int }_{a}^{b}f\left( x\right) \mathrm{d}x = \frac{1}{2}\left( {b - a}\right) \left( {f\left( a\right) + f\left( b\right) }\right) $$

$$ - \frac{1}{2}{\int }_{a}^{b}\left( {x - a + x - b}\right) {f}^{\prime }\left( x\right) \mathrm{d}x $$

$$ = \frac{1}{2}\left( {b - a}\right) \left( {f\left( a\right) + f\left( b\right) }\right) $$

$$ - \frac{1}{2}{\int }_{a}^{b}{f}^{\prime }\left( x\right) \mathrm{d}\left( {x - a}\right) \left( {x - b}\right) $$

$$ = \frac{1}{2}\left( {b - a}\right) \left( {f\left( a\right) + f\left( b\right) }\right) $$

$$ + \frac{1}{2}{\int }_{a}^{b}{f}^{\prime \prime }\left( x\right) \left( {x - a}\right) \left( {x - b}\right) \mathrm{d}x. \tag{3.8} $$

评注 (3.5)式给出梯形公式误差的积分表示, 也就是

$$ {\int }_{a}^{b}f\left( x\right) \mathrm{d}x \approx \frac{1}{2}\left( {b - a}\right) \left( {f\left( a\right) + f\left( b\right) }\right) $$

的误差. 注意到 $\left( {x - a}\right) \left( {x - b}\right) < 0\left( {x \in \left( {a,b}\right) }\right)$ ,如果 ${f}^{\prime \prime }\left( x\right) > 0$ ,即 $f\left( x\right)$ 是凹函数,那么 (3.8) 式给出

$$ {\int }_{a}^{b}f\left( x\right) \mathrm{d}x < \frac{1}{2}\left( {b - a}\right) \left( {f\left( a\right) + f\left( b\right) }\right) . $$

从几何意义上看, 这正表明凹弧下的曲边梯形面积, 小于它所对的弦下的梯形面积.

\subsubsection{三、含定积分的不等式证明}