📝 题目
例 10 设 $f\left( x\right)$ 在 $\left\lbrack {a,b}\right\rbrack$ 上连续且单调增加,求证:
$$ {\int }_{a}^{b}{xf}\left( x\right) \mathrm{d}x \geq \frac{a + b}{2}{\int }_{a}^{b}f\left( x\right) \mathrm{d}x. $$
💡 答案与解析
证法 1 因为 $f\left( x\right)$ 单调增加,所以
$$ \left( {x - \frac{a + b}{2}}\right) \left( {f\left( x\right) - f\left( \frac{a + b}{2}\right) }\right) \geq 0, $$
$$ {\int }_{a}^{b}\left( {x - \frac{a + b}{2}}\right) \left( {f\left( x\right) - f\left( \frac{a + b}{2}\right) }\right) \mathrm{d}x \geq 0. \tag{3.11} $$
又
$$ {\int }_{a}^{b}\left( {x - \frac{a + b}{2}}\right) f\left( \frac{a + b}{2}\right) \mathrm{d}x $$
$$ \frac{t = \frac{a + b}{2} - x}{2}f\left( \frac{a + b}{2}\right) {\int }_{\frac{b - a}{2}}^{-\frac{b - a}{2}}t\mathrm{\;d}t = 0. \tag{3. 12} $$
联立 (3.11) 与 (3.12) 式,解得 $\displaystyle{\int }_{a}^{b}\left( {x - \frac{a + b}{2}}\right) f\left( x\right) \mathrm{d}x \geq 0$ ,即本题得证.
证法 2 用积分第一中值定理.
$$ {\int }_{a}^{b}\left( {x - \frac{a + b}{2}}\right) f\left( x\right) \mathrm{d}x $$
$$ = {\int }_{0}^{\frac{a + b}{2}}\left( {x - \frac{a + b}{2}}\right) f\left( x\right) \mathrm{d}x + {\int }_{\frac{a + b}{2}}^{b}\left( {x - \frac{a + b}{2}}\right) f\left( x\right) \mathrm{d}x $$
$$ = f\left( {\xi }_{1}\right) {\int }_{a}^{\frac{a + b}{2}}\left( {x - \frac{a + b}{2}}\right) \mathrm{d}x + f\left( {\xi }_{2}\right) {\int }_{\frac{a + b}{2}}^{b}\left( {x - \frac{a + b}{2}}\right) \mathrm{d}x $$
$$ \left( {a \leq {\xi }_{1} \leq \frac{a + b}{2} \leq {\xi }_{2} \leq b}\right) $$
$$ = - f\left( {\xi }_{1}\right) \frac{{\left( b - a\right) }^{2}}{2} + f\left( {\xi }_{2}\right) \frac{{\left( b - a\right) }^{2}}{2} $$
$$ = \frac{{\left( b - a\right) }^{2}}{2}\left( {f\left( {\xi }_{2}\right) - f\left( {\xi }_{1}\right) }\right) \geq 0\;\left( {\text{ 因为 }f\left( x\right) \uparrow }\right) . $$
评注 积分第一中值定理的条件要求被积函数中至少有一个因子在积分区间上是不变号的, 本证法中, 对只有一个变号点的因子 $x - \frac{a + b}{2}$ ,通过分段使得在每一段区间中都是不变号的,以便应用积分第一中值定理. 这种“分段”处理问题的技巧是常用的.
证法 3 用积分第二中值定理. 因为 $f\left( x\right)$ 单调增加,所以 $\exists \xi \in$ $\left\lbrack {a,b}\right\rbrack$ ,使得
$$ {\int }_{a}^{b}\left( {x - \frac{a + b}{2}}\right) f\left( x\right) \mathrm{d}x $$
$$ = f\left( a\right) {\int }_{a}^{\xi }\left( {x - \frac{a + b}{2}}\right) \mathrm{d}x + f\left( b\right) {\int }_{\xi }^{b}\left( {x - \frac{a + b}{2}}\right) \mathrm{d}x $$
$$ = f\left( a\right) {\int }_{a}^{b}\left( {x - \frac{a + b}{2}}\right) \mathrm{d}x + \left( {f\left( b\right) - f\left( a\right) }\right) {\int }_{\xi }^{b}\left( {x - \frac{a + b}{2}}\right) \mathrm{d}x $$
$$ = 0 + \left( {f\left( b\right) - f\left( a\right) }\right) \left\lbrack {\frac{{b}^{2} - {\xi }^{2}}{2} - \frac{a + b}{2}\left( {b - \xi }\right) }\right\rbrack $$
$$ = \left( {f\left( b\right) - f\left( a\right) }\right) \frac{b - \xi }{2}\left( {\xi - a}\right) \geq 0. $$
证法 4 用变上限定积分. 令
$$ F\left( x\right) \overset{\text{ 定义 }}{ = }{\int }_{a}^{x}{tf}\left( t\right) \mathrm{d}t - \frac{a + x}{2}{\int }_{a}^{x}f\left( t\right) \mathrm{d}t. $$
则 $F\left( a\right) = 0$ ,且对 $\forall x \in (a,b\rbrack$ ,有
$$ {F}^{\prime }\left( x\right) = {xf}\left( x\right) - \frac{1}{2}{\int }_{a}^{x}f\left( t\right) \mathrm{d}t - \frac{a + x}{2}f\left( x\right) $$
$$ = \frac{x - a}{2}f\left( x\right) - \frac{1}{2}{\int }_{a}^{x}f\left( t\right) \mathrm{d}t $$
$$ = \frac{1}{2}{\int }_{a}^{x}\left\lbrack {f\left( x\right) - f\left( t\right) }\right\rbrack \mathrm{d}t\overset{\text{ 因为 }f\left( x\right) \uparrow }{ \geq }0. $$
由此可见 $F\left( x\right)$ 单调递增,从而 $F\left( b\right) \geq F\left( a\right) = 0$ ,即得
$$ {\int }_{a}^{b}{xf}\left( x\right) \mathrm{d}x - \frac{a + b}{2}{\int }_{a}^{b}f\left( x\right) \mathrm{d}x = F\left( b\right) \geq 0. $$
证法 5 因为 $f\left( x\right)$ 单调增加,所以对 $\forall t,x \in \left\lbrack {a,b}\right\rbrack$ ,有
$$ \left( {t - x}\right) \left( {f\left( t\right) - f\left( x\right) }\right) \geq 0. \tag{3.13} $$
在 (3.13) 式中固定住 $x$ ,对 $t$ 从 $a$ 到 $b$ 积分,得
$$ {\int }_{a}^{b}{tf}\left( t\right) \mathrm{d}t - x{\int }_{a}^{b}f\left( t\right) \mathrm{d}t + {xf}\left( x\right) \left( {b - a}\right) - f\left( x\right) \cdot \frac{{b}^{2} - {a}^{2}}{2} \geq 0. $$
再对 $x$ 从 $a$ 到 $b$ 积分,得
$$ \left( {b - a}\right) {\int }_{a}^{b}{tf}\left( t\right) \mathrm{d}t - \frac{{b}^{2} - {a}^{2}}{2}{\int }_{a}^{b}f\left( t\right) \mathrm{d}t + \left( {b - a}\right) {\int }_{a}^{b}{xf}\left( x\right) \mathrm{d}x $$
$$ - \frac{{b}^{2} - {a}^{2}}{2}{\int }_{a}^{b}f\left( t\right) \mathrm{d}t \geq 0. $$
改写积分变量 $t$ 为 $x$ ,化简即得 $\displaystyle{\int }_{a}^{b}{xf}\left( x\right) \mathrm{d}x \geq \frac{a + b}{2}{\int }_{a}^{b}f\left( x\right) \mathrm{d}x$ .
评注 本题的物理意义是: 如果曲线 $y = f\left( x\right)$ 单调增加,那么密度均匀的曲边梯形
$$ \{ \left( {x,y}\right) \mid a \leq x \leq b;0 \leq y \leq f\left( x\right) \} $$
的重心不可能落在直线 $x = \frac{a + b}{2}$ 的左边.