方企勤 第三章 一元函数积分学 第19题
📝 题目
例 19 设 $f\left( x\right)$ 在 $\left\lbrack {0,1}\right\rbrack$ 上可微,且 $0 < {f}^{\prime }\left( x\right) < 1(\forall x \in$
$\left( {0,1}\right) ),f\left( 0\right) = 0$ . 求证:
$$ {\left( {\int }_{0}^{1}f\left( x\right) \mathrm{d}x\right) }^{2} > {\int }_{0}^{1}{f}^{3}\left( x\right) \mathrm{d}x. \tag{3.29} $$
💡 答案解析
证法 1 问题就是要证明
$$ \frac{{\left( {\int }_{0}^{1}f\left( x\right) \mathrm{d}x\right) }^{2}}{{\int }_{0}^{1}{f}^{3}\left( x\right) \mathrm{d}x} > 1. \tag{3.30} $$
作辅助函数
$$ F\left( x\right) = {\left( {\int }_{0}^{x}f\left( t\right) \mathrm{d}t\right) }^{2},\;G\left( x\right) = {\int }_{0}^{x}{f}^{3}\left( t\right) \mathrm{d}t, $$
则根据柯西中值定理可知 (3.30) 式左端
$$ \frac{{\left( {\int }_{0}^{1}f\left( x\right) \mathrm{d}x\right) }^{2}}{{\int }_{0}^{1}{f}^{3}\left( x\right) \mathrm{d}x} = \frac{F\left( 1\right) - F\left( 0\right) }{G\left( 1\right) - G\left( 0\right) } = \frac{{F}^{\prime }\left( \xi \right) }{{G}^{\prime }\left( \xi \right) } $$
$$ = \frac{{2f}\left( \xi \right) {\int }_{0}^{\xi }f\left( t\right) \mathrm{d}t}{{f}^{3}\left( \xi \right) } = \frac{2{\int }_{0}^{\xi }f\left( t\right) \mathrm{d}t}{{f}^{2}\left( \xi \right) }\left( {0 < \xi < 1}\right) $$
$$ = \frac{2{\int }_{0}^{\xi }f\left( t\right) \mathrm{d}t - 2{\int }_{0}^{0}f\left( t\right) \mathrm{d}t}{{f}^{2}\left( \xi \right) - {f}^{2}\left( 0\right) } = \frac{{2f}\left( \eta \right) }{{2f}\left( \eta \right) {f}^{\prime }\left( \eta \right) } $$
$$ = \frac{1}{{f}^{\prime }\left( \eta \right) } > 1\left( {0 < \eta < \xi < 1}\right) . $$
证法 2 问题就是要证明
$$ {\left( {\int }_{0}^{1}f\left( x\right) \mathrm{d}x\right) }^{2} - {\int }_{0}^{1}{f}^{3}\left( x\right) \mathrm{d}x > 0. \tag{3.31} $$
作辅助函数
$$ F\left( x\right) = {\left( {\int }_{0}^{x}f\left( t\right) \mathrm{d}t\right) }^{2} - {\int }_{0}^{x}{f}^{3}\left( t\right) \mathrm{d}t, $$
因为 $F\left( 0\right) = 0$ ,所以只要证明 ${F}^{\prime }\left( x\right) > 0\left( {\forall x \in \left( {0,1}\right) }\right)$ . 事实上,
$$ {F}^{\prime }\left( x\right) = {2f}\left( x\right) {\int }_{0}^{x}f\left( t\right) \mathrm{d}t - {f}^{3}\left( x\right) $$
$$ = f\left( x\right) \left\lbrack {2{\int }_{0}^{x}f\left( t\right) \mathrm{d}t - {f}^{2}\left( x\right) }\right\rbrack . \tag{3.32} $$
因为 $f\left( 0\right) = 0,0 < {f}^{\prime }\left( x\right) < 1\left( {\forall x \in \left( {0,1}\right) }\right)$ ,所以 $f\left( x\right) > 0(\forall x \in$ $\left( {0,1}\right) )$ .
由此可见, 只要再证明 (3.32) 式中方括弧内的函数大于零, 即有 ${F}^{\prime }\left( x\right) > 0$ . 事实上,记
$$ g\left( x\right) = 2{\int }_{0}^{x}f\left( t\right) \mathrm{d}t - {f}^{2}\left( x\right) , $$
因为 $g\left( 0\right) = 0$ ,所以
$$ {g}^{\prime }\left( x\right) = {2f}\left( x\right) - {2f}\left( x\right) \cdot {f}^{\prime }\left( x\right) = {2f}\left( x\right) \left\lbrack {1 - {f}^{\prime }\left( x\right) }\right\rbrack > 0 $$
$$ \left( {\forall x \in \left( {0,1}\right) }\right) \text{ , } $$
即得 $g\left( x\right) > 0$ ,也就是 (3.32) 式中方括弧内的函数大于零.
\subsubsection{四、含定积分的中值命题}