方企勤 第三章 一元函数积分学 第2题
📝 题目
例 2 (1) 设 $f\left( x\right)$ 为 $\left\lbrack {-1,1}\right\rbrack$ 上的三次多项式,求证:
$$ {\int }_{-1}^{1}f\left( x\right) \mathrm{d}x = \frac{1}{3}\{ f\left( {-1}\right) + {4f}\left( 0\right) + f\left( 1\right) \} . \tag{3.1} $$
(2)设 $f\left( x\right)$ 为 $\left\lbrack {a,b}\right\rbrack$ 上的三次多项式,求证:
$$ {\int }_{a}^{b}f\left( x\right) \mathrm{d}x = \frac{b - a}{6}\left\{ {f\left( a\right) + {4f}\left( \frac{a + b}{2}\right) + f\left( b\right) }\right\} . \tag{3.2} $$
💡 答案解析
证(1)设 $f\left( x\right) = \alpha {x}^{3} + \beta {x}^{2} + {\gamma x} + \delta$ ,则
$$ {\int }_{-1}^{1}f\left( x\right) \mathrm{d}x = {\int }_{-1}^{1}\left( {\beta {x}^{2} + \delta }\right) \mathrm{d}x = 2{\int }_{0}^{1}\left( {\beta {x}^{2} + \delta }\right) \mathrm{d}x $$
$$ = \frac{2\beta }{3} + {2\delta } = \frac{1}{3}\{ f\left( {-1}\right) + {4f}\left( 0\right) + f\left( 1\right) \} . $$
(2)令 $x = \frac{a + b}{2} + \frac{b - a}{2}t$ ,则
$$ {\int }_{a}^{b}f\left( x\right) \mathrm{d}x = {\int }_{-1}^{1}f\left( {\frac{a + b}{2} + \frac{b - a}{2}t}\right) \frac{b - a}{2}\mathrm{\;d}t $$
$$ \text{ 由第 (1) 小题 }\frac{b - a}{6}\left\{ {f\left( a\right) + {4f}\left( \frac{a + b}{2}\right) + f\left( b\right) }\right\} \text{ . } $$
\subsubsection{二、含定积分的等式证明}