关于积分中值定理的专题讨论

$\bf命题:$设$f\left( x \right) \in {C^1}\left[ {0,1} \right]$,则存在$\xi \in (0,1)$,使得$\int_0^1 {f\left( x \right)dx}  = f\left( 0 \right) + \frac{1}{2}f'\left( \xi  \right)$

1

$\bf命题:$设$f\left( x \right) \in C\left[ {0,\pi } \right],\int_0^\pi  {f\left( x \right)dx}  = 0,\int_0^\pi  {f\left( x \right)\cos xdx}  = 0$,则存在${\xi _1},{\xi _2} \in \left( {0,\pi } \right)$,使得$f\left( {{\xi _1}} \right) = f\left( {{\xi _2}} \right) = 0$

1

$\bf命题:$设$f\left( x \right) \in C\left[ {0,1} \right],f\left( x \right) > 0$,则对于$n \in {N_ + }$,存在${\xi _n}$,使得\[\frac{1}{n}\int_0^1 {f\left( x \right)dx}  = \int_0^{{\xi _n}} {f\left( x \right)dx}  + \int_{1 - {\xi _n}}^1 {f\left( x \right)dx} \]\[\mathop {\lim }\limits_{n \to \infty } n{\xi _n} = \int_0^1 {f\left( x \right)dx} /[f\left( 0 \right) + f\left( 1 \right)]\]

1

$\bf命题:$设$f\left( x \right) \in {C^1}\left[ {0,1} \right],f'\left( 0 \right) \ne 0$,则对$\int_0^x {f\left( t \right)dt}  = f\left( {\xi \left( x \right)} \right)x,0 < x < 1$中的$\xi(x)$,有$\xi \left( x \right)/x \to 1/2\left( {x \to {0^ + }} \right)$

1

$\bf命题:$设$f(x)$在$[a,b]$上可积,证明:$$ \lim \limits_{n \to \infty } \int_a^b {f\left( x \right)\left| {\sin nx} \right|dx}  = \frac{2}{\pi }\int_a^b {f\left( x \right)dx} $$

1

$\bf命题:$设$f(x)$在$[-1,1]$上二阶连续可微,则存在$\xi \in(-1,1)$,使得\[\int_{ - 1}^1 {xf\left( x \right)dx}  = \frac{2}{3}f'\left( \xi  \right) + \frac{1}{3}\xi f''\left( \xi  \right)\]

1

$\bf命题:$设$\phi(x)$是以$T$为周期的有界函数,且$\frac{1}{T}\int_0^T {\varphi \left( x \right)dx}  = C$,证明:\[\mathop {\lim }\limits_{n \to \infty } n\int_n^{ + \infty } {\frac{{\varphi \left( t \right)}}{{{t^2}}}dt}  = C\]

1

$\bf命题:$

posted on 2014-04-20 17:31  142857  阅读(544)  评论(0编辑  收藏  举报

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