关于不等式的专题讨论I(微积分基本定理，分部积分法，中值定理，Schwarz不等式)

$\bf命题：$设$f\left( x \right) \in {C^2}\left[ {0,1} \right]$，且$f\left( 0 \right) = f\left( 1 \right) = 0$，则$${\rm{ }}\int_0^1 {\left| {f''\left( x \right)} \right|dx} \ge 4\mathop {\max}\limits_{x \in \left[ {0,1} \right]} \left| {f\left( x \right)} \right|$$

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$\bf命题：$设$f\left( x \right) \in {C^2}\left[ {0,1} \right]$，$f\left( 0 \right) = f\left( 1 \right) = 0,f'\left( 0 \right) = 1,f'\left( 1 \right) = 0$，则\[\int_0^1 {{{\left| {f''\left( x \right)} \right|}^2}dx} \geqslant 4\]

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$\bf命题：$设$f\left( x \right) \in {C^1}\left[ {a,b} \right]$，且$f\left( a \right) = f\left( b \right) = 0$，则$$\left| {\int_a^b {f\left( x \right)dx} } \right| \le \frac{{{{\left( {b - a} \right)}^2}}}{4}\mathop {\max}\limits_{x \in \left[ {a,b} \right]} \left| {f'\left( x \right)} \right|$$

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$\bf命题：$设$f\left( x \right) \in {C^2}\left[ {a,b} \right]$，且$f\left( a \right) = f\left( b \right) = 0$，则$$\left| {\int_a^b {f\left( x \right)dx} } \right| \le \frac{{{{\left( {b - a} \right)}^3}}}{{12}}\mathop {\max}\limits_{x \in \left[ {a,b} \right]} \left| {f''\left( x \right)} \right|$$

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$\bf命题：$ $\bf(内插不等式)$设$f\left( x \right) \in {C^2}\left[ {a,b} \right]$，则$$\mathop {\max}\limits_{x \in \left[ {a,b} \right]} \left| {f'\left( x \right)} \right| \le \int_a^b {\left| {f''\left( x \right)} \right|dx} + \frac{{18}}{{{{\left( {b - a} \right)}^2}}}\int_a^b {\left| {f\left( x \right)} \right|dx} $$

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$\bf命题：$设连续函数$f,g：$$\left[ {0,1} \right] \to \left[ {0,1} \right]$，且$f(x)$单调递增，则$$\int_0^1 {f\left( {g\left( x \right)} \right)dx} \le \int_0^1 {f\left( x \right)dx} + \int_0^1 {g\left( x \right)dx} $$

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$\bf命题：$设$f\left( x \right) \in {C^1}\left[ {a,b} \right]$，且$f\left( a \right) = 0$，则$${\int_a^b {{f^2}\left( x \right)dx} \le \frac{{{{\left( {b - a} \right)}^2}}}{2}\int_a^b {{{\left[ {f'\left( x \right)} \right]}^2}dx} }$$

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$\bf命题：$设$f'\left( x \right) \in C\left[ {a,b} \right],f\left( a \right) = f\left( b \right) = 0$，则\[\int_a^b {{f^2}\left( x \right)dx} \le \frac{{{{\left( {b - a} \right)}^2}}}{4}\int_a^b {{{\left( {f'\left( x \right)} \right)}^2}dx} \]

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$\bf命题：$设$f\left( x \right) \in {C^2}\left[ { - a,a} \right]$，且$f\left( 0 \right) = 0$，则

\[{\left( {\int_{ - a}^a {f\left( x \right)dx} } \right)^2} \le \frac{{{a^5}}}{{10}}\int_{ - a}^a {{{\left[ {f''\left( x \right)} \right]}^2}dx} \]

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$\bf命题：$设$f(x)$在$[-1,1]$上可微，$M = \mathop {Sup}\limits_{x \in [ - 1,1]} \left| {f'\left( x \right)} \right|$，若存在$a \in(0,1)$，使得$\int_{ - a}^a {f\left( x \right)dx} = 0$，则$$\left| {\int_{ - 1}^1 {f\left( x \right)dx} } \right| \leqslant M\left( {1 - {a^2}} \right)$$

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$\bf命题：$设$f\left( x \right) \in {C^2}\left[ { - l,l} \right],f\left( 0 \right) = 0$，则\[{\left( {\int_{ - l}^l {f\left( x \right)dx} } \right)^2} \leqslant \frac{{{l^5}}}{{10}}\int_{ - l}^l {{{\left( {f''\left( x \right)} \right)}^2}dx} \]

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$\bf命题：$

$\bf命题：$设$f(x)$在$(a,b)$上可微，且$\left| {f'\left( x \right)} \right| \leqslant M$，则\[\left| {\int_a^b {f\left( x \right)dx} - \frac{{b - a}}{2}\left( {f\left( a \right) + f\left( b \right)} \right)} \right| \leqslant \frac{{{{\left( {b - a} \right)}^2}M}}{4} - \frac{{{{\left( {f\left( a \right) - f\left( b \right)} \right)}^2}}}{{4M}}\]

$\bf(03川大三)$设$f\left( x \right) \in {C^2}\left[ {a,b} \right],f\left( a \right) = f\left( b \right) = 0$，则\[\left| {\frac{{f\left( x \right)}}{{\left( {x - a} \right)\left( {x - b} \right)}}} \right| \leqslant \frac{1}{{b - a}}\int_a^b {\left| {f''\left( x \right)} \right|dx} ,\forall x \in \left[ {a,b} \right]\]

[b]附录[/b]

$\bf命题：$设$f\left( x \right) \in {C^1}\left[ {0,1} \right]$，且存在$M > 0$，使得对任意$x \in \left( {0,1} \right)$，有$\left| {f'\left( x \right)} \right| \le M$，则$$\left| {\int_0^1 {f\left( x \right)dx} - \frac{1}{n}\sum\limits_{k = 1}^n {f\left( {\frac{k}{n}} \right)} } \right| \le \frac{M}{n}$$

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$\bf命题：$设非负单调递减函数$f(x)$在$\left[ {0,1} \right]$上连续，且$0 < \alpha < \beta < 1$，证明：$\int_0^\alpha {f\left( x \right)dx} \ge \frac{\alpha }{\beta }\int_\alpha ^\beta {f\left( x \right)dx}$

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$\bf命题：$设$f(x)$在$\left[ {a,b} \right]$上可导，且$f'\left( x \right)$在$\left[ {a,b} \right]$上递减，$f'\left( b \right) > 0$，证明：$\left| {\int_a^b {\cos f\left( x \right)dx} } \right| \le \frac{2}{{f'\left( b \right)}}$

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$\bf命题：$设$f(x)$为$\left[ { - \pi ,\pi } \right]$上的凸函数，且$f'\left( x \right)$有界，证明：${\rm{ }}{a_{2n}} = \frac{1}{\pi }\int_{ - \pi }^\pi {f\left( x \right)\cos 2nxdx} \ge 0$

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$\bf命题：$