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)

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

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

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

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

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

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$f命题：$设连续函数$f,g：$$left[ {0,1} ight] o left[ {0,1} ight]$，且$f(x)$单调递增，则$$int_0^1 {fleft( {gleft( x ight)} ight)dx}  le int_0^1 {fleft( x ight)dx}  + int_0^1 {gleft( x ight)dx} $$

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

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

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

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

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

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

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$f命题：$

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

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

附录

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

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$f命题：$设非负单调递减函数$f(x)$在$left[ {0,1} ight]$上连续，且$0 < alpha  < eta  < 1$，证明：$int_0^alpha  {fleft( x ight)dx}  ge frac{alpha }{eta }int_alpha ^eta  {fleft( x ight)dx}$

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

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

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$f命题：$