一.仅知被积函数连续的不等式1.设)(xf在1,0上连续且单调减少,证明:对于任意的1,0,都有100)()(dxxfdxxf。2.设)(xf在ba,上连续且单调减少,证明:babadxxfbadxxxf)()()(2。3.设)(xf在ba,上连续,证明babadxxfabdxxf)()()(22。4.设()fx在ba,上连续,且满足Mxfm)(0,证明:dxxfdxxfbaba)(1)(22)(4)(abmMMm-----------(Kantorovich)康托洛维奇不等式二.被积函数一阶可导的不等式(对)(xf用微分中值公式或对原函数)(xF用泰勒公式)5.设)(xf连续可微,且0)(10dxxf,Mxf)(,证明:任给(0,1)总有8)(0Mdxxf。6.设)(xf在ba,上连续可微,且1)(Mxf,0)(af,证明abMdxxfba21)(2)(。7.设)(xf在ba,上连续可微,且1)(Mxf,)(0)(bfaf,证明abMdxxfba21)(4)(。三.被积函数二阶可导的不等式(用泰勒公式)8.设)(xf在ba,上二次连续可微,且2)(Mxf,)(0)(bfaf,证明abMdxxfba32)(12)(。9.设)(xf在ba,上二次连续可微,且2)(Mxf,0)2(baf,证明abMdxxfba32)(24)(。10.设)(xf在ba,上二次连续可微,且0)(xf,证明2)()()()()2()(bfafabdxxfbafabba。四.具体函数的积分不等式11.证明2202021cos1sindxxxdxxx。12.证明2520sindxex。————————————————————————————————————一.仅知被积函数连续的不等式1.设)(xf在1,0上连续且单调减少,证明:对于任意的1,0,都有100)()(dxxfdxxf。证法1(换元法)令tx,10100)()()(dttfdttfdxxf(因f单减,)()(tftf)。证法2(单调性)设0)()(dxxfF,0)()()()()()()(220ffffdxxffF(单减f,,0)0))()(()()()(2020dxxffdxxffF或所以)(F单减,)1()(FF,即100)()(dxxfdxxf。证法3(利用定积分性质)0)()1()()1()()()1(,1,,,00)()1()()1()()()1()()()()()(1010001010ffdxxfdxxffffdxxfdxxfdxxfdxxfdxxfdxxfdxxf右或左单减右左证法4(函数最值)设1,0,)()()(100xdttfxdttfxFx1,0),()()()()(10fxfdttfxfxFx为函数的驻点,且x时)(,0)(),()(xFxFfxf单增,x时)(,0)(),()(xFxFfxf单减。所以)(F为函数的最大值,而最小值在端点取得。又0)1()0(FF,故0)(xF。(或者10)()()(dttfxfxF为单调递减函数,)(xF为上凸函数,最小值在端点取得。)证法5(微分中值定理)设1,0,)()(0xdttfxFx,则)()(,0)0(xfxFF)1,(),(1)()1(),0(),(0)0()(fFFfFF,因为f单减,所以)()(ff即1)()1()(FFF,也就是)1()(FF,得证。证法6(用定积分定义)ninnindxxfnifnnifndxxf11010)()(1lim)(lim)(2.设)(xf在ba,上连续且单调减少,证明:babadxxfbadxxxf)()()(2。证明作辅助函数xaxadttfxadtttfxF)()()(2)(,0)(aF。xadttfxfaxxxfxF)()()()(2)(0)()()()()(xaxadttfxfdttfxfax(因)(xf单调减少)所以)(xF单调减少,故0)()(aFbF,即不等式成立。3.设)(xf在ba,上连续,证明babadxxfabdxxf)()()(22。证明作辅助函数xaxadttfaxdttfxF)()()()(22,0)(aF)()()()()(2)(22xfaxdttfdttfxfxFxaxadtxfdttfdttfxfxaxaxa)()()()(2220)()(2xadttfxf,所以)(xF单调减少,故0)()(aFbF,即不等式成立。注此题也可用SchwartzCauchy不等式bababadxxgdxxfdxxgxf)()()()(222快速证明。同类的题还有:设)(xf在ba,上连续,且0)(xf,证明2)()(1)(abdxxfdxxfbaba。4.设()fx在ba,上连续,且满足Mxfm)(0,证明:dxxfdxxfbaba)(1)(22)(4)(abmMMm-----------(Kantorovich)康托洛维奇不等式证明:因为0)())()()((xfMxfmxf即MmxfmMxf)()(推得babaabMmdxxfmMdxxf))(()(1)(又babababadxxfdxxfmMdxxfmMdxxf)(1)(2)(1)(故)(2)()(1)(abMmdxxfdxxfmMbaba即dxxfdxxfbaba)(1)(22)(4)(abmMMm。注(特例)设()fx在[0,1]上连续,且满足1()2fx,证明:110019()()8fxdxdxfx。二.被积函数一阶可导的不等式(对)(xf用微分中值公式或对原函数)(xF用泰勒公式)5.设)(xf连续可微,且0)(10dxxf,Mxf)(,证明:任给(0,1)总有8)(0Mdxxf。证法一(对原函数用泰勒公式)记xdttfxF0)()(,1,0x,则0)1()0(FF,而,)(2)())(()()(2xtFxtxFxFtF1,0令1,0t,有,2)()()()0(021xFxxFxFF1,01(1),)1(2)()1)(()()1(022xFxxFxFF1,02(2)(1))1(x(2)x得)1()1)(()(21)(021xxxFxFxF则8)1(21)1()1(21)(MxMxxxxMMxxF。法二记xdttfxF0)()(,1,0x,则0)1()0(FF。若0)(xF,则结论自明。若)(xF不恒为零,则)(xF的最大值必在区间),(10内某点c取得,且该点为极值点,0)(cF。用泰勒公式可得,)(2)())(()()(2cxFcxcFcFxF故2)(2)()(cxMxFcF若2/10c,取0x可得82)(2McMcF,若12/1c,取1x可得8)1(2)(2McMcF。综上有8)(0Mdxxf。6.设)(xf在ba,上连续可微,且1)(Mxf,0)(af,证明abMdxxfba21)(2)(。证明法一dxafxfdxxfdxxfbababa)()()()(211)(2)()()(abMdxaxMdxaxfbaba。法二记xadttfxF)()(,0)()(aFaF,)()(xfxF2))((21)()()(abFaFaFbF所以)(bF212)(2)()(21abMabF。7.设)(xf在ba,上连续可微,且1)(Mxf,)(0)(bfaf,证明abMdxxfba21)(4)(。证记2bac法一仿3题可得2121)(8)(2)(abMacMdxxfca,2121)(8)(2)(abMcbMdxxfbc易知abMdxxfdxxfdxxfbccaba21)(4)()()(。法二令xadttfxF)()(,0)()(aFaF,)()(xfxF2))((21)()()(acFaFaFcF所以)(cF212)(8)()(21abMacF同理令bxdttfxG)()(,0)()(bGbG,)()(xfxG2))((21)()()(acGaGaGcG所以)(cG212)(8)()(21abMacG,从而abMcGcFdxxfba21)(4)()()(法三分部积分(用上条件)(0)(bfaf)babababadxxfcxdxxfcxabxfcxcxdxfdxxf)()()()()()()()()(故abMdxcxMdxxfcxdxxfbababa211)(4)()(三.被积函数二阶可导的不等式(用泰勒公式)8.设)(xf在ba,上二次连续可微,且2)(Mxf,)(0)(bfaf,证明abMdxxfba32)(12)(证法一分部积分可得(用上条件)(0)(bfaf)bababadxxfdxxfxbadxxfxbax)(2)()2()())((进而32)(12))((2)())((21)(abMdxxbaxMdxxfxbaxdxxfbababa。法二记2bac,分部积分可得bababadxxfcxcxdxfdxxf)()()()()(因为))(()()(cxfcfxfbababadxcxfdxcxfcfcxdxxf2))(())(()()()(故32222)(12)()()()(abMdxcxMdxfcxdxxfbababa注(特例)设)(xf在10,上二次连续可微,且2)(Mxf,)1(0)0(ff,证明Mdxxf12)(210。9.设)(xf在ba,上二次连续可微,且2)(Mxf,0)2(baf,证明abMdxxfba32)(24)(证明(涉及到端点外的第三点)记2bac,则0)(cf,在该点用泰勒公式有2))((21))(()(cxfcxcfxf,而bababadxcxfdxcxfcxcfdxxf22))((21))((21))(()(,abMdxcxMdxcxfdxxfbababa32222)(24)(2)()(21)(。10.设)(xf在ba,上二次连续可微,且0)(xf,证明2)()()()()2()(bf