第三章 一元函数积分学§3-1 不定积分不定积分是计算定积分、重积分、线面积分和解微分方程的基础，要求在掌握基本积分法的基础上，更要注重和提高计算的技巧。

一、基本概念与公式1. 原函数与不定积分的概念2. 不定积分与微分的关系（互为逆运算）3. 不定积分的性质 4．基本积分表 2222223122232max{1}d .,1max{1,}1,11,,111max{1,}d d 311max{1,}d 1d 11max{1,}d d .3x x x x x x x x x x x x x x C x x x x x C x x x x x xC ⎧<-⎪=-≤≤⎨⎪>⎩<-==+-≤≤==+>==+⎰⎰⎰⎰⎰⎰⎰1求，因当时；当时；当时例解()()31113211123231lim lim 3,1lim lim 323,232133max{1,}d 1 1.2133x x x x x C x C x C x C C C C C x C x x x x C x x C x -+-+→-→-→→⎧⎛⎫+=+ ⎪⎪⎪⎝⎭⎨⎛⎫⎪+=+ ⎪⎪⎝⎭⎩⎧=-+⎪⎪⎨⎪=+⎪⎩⎧-+<-⎪⎪⎪=+-≤≤⎨⎪⎪++>⎪⎩⎰由原函数的连续性,有得故，，，二、不定积分的基本方法1. 第一类换元法（凑微分法） ()d ()[()]d []d [].f u u F u C f x x x f x x F x C ϕϕϕϕϕ=+'()=()()=()+⎰⎰⎰若，则2. 第二类换元法()10[]()()d []d ()[].x t t x x t t f t t G t f x xf t t t G t CG x C ϕϕϕϕϕϕϕϕϕ-1=()=-''=()()≠()()'()()=+()+⎰⎰令代回若是单调可导函数，且，又具有原函数，则有换元公式3. 分部积分法()()d ()()()()d d d .u x v x x u x v x u x v x xu v uv v u ''=-=-⎰⎰⎰⎰或4. 有理函数的积分法 化有理真分式为部分分式.5. 三角函数有理式的积分(sin cos )d ()tan2R x x x R u v u v xt =⎰对于，(其中，表示关于，的有理函数)，可用“万能代换”化为有理函数的积分.三、题解示例(2222225sin 25sin25sin25sin 25sin 225s .22.sin 4d .sin 4d 2sin 2cos 2d 1sin 2d sin 2d(5+sin 2)212xxxxxx x C e x x ex x e x x xe x x e x e ++++++==+====⎰⎰⎰⎰⎰⎰23求求例解例解2in 2322222..ln(2[ln()].31ln d .(ln )1ln 1ln d d (ln )ln 1xC x x x x C xx x x x x xx x x x x x +==++----=⋅-⎛⎫- ⎪⎝⎭⎰⎰⎰45求求例解例解21ln 1d 1ln ln 11.ln x C x x x x x xC x x⎛⎫=--=+ ⎪⎝⎭⎛⎫-- ⎪⎝⎭=+-⎰sin tan (d ((n nt ax bax b t cx dx a tx a tR x x R x xR x x R x x=++=+==⎛ ⎝⎰⎰⎰⎰令令令令化为有理函数的积分.化为有理函数的积分.化为三角函数有第二类换元法常用来去根号,如式.：理的积分sec (x a tR x x=⎰令化为三角函数有理式的积分.化为三角函数有理式的积分.355323232.1d 6d 6d d 6d 6(1)d 6112366ln |1|1).x t x t tt t t t t t t t t t t t t t t t CC +====-+-+++=-+-++=+⎰⎰⎰⎰6求令，则原式例解2222222222(1)1d 12d d ,11(1)12d 2d 1ln .(1)11x x xI xx xx t t tt x x x t t t t t t t I C t t t t +=+===+---+=⋅==+---⎰⎰⎰法一：令，则，故解222211ln2ln(1).11111ln 22411221ln .2xx C x x C x xI x x Cx x x x C ++=+=+++-+⎛⎫⎛⎫ ⎪==+++-+ ⎪ ⎪⎝⎭⎛⎫⎛⎫⎝⎭+- ⎪ ⎪⎝⎭⎝⎭⎛⎫=++++ ⎪⎝⎭⎰法二：222222222(2)413(2)(2)33tan 23sec d 1cos d 9tan 3sec 9sin 1413.9sin I x x x x x x t t t ttt t tx x C C t ==++++++=-==⋅++=-+=-+⎰⎰⎰⎰8求令原式例解2222222222arcsin d .1arcsin sin d cos d (1sin )d csc d d cot cot d sin 211cot ln |sin |arcsin ln ||(arcsin ).22x I x x xx t x t x t tt t t I t t t t t t t t t t t t x t t t C x x x C =⋅-===+==+=-++-=-+++=-+++⎰⎰⎰⎰⎰9求令，，例解 ()d ()sin d ()cos d ()()d sin d cos d d ()ln ()d ()(arcsin )d ()arctan d ln ()(arcsin )arctan (axm m m ax m m n n m m n n m P x e x P x ax x P x bx x P x m P x u e x ax x bx x v P x ax b x P x x x P x x x ax b x x u P ++⎰⎰⎰⎰⎰⎰一般地，形如，，的积分(其中为次多项式),选取为，，，为应用分部积分降幂；形如，，的积分，选取，，为，)d d x x v 为应用分部积分超越函数代数化.2222111111ln d ln d 122121111ln .21xx x x x I x xx x x x x x x C x -⎛⎫++⎛⎫==-+ ⎪ ⎪--+-⎝⎭⎝⎭-+=++-⎰⎰解3233322233sin ()()d .sin cos sin ().()d d ()()3()d cos sin 3(sin sin )d cos 4sin 6cos .d .(1)d d (1)xx xx xf x x f x x xx x x xf x x x xf x x x f x x f x x f x xx x x x x x x x x x x x x C xe I x e e u x v x e v ''-⎛⎫== ⎪⎝⎭'==-=---=--+=-==-=⎰⎰⎰⎰⎰⎰1112已知是的一个原函数，求求令，，则例解例解322222211d .(1)2(1)11d .2(1)2(1)d 1111d d (1)1(1)1111d ln |1|ln ||11ln |1|1111ln |2(1)21x x x x x x x x x x x x x x e te x e e x x I e e x t t e t t t t t t e e C t t t t e x Ce x I x e e -==---=-⋅+--⎡⎤=====⋅=-⎢⎥-++⎣⎦⎛⎫=---=---++ ⎪+⎝⎭=---++-=-+----⎰⎰⎰⎰⎰⎰令故又所以1|.xe C ⎡⎤-+⎢⎥⎣⎦arcsin arccos d .arcsin arccos d arcsin arccos (arccos arcsin arcsin arccos arcsin )d arcsin arcc I x x x I x x x x x x x x x x x x x x x x x x =⋅=⋅-=⋅+-=⋅-⎫--=⋅⎰⎰⎰13求例解os arcsin )2.x x x x C -++22211sind.1cos12cos sin122d sec d tan d2222cos2tan tan d tan d tan.2222().1(1xx x xx x x xnn nxI e xxx xx xI e x e x e xxx x x xe e x e x e CI nI n xx--+=++==+=-+=+====-⎰⎰⎰⎰⎰⎰⎰⎰1415求求为自然数例解例解(2221(1)(1)().2.1ln..n nn nn x n I InI InI x CI C--=-=-+-=-==+=+移项得而有理函数的不定积分用分解为部分公式之和再分项积分是行之有效的方法，但有时计算比较烦琐，如果针对被积函数的某种特点，可应用更简便的方法。

例如：22(1)tan.(2)111()(3)()d d.(4)d.()n nnmx a x a tR tR x x t x tx n txx x a tx a+==-=-⎰⎰⎰被积函数分母中出现因子时，宜用代换当被积函数分子为，分母为较高次数的多项式时，往往可采用倒代换化为假分式.形如的积分可通过代换化为形如的积分宜用代换23tan423222d.(1)d311cos d sin2sin4(1)84323(53)arctan.88(1)x txxxt t t t t Cxx xx Cx=+====+++++=+++⎰⎰⎰16求例解8289101282282864222753753d .(1)1(1)11101d 1d 11111arctan 75311111arctan .753xI x x A A x A A A x x x x x x x t t I t t t t t t t t t t t t CC x x x x x =++=++++++=⎡⎤=-=--+-+⎢⎥++⎣⎦=-+-+-+=-+-+-+⎰⎰⎰17求若将被积函数分解成部分分式之和，需确定个常数．令，则例解 2100219899100100d .(1)21d d 2d d x txI x x t t I t t t t t t t t -=---=--+-=-+-⎰⎰⎰⎰⎰18令求例解979899979899111974999111.97(1)49(1)99(1)t t t C C x x x---=-++=-++--- 三角函数有理式的不定积分通过万能变换tan2xu = 总可化为有理函数积分，但有时很繁，可用如下代换:(1)(sin cos )(sin cos )cos .(2)(sin cos )(sin cos )sin .(3)(sin cos )(sin cos )tan .R x x R x x t x R x x R x x t x R x x R x x t x -=-=-=-=--==当，，时，可用代换当，，时，可用代换当，，时，可用代换44119d .sin cos I x x x =+⎰例444224421d .sin cos (sin cos )(sin cos )tan sec 111d d d tan 11121.I x x xR x x R x x t x x t I x t t x t t t t t C C =+--==+⎛⎫===- ⎪++⎝⎭⎛⎫-+ ⎪⎝⎭-==⎰⎰⎰⎰19求因，，，故令例解4144242223231d .sin cos (sin cos )(sin cos )sin cos d d d sin (1sin )(1)111111d ln 13221111sin ln .3sin sin 21sin u tI x x xR x x R x x u x x u t I x tx x u u t t t t t t C t t xC x x x ==-=-===----⎛-⎫⎛⎫=-++=-+++ ⎪ ⎪-+⎝⎭⎝⎭-=---++⎰⎰⎰⎰⎰20令求因，，，故令，例解7sin cos d .3sin 4cos x xI x x x +=+⎰21求例sin cos 7sin cos (3sin 4cos )(3sin 4cos )(34)sin (43)cos .3474311.13sin 4cos (3sin 4cos )d ln |3s 3sin 4cos x x x x A x x B x x A B x A B x A B A B A B x x x x I x x x x'+=+++=-++-=⎧⎨-=⎩=⎧⎨=-⎩'+-+==-+⎰被积函数的特点是分子、分母均为和的线性组合，不一定用万能变换，可用待定系数法.令故有，解得所以解in 4cos |.x x C ++ 杂例(222222.ln(1)ln(1)2211d 2d 1111112ln arctan 21n .x x x xI x t x t t t t I t t t t t t t t t C t e C I x -==+--⎡⎤⎡⎤=+=-⎢⎥⎢⎥+--+⎣⎦⎣⎦⎡+⎤=-+⎢⎥-⎣⎦=-==⎰⎰22求法一：令，，则法二：例解ln arcsin .x x e e C -=++ln ln (1ln )d .d(ln ).xx x x x x I x x x I e x x e C x C =+==+=+⎰⎰23求例解23222322()()()d .()()()()()()()()()()d d ()()()()()1()d .()()2()f x f x f x I x f x f x f x f x f x f x f x f x f x f x I x xf x f x f x f x f x f x C f x f x f x ''⎛⎫=- ⎪''⎝⎭'''''--==⋅'''⎛⎫⎛⎫==+ ⎪ ⎪'''⎝⎭⎝⎭⎰⎰⎰⎰24求例解22222222322(sin )cos 2tan ()(01)(sin )(sin )d(sin )(cos 2tan )d(sin )(cos 2tan )sin d 1sin cos 2d(cos 2)2d 2cos 11cos 22d cos 2cos d cos 4cos 1cos 22ln |cos 4f x x x f x x f x f x x x x x x x x xxx x xx x x x xx x x '=+<<'==+=+=-+=--+=--⎰⎰⎰⎰⎰⎰⎰25设，求法一：因例解2242212222|cos 3ln(1sin )sin .43()ln(1)ln(1).4sin (sin )12sin 1sin 1()122.11x Cx x C f x x x C x x C xf x x xx f x x x x x++=---+=---+=---+'=-+-'=-+=-+--所以法二：因故21()2d ln(1).1f x x x x x C x ⎛⎫=-+=---+ ⎪-⎝⎭⎰从而§3-2 定积分一、重要概念及公式1. 定积分定义11101(),()d lim ().()[,].[,]()d lim .0,1,1()d lim .nb i i ai i i nb an i nn i f x f x x f x f x a b a b n b a b a f x x f a i n n a b i f x x f n n λξξξ→=→∞=→∞==∆--⎛⎫=+ ⎪⎝⎭==⎛⎫= ⎪⎝⎭∑⎰∑⎰∑⎰若可积则注：定积分的存在与区间的分法和的取法无关,仅与被积函数与区间有关如果把分成等份,取每个小区间的右端点,则有特别地,若则有2. 定积分的性质(略)()[,],()[,][,]()()d ()()d .()1,()[,],[,]()d ()().b baab af x C a bg x C a b a b f x g x x f g x x g x f x C a b a b f x x f b a ξξξξ∈∈∃∈=≡∈∃∈=-⎰⎰⎰积若且不变号,则使特别地,取则有积若则使分第一中值定理分中值定理3. 积分上限函数的导数d [,],()d ()().d xa f C ab f t t f x a x b x ∈=≤≤⎰若则()()d ()d ()[()]()[()].d u x v x f t t u x f u x v x f v x x ''=-⎰一般地, 4. Newton-Leibniz 公式[,],()()[,],()d ()()().b ba af C a b F x f x a b f x x F x F b F a ∈==-⎰若是在上的原函数则5. 定积分的换元公式和分部积分公式[,],[,]([,])(),(),()d [()]()d .()d ()()()()d ().b a b b ba aaf C a b C a b f x x f t t t u x v x u x v x v x u x βαϕαββαϕαϕβϕϕ∈∈=='==-⎰⎰⎰⎰若或且则6. 反常积分 定义略.反常积分是常积分的极限问题. 7. 一些常用的公式2202020[,],2()d ,()()d 0,().[0,1],(sin )d (cos )d (sin )d 2(sin )d (sin )d (sin )d (sin )d .2(,),()()(0)a a af C a a f x x f x f x x f x f C f x x f x x f x x f x x x f x x f x x f x x f C f x l f x l πππππππππ-∈-⎧⎪=⎨⎪⎩∈====∈-∞+∞+=>⎰⎰⎰⎰⎰⎰⎰⎰⎰若则当是偶函数；当是奇函数若则；；若0,()d ()d ()d ()d Cauchy-Schwarz ,[,],a l la a nl laf x x f x x a f x x n f x x a k Z fg C a b +++==∈∈⎰⎰⎰⎰则(为常数).(为常数,).不等式若则222()()d ()d ()d .b a a a b b f x g x x f x x g x x ⎡⎤≤⋅⎢⎥⎣⎦⎰⎰⎰二、利用定积分的定义计算数列的极限[]111001ln 1ln(1)d (1)ln(1)2ln 212)(2).1lim e 4e ee .e ni n i n n n n x xx x x n n =→∞⎛⎫+ ⎪⎝⎭→∞+++--∑⎫⎛⎫=+=⎪⎪⎭⎝⎭⎰====1求极限原式例解111100112sin sin sin lim .11122sin sin sin 11sin sin ,11112112lim sin sin d cos ,1lim sin lim 1n nn i i n n i n n n i n n n n n n n n n n i i n n n n n n n n n n n ni x x xn n n i n n n n ππππππππππππππ→∞==→∞=→∞→∞=⎛⎫ ⎪+++ ⎪+ ⎪++⎝⎭⋅≤+++≤++++==-=⋅=+∑∑∑⎰∑2求极限因又例解112lim sin ,12sin sin sin2lim .1112n n i n i n n n n n n n n n n n ππππππ→∞=→∞⋅=+⎛⎫ ⎪+++= ⎪+ ⎪++⎝⎭∑由夹逼准则知三、利用定积分的几何意义和定积分的性质[]123[,]()0,()0,()0.()d ,1()(),()()(),2baa b f x f x f x S f x x S f b b a S f a f b b a '''><>==-=+-⎰3设在闭区间上记则()例213()[,][,]()()()()()(),()()()d ()d ()()d ,,(D).bb b a a a y f x a b x a b f b f a f b f x f a x a b af b f a f b x f x x f a x a x b a S S S =-≤≤+---⎡⎤≤≤+-⎢⎥-⎣⎦<<⎰⎰⎰由已知条件,曲线在闭区间上是位于由上方的单调下降的(向上)凹弧,故在上从而有即故选解()2sin 222sin sin sin 20002sin 2sin 20sin sin sin ()e sin d ,(A)(B)(C)(D)()e sin d e cos e cos d e cos d 0[0,2]ecos 0()e sin d e sin d e sin d x t xt t t t tt t t F x t t F x t t t t tt t t F x t t t t t πππππππππ+--===-+=>≥==+⎰⎰⎰⎰⎰⎰4设则为正常数；为负常数；恒为零；不为常数；因因在上；或例解()()sin sin 0sin sin 0sin sin e sin d e sin d ee sin d 0[0,]ee ,sin 0(A).u tu t tt tt tu u t tt t t πππππ=----=======+=->≥≥⎰⎰⎰⎰前一式令因在上,故选44120012122121tan d d tan (A)1(B)1(C)1(D)1.x xI x I x xx I I I I I I I I ππ==>>>>>>>>⎰⎰5设,,则()；；；例()122222tan 0,,tan ,,.4tan tan sec tan sin cos 00sin ,cos x x x x x I I x x x x x x x x xx x x x x x x π⎛⎫∈>>> ⎪⎝⎭'--⎛⎫==>>> ⎪⎝⎭当时因故从而又因当时故解444100tan tan 4,0,,4tan 4d d 1.(B).x x xx x xx I x x xππππππ=⎛⎫<=∈ ⎪⎝⎭=<=⎰⎰从而故选四、积分上限函数的导数的应用20(1)22202000020()d ,(0)0,(0)0,lim.()d 2()2()limlim2()d ()2()d ()4()4()limlim()3()()3()lim ()lim ()(0)x xx xx x x x x x x f t tf C f f xf t tx f x f x x f t t x f x f t t x f x x f x f x f x f x x f x f x xf x f x f →→→→→→→'∈=≠==++''=='+'+'''==⎰⎰⎰⎰6设求原式因例解000,()limlim ()(0)04(0)1.3(0)(0)x x f x f x f x f f f →→≠''==≠'==''+故原式1ln 1d .lim lim lim 2,n n tt t t x →+∞→+∞⎛⎝===7求极限因故例解1ln 1d 2.n n x ⎛= ⎝()22222224500032054020024300,1lim e d ,e d 31e limlim ,5lim 31e 0,1,31e62e limlim 520xt x xt xx x xx x x x x a b ab t x x x ax x b tax b xx ax b b ax ax x x x -→--→→-→--→→⎛⎫++ ⎪⎝⎭++++====++==-+-+=====⎰⎰8型型确定的取值,使得极限存在并求出极限值.原式由原极限存在知故得于是原式例解()222200222003e lim ,10lim 3e 0,1,3e 11lim lim .101010xx xx x x x a x a a x x x -→-→-→→++==---===-由原极限存在知故得且原式2222(),(0)1,()||()d ().0,()()()d ()()d ()d ()d ()d ()d ,()()d ()()x xx x xxx x xxxg x T g f x x t g t t f T x f x x t g t t t x g t tx g t t tg t t tg t t x g t t f x g t t xg x xg x ==-'>=-+-=-+-'=+-⎰⎰⎰⎰⎰⎰⎰⎰90000设是以为周期的连续函数,求设因故例解22204(2)()()d 2(2)()()d ()d 2(2),()()d ()d 2(2)()d ()d 2(0)2.x xxx xTT T T Txg x xg x g t t xg x xg x g t t g t t xg x f T g t t g t t Tg T g t t g t t Tg T +---+=-+'=-+=-+=⎰⎰⎰⎰⎰⎰⎰000所以1012010()(,),lim2.()()d ,()().01()()d ()d ,1()()()d .()(0)lim ()lim 0,(0)(0)d 0,(0)limx xt uxx x x x f x f C xx f xt t x x x x f xt t f u u x f x x f u u x xf x f f x x xf t ϕϕϕϕϕϕϕ→=→→→∈-∞+∞==''≠====='=-+==⋅==='=⎰⎰⎰⎰⎰10令已知且设试求并讨论的连续性当时故又故例解02002200()d ()(0)1()limlim1,21()()d ,0().1,1()1()lim ()lim ()d lim 21(0),2x x x x x x x x f u u x f x xxxf x f u u x x x xx f x f x x f u u xx x ϕϕϕϕϕ→→→→→-===⎧-+≠⎪'=⎨⎪=⎩⎡⎤''=-+=-+==⎢⎥⎣⎦⎰⎰⎰所以因()(,).x ϕ'-∞+∞故在上连续1223012231123002223002()()()d ()d ,().()d ,()d ,(),111()d ()d ,2348()d ()d 24,33,1,83()8f x f x x x f x x x f x x f x A f x x B f x x f x x Ax Bx A f x x x Ax Bx x A B B f x x x Ax Bx x A B A B f x x x x =++===++==++=++==++=++==-=+-⎰⎰⎰⎰⎰⎰⎰⎰11设连续函数满足求设则从而有解得故例解312.()d ()d f x x f x x ⎰⎰注：和都是常数.五、定积分和反常积分的计算2012101111010101011,01(),1,01e (1)d .(1)d ()d ()d ()d 11e d d d ln(1)1e 11e ln(1e )ln 2ln(1).x t x tt tt x xf x x f x x f x x f t t f t t f t tt t t t t e =---------⎧≥⎪⎪+=⎨⎪<⎪+⎩--======+=+=+++++=-++=+⎰⎰⎰⎰⎰⎰⎰⎰12令设求例解222212220011min ,d .||1122min ,d 2d d 2ln 2.3I x x x I x x x x x x x -⎧⎫=⎨⎬⎩⎭⎧⎫⎡⎤==+=+⎨⎬⎢⎥⎩⎭⎣⎦⎰⎰⎰⎰13求易见被积函数为偶函数,故例解404,.()|sin cos |d (cos sin )d (sin cos )d .n I x n Z f x x I n x n xn x x x n x x x x x x ππππππππ+=∈===⎡⎤=-=-+-⎢⎥⎣⎦=⎰⎰⎰⎰⎰⎰14计算因是周期为的周期函数,故例解121211222001122220011222201(cos )lnd .11ln112ln d [ln(1)ln(1)]d()111[ln(1)ln(1)]d 11111ln 32d ln 32ln 414131ln 3.4xI x x x xxxx I x x x x x xx x x x x x x x x x x x x -+=+-+-+==+---⎛⎫=+---+ ⎪+-⎝⎭+⎡⎤=-=--+⎢⎥--⎣⎦=-⎰⎰⎰⎰⎰15计算因为奇函数,故例解2020,2I x t I ππππ==-===⎰⎰⎰16计算令有例解2200211d d.2240,sind.sin cos4pp pI x xpxxx xπππππ⎡⎤===>=+⎰⎰⎰所以类似地,对任意常数有12404444004400ln(1)d.1tan,ln(1tan)d ln1tan d41tan2ln1d ln d.1tan1tanln2d ln(1tan)d ln2,4ln2.8u txI xxx tI t t t uuu uu uu u u IIππππππππππ=-+=+=⎡⎤⎛⎫=+====-+-⎪⎢⎥⎝⎭⎣⎦-⎡⎤=+=⎢⎥++⎣⎦=-+=-=⎰⎰⎰⎰⎰⎰⎰17求令则所以例解111100010,d ln.ln1d,ln ln11d d d d d d ln.ln11b aby b ab yaab ab b by ya a ax x ba b xx ax x xx yx xx x b x x x y y x x yx y a-+<<=+-==-+==========++⎰⎰⎰⎰⎰⎰⎰⎰18交换积分次序设证明因故例解121111000()arctan(1)(0)0,()d.()d()d(1)(1)()(1)()dy x x y y x xy x x y x x x y x x y x x'=-='=-=---⎰⎰⎰⎰19若,计算例解2(1)1120011(1)arctan(1)d arctan d ln2.284x tx x x t tπ-==---=======-⎰⎰令)(32123121321121222ln ln2.2xπ==+==+=++⎰⎰⎰20计算因为无穷间断点(瑕点),故原式例解2122211d.(1)11,(1)11d ln ln2.1xIx xxx x x xxI x xx x+∞+∞+∞+∞=+=-++⎛⎫⎡=-=-==⎪⎣+⎝⎭⎰⎰21计算因故例解151011101,111555111ln525utIt xIu+∞=+∞====⎛⎫=++=⎪⎝⎭⎰⎰⎰⎰22令求令则有例解六、定积分的证明题4tan d,1,nnI x x nπ⎰23设=为大于的整数试证：例11(1)111(2).2222n nnI InIn n-+=+<<+-；222442001422222(1)(tan tan)d tan sec d11tan.11(2)0,,tan1,tan tan(1),,412,11.2212,11.2212n n nn nnn nn nn n nnn nn n nnI I x x x x x xxn nx x x x n I II I InInI II I InInππππ-------+++=+=⋅==--⎡⎤∈≤≤≥≤⎢⎥⎣⎦≤+=-≤-≥≥+=+≥+⎰⎰当时故从而故即同理,由得即所以证1.222nIn n≤≤+-[]202220002020()[0,2],()0.2()sin d (2)(0).11()sin d ()cos ()cos d 11(2)(0)()cos d 11(2)(0)()cos d .()0,f x f x n f x nx x f f n f x x x f x nx f x nx xn n f f f x nx x n n f f f x nx x n n f x ππππππππππ'≥≤-⎡⎤'=-+⎢⎥⎣⎦'≤-+'≤-+'≥⎰⎰⎰⎰⎰24设函数在上导数连续求证对任意正数有因故例证()()(0)0,f x f f π2-≥为单调增函数,从而又[][]2200()cos ()(),112()sin d (2)(0)()d (2)(0).f x nx f x f x f x x x f f f x x f f n n n ππππ'''≤='≤-+=-⎰⎰所以0000000000000()()d ()d d ,().()()d ()d ()d ()()d ()d d ()d d .()()()d ()d x x ux x uxu x ux ux uf u x u u f t t u f x f u x u u x u f t t x u f t t f t t u f t t u F x f u x u u f t t ⎡⎤-=⎢⎥⎣⎦⎡⎤-=-⎢⎥⎣⎦⎡⎤⎡⎤=-+⎢⎥⎢⎥⎣⎦⎣⎦⎡⎤=⎢⎥⎣⎦⎡⎤=--⎢⎥⎣⎦⎰⎰⎰⎰⎰⎰⎰⎰⎰⎰⎰⎰⎰25证明：其中为连续函数法一法二令例证00000000d ,()()d ()d ()d d ,()()d ()()()d 0,().(0)0,0,()0,()()d ()d d .x x x x uxxx x uu F x x f u u u f u u f t t u F x f u u x f x x f x f t t F x C F C F x f u x u u f t t u ⎡⎤=--⎢⎥⎣⎦'=+--≡====⎡⎤-=⎢⎥⎣⎦⎰⎰⎰⎰⎰⎰⎰⎰⎰⎰由于故所以又故因此有即1100111[0,1]()0,ln ()d ln ()d ..111ln ln ,n i nni i f C f x f x x f x x i f n n i i f f n n n n n ===⎡⎤∈>≥⎢⎥⎣⎦⎛⎫≥ ⎪⎝⎭⎡⎤⎛⎫⎛⎫≥ ⎪ ⎪⎢⎥⎝⎭⎝⎭⎣⎦→∞⎰⎰∑∑∑26设且证明：法一用定积分的定义由于两边取对数,有令取极限,注意到例证10111011100111lim ln ln lim ln ()d ,1lim ln ln ()d ,ln ()d ln ()d .()d ,0,()()lnln 11nn n n i i nn i i i f f f x x n n n n i f f x x n n f x x f x x a f x x a f x f x a a →∞→∞==→∞=⎡⎤⎡⎤⎛⎫⎛⎫⎡⎤== ⎪ ⎪⎢⎥⎢⎥⎢⎥⎣⎦⎝⎭⎝⎭⎣⎦⎣⎦⎛⎫= ⎪⎝⎭⎡⎤≥⎢⎥⎣⎦=>⎡⎛⎫=+- ⎪⎝⎭⎣∑∑⎰∑⎰⎰⎰⎰即得法二令由题设有故11110000111000()1,()1ln ()d ln d 1d ()d 10,ln ()d ln d ln ln ()d .f x a f x f x x a x x f x x a a f x x a x a f x x ⎤≤-⎢⎥⎦⎛⎫-≤-=-= ⎪⎝⎭⎡⎤≤==⎢⎥⎣⎦⎰⎰⎰⎰⎰⎰⎰两边积分得所以110111000111000[0,1],()d 0,()d 1.[0,1]() 4..[0,1]() 4.11()d ()d ()d 1,221111()d ()d 4d 1,222[0,f C f x x x f x x f x f x x f x x x f x x f x x x f x x x f x x x x ξξξ∈==∃∈≥∀∈<⎛⎫-=-= ⎪⎝⎭⎛⎫=-≤-<-= ⎪⎝⎭∃∈⎰⎰⎰⎰⎰⎰⎰⎰27设且证明：使用反证明法设有由题设从而有矛盾.所以假设不成立,即例证1]() 4.f ξ≥使(1)2[,][,],()()0,4max |()||()|d .()(,),()[,][,]Lagrange b ax a b f C a b f a f b f x f x x b a x a b f x a x x b ∈∈=='≥-∈⎰28设且证明设因在区间和上满足中值定理条例证1122[,]222()()()()((,)),()()()()((,)),max |()|,()()0,|()|(),|()|(),|()|d |()|d |()|d ()d ()d x a b a b b ba b aaa b af x f a f x a a x f b f x f b x x b M f x f a f b f x M x a f x M b x f x x f x x f x xM x a x M b x x ξξξξ∈+++'-=-∈'-=-∈'===≤-≤-=+≤-+-⎰⎰⎰⎰件,故记并注意到得于是222[,](),44max |()||()|d .()ba b b ax a b b a M f x f x x b a +∈-='≥-⎰⎰即2(2)0[0,2]222222201[0,2],(1)0.()d 3max |()|.Taylor ()()()(1)(1)(1)(2)(1)(1)(2),2!2()1()d (1)(1)d (2)d ()(2)d 22x f C f f x x M M f x f f f x f f x x f x x f f x x f x x x x f x x ξξξξ∈∈=≤''=''''''=+-+-=-+-'''''=-+-=-⎰⎰⎰⎰29已知证明：，其中由一阶公式故例证222222000220,11()d ()(2)d ()(2)d 2211(2)d .23f x x f x x f x x M x x M ξξ''''=-≤-≤-=⎰⎰⎰⎰⎰从而1100d 8[0,1],1()2,()d .()91()2,x f C f x f x x f x f x ∈≤≤≤≤≤⎰⎰30设且证明：因有例证[][]21101101100()1()20,()3()20.()0,2() 3.()d ()d 2 3.()d ()d 2()d 8()d .()9f x f x f x f x f x f x f x xf x x f x xf x x f x x f x x f x --≤-+≤>+≤+≤+≥≤⎰⎰⎰⎰⎰⎰即又因故有积分得由于故2[,],()[,],0,1()()d d .()4b b aaf C a b f x a b M m m M m f x x x f x M m∈>+≤⎰⎰推广：设在上的最大值和最小值分别为和且则有(1)2113002300300[,],(0,1),0()1,(0)0.()d ()d .()0,()(0)0,(0,1).()()d ()d ,[0,1],(0)0,()2()()d ()()2()d x xx xf C a b x f x f f x x f x x f x f x f x F x f t t f t t F C F F x f x f t t f x f x f t t '∈∈<<=⎡⎤>⎢⎥⎣⎦'>>=∈⎡⎤=-∈=⎢⎥⎣⎦'=-=-⎰⎰⎰⎰⎰⎰31设当时证明：因故有设则且例证[]220().()2()d (),[0,1],(0)0,()2()2()()2()1()0,(0,1),xf x x f t t f x C x f x f x f x f x f x x ϕϕϕϕ⎡⎤⎢⎥⎣⎦=-∈='''=-=->∈⎰令则且211300(0,1)()(0)0,()0((0,1)),()[0,1](1)(0)0,()d ()d .x x F x x F x F F f x x f x x ϕϕ∈>='>∈>=⎡⎤<⎢⎥⎣⎦⎰⎰从而当时于是即在上单调增加,因此有即21201()[0,],()cos d ()sin d 0,(0,),,()()0.()[0,](0,)sin 0()sin d 0()(0,).()[0,](0,),()0f x f x x x f x x x f f f x x f x x x f x f x f πππππξξξξπππππξξ====>==⎰⎰⎰3211设在上连续且证明：在内存在相异的两点使不妨设在上不恒为零.由在上及条件可知在内必定变号又由在上的连续性可知,在内至少有一点使例证1111111101110.()(0,),()(0,)(,)sin()(0,)(,)()sin()(0,)()sin()d 0.()sin()d cos ()sin d sin ()cos d 0,.()(0,f x f x x f x x f x x x f x x x f x x x f x x x f x ππππξπξξπξξξπξπξξξξπ---≠-=-=⎰⎰⎰⎰假设是在内的唯一零点则在和上异号.又因在和上异号,故在内不变号.于是另一方面矛盾因此在212)(),()0f ξξξξ≠=2内至少存在一点使.0000()()0,(),11[()]d ()d (0).1()d ,()d .()0,Taylor a a aay f x f x u u t f u t t f u t t a a a b u t t a ab u t t f x ''=≥=⎡⎤≥>⎢⎥⎣⎦==''≥⎰⎰⎰⎰33设二次可微,且求证：对任意连续函数必有记即由于由公式知例证20000()[()]()()[()][()]2!()()[()],[()]d ()()[()]d ()()()d ().0,11[()]d ()()d .a aaa a f f u t fb f b u t b u t b f b f b u t b f u t t a f b f b u t b ta fb f b u t t ab a f b a f u t t f b f u t t a a ξ'''=+-+-'≥+-'≥+-⎡⎤'=+-=⎢⎥⎣⎦>⎡⎤≥=⎢⎥⎣⎦⎰⎰⎰⎰⎰于是又因故有01010100011()[0,1](0)(1)0,()0,|()|d 4max |()|.()0,max |()|0.(0)(1)0,max |()|(0,1),(0,1)|()|max |()|.Lagrange ()x x x x f x f f f x f x x f x f x f x f f f x x f x f x f ξ≤≤≤≤≤≤≤≤==≡/''≥≡>==/∃∈='⎰341设在上具有二阶连续导数,且证明：因故又因故在内取得即使由中值定理,有例证221100100000220002100001000000()(0)(),(0,),(1)()()(),(,1).11|()|d |()|d ()d |()()|()()11|()|max |()|,1(1)(1)x f x f f x x x x f f x f x f x x x f x x f x x f x x f f f x f x f x f x x x x x x x x ξξξξξξξξξ≤≤-==∈--'==∈--''''''''≥≥=--=-=⋅=---⎰⎰⎰10因此而200001111(1),424|()|d 4max |()|.x x x f x x f x ≤≤⎛⎫-=--≤ ⎪⎝⎭''≥⎰10所以§3-3 定积分的应用本节最重要的知识是元素法。