第一節(jié)定積分的概baxdx(0a將區(qū)間[ab]等分為n份分點(diǎn)為ax0x1x2xn其中xaba.小區(qū)間長(zhǎng)度為xx b n(ababa那么xixi bxdxI

na

i1 2nab ban2n b2lim

2ni1 b

akdx(0ab)k解：將區(qū)間[ab]等分為n份分點(diǎn)為ax0x1x2xn其中xaba.小區(qū)間長(zhǎng)度為xx b bn那么kxi ,n

bxdxIlim n2

kbalimk(ba)k(ba). n2(1) 2(1) n3 in1n 2 2x2dxlim313i

lim316i9i22 ni1n

n ni1n

nlim

27i2

27n(n1)(2n1)

nn2

n33

i1 399x0a

n1nn

an那a1n1a1/nn n

aaxdxlim an

ni1n 1a1/n

n1ln ln設(shè)f(x)在區(qū)間[acbc]上可積證明：f(xc)在[ab]上可積 af(xc)dxacf證明：設(shè)acf(x)dxI,即將區(qū)間[acbc]acx0cx1cx2cxncb

I,

n

0于是可知：將區(qū)間[a,b]被分ax0x1x2xn任意分成n個(gè)小區(qū)間,小區(qū)間長(zhǎng)度為xixixi1i12,n記maxxi在每個(gè)小 上任取一點(diǎn)i[xi1xi做和式f(ic)xi滿足limf(ic)xiI,b

0nf(xc)dxlimf(ic)xi因此

0 af(xc)dxacf設(shè)b

f(x)00

x ,x[a,c)(c,求證:af(x)dx

ax0x1x2xn任意分成n個(gè)小區(qū)間,小區(qū)間長(zhǎng)度為xixixi1i12,n記maxxin上任取一點(diǎn)ii[xi1xi做和式f(i)xin0f(i

f(c)n于是當(dāng)趨近于零時(shí),根據(jù)夾迫性原理可知 f(i)xi0即baf(x)dx成另一個(gè)函數(shù)f*(x),證明：f*(x)也在區(qū)間[a,b]可積,其積分也是I. 證明：首先設(shè)將函數(shù)f(x)改變m個(gè)點(diǎn)x x后得到函數(shù)f*( mf(xij)Mj由于f(x)在區(qū)間[a,b]可積且積分為I我們知道區(qū)間[a,b]被分ax0x1x2xn任意分成n個(gè)小區(qū)間,小區(qū)間長(zhǎng)度為xixixi1i12,n記maxxi在每個(gè)小區(qū)nnnlimf(i)xi0i1 那么對(duì)任意的0,0,當(dāng) ,)時(shí)有f(i)xi

nf*(i)xi

nf(i)xi

mfj

nf(i)xi因此函數(shù)f*(x)也在區(qū)間[a,b]可積,其積分也是

M 第二節(jié)定積分的基本性設(shè)f(x)在[ab]連續(xù),f(x0,f(x)不恒為零證明a

f(x)dx證明：設(shè)有一點(diǎn)x0[ab]使得f(x00,那么有函數(shù)極限的局部保號(hào)性原理可知當(dāng)0x

時(shí)f(x0.于是有xxf(x)dx dx0x0 x0又由于函數(shù)f(x)在區(qū)間[ab]上滿足f(x)0, x0f(x)dx x0則

f(x)dx bf(x)dxx0f(x)dxx0f(x x0 x0

f(x)dx0設(shè)f(x)在[ab]上連續(xù),bf2x)dx0,證明f(x)在區(qū)間[ab]a證明：由于函數(shù)f(x)在[ab]上連續(xù)可知函數(shù)f2x)在[ab]上連續(xù),又因?yàn)楸赜衒2x0,如果函數(shù)f2(x)不恒為零,則由上題可知bf2(x)dx0,這與題意a因此必有f2x0,于是f(x)舉例說明f2(x)在[a,b]上可積,f(x)在[a,b]上不可積。 D(x)1

.顯然有D2(x)1,在任意區(qū)間上可積;而函數(shù)D(x)在任意區(qū)間上不可積001xdx,100 解：在區(qū)間[0,1]上必有xx2成立,1xdx1 2xdx,2sin 22解：在區(qū)間[0,]上必有xsinx成立,因此有xdxsin22 1(1)xdx,13x3 3解：由上一節(jié)T可知()dx

12

dx

1911 ( 2 22

0

3x成立,因此有()dx11

dx

3x 2 0 11ex2dx0 證明：在區(qū)間[0,1]上顯然有1ex2e,11dx1ex2dx1edx,即11ex2 sin 12 dx;證明：在區(qū)間 sin sin [0,]上 ,于 sin dx dx20 sin 2 dx. 11sin2211sin22證明：在區(qū)間上 單調(diào)遞增11sin2211sin2222 2

因此

e4elnxdxee lnx lne2 上 e

ln 上 .因有 4elnxdxe

4e1dx4elnxdx4e2 e2lim1xndx0;(2)limsinnxdx2n01 n設(shè)f(xg(x)在區(qū)間[ab]連續(xù) limf(i)g(i)x

f

其中ax0x1x2xnbxixixi1,i,i[xixi1i12,maxxi證明：由f(x)在區(qū)間[ab]連續(xù),可知其在此閉區(qū)間上有界,即存在M0,使得對(duì)x[a,b],有f(x)M.又由于g(x)在區(qū)間[ab]連續(xù),因此在區(qū)間[ab]上一致連續(xù);于是對(duì)10,10,對(duì)x1x2[ab],x1x21時(shí),有g(shù)(x1)g(x2) .2M(b由于f(xg(x)在區(qū)間[ab]連續(xù)那么f(x)g(x)在區(qū)間[ab] 00

fn即對(duì)0,0,當(dāng)limn

f()g()x

bf(x)g(x)dx

那么我們?nèi)?此時(shí)由于i,i[xi,xi1因此ig(ig(i) 2M(b

xi

f(i)g(i)xi

fnf(i)g(i)xi

f(x)g(x)dxf(i)[g(i)g(i f(i)g(i)xi

f(x)g(x)dxf(i)[g(i)g(i

2M(bn

b

(ba) 2M(b

f(x)g(x)dx成立0 設(shè)01,

1(1t2)n n1(1t2)n0證明：由于在區(qū)間[0,1]上函數(shù)(1t2)n是非負(fù)且單調(diào)下降的,因 0lim1(1t2)n

1(1t2)n n1(1t2)n0

2(1t2)n1(12)n 2lim (1lim n 2n2(1)n [1()]

n1 nn 1(

1(1t2)n n1(1t2)n0b9.(1)設(shè)f(x)在[a,b]上連續(xù),且對(duì)[a,b]上任一連續(xù)函數(shù)g(x)均有 f(x)g(x)dx0,證明bf(x)0,x[a,證明：取連續(xù)函數(shù)g(x)f(x),bf2x)dx0,于是由T可知f(x0,x[a 設(shè)f(x)在[ab]上連續(xù)且對(duì)所有那些在[ab]上滿足附加條件g(a)g(b)0的連續(xù)函數(shù)b有af(x)g(x)dx0,證明在區(qū)間[ab]上同樣有f(x)證明：由于函數(shù)f(x)在閉區(qū)間[a,b]上是連續(xù)函數(shù),因此在該區(qū)間上有界,即存在 0,對(duì)x[ab]有f(xM對(duì)于0,我們?nèi)缦聵?gòu)造函數(shù)g(x取min(ba ),當(dāng)sa,tb時(shí),構(gòu)造函 2M s f(s)(xs)f x s 顯然

g(x)ff(t)(x f(t) t

x[s,t]x(t, sf(x)g(x)dxaf(x)g(x)dxaf(x)g(x)dxtf af(x)g(x)dxaf(x)g(x)dx

f0af(x)f(s)dxtf(x)fat0sf2(s)dxbf2(t)dxf2(s)asf2(t)bat[f2(s)f2(t)]2M 2M s s s于是有 f(x)f(x)dx f(x)g(x)dx t tt由于函數(shù)F(stsf(xf(x)dx是連續(xù)函數(shù) 因此可得f(x)0,x[a

f(x)f(x)dxt

f(x)f(x)dx設(shè)f(xg(x)在[ab]上連續(xù)求證 bf(x)g(x)dx bf2 bg2(x)dx 而且等號(hào)成立當(dāng)且僅當(dāng)g(x)f(x)(或f(x)g(x其中為常數(shù)證明：若f(xg(x)在[ab]上連續(xù)則f2xg2x),f(xg(x)在[ab]上都連續(xù),故可積那么對(duì)于任意的實(shí)數(shù)有[f(x)g(x)]2在[ab]上也可積,且有b[f(x)g(x)]2dxa f2(x)dx f(x)g(x)dx g2(x)dx 那么關(guān)于實(shí)數(shù)的二次三項(xiàng)式的判別式非正,即 4[bf(x)g(x)dx]24bf2(x)dxbg2(x)

bf(x)g(x)dx bf2 bg2 顯然當(dāng)且僅當(dāng)g(x)f(x)b[f(xg(x)]2dx0此時(shí)判別式0,a設(shè)f(xg(x)在[ab]上連續(xù)

b[f(x)g(x)]b[f(x)g(x)]2aa而且等號(hào)成立當(dāng)且僅當(dāng)g(x)f(x)(0為常數(shù)b[f(x)g(x)]2b[f(x)g(x)]2aab[f(x)g(x)]2a2b[f(x)g(x)]2a 即 b[f(x)g(x)]2dxbf2(x)dxbg2(x)dx bf2(x)g2 由上題可知bf(x)g(x)dx bf2 b[f(x)g(x)]2dxbf2(x)dxbg2(x)dx2bfa

bf2(x)dxbg2(x)dx

bf2(x)g2a設(shè)f(x)在[0,1]上連續(xù)f(x0 dx 證明：由T10

0f

10f dx1f(x)dx

f

0f 即 dx1f(x)dx1.于是0f

f dx 0f

10f設(shè)y(x)(x0)是嚴(yán)格單調(diào)增加的連續(xù)函數(shù),(0)0,xy)是它的反函數(shù)證 0(x)dx0(y)dy (a0,b證明：i我們首先

(a0(x)dx (y)dy由于(x)是嚴(yán)格單調(diào)增加函數(shù),因此它的反函數(shù)y)也是嚴(yán)格單調(diào)增加函數(shù);因此它們分別在區(qū)間[0,a],[0,(a)]上可積。將區(qū)間[0,a]分為n份其各分點(diǎn)為0x0x1x2xn那么對(duì)應(yīng)該分點(diǎn)有yi(xi)(i0,12,又因?yàn)?x)在[0,a]上連續(xù),那么就一致連續(xù)。于是對(duì)0,10,maxximax(xixi1) 時(shí),(xi1(xi那

a(x)dx(a)(0 lim(x)x( i

i limy(x )

(y n

i1lim[yi(xixi1)xi1(yiyi1

lim[yixiyixi1xi1yixi1yi1lim[yx

)]lim(y

yxin

n 0 那么當(dāng)b(a)時(shí),有0(x)dx0y)dy 當(dāng)0b(a)時(shí),由介值定理可知x0(0a)使得(x0)b,于是由于(x)是嚴(yán)格單 0(x)dx0( x0(x)dxa(x)dx(x0)(0a0x0(x)dxx(x0)(y)dy (x0a0 0x0(x0)(ax0)(x0)a(x0 0(x)dx0( (y0)(x)dxb(y)dxb(0(y0)(x)dx

y0(y)dxb

(b

0b(y0)(by0)(y0)b(y00 0(x)dx0(y)dy

(a0,bf(x3x在[0,1]f(xsinx在()上是一致連續(xù)的f(xx2在[ab]上一致連續(xù)但在()f(xsinx2在()上非一致連續(xù)。(1)對(duì)于任意的x1x2[0,1],由于

x1333

x2/3x1/3x1/3

1/3 x1x13x1x13

那么對(duì)于0,3,x3x13x1

因此f(x3x在[0,1](2).對(duì)于x1x2x12sinxsinx2sinx1x2 2 2xx12

xx 于是對(duì)于0,,x1

時(shí),有sinx1sin

成立因此在(,)上sinx一致連i.對(duì)于x1x2[ab 12 x2x2x xxxxxx2bxx 12 于是對(duì)于0,,xxx2x22bxx成立 因此在[a,b]上x2一致連ii.取1,無(wú)論多么小,總存在x1x1xx x1x2x1x2x1x21 14因此在(,)上x2非一致連取x ,x 2n sinx2sinx21,但此 22n222n2

0(n于是存在01,對(duì)于任意的0,x1sinx2sinx

1 因此在(,)上sinx2非一致連 1

第三節(jié)微積分基本定 0cosxdx2x2sin2x 2 2222

axdx

2a

a3 1sinxdx

cosxdx2sinx201 xx21xxx21x

d

1x1x2

1 22x 2x 2ln ln2 xdx1lnxdlnx 21

lnxdx lnxdxelnxdxxlnxx1xlnxxe22 1 11e 1e n limsin

sin(x)dx cos( nk1

lim1 1lim lim 1lim n 2n ni1n ni1ni ni11i 11dxln(1x)1ln01 n1 n1k limn2limnn0xdx 2nklim1

nk

lim nn n 1111cos(x)dx1 cos(x)dx1 sin 11 2 若f(x)連續(xù)求F'(xF(x)0f解：我們首先設(shè)G'(xf(x),( F'(x)limF(xx)F(x)lim f(t)dt

f (lim f

22

lim [G(x)]'G(x)x'2xf(x bF(x)xfb解：我們首先設(shè)G'(xf(x), F'(x)limF(xx)F(x)limxxf(t)dtxf

f

G(x) limG(x)G(xx)limG(xx)G(x)G'(x)f x3F(x)xe

解：我們首先設(shè)G'(xex2則F'(x)limF(xxF (xx)3

( lim edtxf

G((xx)3)G(xx)[G(x3)limG((xx)3)G(x3)limG(xx) [G(x3)]'G'(x)ex6x3'ex23x2ex6ex2 法 cos et 法et 法0

eedt1lim

lim lim0

lim 0

e2t2

e2

xx求函數(shù)F(x1tdtxx

解：當(dāng)x[1,0]時(shí),F(x1tdt1tdt

t2 t2

當(dāng)x(0,1]時(shí),F(x綜上

tdt

tdt

tdt 1

1F(x)

x[1,. 1x x設(shè)f(x)在[0,)F(x)1xf(t)dtx0在區(qū)間(0,)上連續(xù)且單調(diào)遞1xf(t)dt在區(qū)間(0)內(nèi)都連續(xù),因此函數(shù)F(x1xf(t)dtx x連續(xù)?？梢郧蟮煤瘮?shù)F(x)的導(dǎo)函數(shù)為F'(x

xf(x)

xf, xf(x)

f xf(x)

f

xf(x)xf由于函數(shù)f(t)單調(diào)遞增,因此有F'(x)

于是可知函數(shù)F(x)單調(diào)遞設(shè)f'(x)在(ab)連續(xù)且f(a0

(b2af(x)dx maxf'(x)2a 證明：取Ma

f'(x由于f(xaf'(t)dtf(aaf b baf(x)dxaaf'(t)dtdxaabMaadtdxbb

a(xa)dxb(bb

(b(x(x22a綜上即 f(x)dx a

f'(x)第四節(jié)定積分的計(jì)2

12(x dx2x11

13 x

x 9ln214011ln 2 1x2 11 0x41dx20x2 2x1x22x1 2 2 2 2x2

2x2 2 2 25x 2t2 55

2

td

12t2t 14t22t 3 dt t3 3

31319

1dx2x32

9186

4

444 14x2dx 4arcsinx1 2arcsin10 3 2 a xasint

a2x2

2

sint

a

tdasint 2a4

tcos2tdta421

a4t1sin4t 0 4 a2.

2sinmxcos

1sin(mxnx)sin(mx211 0211 2sin(mxnx)dx

2sin(mx2 2

cos(m

2 2(m

cos(m0

cos(mn)

cos(mn)2(m 2(m 2(m 2(m mcosmcosnnsinmsin m2

2m21114

2

13/1

32 3/3

3xx

0 1

4 2 23 415 11 2

xx xdx x24 420

cos

dx

2dsinxarctansinx2arctan11sin2 11sin2 x xt tx

edx

2tedt2te

(2e2e)(02)0 x2 0xarctanxdx0arctanxd

arctan

1

darctan2

0 1 1xarctanx118201x2dx

2x2cos2xdx1x31x2sin2x1xcos 1sin2x24

x2cos2xdx

x3

x2sin2x

xcos2x

sin

ln

2

2x2t

lnt

tln

x3ex2dx1ln2x

dx

tedt

te

ln21 2 2 1ln2

sinmxcosnxdx

cos(mn)x

cos(m

a0

a aa

2(m 2(m ax2(a0

aa(x2a2)a3dx a2 2aaa2x2dxa3 1a2 2aa2adxa31002a(a2x2aa2adxa31002

a2a2a2 4a3a2xax22x3

a 1a3(3

aa

x1x 2a a2 1

dx

xdx

11d

3

a a 1

a

13a2x2

3

4a2

8a2 a

5x3(15x2)10dx2010

5x2(15x2)10dx2 50

5(15x2)11d15x2 50

5(15x2)10d15x2150 150 (15x2)12 (15x2)11 2sin9

解：I2sin0xdx

解：I0sin0xdx

I1sin1xdxcosxI2sin1xdxcosx2

I2sinnxdx2sinn1xdcos Insinnxdxsinn1xdcos

cosxsinn1x22cosxdsinn1

(n1)sinn2xcos22(n1)sinn2xcos20

(n1)sinn2x(1sin22(n1)sinn2x(1sin2

(n

(n1)I0n(n1)In2(n1)Inn

nn 于是有In In2,那于是有In In2,那 I8642I384

I5531I115215128x29 9711 945 3x2

0

xdx

dx

2sin6

1 1 I0 2sin0tdt2,I 2sin1tdt

I66

2cos7xdx2sin7xdxsin7 2 I sin0tdt3,I 2 2

tdt

2cos7xdxI

I135 xa (a2x2)n a2n2(1cos2t)nd(acost)a2n1sin2n1tdta2n12sin2n 2

I0

2sin0tdt

2,I1

2sin1tdt0a(a2x2)ndxa2n1I0

a2n1 (2k)(2kI(2k1)(2k1

a2n1(2k)!!(2k xcost (1x2)6dx2(1cos2t)6d(cost)sin13tdt2 2

I0

2sin0tdt

2,I1

2sin1tdt1(12)6dxI 2I (12)!!1024. 證明：i.若函數(shù) (x)為奇函數(shù),則其所有原函數(shù)可以表示xF(x)0f(t)dt 由于f(t)是奇函數(shù)那么有xf(t)dt0f(t)dt, F(x) f(t)dtCxf(t)dtC0f(t)dtCF因此奇函數(shù)的所有原函數(shù)都是偶函若函數(shù)g(x)為偶函數(shù),xG(x)0g(t)dtx 由于g(t)是偶函數(shù)那么有xg(t)dt0g(t)dt, G(x)0g(t)dtCxg(t)dtC0f(t)dt若要使得G(xG(x那么必有C因此偶函數(shù)的所有原函數(shù)中只有一個(gè)是奇函數(shù)。設(shè)f(x)在所示區(qū)間是連續(xù)函數(shù) 2f(sinx)dx 2f(cos

x

0f(sin

)dt f(cost)dt2f(cost)dt2 2000 xf(sinx)dxf(sin 證明：xf(sinx)dx(tf(sint)dttf(sint)dtf(sin 0tf(sint)dt0f(sin

xf(sinx)dxf(sin2 22aa

a2f(x2x2)x

af(x

a)dx x2t

dt a

a

f(x2a)

f(t

f(ta

f(t 1

x

t2af(x2t21

2 aa) a)

f(x

a)dxa 1

0xfa

)dx 212

xf00 x2t1 100證明:

x3f(x2)dx2

x2f(x2)dx22

tf(t)dt2

xf(x)dx計(jì)算積分0

sincosxsin

I sin cosxsinx cosxJ cosxsinIJsinxcosxdxx

cosxsinxcosxsinxcosxsin

dx

dcosxsincosxsin

lnsinxcosxI sinx dxxlnsinxcosxC,cosxsinx 那么 可20

sincosxsin

.xsinx.xsinxcos240 xf(u)(xu)duxuf(t)dt 證明：xuf(t)dtduuuf(t)dtxuduf(t)dt 0uf(t)dt0uf(u)du0xf(u)du0uf(u)du0(xu) 設(shè)f''(x)在[ab]上連續(xù)且f(af(b求證：(1)bf(x)dx1bf''(x)(xa)(x 2 (b aaf(x ax

f''(x)b證明：可以知道if(x)af'(t)dtiif(x

ff'(x)af''(t)dtf'(a),iv.f'(x)f'(b)xfb 1af(x)dxaaf'(t)b xxf'(t)dtbbxdxf'(t)dtbbf'(t)dtbxf b

b(xb)f'(x)dxiii

(xb)

f''(t)dtf'(a) 1(xb)2xf

1bxb2 xf 1(xb)2f 2 1b(xb)2f''(x)dx1(ab)2f2 1b(x22bxb2)f''(x)dx1(ab)2f 2 f(x)dx f xbf'(t)dtbbxdbf'(t)dtabf'(t)dtbxfb b '(x)dx

(x

f''(t)dtf'(a)a(xa)

1(xa)2xf''(t)dtb1bxa2dxf''(t)dt1(xa)2f'(a) 2 1(ba)2bf''(t)dt1bxa2f''(x)dx1(ab)2f1 2 bxa2(ba)2f''(x)dx (ab)2f2a 1b(x22ax2abb2)f''(x)dx1(ab)2f2 將12b2f(x)dxb

1

2bx

2ax2ab

)f 2 b(x2bxaxab)f''(x)dxb(xa)(xb)f 于是有bf(x)dx1bf''(x)(xa)(xb)dx 2(2)取Ma

f''(xbf(x)dxa

1bf''(x)(xa)(xb)dx22

b(xa)(xa

M (xa)(ba)dx a(xM(ba)3M(ba)3

2 bf

(ba)3

f''(x)

a設(shè)f(x)在x0時(shí)連續(xù),對(duì)任意ab0積分abf(x)dx與a無(wú)關(guān)求證:f(xc(c為常數(shù) 那么由于F(a)

F(a)afdF(a)bf(ab)f(a)于是有f(abf(a令a1可得