微積分早期復(fù)習(xí)注意修改時間微二第8周習(xí)題課答案

af(x區(qū)間[abF(x)（注釋：在區(qū)間[ab可積的函數(shù)未必有原函數(shù)）則有bf(x)dxF(b)F(a)．a(chǎn)證明：對于區(qū)間[ab的任意分割Tax0x1·xnb F(b)F(a)[F(xi)F(xi1)]F(i)xi，i(xi1,xi) b當(dāng)分割的直徑af(x)dxbfx在[a,b]f(x0(ba)f(a)bf(x)dx(ba)f(a)f(b) f(x)在[ab]上嚴(yán)格單增，xabf(x)f(a)令g(x)f(bf(a(xaf(a)，易證bg(x)dx(ba)f(afb 而f"(x)0f(xf(a)x

f(b)f(a)b

f(b)f b 易知xabf(xg(x)f(x在[ab上可積，則0，存在區(qū)間[abg(xab|f(x)g(x)|dxa證明：0，由定理（定理2.1.4）推出，存在0，使得對于直徑的任n劃分Tx0x1x2·xn}(Mkmkk

今取一個滿足直徑的確定的劃分Tx0x1x2·xn}g(x)mk,x[xk1,xk)(k1,2,·,n)n |f(x)g(x)|dx(f(x)g(x))dxx

(f(x)m kkk

(Mkmk)dxk1kf(x在[ab上可積，求證函數(shù)expf(x在[abf(x在[abMsup{|f(x|:axb}．對于區(qū)間[abT{x0,x1,x2,·,xnMisup{f(x)|xi1xxi}，miinf{f(x)|xi1xxi}，iMimiuv[xi1xi]|exp[f(uexpf(v|exp()|f(uf(v|Mi，其中介于f(u),f(vMexp(M)．i對于上述劃分Tx0x1x2·xn}M*sup{exp[f(x|im*inf{exp[f(x| xx}*M*m* n

xxin*x

(M*m*

nsup{exp[f(u)]exp[f(v)]: xx}

nnMsup{f(u)f(v):xi1xxi}

Mixi f(x可積，當(dāng)劃分直徑趨向于零時，ixi0

ixi0即函數(shù)exp[f(x在[abf(x在區(qū)間[abf([abAB]g(u在區(qū)間AB可積。能否斷定函數(shù)g(f(x))在區(qū)間[ab]可積？試研究函數(shù)f(x)

若x

，g(u)

u若u0n n解：不能斷定函數(shù)gf(x))在區(qū)間[ab]1

n所以其振幅為1，則limkk

nnk

1１．lim1[(n1)(n2)·2n]nnS

[(n1)(n2)·2n]n[(1

2)·1

AlnS1[ln(11)ln(12)·ln(1n)]1ln(1x)dx2ln21 lim

4e２．lim1x2sin2nxdxn 將[0,1]n等分

x2sin2n

k

n

x2sin2n2積分中值定理 2 k1

n

sin2n換元積分 2 11 nk

sintdt

2n

2

xdx 6k

(k

k nx,0x1 ３．設(shè)

(x)nn

gn(x)(xk

)．求極限 0

exgn(x)dx1

knn

(x)dx

e(x

nk1 k )dx nekk1(x n

nk

k 1 eknnxdx

ek

exdx

(e1k

2n 2 ff f() 1 i 據(jù)．設(shè)定積分f(x)dx存在，則當(dāng)n時，兩個和式：Sn f

n)) 1nf(2i)

1f(x)dx n （1）|0f(x)dxSn|2nM （2）|0f(x)dxn|4nMMmax{|f(x)|ax1

1 k kn|n

f(x)dx

f(x)dx f

)|

|f(x)f )|nnk1n

k

k1 k k k

k Mn1 2kn|f

)(x )|dx

n(x

)dx

k1

k

k

n 2k（2）|

f(x)dxn|n1|f(x)f(2n)|kk1k 2k

|f(k)(x )|k k1 k

2kMk

n

|x

| 2k n1 22Mn1|x |dx2M 2

k1 k12

k1

2 １．lnxdx(p0) ２．lnxdx(p0) x xln(1 x1３． x dx(p0) ４．1 )dx(p0) 1x3x(x2)2(x103x(x2)2(x100５．2(lnsinxlnsin0cos(ln 1 dx 1(1 0p1p1時收斂。３．1p2 x1 ４．1ln )dx1ln(1 )dxp1p1 1x 1x0 2

1limx2lnsin2

00lnsin

．因?yàn)?xlnsin2

0101

dx

dxx0,x2x4 33x(x2)2(x

x3(x2)3(x4) 13limx13

11

x(x2)2(x4)dx3x(x2)2(3x(x2)2(x0

x3(x2)3(x4) cos(ln cos ７．解 dx 1 01

etdt 2n 12n因?yàn)閘im 1，所以當(dāng)n充分大時，有 122n21

etdt122n 2散．８．解：因?yàn)?cosusinu

1u21u4o(u4)1u2o(u4) u4o(u4， 2 ， 1x所以1x

sin2x24x2o(x2)limx2(1cos1sin11x x１．x3ex2dx ２．a(chǎn)rctanxdx ３． xln dx x (1x2/1

lnsinxdx１．，換元法、分部積分法；21２．(ln4)4因?yàn)閤0是其唯一奇點(diǎn)，而

3xlnsinxlim2x2cosxlim2xcos

sin sin yx 2lnsinxdx2ln2sincosdx 2ln2dx2ln dx2lncos

0y ln22lnsinxdxlnsinxdxln2lnsin 2ln222lnsin x2lnsin0

2ln2

三、證明題（１）舉例說明：f(x)dx收斂未必有

f(x0 a（２）f(x在[a,a

f(x)dx

f(x)01,nxn（１）解：例如f(x

12,則f lim

f(x)0

Nn1（２）反證：若

f(x)0不成立