微积分早期复习注意修改时间微二第8周习题课答案

af(x区间[abF(x)（注释：在区间[ab可积的函数未必有原函数）则有bf(x)dxF(b)F(a)．a证明：对于区间[ab的任意分割Tax0x1·xnb F(b)F(a)[F(xi)F(xi1)]F(i)xi，i(xi1,xi) b当分割的直径af(x)dxbfx在[a,b]f(x0(ba)f(a)bf(x)dx(ba)f(a)f(b) f(x)在[ab]上严格单增，xabf(x)f(a)令g(x)f(bf(a(xaf(a)，易证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上可积，则0，存在区间[abg(xab|f(x)g(x)|dxa证明：0，由定理（定理2.1.4）推出，存在0，使得对于直径的任n划分Tx0x1x2·xn}(Mkmkk

今取一个满足直径的确定的划分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上可积，求证函数expf(x在[abf(x在[abMsup{|f(x|:axb}．对于区间[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对于上述划分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可积，当划分直径趋向于零时，ixi0

ixi0即函数exp[f(x在[abf(x在区间[abf([abAB]g(u在区间AB可积。能否断定函数g(f(x))在区间[ab]可积？试研究函数f(x)

若x

，g(u)

u若u0n n解：不能断定函数gf(x))在区间[ab]1

n所以其振幅为1，则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 将[0,1]n等分

x2sin2n

k

n

x2sin2n2积分中值定理 2 k1

n

sin2n换元积分 2 11 nk

sintdt

2n

2

xdx 6k

(k

k nx,0x1 ３．设

(x)nn

gn(x)(xk

)．求极限 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 据．设定积分f(x)dx存在，则当n时，两个和式：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时收敛。３．1p2 x1 ４．1ln )dx1ln(1 )dxp1p1 1x 1x0 2

1limx2lnsin2

00lnsin

．因为2xlnsin2

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因为lim 1，所以当n充分大时，有 122n21

etdt122n 2散．８．解：因为1cosusinu

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

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

lnsinxdx１．，换元法、分部积分法；21２．(ln4)4因为x0是其唯一奇点，而

3xlnsinxlim2x2cosxlim2xcos

sin sin yx 2lnsinxdx2ln2sincosdx 2ln2dx2ln dx2lncos

0y ln22lnsinxdxlnsinxdxln2lnsin 2ln222lnsin x2lnsin0

2ln2

三、证明题（１）举例说明：f(x)dx收敛未必有

f(x0 a（２）f(x在[a,a

f(x)dx

f(x)01,nxn（１）解：例如f(x

12,则f lim

f(x)0

Nn1（２）反证：若

f(x)0不成立