微积分答案
高等数学教材微积分课后答案
高等数学教材微积分课后答案第一章微积分基本概念1. 第一节课后习题答案1.1 单项选择题1. A2. B3. C4. D5. A1.2 填空题1. 42. 273. 184. 05. 21.3 解答题1. (a) 首先将函数对x求导,得到f'(x) = 6x^2 + 12x - 8。
令f'(x) = 0,解得x = -2和x = 2/3。
然后再带入原函数,得到f(-2) = 0和f(2/3) = -1/27。
因此,函数在x = -2和x = 2/3处取得极值,极大值为0,极小值为-1/27。
(b) 由于f'(x) = 6x^2 + 12x - 8 > 0,说明函数在(-∞, -2)和(2/3, +∞)上为增函数;当-2 < x < 2/3时,f'(x) < 0,说明函数在(-2, 2/3)上为减函数。
结合图像,可以得到函数的单调性为:在(-∞, -2)上递增,在(-2, 2/3)上递减,在(2/3, +∞)上递增。
2. 第二节课后习题答案2.1 单项选择题1. C2. A3. D4. B5. C2.2 填空题1. 82. 123. 04. -∞5. +∞2.3 解答题1. (a) 首先求函数的导数,得到f'(x) = 2e^x - 12x。
令f'(x) = 0,解得x = ln6。
然后带入原函数,得到f(ln6) = 4ln6 - 6ln^2(6)。
因此,函数在x = ln6处取得极值。
(b) 由于f'(x) = 2e^x - 12x > 0,说明函数在(-∞, ln6)上为增函数;当x > ln6时,f'(x) < 0,说明函数在(ln6, +∞)上为减函数。
结合图像,可以得到函数的单调性为:在(-∞, ln6)上递增,在(ln6, +∞)上递减。
第二章微分学中值定理1. 第三节课后习题答案1.1 单项选择题1. B2. D3. C4. A5. D1.2 填空题1. 42. 53. π/24. √35. 01.3 解答题1. 根据罗尔定理,首先证明f(x)在区间[0, 1]上连续。
《微积分》课后习题答案
习题五 (A )1.求函数)(x f ,使)3)(2()(x x x f --=',且0)1(=f .解:6x 5x )(f 2++-='xC x x x x f +++-=⇒62531)(236230625310)1(=⇒=+++-⇒=C C f 62362531)(23+++-=x x x x f2.一曲线)(x f y =过点(0,2),且其上任意点的斜率为x x e 321+,求)(x f .解:x e x x f 321)(+=C e x x f x ++=⇒341)(21232)0(-=⇒=+⇒=C C f1341)(2-+=⇒x e x x f 3.已知)(x f 的一个原函数为2e x,求⎰'x x f d )(.解:222)()(x x xe e x f ='=⎰+=+='C xe C x f dx x f x 22)()(4.一质点作直线运动,如果已知其速度为t t dtdxsin 32-=,初始位移为20=s ,求s 和t 的函数关系.解:t t t S sin 3)(2-=C t t t S ++=⇒cos )(31212)0(=⇒=+⇒=C C S1cos )(3++=⇒t t t S5.设[]211)(ln x x f +=',求)(x f .解:[]1arctan )(ln 11)(ln C x x f x x f +=⇒+=')0()(arctan arctan 1>==⇒+C Ce e x f x C x6.求函数)(x f ,使5e 1111)(22+--++='x x x x f 且0)0(=f .解:C x e x x x f e x x x f x x ++-++=⇒--++=+521arcsin 1ln )(1111)(252 21002100)0(=⇒=++-+=C C f 21521arcsin 1ln )(2++-++=⇒x e x x x f x7.求下列函数的不定积分 (1)⎰-x xx x d 2(2)⎰-)1(t a dt(3)⎰mnx x d (4)⎰+-x xx d 1122(5)⎰++x x x d 114 (6)⎰++x xx xd cos sin 2sin 1(7)⎰+x x x x d cos sin 2cos (8)⎰++x xxd 2cos 1cos 12(9)⎰x x x xd cos sin 2cos 22 (10)x x x d sin 2cos 22⎰⎪⎭⎫⎝⎛+ (11)⎰-x xx x d cos sin12cos 22(12)⎰+-x xx d 1e 1e 2 (13)⎰⨯-⨯x xxx d 85382 (14)x xx x d 105211⎰-+-(15)⎰-x xx -x x d )e (e (16)⎰++x xx x d )31)(2e ( (17)x x x xx d 1111⎰⎪⎪⎭⎫⎝⎛+-+-+ (18)⎰----x x x x x x d 151)1(222(19)x xx d 1142⎰-+ (20)⎰-+-x xx xd sincos 1cos 1222(21)⎰+-+x x x x x d )1(1223 (22)⎰+-x x x x d 1224解:(1)=⎰+-=-C x x dxx x 252323215232)( (2)=⎰+-=--C tatt d a2121)1(2)1()1(.1(3)=⎪⎪⎪⎪⎩⎪⎪⎪⎪⎨⎧=+=-=+=≠-≠++=⎰⎰⎰+0 0, m C x dx n m C x In dx x m n m C x m n m dx x m n m m n m n(4)=⎰+-=⎪⎪⎭⎫⎝⎛+-C x x dx x arctan 2 121(5)=C x x x dx x x x x ++-=++-+⎰arctan 2311)1(32222(6)=⎰⎰++=+++dx xx x x dx xx xx x x cos sin )cos (sin cos sin cos sin 2cos sin 222=⎰+-=+C x x dx x x cos sin )cos (sin(7)=⎰⎰-=+-dx x x dx xx xx )sin (cos cos sin sin cos 22=C x x ++cos sin (8)=⎰⎰++=⎪⎪⎭⎫⎝⎛+=+C x x dx x dx xx2tan 21 1cos 121cos 2cos 1222 (9)=⎰⎰+--=⎪⎪⎭⎫ ⎝⎛-=-C x x dx x x dx x x xx tan cot cos 1sin 1cos sin sin cos 222222 (10)=⎰⎰⎪⎭⎫ ⎝⎛+-=-++dx x x dx x x 122cos 2cos 22cos 121cos =C x x x +-+2sin 41sin 21(11)=⎰⎰+-=-=---C x dx x dx xx xx x x tan 2cos 12cos sin sin cos sin cos 2222(12)=()⎰+-=-C x e dx e x x 1(13)=⎰⎰+⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-=⎪⎭⎫⎝⎛-C x dx dx xx85ln 85328532(14)=⎰⎰++-=⎪⎭⎫ ⎝⎛-⎪⎭⎫⎝⎛--C dx dx x x xx22ln 5155ln 22151512(15)=⎰+-=⎪⎭⎫⎝⎛-C x e dx x e x x ln 1(16)=[]⎰+++++=+++C e e dx e e xx x xxxxx6ln 63ln l )3(2ln 2)3(26(17)=⎰⎰+=-=--++C x dx xdx xx x arcsin 211211122(18)=⎰+--=⎪⎪⎭⎫⎝⎛---C x x x dx x xx arcsin 5ln 21151222 (19)=⎰+=-C x dx xarcsin 112(20)=⎰⎰+-=⎪⎪⎭⎫⎝⎛-=-C x x dx x dx xx2tan 211cos 121cos 2cos 1222 (21)=⎰⎰+++=⎪⎪⎭⎫ ⎝⎛++-=+-+C x x x dx x x x dx x x x x arctan 1ln 1111)1(1)1(22222 (22)=⎰⎰++-=⎪⎪⎭⎫ ⎝⎛++-=+++--C x x x dx x x dx x x x arctan 22312212)1(13222248.用换元积分法计算下列各题. (1)⎰+-x x x d 24 (2)⎰-x x d )23(8(3)x xxd e 3e 42⎰+ (4)⎰⎪⎭⎫ ⎝⎛+32cos d 2πx x(5)⎰-x xx d 432 (6)⎰+-52xd 2x x(7)⎰-+xxxe ed (8)⎰--xxxe e d(9)⎰-1tan cos d 2x xx(10)⎰)ln -(1d x x x(11)⎰-xx x2ln 1d (12)⎰-x xx d e9e 2(13)⎰+x xxx d sin2cos sin (14)⎰-x x x d 212(15)x xx x d 1arctan 2⎰++ (16)⎰+xxe1d(17)x x x d 11arctan2⎰+ (18)⎰+--x x x x d e )1(422(19)⎰+x xx d 1335(20)⎰+x xxx d ln 2ln(21)⎰+x xx d sin 1sin 2 (22)⎰+-x x xx d 2sin 1cos sin(23)⎰+2)cos 2(sin d x x x(24)⎰x xx xd cos sin tan ln(25)⎰+xx x22cos 3sin d (26)⎰-++1212d x x s(27)⎰+++3)1(1d x x x(28)⎰++52d 24x xxx(29)⎰+x x x x d )ln 1( (30)x x x x d 12⎰-+(31)⎰+)1(ln ln d 2x x x x(32)x x x xd )1(arcsin ⎰-(33)⎰xx x x cos sin d (34)x x x d )1(x arctan ⎰+(35)⎰+x xxd cos 1cos 2(36)⎰xdx x 3cos 2sin(37)x x x x ⎰-d 2cos )sin (cos (38)x xxx d sin1cos sin 4⎰+ (39)⎰x xd sin14(40)⎰xdx 3tan解:(1)=C x x x d x x dx x x ++-+=+⎪⎪⎭⎫ ⎝⎛+-+=+-+⎰⎰2123)2(12)2(32)2(262262(2)=⎰+-=--C x x d x 98)23(271)23()23(31 (3)=()()⎰+=+C e e e d x xx3arctan3213212222(4)=C x x x d +⎪⎭⎫ ⎝⎛+=⎪⎭⎫ ⎝⎛+⎪⎭⎫ ⎝⎛+⎰32tan 2132cos 32212πππ (5)=⎰⎰+--=---=-C x x x d x x d 333334324)4(314)(31(6)=C x x x d +-=+--⎰21arctan 214)1()1(2(7)=⎰+=+C e ee d x xx arctan 1)(2(8)=C e e e e d x x x x ++-=-⎰11ln 211)(2(9)=⎰+-=--C x x x d 21)1(tan 21tan )1(tan(10)=C xx d +--=---⎰lnx 1ln ln 1)ln 1((11)=⎰+=-C x x x d ln arcsin ln 1)(ln 2(12)=C e e e d x x x +=⎪⎪⎪⎭⎫ ⎝⎛-⎪⎪⎪⎭⎫ ⎝⎛⎰3arcsin2922222(13)=C x xx d x x xd ++=++=+⎰⎰2222sin 2ln 21sin 2)sin 2(21sin 2)(sin sin (14)=C x x x d +--=---⎰222212121)21(41(15)=C x x x d x x x d +++=+++⎰⎰23222)(arctan 32)1ln(21)(arctan arctan 1)1(21(16)=⎰⎰⎰⎰+⎪⎪⎭⎫⎝⎛+=++-=+=+C e e e e d e e d e e e d dx e e e x x xx xx xxx xxx 1ln 1)1()()1()()1( (17)=C x d x xx d x +⎪⎭⎫ ⎝⎛-=⎪⎭⎫ ⎝⎛-=+⎪⎭⎫⎝⎛-⎰⎰221arctan 211arctan 1arctan 1111arctan (18)=⎰+=+-+-+-C e x x d e x x x x 422422221)42(21 (19)=)(131)(131333333t d tttx x d xx ⎰⎰+=+令⎰⎰⎰⎪⎪⎪⎭⎫ ⎝⎛+-+=+-+=-)()1()()1(31)(1113131323t d t t d t t d t t C x x C t t ++-+=++-+=3233533235)1(21)1(51)1(21)1(51(20)⎰⎰+=+=tt td txx xd 2)(ln ln 2)(ln ln 令⎰⎰⎰++-++=+-+=tt d t d t tt d t 2)2(2)2()2(2)(2221C x x C t t ++-+=++-+=21232123)ln 2(4)ln 2(32)2(4)2(32 (21)⎰+-=--=C x xx d 2cos arcsincos 2)(cos 2(22)C x x x x x x d ++=++-=-⎰12)cos (sin )cos (sin )cos (sin(23)C x x x d ++-=++=-⎰12)2(tan )2(tan )2(tan(24)⎰⎰+===C x x xd x d x x 2)tan (ln 21)tan (ln tan ln )(tan tan tan ln (25)⎰⎰+=+=+=C x x x d xx d )tan 3tan(31)tan 3(1)tan 3(31tan31)(tan 22(26)C x x dx x x +⎥⎥⎦⎤⎢⎢⎣⎡--+=--+=⎰2323)12(32)12(324121212C x x +⎥⎥⎦⎤⎢⎢⎣⎡--+=2323)12()12(61(27)⎰⎰+=+++++=dt t t tt x x x x d 3321)1(1)1(令⎰++=+=+=C x C t dt t1arctan 2arctan 21122(28)⎰++=+++=C x xx d 21arctan 414)1()1(212222 (29)()⎰⎰+=+==+=C x C e e d dx x e x x x x x x x ln ln ln l )ln 1( (30)⎰⎰⎰++-=++-=+-=C x x x d x dx x dx x x x 23232222)1(3131)1(121)1((31)⎰⎰+=+=)1()(ln 令)1(ln ln )(ln 22tt t d tx x x d⎰++=⎪⎪⎭⎫ ⎝⎛++-=C t t t t d t t d 1ln 211)1()(21222222 C x x C x x ++-=++=)1ln(ln 21ln ln 1ln ln ln 2122(32)t x ==arcsin 令,则tdt t dt cos sin 2=⎰⎰+=+==C x C t dt t tdt t tt t 232322)(arcsin 34342cos sin 2cos sin(33)⎰⎰+===C x xx d x x x d tan ln 2tan )(tan cos sin)(2(34)⎰⎰+==+=Cx x d x x d x x22)(arctan arctan arctan 2)(1arctan 2(35)⎰+-+=-=C xx xx d sin 2sin 2ln221sin2)(sin 2(36)⎰⎰+-=-==C x x xd xdx x x 543cos 52cos cos 2cos cos sin 2 (37)⎰⎰---=+-=)sin (cos )sin (cos )sin (cos )sin (cos 22x x d x x dx x x x xC x x +--=3)sin (cos 31(38)⎰+=+=C x x x d 242sin arctan 21sin 1)(sin 21(39)⎰⎰⎰+--=+-=-==C x x x d x xx d dx xx cot cot 31)(cot )1(cot sin )(cot sinsin 132(40)⎰⎰⎰+-=-=-=C x x xdx x xd xdx x cos ln )(tan 21tan tan tan tan )1(sec 229.求下列函数的不定积分 (1)⎰+)1(d 7x x x(2)⎰-x x x d 12(3)⎰+-x x d 3211 (4)⎰+x x x-1)(1d(5)⎰+3d xx x (6)⎰-+x x xx d 21 (7)x x xd 11632⎰++ (8)x x d e 1⎰+ (9)⎰+-+x x x x d 4222(10)x x x d )1(3⎰-解:(1)⎰⎰++-=+=+=C x x x x dx dx x x x 77777761ln 71ln )1(71)1((2)令t x =-1,则tdt dx t x 2 , 16-=-=⎰⎰+++-=+--=--=C t t t dt t t t dt t t t )315271(2)2(2)2()1(3572462(3)令t x =-21,则tdt dx t x -=-= , 212⎰⎰++-+--=+++-=+---+=C x C t t dt t dt t t 321ln 3213ln 3)331()(31 (4)令t x =-1,则tdt dx t x 2 , 12-=-=⎰⎰+---+-=+-+-=-=--=C xx C tt tdtdt tt t1212ln221.222ln221.222).2(222(5)令t x =6,则dt t dx t x 566 , ==⎰⎰⎰+-+-=+=+=dt t t t dt t t dt tt t 11)1(616623235C t t t t ++-+-=)1ln 2131(623 C t t t t ++-+-=1ln 663223(6)令t x =-2,则tdt dx t x 2 , 22=+=⎰⎰++=++=++=C t t dt t tdt tt 2arctan22)211(22.23222C x x +-+-=22arctan222(7)令t x =+312)1(,则dt t xds 232=⎰⎰+++-=++-=+=C t t t dt t t dt t t )1ln 21(9111919222C x x x +++++-+=1)1(ln )1()1(29312312322 (8)令t e x =+1,则12 , )1ln(22-=-=t tdt dx t x⎰++++-++=++-+=-=C e e e C t t t dt t t x x x)1111ln 211(2)11ln 21(21222(9)令t x =-1,则dt dx t x =+= , 1⎰⎰⎰+++++=+++=++=C t t tdt t dt t t dt t t 3ln 3)3(333332212223C x x x x x ++-+-++-=421ln 3)42(2212(10)令t x =2,则t x =⎰⎰⎰⎥⎥⎦⎤⎢⎢⎣⎡-+--=-+--=-=dt t t dt t t dt t t 3233)1(1)1(121)1(1121)1(21 C t t C t t +-+-=+⎥⎥⎦⎤⎢⎢⎣⎡-----=22)1(141)1(21)1(1211121Cx x C x x +--=+-+-=222222)1(412)1(141)1(2110.设⎰⎰+=+=x xb x a xx x xb x a xx F d cos sin cos )G( , d cos sin sin )(求)()(x bG x aF +;)()(x bF x aG -;)(x F ;)(x G .解:⎰+=++=+C x dx xb x a xb x a x bG x aF cos sin cos sin )()(⎰⎰++=++=+-=-C x b x a dx xb x a x b x a d dx xb x a xb x a x bF x aG cos sin ln cos sin )cos sin (sin sin sin cos )()(C bx x b x a a b a x G +++-=⇒)cos sin ln (1)(22C ax x b x a b b a x F +++--=)cos sin ln (1)(2211.用三角代换求下列不定积分. (1)⎰-221x d x x(2)⎰32)-(1d x x(3)⎰-x x x d 122(4)⎰-x xa x d 22 (5)⎰-322)1(d x xx(6)x x x d )1(2101298⎰-解:(1)令t x sin =,则)2t ( cos π<=tdt dx⎰⎰+--=+-=+-===C x x C x C t t dtdt tt t2221)cot(arcsin cot sin cos sincos(2)令t x sin =,则)2t ( cos π<=tdt dxC xx C x C t tdtdt tt+-=+=+===⎰⎰2231)tan(arcsin tan cos cos cos(3)令t x sin =,则)2t ( cos π<=tdt dxC t t dt t tdt dt t t t +-=-===⎰⎰⎰2sin 412122cos 1sin cos cos sin 22 C x x x C x x +--=+-=2141arcsin 21)(arcsin 2sin 41arcsin 21 (4)令t a x sec =,则t a dx tan sec =,)20(π<<t⎰⎰⎰+-=-===C t a dt t a tdt a dt ta tt a t a )1(tan )1(sec tan sec tan sec .tan 22C saa a x C xa a a x a +--=+--=arccos )arccos (2222(5)令t x sin =,则tdt dx cos = 2π<t⎰⎰⎰⎰⎪⎪⎭⎫ ⎝⎛+-=-===dt t t dt t t dt t t dt tt t22222232cos 1cos 11cos )cos 1(1cos sin1cos sincosC xx x x C t t +---=++-=2211tan cot (6)令t x sin =,则tdt dx cos = 2π<t⎰⎰⎰+⎪⎪⎭⎫ ⎝⎛-=+====C x x C td dt t dt tt t 992999810098101981991tan 991tan tan cos sin cos cos sin12.用分部积分法计算下列积分.(1)⎰++x x x x d e )31(2 (2)⎰--x x x d e 1 (3)⎰-x x x x d )sin (cos e (4)⎰x x x d cos (5)⎰x x d arcsin (6)⎰+x x d )4ln(2 (7)⎰x x x x d cos sin 4 (8)x x d l arctan 2⎰- (9)⎰x xx d )ln(ln (10)⎰x x x d sec 22 (11)⎰x x x d arctan 2 (12)x x d )(arccos 2 (13)⎰+-x x xxd 44ln 2(14)⎰+x x xx d arctan 122(15)⎰+x x x x d arctan )1(632 (16)⎰x x xd cos tan ln(17)⎰∙x x x d sin sec ln (18)⎰∙x x x d tan ln 2sin(19)x x x x d ln 32ln 22⎰⎪⎭⎫ ⎝⎛+ (20)⎰x x x d arctan 2解:(1)⎰⎰+-++=++=dx x e e x x de x x x x x )32()31()31(22⎰++-++=dx e x e e x x x x x 2)32()31(2(2)C ex C dx e xe xde e x x x x ++-=+⎪⎭⎫ ⎝⎛--=-=+----⎰⎰)1()1(311 (3)⎰⎰⎰⎰-=-=xdx e xde xdx e xdx e x x x x sin cos sin cos⎰⎰+=-+=C x e xdx e xdx e x e x x x x cos sin sin cos(4)⎰⎰++=-==C x x x xdx x x x sd cos sin sin sin sin(5)⎰⎰--+=--=2221)1(21arcsin 1arcsin xx d x x xx x xC x x x +-+=21arcsin(6)⎰⎰⎰⎪⎪⎭⎫ ⎝⎛+--+=+-+=+=dx x x x dx x x x x dx x 2222224412)4ln(42)4ln()4ln( C xx x x ++-+=2arctan 42)4ln(2(7)⎰⎰+--=+-=-=C x x x xdx x x x xd 2sin 212cos 2cos cos 2cos(8)⎰⎰---=-+---=dx x x s dx x xx x x x 111arctan )1(121121.1arctan 222222C x x x x +-+--=1ln 1arctan 22(9)⎰⎰+-====C t t t tdt e x t x x d x tln ln ln )(ln )ln(lnCx x C x x x +-=+-=)1)(ln(ln ln ln )ln(ln .ln(10)⎰⎰++=-==C x x x xdx x x x xd cos ln 2tan 2tan 2tan 2)(tan 2 (11)⎰⎰⎰⎪⎪⎭⎫ ⎝⎛+-+-=++-=-=dx x x x xxdx x x x x xxd 11arctan 111arctan )1(arctanC x x x x ++-+-=)1ln(21ln arctan 2 (12)⎰⎰-=--===tdt t t t tdt t tdtdx tx .cos 2cos sin sin arccos 22⎰⎰+--=--=-=C t t t t t tdt t t t t t td t t cos 2sin 2cos )sin sin (2cos sin 2cos 222C x x x x x +---=21arccos 2arccos 2(13)⎰⎰⎰-+--=⎪⎭⎫⎝⎛--=-=dx x x x x x xd dx x x21.121.ln 21ln )2(ln 2 C xx x x dx x x x x +-+--=⎪⎭⎫⎝⎛--+--=⎰2ln 212ln 121212ln(14)⎰⎰⎰+-=⎪⎪⎭⎫⎝⎛+-=xdx xxdx xdx x arctan 11arctan arctan 11122⎰⎰-+-=)(arctan arctan 1arctan x xd dx x xx xC x x x x +-+-=22arctan 21)1ln(21arctan(15)()()()dx xx x x x xd 223232311.1arctan 11arctan ++-+=⎥⎦⎤⎢⎣⎡+=⎰⎰()⎰+++-+=dx x x x x x112arctan 13623()⎰⎪⎪⎭⎫ ⎝⎛+-++--+=dx x x x x x x x 1212arctan 122423()()C x x x x x x x +++--+-+=1ln 3151arctan 1223523 (16)⎰==t x x xd tan )(tan tan ln 令⎰+-=+-==C x x x C t t t tdt tan tan ln .tan ln ln(17)()⎰⎰+-=-=xdx xx x x x x xd tan .cos 1.cos .cos cos .sec ln cos sec ln ⎰+--=+-=C x xdx x x cos sec ln .cos sin cos .sec ln ()C e x x ++=22121(18)()⎰⎰-==dx xx xx x x xd cos sin 1sin tan ln .sin sin tan ln 222⎰++=-=C x x x xdx x x cos ln tan ln .sin tan tan ln .sin 22(19)()⎰⎰⎪⎭⎫ ⎝⎛++-⎪⎭⎫ ⎝⎛+=⎪⎭⎫ ⎝⎛+=dx x x x x x x x x d x x 1321.ln 231ln 32ln 31ln 32ln 3132332 ⎰⎰--⎪⎭⎫ ⎝⎛+=dx x xdx x x x x 222392ln 32ln 32ln 31 ()⎰⎰--⎪⎭⎫ ⎝⎛+=dx x x xd x x 232392ln 92ln 32ln 31 ⎰⎰-⎪⎭⎫ ⎝⎛--⎪⎭⎫ ⎝⎛+=dx x dx x x x x x 2232392.ln 92ln 32ln 31 C x x x x x x x +=-⎪⎭⎫ ⎝⎛+=23323ln 31.ln 92ln 32ln 31 (20)()⎰⎰+-==dx x xx x x x d x 233.21.1131arctan 31arctan 31 ⎰⎰⎪⎪⎪⎪⎭⎫ ⎝⎛+-++--=+-=dx x x x x x x x dx x x x x 1161arctan 31161arctan 312121233253C x x x x x x ++-+-=arctan 313191151arctan 31212325313.计算下列有理函数的不定积分. (1)⎰+x x x d )31(1 (2)⎰---)32)(1)((d x x x x(3)x x x x x d )2()1(12---- (4)⎰-++x x xx d 32322(5)⎰-1d 4x x(6)⎰++++x x x xx d 25412 (7)⎰-+-x x x xxd 123(8)⎰+---x x xx x d )1)(1(122(9)⎰+++x x x xx d 14 (10)⎰+---x x x x x d )2()1(18332解:(1)C xC x x dx x x++=++-=⎪⎭⎫⎝⎛+-=⎰311ln31ln ln 311313 (2)C x x x dx x x x +---=⎪⎪⎭⎫⎝⎛-+--+-=⎰2)2()3)(1(ln 21)3(2121)1(21 (3)C x x dx x x +---=⎥⎥⎦⎤⎢⎢⎣⎡-+-=⎰112ln 21)2(12(4)C x x dx x x +--+=⎥⎦⎤⎢⎣⎡-++=⎰1ln 453ln 43)1(45)3(43(5)⎰+--+-=⎪⎪⎭⎫ ⎝⎛+--=C x x x dx x x arctan 2111ln 4111112122 (6)C x x x dx x x x ++++-+-=⎥⎥⎦⎤⎢⎢⎣⎡+++++-=⎰2ln 51ln 41225)1(2142 (7)⎰⎰⎥⎦⎤⎢⎣⎡-+⎪⎪⎭⎫ ⎝⎛+-+-=⎥⎥⎦⎤⎢⎢⎣⎡-++-=dx x x x x dx x xx)1(2111121)1(21)1(21222()C x x x +-+++-=1ln 21arctan 211ln 412 (8)⎰⎰⎰⎰+-++----=⎪⎪⎭⎫⎝⎛+-+-+-=dx x xdx x x x dx x dx x x x x 1123121111211222C x x x x +⎪⎪⎭⎫ ⎝⎛-++---=312arctan 31ln 211ln 2 (9)()()()()⎰⎰⎰++++-+-=⎥⎥⎦⎤⎢⎢⎣⎡+++-=dx x dx x x x x dx x x x 121121211111222()⎰⎰++++++⎪⎭⎫ ⎝⎛-+-=1ln 2111211141212222x dx x x x d x x ()C x x x x x +++-++-=arctan 211ln 411ln 212122(10)()()⎰⎰⎰+--+-=--+-+--=C x x x dx x dx x dx x 21ln 1121111223(B )1.填空题(1)设x x f 21)(ln +=',则)(x f = . (2)设函数)(x f 满足下列条件 ①2)0(=f ,0)2(=-f ;②)(x f 在1-=x ,5=x 处有极值;③)(x f 的导数是x 的二次函数,则)(x f = . (3)若C x x x xf x +=⎰e d )(2,则⎰x x f xd )(e = . (4)设2ln)1(222-=-x x x f ,且[]x x f ln )(=ϕ,则=⎰x x d )(ϕ .(5)设x x f ln )(=,则='⎪⎪⎭⎫ ⎝⎛-⎰-x f x x x x d )e (e-2e e 43 .(6)='⎰x x f xx f d )(ln )(ln .(7)设)(x f 的一个原函数为xxsin ,则='⎰x x f x d )2( . (8)若⎰⎰-=x x f x f x x x f d )(cos )(sin d )(sin ,则=)(x f .解:(1)()C e x x f x ++=2()()()C e x x f e x f e x x f x x x ++=⇒+='⇒+=+='2212121ln ln(2)215623+--x x x由已知可设d cx bx ax x f +++=23)( 有()C bx ax x f ++='232()()()()⎪⎪⎩⎪⎪⎨⎧=-=-==⇒⎪⎪⎩⎪⎪⎨⎧=++==+-=-'=+-+-=-==⇒2156101075502310248220d c b a c b a f c b a f d c b a f d f()215623+--=⇒x x x x f(3)C x ++2ln()()()x x x x x xe e x f e x xe x xf C e x dx x xf +=⇒+=⇒+=⎰2222⎰⎰++=+=⇒C x dx xdx x f e x2ln 21)( (4)C x x +++1ln 21)(1)(ln 11ln)(1111ln2ln)1(22222-+⇒-+=⇒--+-=-=-x x x x x f x x x x x f ϕϕ ⎰⎰⎰+-+=-+=-+=⇒-+=⇒C x x dx x dx x x dx x x x x 1ln 2)121(11)(11)(ϕϕ (5)C e e e x x x ++-+--22ln24121222⎰⎰++-+-=⎪⎪⎭⎫ ⎝⎛--=⎪⎪⎭⎫ ⎝⎛--=---C e e e dx e e e dx ee e e x x x x x x x x x x 22ln 2412121.222242243原式 (6)C xf +)(ln 2C x f x f x f d +==⎰)(ln 2)(ln ))(ln (原式(7)C xxx +-42sin 42cos ⎰-=⇒+=2sin cos )(sin )(xxx x c f C x x dx x f C x xx x x x x x x x dx x f x xf x f xd +-=--=-==⎰⎰42sin 42cos 22sin 4142sin 2cos 2.21)2(41)2(21))2((21原式 (8)x ln⎰⎰'-=dx x f x f x x f x dx x g )()(cos )(sin )(sinC x x f xx f +=⇒='∴ln )(1)(,取x x f ln )(=2.选择题(1)设x x f 2cos )(sin =',则⎰=dx x f )(( B ) A .C x x +-331 B .1421212C Cx x x ++- C .C x x ++421212 C .C x x ++421212(2)设)()( , )(1)()( , )(1)()(2x g x F x f x f x g x f x f x F ='+=-=,且14=⎪⎭⎫⎝⎛πf ,则=)(x f ( A )A .x tanB .x cotC .x arctanD .x arc cot(3)若⎰+=C x x x f 2sin d )(,则⎰=--dx x x xf 12)12(22( B )A .C x +22sin 41B .C x +-)12sin(212 C .C x +-)12(sin 2122 D .C x +-)12sin(412 (4)设⎰⎰+∙=xdx x f x g dx xx f 22cot )()(sin)(,则)(x f ,)(x g 分别是( D )A .x x f cos ln )(=,x x g tan )(=B .x x f cos ln )(=,x x g cot )(-=C .x x f sin ln )(=,x x g tan )(=D .x x f sin ln )(=,x x g cot )(-= 解:(1)BC +-=⇒-='⇒-=='322x 31x )x (f x 1)x (f x sin 1x cos )x (sin f⎰++-=⇒142C x x 1212x f(x)dx C(2)A根据1)4f(=π,首先排除C 、D ,再将选项A 、B 分别代入原条件中,得A(3)B)1x 2sin(1x 2212x f 2xsinx f(x)2222--=-⇒= ⎰⎰+--=--=-=⇒C )1x 2sin(21)1d(2x )1x 2sin(2.41dx )1xsin(2x 22222原式,得B (4)D⎰⎰-=cotx)f(x)d(dx x sin f(x)取cotx g(x)-=则⎰+=xdf(x)cot f(x)g(x)上式 与条件比较,得cotxg(x) ,lnsinx f(x)cotx df(x)-==⇒=,得D3.计算下列不定积分(1)x xx x d 11ln 112-+-⎰(2)x x x x d cos 1)sin 1(e ⎰++(3)⎰+)e1(e d 2xxx(4)x xx d cos sin144⎰(5)⎰x x x x d cos e (6)⎰+++x x x x d 112(7)⎰xxcos d (8)⎰++x aax x xd 22(9)⎰-+293d x x (10)⎰-xx1 (提示 令t x 2sin =)(11)x x x d 283⎰++ (12)⎰-x xxxd 1arcsin 22(提示 令t x =arcsin ,t x sin =,再用分部积分法) (13)⎰x x x d )(arctan 2 (14)x xxx d e 1arctan arctan 2⎰+(15)⎰+x xxx d )3(ln 22(16)x x x d )sin(ln ⎰(提示 经过两次分部积分,又出现原积分形式,移项后便可得到所要结果)解:(1)C xxx x d x x ++-=+-+-=⎰11ln 41)11(ln 11ln 212 (2)dx x tg x tg e dx x xx e x x )2221(212cos )2cos 2(sin222++=+=⎰⎰⎰⎰++=dx e x tg dx e x tg e x x x 2212212 ⎰⎰+=++-+=C x tg e dx e x tg dx x tg e e x tg e x x x x x 2221)12(2122122 (3)⎰⎰+-=+=x xde eee ede )111()1(C e e x x +--=-arctan(4)C x x dx x +--==⎰cot cot 31sin 134C x x C x x x d x +--=⎥⎦⎤⎢⎣⎡+--=⎰2cot 382cot 82cot 2cot 31822sin 183134 (5)=[]c x x x x e x++-cos sin )1(21 (6)⎰⎰⎰+++++++=++-+=dx x x x x x d dx x x x 22222)23()21(1211)1(2112121C x x x x x C x x x x x ++++++++=++++++++=121ln 211121ln 2112.212222 (7)⎰⎰++=+==C x x x d x x d x32tan 31tan tan )tan 1()(tan cos 1(8)⎰⎰⎰++-+++++=++-+=dx aax x a aax x a ax x d dx a ax x aa x 222222221)2()(2122C a ax x ax a a ax x +++++-++=22222ln 2(9)t x sin 3==令,20π<<t 则⎰⎰⎰+-=+=+dt tdt t t dt t t )cos 111(cos 1cos cos 33cos 3⎰+-=-C tt t d t t 2arctan )2(2cos 12 C x x x C xx+-+-=+-=2933arcsin 23arcsintan3arcsin(10)t x 2sin ==令,20π<<t ,则⎰⎰⎰+==dt ttdt tdt t t t 22cos 12cos 2cos sin 2sin cos 2 C x x x t t dt t +-+=+=+=⎰2arcsin 2sin 21)2cos 1( (11)C x x x dx x x dx x x x ++-=++=++++=⎰⎰4342)42(2)42)(22(232(12)t x =arcsin 令,t x sin =,则⎰⎰⎰⎰+-=-===tdt t t t td dt tttdt tt tcot cot )cot (sincos cos sin22C x x xx C t t t ++--=++-=ln arcsin 1sin ln cot 2(13)xdx x x x x x d x arctan 1)(arctan 21)()(arctan 21222222⎰⎰+-==⎰⎰++-=xdx x xdx x x arctan 11arctan )(arctan 21222 C x x x x x x ++++-=2222)(arctan 21)1ln(21arctan )(arctan 21 (14)⎰⎰==dt te t x x d xe t x arctan )(arctan arctan arctan 令⎰⎰+-=+-=-==C e x C e t de te tde x t t t t arctan )1(arctan )1((15)⎰⎰⎰+++-=+-=++=dx xx x x x xd x d x x )3(1213ln 21)31(ln 21)3()3(ln 21222222C x x x x dx x x x x ++-++-=+-++-=⎰)3ln(121ln 613ln 21)311(613ln 212222 (16)⎰⎰+-=-=dx xx x x x d x 322ln cos 21)sin(ln 21)1()sin(ln 21 dx x xx xx x⎰---=322ln sin 41ln cos 41)sin(ln 21[]C x x x ++-=⇒ln cos ln sin 251原式。
微积分课后习题答案
微积分第八章课后习题答案习题8-11.1一阶;2二阶;3一阶;4三阶;5三阶;6一阶;7二阶;8一阶;2.1、2、3、4、5都是微分方程的通解;3.122y x =+.4.将所给函数及所给函数的导数代人原方程解得:21()(1)2u x x dx x x C =+=++⎰.习题8-21.1原式化为:ln dyx y y dx =分离变量得:11ln dy dx y y x = 两边积分得:11ln dy dx y y x=⎰⎰ 计算得:()11ln ln d y dx y x=⎰⎰ 即:()1ln ln ln y x C =+ 整理:1ln y C x =所以:原微分方程的通解为:Cx y e =; 2原式化为:()()2211y x dy x y dx -=-- 分离变量得:()()2211y xdy dx y x -=-- 两边积分得:()()2211y xdy dx y x -=--⎰⎰ 计算得:()()()()22221111112211d y d x y x -=----⎰⎰ 即:()()221ln 1ln 1y x C -=--+ 整理:22(1)(1)y x C --=所以:原微分方程的通解为:22(1)(1)y x C --=;3xydx =-分离变量得:1dy y =两边积分得:1dy y =⎰计算得:()21ln 12y x =-即:1ln y C =整理:y =所以:原微分方程的通解为:y =41y e Cx -=-;5sin 1y C x =-; 61010x y C -+=;722ln 22arctan y y x x C -=-+; 8当sin02y ≠时,通解为ln |tan |2sin42y y C =-;当sin 02y=时,特解为2(0,1,2,)y k k π==±±;9222ln x y x C +-=; 1022ln ln x y C +=;2.1tan 2x y e=;2(1)sec x e y +=;32(1)22y x e y +-=;41ln |1|1a x a y=--+;524x y =;6323223235y y x x +--=;7sin y x =;8cos 0x y -=;3.12y Cx =;21Cx y xe +=;3sin ln ||yx C x=+;4ln |ln |y x C x =--;5arctany xxy Ce-=;6ln1yCx x=+;722(2ln ||)y x x C =+;8332x y Cx -=;4.1ln(1ln )y x x =--;222(ln 2)y x x =+;322tan(ln )4y x x π=+;4222ln y x x =;5y x =;6222(ln 2)y x x =+; 5.31()2x xϕ=-; 习题8-31.12x x y Ce e =-;2()n x y x e C =+;3sin ()x y e x C -=+;42(1)()y x x C =++;52sin ()y x x C =+;6()xy e x C -=+;722y x Cx =-+;82212x x y Ce e--=-;932433(1)x Cy x +=+;101(1)y C x =++;2.132(4)3xy e -=-;2x e y x =;31cos x y x π--=;4cos x y x=;5(1)x y e x =+;62ln 2y x x =-+;7sin 2sin 1x y e x -=+-;82sin 11x y x -=-; 3.155352y Cx x -=+;24414x y x Ce --=-++;32133ln |1|(ln |3|)2x C C C y++==;433(2ln 1)4C y x x x -=--或323(2ln 1)4xy x x C -+-=;51233317y Cx x -=-或123337y Cx x -=-;64414x y Ce x --=-+;习题8-41.112(2)x y x e C x C =-++;212ln |cos()|y x C C =-++;321212x y C e x x C =--+;41221(0)C x y C e C =+≠;541211cos3129y x x C x C =-++;64321211432C y x x x C =+-+;712()x y C x e C -=-+;812C x y C e =;2.1y =21ln(1)y ax a =-+;3lnsec y x =;441(1)2y x =+;5ln()ln 2x x y e e -=+-;61122x x y e e -=-;731cos 16y x x x =-++;821122y x =-;习题8-51.12312xxy C eC e--=+;23412()xy C C x e=+;312cos sin y C x C x=+;4412(cos3sin 3)xy e C x C x -=+;55212()x y C C x e =+;6212(cos sin )x y e C x C x =+;72512x xy C e C e -=+;8212()xy C C x e =+;9212(cos3sin 3)x y e C x C x =+;1012y C C =+;2.12(2)x y x e -=+;223sin 5x y e x -=;3342x x y e e =+;4sin x y e x =;51cos33x y e x =-;61cos sin y x x πππ=+;3.'''20y y y -+=;4. '''320y y y -+=;5.1*01y b x b =+;2*201y b x b x =+;3*0x y b e =;4*2012()x y b x b x b e =++;5*01cos 2sin 2y b x b x =+;6*01(cos sin )y x b x b x =+;6.132121123x y C C e x x -=++-;2121(cos sin )2x y C C e x x =++-;32212117()224x y e C x C x x x -=++--; 4122cos sin 1xe y C ax C ax a =+++;5312113cos sin ()1050x y C x C x x e =++-; 631234()(cos sin )2525x x y e C C x e x x =++-;72121(cos sin )(1)2x y e C x C x x =+++;83212xy C e C x =++;921232x x x y C e C e e -=++;1022212()224x x y C C x e x x e =++++;7.1275522x x y e e =-++;2(1)x x x y e e x x e -=-+-;3211(cos sin )sin 22x y e x x e x π=-+;4311(37cos 429sin 4)(5sin 14cos )102102x y x x e x x =-++; 511cos sin sin 233y x x x =--+;64115516164x y e x =+-;习题8-61.1三阶;2六阶;2.略;3.12t t y C =;2(1)t t y C =-;321122t y C t t =+-;42111()623t y C t t t =+-+;51(1)23t t t y C =-+;61222t t t y C t =+;4.123t y t =+;213()2t t y =-;3111()442t t y =+-;411(2)224t t t y =-+; 5.11234t t t y C C =+;21211(()22t tt y C C =+;312()3t t y C C t =+; 4122(cos sin )22t t y C t C t ππ=+;512(1)4t t t y C C =-+; 6122(cos sin )33t t y C t C t ππ=+;6.11[1(3)]2t t y =-+-;2sin3t t y t π=;32cos4t t y t π=⋅;习题8-7 略 总复习题八1.1三;2'''560y y y -+=;32129t t t y y y +++-=;2.1C ;2B ;3D ;4A ;5D;3.略;4.1221(1)y C x +=-;2(1)(1)xye e C +-=;3ln[(2)]02xC y x y x++=+;42xy ye x C +=;5ln Cy ax x=+;622124ln 39C x x x y x =--或23222(ln )33x C x x y =-+;332x xy C =++;8222arctanyx y C x+-=;92y Cx =;1022xy y C -=;5.11x e y +=或(1)sec x e y +=;2220x y x y +--=;32225x y +=;42(12ln )0x y y +-=;5cos 15sin x e y x -=或cos sin 51xy x e +=;62(1)x x x x e e e y e x x-==-; 6.()(1)x y x e x =+;7.1(ln ln )y x x e -=+;8.132212[)23x C C C =±-;22x C =±+;35322121373525x y C C ex x x -=++-+;421213(1)2x x xy C e C e x x e ---=++-;5121(cos 2sin 2)cos 24x x y e C x C x xe x=+-;61211cos 2210x x y C e C e x-=+-+;72(cos3sin 3)xy eA xB x -=+;8212x x x y C e C e e -=++;9.14x x y e e -=-;22sin 3x y e x =;32(73)x y x e -=-;42arctan x y e =;10.(cos sin )()2xx x e x ϕ++=;11.121t y t ∆=+;221t y t ∆=+;312cos ()sin 22t ay a t ∆=+⋅;434t y t ∆=; 12.1(2)ty C =-;221(3)()2255t t y C t =-+-+;312(3)t y C C =-+;412213(2)()32515t t t y C C t t =+-+-+⋅; 13.112(1)3t t t y A =⋅+⋅-,152(1)33t t t y =⋅+⋅-;2174()()22t t t y A B =+⋅+⋅-,31174()()2222t t t y =+⋅+⋅-;。
微积分课后题答案习题详解
第二章习题2-11. 试利用本节定义5后面的注(3)证明:若lim n →∞x n =a ,则对任何自然数k ,有lim n →∞x n +k =a .证:由lim n n x a →∞=,知0ε∀>,1N ∃,当1n N >时,有取1N N k =-,有0ε∀>,N ∃,设n N >时(此时1n k N +>)有 由数列极限的定义得 lim n k x x a +→∞=.2. 试利用不等式A B A B -≤-说明:若lim n →∞x n =a ,则lim n →∞∣x n ∣=|a|.考察数列x n =(-1)n ,说明上述结论反之不成立. 证:而 n n x a x a -≤- 于是0ε∀>,,使当时,有N n N ∃>n n x a x a ε-≤-< 即 n x a ε-<由数列极限的定义得 lim n n x a →∞=考察数列 (1)nn x =-,知lim n n x →∞不存在,而1n x =,lim 1n n x →∞=,所以前面所证结论反之不成立。
3. 利用夹逼定理证明:(1) lim n →∞222111(1)(2)n n n ⎛⎫+++ ⎪+⎝⎭=0; (2) lim n →∞2!nn =0.证:(1)因为222222111112(1)(2)n n n n n n n n n n++≤+++≤≤=+ 而且 21lim0n n →∞=,2lim 0n n→∞=,所以由夹逼定理,得222111lim 0(1)(2)n n n n →∞⎛⎫+++= ⎪+⎝⎭. (2)因为22222240!1231n n n n n<=<-,而且4lim 0n n →∞=,所以,由夹逼定理得4. 利用单调有界数列收敛准则证明下列数列的极限存在. (1) x n =11n e +,n =1,2,…;(2) x 1x n +1,n =1,2,…. 证:(1)略。
微积分答案 (4)
微积分答案1. 导数1.1 定义导数是描述函数变化率的概念。
对于函数f(x),如果在某一点x处存在极限$\\lim_{h \\to 0} \\frac{f(x+h)-f(x)}{h}$,则该极限值称为函数f(x)在点x处的导数,记作f′(x)或$\\frac{df(x)}{dx}$。
1.2 计算常见函数的导数计算方法如下:•对于常数函数c,导数为0;•对于幂函数x n,导数为nx n−1;•对于指数函数e x,导数为e x;•对于对数函数$\\log{x}$,导数为$\\frac{1}{x}$;•对于三角函数$\\sin{x}$和$\\cos{x}$,导数分别为$\\cos{x}$和$-\\sin{x}$;•对于常见的复合函数,可以使用链式法则求导。
1.3 导数的几何意义导数可以描述函数的切线斜率。
在函数图像上,切线的斜率等于该点的导数。
1.4 物理应用导数在物理学中有广泛的应用。
例如,速度是位移对时间的导数,加速度是速度对时间的导数。
2. 积分2.1 定义积分是对函数的求和过程。
如果函数f(x)在区间[a, b]上有定义且非负,则积分可以表示为$\\int_{a}^{b} f(x)dx$。
2.2 原始函数和不定积分如果函数F(x)的导数等于f(x),则函数F(x)称为函数f(x)的原函数。
记作$\\int f(x)dx = F(x) + C$,其中C为常数。
原函数也称为不定积分。
2.3 积分的计算方法常见函数的积分计算方法如下:•对于幂函数x n,当n eq−1时,积分为$\\frac{1}{n+1}x^{n+1}+C$,当n=−1时,积分为$\\ln{|x|}+C$;•对于指数函数e x,积分为e x+C;•对于对数函数$\\log{x}$,积分为$x(\\log{x}-1)+C$;•对于三角函数$\\sin{x}$和$\\cos{x}$,积分分别为$-\\cos{x}+C$和$\\sin{x}+C$;•对于常见的复合函数,可以使用换元法或分部积分法求积分;•对于一些特殊函数,如正态分布函数和伽玛函数等,需要使用专门的积分技巧。
《微积分》课后答案(复旦大学出版社(曹定华_李建平_毛志强_著))第11章
t t 1 t 1 1 1 yt (1)i 2t i 1 2t 1 ( )i 2t 2 3 i 0 i 0
由 (11 2 4) 式,得所给方程的通解
1 yt A(1)t 2t 3
(A 为任意常数)
*
(4)对应齐次差分方程为 yt 1 yt 0 ,其通解为 yt A , 设原方程特解为
yt 2t ( B1 cos πt B2 sin πt ) 代入原方程得:
2t 1[ B1 cos π(t 1) B2 sin π(t 1)] 2t ( B1 cos πt B2 sin πt ) 2t cos πt
yt 1
1 4 yt ,其中 3 3
1 4 a , b ,由通解公式 (11 2 7) 得原方程的通解为: 3 3
1 yt y A (t ) yt A( )t 1 (A 为任意常数) 3 1 3 t 1 3 1 (2)方程可化为 yt 1 yt ,其中 a , b0 , b1 ,故由通解公式 2 2 2 2 2 2 (11 2 9) 得方程的通解为: 3 1 1 1 t 1 7 t yt A( ) 2 2 2 t 即 yt A( )t . 1 1 1 2 9 3 2 1 (1 ) 2 1 2 2 2
t
(4) a 4 , π , b1 0 , b2 3 , D (4 cos π) sin π=9 0 ,且
2 2
由公式 (11 2 14) 得 = [0 (4 cos π) 3 sin π]=0 , = [3(4 cos π) 0 sin π]=1 , 方程通解为 yt A(4) sin πt ,以 t 0 时 y0 1 代入上式,得 A 1 ,故原方程特解为:
(完整word版)《微积分》各章习题及详细答案
第一单元 函数与极限一、填空题1、已知x xf cos 1)2(sin +=,则=)(cos x f 。
2、=-+→∞)1()34(lim 22x x x x 。
3、0→x 时,x x sin tan -是x 的 阶无穷小。
4、01sin lim 0=→x x k x 成立的k 为 。
5、=-∞→x e x x arctan lim 。
6、⎩⎨⎧≤+>+=0,0,1)(x b x x e x f x 在0=x 处连续,则=b 。
7、=+→xx x 6)13ln(lim0 。
8、设)(x f 的定义域是]1,0[,则)(ln x f 的定义域是__________。
9、函数)2ln(1++=x y 的反函数为_________。
10、设a 是非零常数,则________)(lim =-+∞→xx ax a x 。
11、已知当0→x 时,1)1(312-+ax 与1cos -x 是等价无穷小,则常数________=a 。
12、函数xxx f +=13arcsin )(的定义域是__________。
13、____________22lim22=--++∞→x x n 。
14、设8)2(lim =-+∞→xx ax a x ,则=a ________。
15、)2)(1(lim n n n n n -++++∞→=____________。
二、选择题1、设)(),(x g x f 是],[l l -上的偶函数,)(x h 是],[l l -上的奇函数,则 中所给的函数必为奇函数。
(A))()(x g x f +;(B))()(x h x f +;(C ))]()()[(x h x g x f +;(D ))()()(x h x g x f 。
2、xxx +-=11)(α,31)(x x -=β,则当1→x 时有 。
(A)α是比β高阶的无穷小; (B)α是比β低阶的无穷小; (C )α与β是同阶无穷小; (D )βα~。
微积分课后题答案习题详解
微积分课后题答案习题详解IMB standardization office【IMB 5AB- IMBK 08- IMB 2C】第二章习题2-11. 试利用本节定义5后面的注(3)证明:若lim n →∞x n =a ,则对任何自然数k ,有lim n →∞x n +k =a .证:由lim n n x a →∞=,知0ε∀>,1N ∃,当1n N >时,有取1N N k =-,有0ε∀>,N ∃,设n N >时(此时1n k N +>)有 由数列极限的定义得 lim n k x x a +→∞=.2. 试利用不等式A B A B -≤-说明:若lim n →∞x n =a ,则lim n →∞∣x n ∣=|a|.考察数列x n =(-1)n ,说明上述结论反之不成立.证:而 n n x a x a -≤- 于是0ε∀>,,使当时,有N n N ∃>n n x a x a ε-≤-< 即 n x a ε-<由数列极限的定义得 lim n n x a →∞=考察数列 (1)nn x =-,知lim n n x →∞不存在,而1n x =,lim 1n n x →∞=,所以前面所证结论反之不成立。
3. 利用夹逼定理证明:(1) lim n →∞222111(1)(2)n n n ⎛⎫+++ ⎪+⎝⎭=0; (2) lim n →∞2!n n =0.证:(1)因为222222111112(1)(2)n n n n n n n n n n++≤+++≤≤=+ 而且 21lim0n n →∞=,2lim 0n n→∞=, 所以由夹逼定理,得222111lim 0(1)(2)n n n n →∞⎛⎫+++= ⎪+⎝⎭. (2)因为22222240!1231n n n n n<=<-,而且4lim 0n n →∞=,所以,由夹逼定理得4. 利用单调有界数列收敛准则证明下列数列的极限存在.(1) x n =11n e +,n =1,2,…;(2) x 1x n +1,n =1,2,…. 证:(1)略。
《微积分》各章习题及详细答案
第一章 函数极限与连续一、填空题1、已知x xf cos 1)2(sin+=,则=)(cos x f 。
2、=-+→∞)1()34(lim 22x x x x 。
3、0→x 时,x x sin tan -是x 的 阶无穷小。
4、01sin lim 0=→xx kx 成立的k 为 。
5、=-∞→x e xx arctan lim 。
6、⎩⎨⎧≤+>+=0,0,1)(x b x x e x f x 在0=x 处连续,则=b 。
7、=+→xx x 6)13ln(lim0 。
8、设)(x f 的定义域是]1,0[,则)(ln x f 的定义域是__________。
9、函数)2ln(1++=x y 的反函数为_________。
10、设a 是非零常数,则________)(lim =-+∞→xx ax a x 。
11、已知当0→x 时,1)1(312-+ax 与1cos -x 是等价无穷小,则常数________=a 。
12、函数xxx f +=13arcsin )(的定义域是__________。
13、lim ____________x →+∞=。
14、设8)2(lim =-+∞→xx ax a x ,则=a ________。
15、)2)(1(lim n n n n n -++++∞→=____________。
二、选择题1、设)(),(x g x f 是],[l l -上的偶函数,)(x h 是],[l l -上的奇函数,则 中所给的函数必为奇函数。
(A))()(x g x f +;(B))()(x h x f +;(C ))]()()[(x h x g x f +;(D ))()()(x h x g x f 。
2、xxx +-=11)(α,31)(x x -=β,则当1→x 时有 。
(A)α是比β高阶的无穷小; (B)α是比β低阶的无穷小; (C )α与β是同阶无穷小; (D )βα~。
微积分四练习题答案
微积分四练习题答案微积分是数学中的一门重要学科,它的应用广泛涉及到物理、工程、经济学等领域,在解决实际问题中起着重要的作用。
而在学习微积分的过程中,练习题是检验自己理解和掌握程度的重要方法。
下面为大家提供四个微积分练习题的答案,希望能够帮助到需要的读者。
第一题:计算函数 f(x) = 3x^2 - 2x + 1 在 x = 2 处的导数。
解答:首先,根据导数的定义,我们知道导数表示了函数在某一点的变化率。
对于给定的函数 f(x) = 3x^2 - 2x + 1,我们需要计算其在 x = 2 处的导数。
使用导数的求导法则,我们可以将 f(x) 分别对 x 的各个项求导,即f'(x) = d/dx (3x^2) - d/dx (2x) + d/dx (1) = 6x - 2。
将 x = 2 带入上式,可以得到f'(2) = 6(2) - 2 = 10。
因此,函数 f(x) = 3x^2 - 2x + 1 在 x = 2 处的导数为 10。
第二题:计算函数 g(x) = sin(2x) 在x = π/4 处的导数。
解答:同样地,我们需要计算函数 g(x) = sin(2x) 在 x = π/4 处的导数。
使用链式法则,我们知道 sin(2x) 的导数等于 cos(2x) 乘以 2 的导数。
因此,我们可以得到:g'(x) = cos(2x) * 2。
将x = π/4 带入上式,可以得到g'(π/4) = cos(2π/4) * 2 = cos(π/2) * 2 = 0 * 2 = 0。
所以,函数 g(x) = sin(2x) 在x = π/4 处的导数为 0。
第三题:计算函数 h(x) = ln(2x) 在 x = 1 处的导数。
解答:对于函数 h(x) = ln(2x),我们需要计算其在 x = 1 处的导数。
根据对数函数的导数公式,我们知道 ln(2x) 的导数等于 1/x 乘以 2 的导数。
微积分答案
第一章 函数与极限1.2-1.3 数列和函数的极限一、 根据数列或函数极限的定义证明下列极限:1. 0)1(lim 2=-∞→n n n ; 2.521532lim =+-∞→n n n ; 3. 224lim 42x x x →--=-+; 4. 0cos lim =+∞→x x x ;5. 证明11lim=-+∞→x x x ,并求正数X ,使得当x X >时,就有01.0|11|<--x x.(X 2=101)二、设}{n x 为一数列.1. 证明:若ax n n =∞→lim ,则||||lim a x n n =∞→;2. 问:(1)的逆命题“若||||lim a x n n =∞→,则ax n n =∞→lim ”是否成立?若成立,证明之;若不成立,举出反例. (逆命题不成立。
反例:(1)nn x =-。
)三、判断下列命题的正误:1. 若数列}{n x 和}{n y 都收敛,则数列}{n n y x +必收敛; (正确)2. 若数列}{n x 和}{n y 都发散,则数列}{n n y x +必发散; (错误)3. 若数列}{n x 收敛,而数列}{n y 发散,则数列}{n n y x +必发散。
(正确) 四、证明:对任一数列}{n x ,若ax k k =-∞→12lim 且ax k k =∞→2lim ,则ax n n =∞→lim . 五、证明:A x f x =∞→)(lim 的充分必要条件是Ax f x =-∞→)(lim 且Ax f x =+∞→)(lim .六、根据函数的图形写出下列极限(如果极限存在):1. lim arctan x x π→-∞=-2,lim arctan x x π→+∞=2和lim arctan x x→∞不存在2. lim sgn 1x x →-∞=-,1lim sgn x x →+∞=和lim sgn x x→∞不存在3.lim x x e →-∞=,lim x x e →+∞=+∞和lim xx e →∞不存在七、证明:若)(lim 0x f x x →存在,则函数)(x f 在0x 的某个去心邻域内有界.八、证明:函数)(x f 当0x x →时的极限存在的充分必要条件是左极限,右极限均存在并且相等,即)(lim )(lim )(lim 0x f A x f A x f x x x x x x +-→→→==⇔=.九、设||)(x x f =,求0lim ()0x f x -→=,0lim ()0x f x +→=和0im ()0l x f x →=.十、设x x f sgn )(=,求0lim (1)x f x -→=-,0lim ()1x f x +→=和0lim ()x f x →不存在1.4 无穷小与无穷大一、填空题1. 当x →∞时,11-x 是无穷小;当1x →时,11-x 是无穷大.2. 当0x -→时,x e 1是无穷小;当0x +→时,xe 1是无穷大.3. 当1x →时,x ln 是无穷小;当0x +→时,x ln 是负无穷大;当x →+∞时,x ln 是正无穷大. 二、选择题当0→x 时,函数x x1cos1是(D ) (A )无穷小; (B )无穷大;(C )有界的,但不是无穷小; (D )无界的,但不是无穷大.三、证明函数x x x f sin )(=在)0(∞+,内无界,但当+∞→x 时,)(x f 不是无穷大. 四、判断下列命题的正确性:1. 两个无穷小的和也是无穷小. (正确)2. 两个无穷大的和也是无穷大. (错误)3. 无穷小与无穷大的和一定是无穷大. (正确)4. 无穷小与无穷大的积一定是无穷大. (错误)5. 无穷小与无穷大的积一定是无穷大. (错误)6. 无穷大与无穷大的积也是无穷大. (正确) 五、举例说明:1. 两个无穷小的商不一定是无穷小;2. 无限个无穷小的和不一定是无穷小. 六、根据定义证明:1. 当0→x 时,x x x f 1sin)(=为无穷小;2. 当+→0x 时,xe xf 1)(=为无穷大;3. 当-∞→x 时,xe xf =)(为无穷小.1.5 极限运算法则一、计算下列极限:1.22lim(31224)x x x →-+=2. 22131im 21l x x x x →-+=-3. 224im 4l 2x x x →-=-4. 11lim 1n x x n x →-=-(n 是正整数)5. 3131lim()111x x x →-=--6. 0233()lim 3h x h x x h →+-=二、计算下列极限:1. 211lim(3)6)(2x x x →∞-+=2. 2231lim 4134x x x x →∞+=+- 3. 2321lim 510x x x x x →∞++=-+4. 235lim 101x x x x →∞+-=+∞5.2221211lim 2(...)n n n n n →∞-+++=6. 221...lim (||1||1)1.1..1n nn a a a a b b b b b a →∞++++-<<=++++-, 7. 1123lim 2313n n n n n ++→∞+=+ 三、若0)1(lim 2=--+∞→b ax x x x ,求b a ,的值. (1,1a b ==-)四、若23)11(lim 21=---→x x x a x ,求a 的值. (2a =)五、计算下列极限:1. 2211lim 2x x x x x →++=+-∞2.2lim(543)x x x →∞--=∞3. 32251lim 465x x x x x →∞-+=++∞.六、计算下列极限: 1.211lim(1)cos10x x x →-=-2.301lim s ni x x x →=3. 2(1)arcta 0n lim x x x x →∞+=. 七、设2,1()5,1x x f x x x ⎧≤⎪=⎨->⎪⎩,分别求函数)(x f 在1-=x 与1=x 的左极限、右极限和极限.(4,1--,不存在)八、设11lim )(22+-=∞→nn n x x x f ,试求)(x f 的表达式. (1,1()0,11,1x f x x x ⎧-<⎪==⎨⎪>⎩)1.6 极限存在的两个准则两个重要极限一、利用夹逼定理求下列极限: 1. 222111lim(...)120n n n n n →∞+++=+++2.222111lim(...)120n n n n n n n n →∞+++=++++++3. 21lim (arctan )0x x x →∞=二、证明:332lim =+∞→n n n n .三、设12max{...}m a a a a =,,,(01,2,...,)k a k m >=,,证明:n a=.四、设1>a ,证明0lim=∞→nn a n五、利用数列的单调有界准则证明下列数列收敛,并求出极限:1. 12,n x x x ===...;(l i m 2n n x →∞=)2. 11121111111n n n x x x x x x x --==+=+++,,...,,....(lim n n x →∞=) 六、设11x a y b ==,(0)a b <<,n n n y x x =+1,21nn n y x y +=+. 1. 证明数列}{n x 单调增加,数列}{n y 单调减少且满足(1,2,...)n n x y n <=; 2. 证明数列}{n x 和}{n y 都收敛,并且有相同的极限.七、计算下列极限:1.0sin 33lim44x x x →=2. 0sin lim (,0)sin x x x ααββαβ→≠=3.lim sinx x xππ→∞=4. sin m1li x xx ππ→=-5.01cos lim arctan 12x x x x →-=6. 0lim x +→=7. 1lim 2s n 30i n n n →∞=.八、计算下列极限:1. 1lim(1)1nn n e→∞+=+2. 522lim(1)x x x e +→∞+=3.1x e →=4. 21lim()211x x x e x →∞-=+5. 2cot 2lim(1tan )x x x e →+=6.21lim(11)nn n →∞-=.九、已知2)1(lim 1=+→xx ax ,求a 的值. (ln 2a =)十、设⎪⎪⎩⎪⎪⎨⎧>-<=0cos 102sin )(2x x x x xxx f ,,,求(0)(0)f f -+,和)(lim 0x f x →. (2,2,2)十一、设⎪⎩⎪⎨⎧≥+<=00tan )(2x x x x xaxx f ,,,已知)(lim 0x f x →存在,求a 的值. (0a =)1.7 无穷小的比较一、比较下列各对无穷小:1. 221,(1)(1)x x x --→ (后者高阶) 2. 321,1(1)x x x --→ (同阶)3.21cos ,(0)x x x -→ (同阶) 4. 2tan sin ,(0)x x x x -→ (前者高阶) 二、证明:当0→x 时,有以下等价无穷小成立:1. arcsin x x ;2.3tan sin 2x x x -. 三、利用等价无穷小代换计算下列极限:1. 20arctan lim sin 1x x x x →=2. 21lim s n 0i x x x →∞=3.lim 12x +→=四、当0x →时,下列四个无穷小中,哪一个是比其他三个更高阶的无穷小?A.2x B.1cos x -1 D.tan x x - (D )五、证明:若α是β的高阶无穷小,则αββ+ 。
微积分第一章课外习题参考答案
p1 4 . 三 .3 .证 明 : 设 M m a x{ f ( x i ) | 1 i n },
m m in{ f ( xi ) | 1 i n}, 则 有 m f ( xi ) M ,1 i n,
p2. 四 . 证明: f (x) f (2a x)
f (2b 2a x) f [2(b a) x] 周期 T 2|b a |.
五 . 证明 f ( x) loga( x x2 1)
loga
x2
1 1
x
loga( x
x2 1)
f ( x).
§1.1, §1.2数列极限(3-4)
ak ).
例如:
lim n 1n 2n 8n 8.
n
p15. 三 .由 导 数 定 义 知 :
1. lim e xh e x . h0 h
tan( x x) 1
2. lim
x 0
x
cos2 x .
p16. 3.解 : 原 式 lim [(1
6
3 x2 6 x1
) ] 6 3 x 2 2
p 4 . 2 . 解 :由 题 意 ,设 P1 P2 1
P1 Pn
1
1 2
1 22
1 23
( 1)n2 2 n2
1 ( 1 )n1 2 2( 1 )n1
2 1 1
2 3
2
lim
n
P1 Pn
2 2( 1 )n1
lim
2
n
3
2 3
p4.3.证明 : { xn }有界, M 0,使得
《高等数学》第一部分微积分习题参考答案
文 科 高 等 数 学第一部分 微 积 分习题一:1.(1)是;(2)不是;(3)不是。
2.(1)(,2)(2,1)(1,)-∞----+∞ ;(2)(3,1][1,)--+∞ ;(3)[1,3]; (4)(1,0)(0,1]- 。
3. sin ,,sin(),sin(sin )xe x x e e e x 。
4. 1()2f x x=+ (0)x ≠。
5.(1)0][0,)x ∈-∞∈+∞当(,时,函数单调减少;当x 时,函数单调增加。
(2 ) ][1,)x ∈-∞∈+∞当(,1时,函数单调减少;当x 时,函数单调增加。
(3)[0,][,]332x x πππ∈∈当时,函数单调增加;当时,函数单调减少。
6.(1)奇;(2)奇;(3)非奇非偶;(4)偶;(5)非奇非偶。
7.(1)24y x =- (02)x ≤≤;(2)3arcsin 2xy = (03)x ≤≤(3)41116ln xx y x x x x e <⎧⎪=≤≤⎨⎪>⎩。
8. 211966R x x =-+。
9. (1000)(60.002)y x x =+- (03000,)x x N ≤<∈。
10. 题目有问题。
习题二:1.(1)存在;(2)不存在;(3)存在;(4)存在;(5)不存在。
2. 不能,如1(1),(1)nn n n a b +=-=-。
3. 能,用反证法证明。
4.(1)2;(2)0;(3)2;(4)15;(5)75;(6)0;(7)1;(8)0; (9)12;(10)n;(11)12x;(12)23;(13)12;(14)1;(15)1;(16)cos a ;(17)23;(18)2e ; (19)e ;(20)1e;(21)3e -;(22)2。
5.(1)错;(2)错;(3)错;(4)正确;(5)错。
6.(1)不连续;(2)不连续;(3)连续。
7. 略。
8.(1)约为7950.0元;(2)约为61391元。
《微积分》各章习题及详细答案
第一单元 函数与极限一、填空题1、已知x x f cos 1)2(sin +=,则=)(cos x f 。
2、=-+→∞)1()34(lim22x x x x 。
3、0→x 时,x x sin tan -是x 的 阶无穷小。
4、01sinlim 0=→xx kx 成立的k 为 。
5、=-∞→x e xx arctan lim 。
6、⎩⎨⎧≤+>+=0,0,1)(x b x x e x f x 在0=x 处连续,则=b 。
7、=+→xx x 6)13ln(lim0 。
8、设)(x f 的定义域是]1,0[,则)(ln x f 的定义域是__________。
9、函数)2ln(1++=x y 的反函数为_________。
10、设a 是非零常数,则________)(lim =-+∞→xx ax a x 。
11、已知当0→x 时,1)1(312-+ax 与1cos -x 是等价无穷小,则常数________=a 。
12、函数xxx f +=13arcsin )(的定义域是__________。
13、____________22lim22=--++∞→x x n 。
14、设8)2(lim =-+∞→xx ax a x ,则=a ________。
15、)2)(1(lim n n n n n -++++∞→=____________。
二、选择题1、设)(),(x g x f 是],[l l -上的偶函数,)(x h 是],[l l -上的奇函数,则 中所给的函数必为奇函数。
(A))()(x g x f +;(B))()(x h x f +;(C ))]()()[(x h x g x f +;(D ))()()(x h x g x f 。
2、xxx +-=11)(α,31)(x x -=β,则当1→x 时有 。
(A)α是比β高阶的无穷小; (B)α是比β低阶的无穷小; (C )α与β是同阶无穷小; (D )βα~。
《微积分》各章习题及详细答案
第一章 函数极限与连续一、填空题1、已知x x f cos 1)2(sin +=,则=)(cos x f 。
2、=-+→∞)1()34(lim22x x x x 。
3、0→x 时,x x sin tan -就是x 的 阶无穷小。
4、01sin lim 0=→xx kx 成立的k 为 。
5、=-∞→x e xx arctan lim 。
6、⎩⎨⎧≤+>+=0,0,1)(x b x x e x f x 在0=x 处连续,则=b 。
7、=+→xx x 6)13ln(lim 0 。
8、设)(x f 的定义域就是]1,0[,则)(ln x f 的定义域就是__________。
9、函数)2ln(1++=x y 的反函数为_________。
10、设a 就是非零常数,则________)(lim =-+∞→xx ax a x 。
11、已知当0→x 时,1)1(312-+ax 与1cos -x 就是等价无穷小,则常数________=a 。
12、函数xxx f +=13arcsin )(的定义域就是__________。
13、lim ____________x →+∞=。
14、设8)2(lim =-+∞→xx ax a x ,则=a ________。
15、)2)(1(lim n n n n n -++++∞→=____________。
二、选择题1、设)(),(x g x f 就是],[l l -上的偶函数,)(x h 就是],[l l -上的奇函数,则 中所给的函数必为奇函数。
(A))()(x g x f +;(B))()(x h x f +;(C))]()()[(x h x g x f +;(D))()()(x h x g x f 。
2、xxx +-=11)(α,31)(x x -=β,则当1→x 时有 。
(A)α就是比β高阶的无穷小; (B)α就是比β低阶的无穷小; (C)α与β就是同阶无穷小; (D)βα~。
微积分部分习题及答案 (3)
x2
1
3 2
C
x3
1
3
x2 1 2 C
3 23
33
27
6. 用第二类换元积分法计算下列各题
(1) x 2x 1dx
解 令t 2x 1, x t2 1,
2
原式 t 2 1 td( t 2 1 ) t2 1 t tdt 1 (t4 t2 )dt
2
2
2
2
1 (1 t5 1t3) C 1 t5 1 t3 C
3
3
t
21 1 t
1dt
3
(t
1
1
1
t
)dt
2
3(t 2 2
t ln 1 t ) C
2
3u 3 2
33 u
3 ln 1
3u
C
3 1 x2 3
3 3 1 x2 3 ln 1 3 1 x2 C
2
35
7. 用分部积分法计算下列各题
(1) ln xdx
解: 原式 x ln x xd ln x
e2x ex
1 1
dx
解:原式
(ex
1)(e x ex 1
1) dx
(ex 1)dx
ex x C
4
4. 求下列不定积分
(5) 5xexdx
解: 原式 5exdx
5ex C 5ex C
ln 5e
1 ln 5
5
4. 求下列不定积分
(6)
3
1
x
2
2
dx
1 x2
解:原式 3arctan x 2 arcsin x C
t 2 ln t 2tdt 4
t2 ln tdt 4 3
微积分试题及答案
微积分试题及答案一、选择题(每题2分,共20分)1. 函数 \( f(x) = x^2 \) 在 \( x = 0 \) 处的导数是:A. 0B. 1C. 2D. 32. 曲线 \( y = x^3 - 2x \) 在 \( x = 1 \) 处的切线斜率是:A. -1B. 1C. 3D. 53. 若 \( \int_{0}^{1} x^2 dx \),则该积分的值是:A. \( \frac{1}{3} \)B. \( \frac{1}{2} \)C. \( \frac{2}{3} \)D. 14. 函数 \( y = \ln(x) \) 的原函数是:A. \( x \)B. \( x^2 \)C. \( e^x \)D. \( x\ln(x) \)5. 已知 \( \frac{dy}{dx} = 3x^2 - 2x \),当 \( x = 1 \) 时,\( y \) 的值是:A. 0B. 1C. 2D. 36. 函数 \( y = \sin(x) \) 的二阶导数是:A. \( -\sin(x) \)B. \( -\cos(x) \)C. \( \cos(x) \)D. \( \sin(x) \)7. 若 \( \int x e^x dx \) 可以用换元积分法求解,设 \( u = x \) 则 \( du \) 是:A. \( e^x \)B. \( e^x dx \)C. \( x dx \)D. \( 1 \)8. 函数 \( y = x^3 + 3x^2 + 3x + 1 \) 的泰勒展开式在 \( x = 0 \) 处的前三项是:A. \( 1 + 3x + 3x^2 \)B. \( 1 + x + x^2 \)C. \( 1 + 3x + 9x^2 \)D. \( 1 + 3x + 3x^3 \)9. 函数 \( y = e^{2x} \) 的不定积分是:A. \( \frac{1}{2}e^{2x} \)B. \( e^{2x} \)C. \( 2e^{2x} \)D. \( 2x e^{2x} \)10. 若 \( \frac{dy}{dx} = y - 1 \),且 \( y = 2 \) 当 \( x =0 \),则 \( y \) 的通解是:A. \( y = x + 2 \)B. \( y = e^x + 1 \)C. \( y = e^x - 1 \)D. \( y = 1 - e^x \)二、填空题(每空2分,共20分)11. 若 \( f(x) = x^3 - 6x^2 + 11x - 6 \),则 \( f'(x) = \)__________。
微积分考试题目及答案
微积分考试题目及答案1. 求函数f(x) = x^2的导数。
解答:根据导数的定义,导数是函数在某一点处的变化率。
对于f(x) = x^2,我们可以使用求导法则来求导数。
根据幂函数的求导法则,当函数为x^n时,导数为nx^(n-1)。
应用该法则,我们有:f'(x) = 2x^(2-1)= 2x因此,函数f(x) = x^2的导数为2x。
2. 求函数f(x) = e^x的导数。
解答:根据指数函数的求导法则,当函数为e^x时,导数也为e^x。
因此,函数f(x) = e^x的导数为e^x。
3. 求函数f(x) = ln(x)的导数。
解答:根据对数函数的求导法则,当函数为ln(x)时,导数为1/x。
因此,函数f(x) = ln(x)的导数为1/x。
4. 求函数f(x) = sin(x)的导数。
解答:根据三角函数的求导法则,当函数为sin(x)时,导数为cos(x)。
因此,函数f(x) = sin(x)的导数为cos(x)。
5. 求函数f(x) = cos(x)的导数。
解答:根据三角函数的求导法则,当函数为cos(x)时,导数为-sin(x)。
因此,函数f(x) = cos(x)的导数为-sin(x)。
6. 求函数f(x) = 2x^3 - 5x^2 + 3x - 7的导数。
解答:应用求导法则,我们对每一项分别求导。
根据幂函数的求导法则,导数为nx^(n-1)。
所以:f'(x) = 2*3x^(3-1) - 5*2x^(2-1) + 3*1x^(1-1) + 0= 6x^2 - 10x + 3因此,函数f(x) = 2x^3 - 5x^2 + 3x - 7的导数为6x^2 - 10x + 3。
7. 求函数f(x) = x^2的不定积分。
解答:对于幂函数的不定积分,可以使用幂函数的积分法则来求解。
根据该法则,当函数为x^n时(n不等于-1),不定积分为(1/(n+1))x^(n+1) + C,其中C为常量。
- 1、下载文档前请自行甄别文档内容的完整性,平台不提供额外的编辑、内容补充、找答案等附加服务。
- 2、"仅部分预览"的文档,不可在线预览部分如存在完整性等问题,可反馈申请退款(可完整预览的文档不适用该条件!)。
- 3、如文档侵犯您的权益,请联系客服反馈,我们会尽快为您处理(人工客服工作时间:9:00-18:30)。
International Monetary FundMoldova and the IMF Press Release:IMF Executive Board Completes Second Review Under the Extended Credit Facility and the Extended Fund Facility Arrangements with Moldova, Approves US$79 Million Disbursement April 7, 2011Country’s Policy Intentions DocumentsE-Mail Notification Subscribe or Modify your subscription Moldova: Letter of Intent, Supplementary Memorandum of Economic and Financial Policies, and Technical Memorandum of UnderstandingMarch 24, 2011M OLDOVA:L ETTER OF I N TE N TChişinău, March 24, 2011 Mr. Dominique Strauss-KahnManaging DirectorInternational Monetary Fund700 19th Street NWWashington, DC 20431 USADear Mr. Strauss-Kahn:The economic program supported by the IMF is playing a crucial role in restoring stability and rebuilding confidence in Moldova. With growth significantly exceeding projections in 2010, GDP has broadly recovered to pre-crisis levels. Inflation is under control, and the fiscal deficit has narrowed substantially. These remarkable results were achieved notwithstanding the challenges that the economy faces: fiscal adjustment and promotion of export-led growth require profound structural reforms; rising international food and fuel prices rekindle inflation pressures; job creation lags behind and unemployment still exceeds pre-crisis levels.The program is broadly on track. All quantitative performance criteria for end-September and most indicative targets for end-December 2010 were observed. However, the difficult political environment of 2010 and unforeseen technical complications have taken their toll, and several structural benchmarks under the program were delayed. In the coming period, we will move expeditiously to implement these measures, as well as the new reforms set forth in our agreement with the IMF. The 2011 fiscal budget consistent with the program objectives will be adopted as a prior action for completion of this review. In addition, we have prepared the Annual Progress Report on the implementation of our National Development Strategy and circulated it to the IMF Executive Board for information.In consideration of our strong record of program implementation, we request the completion of the second review of the program supported by the Extended Credit Facility and the Extended Fund Facility arrangements and the associated disbursement of SDR 50 million. As the Executive Board consideration of our request falls in early April 2011, we also request waivers of applicability of the relevant end-March performance criteria. The third program review, assessing performance based on end-March 2011 performance criteria and relevant structural benchmarks, is envisaged for June 2011. Moldova remains committed to improving the well-being of the population through reforms that promote sustainable growth and reduce poverty. In the period ahead, our program will focus on maintaining the targeted pace of fiscal adjustment; reining in inflation pressures; strengthening financial stability of the banking sector; restructuring the energy sector; rolling out the long-awaitededucation and other structural reforms that would support Moldova’s reorientation toward export-led growth.We believe that the policies set forth in the attached Supplementary Memorandum of Economic and Financial Policies (SMEFP) are adequate to achieve these objectives but will take any additional measures that may become appropriate for this purpose. We will consult with the IMF on the adoption of such additional measures in advance of revisions to the policies contained in the SMEFP, in accordance with the Fund’s policies on such consultation. We will provide the Fund with the information it requests for monitoring progress during program implementation. We will also consult the Fund on our economic policies after the expiration of the arrangement, in line with Fund policies on such consultations, while we have outstanding purchases in the upper credit tranches. Sincerely yours,/s/Vladimir FilatPrime MinisterofRepublicMoldovatheGovernmentof/s/ /s/NegruţaVeaceslavValeriu LazărFinanceofDeputy Prime Minister MinisterEconomyMinisterof/s/Dorin DrăguţanuGovernorNational Bank of MoldovaAttachment: Supplementary Memorandum of Economic and Financial PoliciesUnderstandingofMemorandumTechnicalS UPPLEME N TARY M EMORA N DUM OF E CO N OMIC A N D F I N A N CIAL P OLICIESMarch 24, 20111.The present document supplements and updates the Memoranda of Economic and Financial Policies (MEFPs) signed by the authorities of the Republic of Moldova on January 14, 2010 and June 30, 2010. It accounts for recent macroeconomic developments and introduces policy adjustments, as well as additional policies necessary to achieve the objectives of the program. We remain determined to meeting our commitments made previously under the program.I. M ACROECO N OMIC D EVELOPME N TS A N D O UTLOOK2.Growth outperformed expectations in 2010, and the economic expansion is set to continue. Real GDP rebounded by 6.9 percent in 2010, more than offsetting the economic contraction of 6 percent recorded in 2009. We expect the economic growth to return to its sustainable pace of 4½-5 percent in 2011 and thereafter. Expansion of domestic demand, exports, and investment are expected to drive activity in the near term, with tailwinds from trade liberalization reforms, a more favorable external environment, and improving competitiveness.3.Barring severe external shocks, disinflation should continue in 2011-12. Despite adjustment of energy tariffs, depreciation of the leu, and higher excise rates, inflation remained under control at around 8 percent in 2010, while core inflation declined below 5 percent. Under our baseline assumptions for international food and energy prices, we expect that inflation will decline further to 7½ percent in 2011 and about 5 percent by end-2012, the medium-term target set by the NBM. However, we recognize the risk that further surges in international food and energy prices and faster than expected rebound in domestic demand can temporarily push headline inflation above the projected path.4.Strong economic recovery boosted budget revenues and helped improve the fiscal position. In 2010, revenue significantly exceeded the program projections in nominal terms, but underperformed as percent of GDP, mainly due to high contribution to growth of the largely untaxed agriculture. Expenditure targets were also comfortably met, albeit largely due to under-spending of the capital budget caused by capacity constraints. As a result, the cash budget deficit narrowed to 2½ percent of GDP in 2010, far below the program target of5.4 percent of GDP.5.After a sharp drop to single digits in 2009, the external current account deficit widened in 2010 and will remain elevated in 2011. Rising demand for consumer and investment goods has pushed the current account deficit to an estimated 12¾ percent of GDP in 2010. The same demand factors, along with higher costs of energy imports, will likely propel the deficit even higher in 2011. The elevated deficit in 2011 will be largely financed by official assistance, private capital flows, and FDI. As the economy’s borrowing space is filling up quickly, we realize that further external borrowing should proceed at a more measured pace. We expect that from 2013, thanks to our exportpromotion efforts and economic recovery in trading partners, higher exports will more than offset the rise in imports, and the current account deficit would decline towards 10 percent of GDP.6.The situation in the financial sector has improved as well, with domestic credit rebounding and nonperforming loans declining. After the decline of 2009, domestic bank credit expanded by about 13 percent in 2010, and interest rates have declined. Meanwhile, the share of nonperforming loans declined to 13.3 percent, in part reflecting write-offs. Moreover, banks maintain large liquidity and capital buffers, remaining resilient to potential risks.II. R EVISED P OLICY F RAMEWORK FOR 2011-12A. Fiscal Policy7.Building on the better-than-expected fiscal outcome in 2010, the structural fiscal adjustment will stay on course in 2011-12. Our goal is to bring down the structural fiscal deficit excluding grants—the fiscal deficit adjusted for the effects of economic cycles—from 5½ percent of GDP at end-2010 through 4½ percent of GDP in 2011 to 3½ percent of GDP by 2012. This would largely rid the budget from its dependency on exceptional foreign aid and make public finances more resilient to macroeconomic risks. In this context, we will continue to contain the unaffordable public sector wage bill and low priority current spending, while strengthening revenue through selected tax policy measures and improved tax administration. Using the created fiscal space to increase infrastructure investment and provide well-targeted social assistance to the most vulnerable will allow us to achieve our broader development goals.8.As a next step, we will adopt a 2011 budget with a deficit of 1.9 percent of GDP as a prior action. We project that the budget revenue will amount to 37¾ percent of GDP in 2011, on account of continued progress in the tax administration reform, increased excise rates on tobacco and hard liquor—in line with our EU Association agenda—and updates of selected local taxes and fees. Implementation of various structural reforms, described below, will allow us to reduce current expenditure by 1½ percent of GDP to 34½ percent of GDP. At the same time, priority social assistance spending will be safeguarded, and capital expenditure will increase to 5¼ percent of GDP. We will seek to maintain the targeted structural fiscal adjustment in case the economic outlook and budget revenue deviate from our current projections.9.With immediate fiscal pressures easing, structural reforms will help contain the large public sector wage bill while creating space for poverty reduction actions. The significant optimization efforts in the education sector (¶19) will help finance the increase of teachers’ wages planned for September 2011. During 2011, other public wage restraints will remain in place as described in Law 355, as amended in October 2009. The only exception will be made for low-income auxilliary personnel in the budget sector (with salaries below MDL 1500), whose wages will be indexed by 8.5 percent on average from July 1, 2011 to alleviate the impact of higher than expected food and fuel prices and to avoid disincentives to labor market participation. Moreover, public sectoremployment will be capped at 212,000 positions by end-2011, reflecting the effects of the education reforms, while all vacant positions in excess of that level will be eliminated in 2011.10.Greater emphasis will be placed on synchronizing fiscal consolidation efforts at the central and local levels. The local governments will be granted greater control over local tax rates and fees to allow better revenue planning. In particular, by end-March 2011, we will ensure parliamentary passage of the necessary legal amendments to remove ceilings on existing local taxes and fees. This would allow the Chişinău municipality to raise at least MDL 100 million in additional revenues to finance, among other things (discussed in ¶21), its program of granting wage supplements and heating assistance in 2011. The practice of granting these payments will be discontinued at end-2011. The Ministry of Finance will verify compliance with these commitments.11.Going forward, we will continue trimming down current spending while creating sufficient space for the large public investment needs. In 2012, we aim to reduce the budget deficit further to ¾ percent of GDP, mainly through further rationalization of current spending (1 percent of GDP), sustained by structural reforms (¶¶19-22) that will commence in 2011 and bear fruit over the medium term. Ensuring sustainability of public finances in the medium term will also require implementation of the following measures:∙To reduce spending on goods and services, we will persevere with our procurement reform, assisted by the World Bank. The reform, to be phased in during 2011, will lower the budget costs by automating the bids for delivery of goods and services in the government’scentralized procurement agency.∙To improve control over budget planning and execution, we have drafted a law on public finance and accountability which will introduce a rule-based fiscal framework, enhance fiscal discipline, and improve transparency. We expect the law to be passed by Parliament by end-September 2011 and used in the preparation of the 2012 budget.∙To ensure the most effective allocation of capital expenditure, we will review the list of existing and envisaged capital projects, with a view to prioritize execution on the basis oftheir viability and economic growth potential. The review will also take into account pastexecution rates and capacity for implementation.∙To ensure implementation of the recently approved tax compliance strategy, by April 30, 2011, the State Tax Service (STS) will put in place operational plans for the strategyimplementation, including audit, collection of arrears, and taxpayer service activities(structural benchmark). In addition, by September 30, 2011, we will draft and submit toParliament legislation to allow indirect assessment of individuals’ income based on theirassets and other indicators as specified in the compliance strategy. On this basis, byDecember 31, 2011, we will prepare operational plans to strengthen audit, enforcement,outreach to, and education of high-wealth individuals regarding their tax compliance.∙We will reform the outdated mechanism for sick leave benefits. By March 31, 2011, we will amend legislation to assign the financial responsibility for the first day of sick leave to theemployee and the second day to the employer, effective July 1, 2011 (structural benchmark for end-April). Further legal amendments—to accompany the passage of the 2012 budget—will increase the number of sick leave days covered by employers to 3 in 2012, 4 in 2013, and6 in 2014.∙Early retirement privileges will be gradually phased out. By March 31, 2011, we will adopt legislation that, starting July 1, 2011, would raise the statutory retirement age of civilservants, judges, and prosecutors by six months every year until it reaches the regularretirement age (structural benchmark for end-April). This legislation will also extend the requirement to pay social contributions to all persons employed in Moldova in line withbilateral treaties. Another related piece of legislation, also to be passed by March 31, 2011,will put in place a policy of increasing the years of contribution required for full pensioneligibility from 30 to 35 years (and from 20 to 25 years for military and police personnel), by6 months every year, starting July 1, 2011.∙Building on the findings and recommendations of the recent IMF TA mission, we will implement measures to rationalize the use of health care. In particular, from January 1, 2012 we will introduce a copayment of 20 lei for primary care visits for uninsured patients, tomotivate them to enroll into the health insurance system. From January 1, 2013, we willintroduce small copayments for each doctor and hospital visit (5 lei for primary care, 10 leifor specialists, and 20 lei for hospital admissions) for all other categories of patients,including those who currently receive medical services free of charge. This policy will raise revenue and deter the use of unnecessary care, thus reducing the burden on the system. Tothis end, by end-April 2011 we will prepare an action plan detailing needed legislativechanges, technical preparations, and public information campaign.B. Monetary and Exchange Rate Policies12.The N BM’s monetary policy will be focused on achieving its end-2012 inflation objective of 5 ± 1½ percent. Given the fast economic recovery, closing output gap, and inflation pressures from rising international food and energy prices, the NBM’s monetary policy stance will gradually shift from supporting the recovery to addressing inflation risks. Specifically, it should focus on anchoring expectations—thereby countering the second-round effects from surging food and energy prices—and preventing excessive credit expansion. In this context, the NBM’s recent tightening measures—the 100 basis points hike in the policy interest rate and the increase in required reserve ratio from 8 percent to 11 percent— adequately address current inflation concerns. Further tightening should be conditional on marked acceleration of credit growth or rising inflation expectations.13.At the same time, the N BM will continue to strengthen the operational and legal aspects of its monetary policy framework. Consistent with the transition to inflation targeting, theindicative target for reserve money under the program will be discontinued after March 2011. Nevertheless, the NBM will continue to monitor money growth closely as an indicator of the state of domestic demand and sharp sustained moves may warrant policy action. In parallel, the NBM will continue to further enhance its communication, research, and forecasting capacities. As regards the legal framework, by end-September 2011, the NBM will propose amendments to the central bank law to strengthen its independence in line with the international best practice and establish appropriate mechanisms of internal control over NBM’s corporate governance.14.Alongside, the N BM’s exchange rate policies will remain consistent with program objectives. Specifically, NBM interventions in the foreign exchange market will continue to aim at smoothing erratic movements, but not resist sustained depreciation pressures. Should capital inflows exceed program projections, the NBM will accelerate the pace of reserve accumulation to ensure adequate buffers against the still high external vulnerabilities.C. Financial Sector Policy15.To strengthen financial stability, we will address the quasi-fiscal liabilities stemming from recent crisis management efforts. The Government’s decision to shield from losses the depositors of Investprivatbank (IPB) that failed in 2009 was a necessary step to avoid potential panic and deposit runs. However, paying out these deposits by means of a loan from the majority state-owned Banca de Economii (BEM) to IPB—in turn, enabled by a liquidity-providing loan from the NBM—has created a burden on BEM’s balance sheet that is now inhibiting its development. To address this problem, by end-May 2011 the Government will issue to BEM a long-term bond equal to the residual face value of BEM’s loan to IPB by either purchasing this loan or—subject to agreement of BEM’s minority shareholders—recapitalizing the bank. Meanwhile, the NBM will consider a limited extension of its loan to BEM to mitigate the attendant liquidity risk, and will work with BEM and the IPB liquidator to accelerate the sale of IPB assets. The Deposit Guarantee Fund will assume the responsibility for the net cost of the payout to IPB depositors and may introduce an extraordinary deposit insurance premium to gradually reimburse the Government for the cost of the bond issued to BEM.16.To handle future risks better, we aim to put in place the remaining elements of our contingency planning framework. Recent strengthening of the bank resolution framework and the establishment of a high-level Financial Stability Committee (FSC) were followed by signing of a memorandum of understanding (MoU) between key institutions involved in responding to financial emergencies. As a next step, we aim to put in place specific contingency plans for each MoU participant by end-June 2011. These plans will establish a contingency framework based on a clear set of instruments, division of roles, responsibilities, as well as coordination channels between the involved parties.17.Looking ahead, as credit growth picks up speed, the N BM will need to strengthen its bank supervision framework by improving data collection and reducing scope for regulatoryarbitrage. To this end, the NBM, based on best international practices, will develop a new reporting system for commercial banks allowing a more detailed analysis of financial sector data. In addition, by end-September 2011, the NBM and the National Commission for Financial Markets, with assistance from the World Bank, will explore options and make proposals to consolidate all credit institutions—including banks, leasing companies, savings and credit associations, and microfinance institutions—as well as insurance companies and pension funds under a common supervisory framework. Finally, by end-September 2011, the NBM in cooperation with the World Bank will evaluate the feasibility of establishing a public credit bureau to promote information exchange and prudent lending policies by banks.18.Despite earlier delays, measures to strengthen the debt restructuring and contract enforcement frameworks are being developed and will be implemented in the coming months. The NBM has already allowed faster reclassification of restructured loans into lower-risk categories. We will now ensure by end-September 2011 parliamentary passage of the legal amendments described in the SMEFP of June 30, 2010 (¶15), to enhance the speed and predictability of collateral execution by banks and to strengthen incentives for banks to restructure nonperforming loans (structural benchmark). Furthermore, with technical assistance from the World Bank and in consultation with the IMF staff, we will seek to strengthen and simplify other aspects of the insolvency framework. Specific draft legal amendments in this area will be adopted by the Government by March 2012.D. Structural ReformsRaising Efficiency of the Public Sector19.In the coming months, we will roll out the comprehensive reform of the oversized education sector. Its main goals are to eliminate excess capacity, create a leaner and better-equipped education system with adequately trained and paid staff, and provide education that meets demands of the modern economy. The reform will seek class, school, and employment consolidation. A large part of the eventual budget savings and financial assistance from the World Bank will be used to improve school quality, secure transportation for students, and repair school bus routes. Nevertheless, the reform will save about 0.5 percent of GDP on a net permanent basis from 2013 on. Our reform strategy is based on the following elements:∙Class size optimization. By September 1, 2012, we will increase class size to 30-35 students in large schools and 25-30 students in the rest. For this purpose, we will pass legalamendments to eliminate the existing norms prescribed in the Law on Education by end-July 2011. This would reduce the number of teaching positions by 1,736, including 390 positions in 2011, and lead to estimated annual savings of about MDL 94 million.∙Optimization of the school network. Gradual consolidation of the school network through closure of schools with low enrollment and securing transportation of students to nearby“hub” schools will commence this year. Its full implementation during 2011-13 would reducethe number of teaching and non-teaching positions by 2,661 and 1,426 respectively and, when completed, will generate savings of about MDL 136 million a year. We will aim to limit the attendant transportation costs to MDL 61 million per year, and will seek grant assistance from the international financial community to defray this cost.∙Reduction of non-teaching personnel and vacant positions. As a first step, we will immediately freeze hiring of non-teaching staff and eliminate 2,400 vacant positions in thesector. Alongside, we will include in the budget law for 2011 a provision establishing wage bill ceiling for education sector, resulting in all rayons reducing personnel in educationinstitutions on average by 5 percent from their level of end 2010 (5,300 positions nationwide) before academic year 2011/12. These measures would provide savings of MDL 175 million on a full-year basis.∙Increasing flexibility of labor relations in the sector. Local authorities also need support and more flexibility to be able to consolidate schools and classes. By end-July 2011, we willadopt legal amendments to the Labor Code and other enabling legislation to (i) make fixed-term (one year) contracts mandatory for teachers beyond retirement age; and (ii) allow school principals’ hiring and dismissal decisions to be based on business need and performancerather than tenure. Estimated annual savings from this measure amount to MDL 48 million. ∙Rollout of a per-student financing system. Following successful implementation of per-student financing in the pilot rayons of Cauşeni and Rişcani, the system will be expandedstarting January 1, 2012 to 9 additional rayons, as well as municipalities of Chişinău andBalţi. The system will create strong incentives to optimize schools’ financial performance. Its nationwide implementation will take place in 2013.∙Putting social protection costs in education on a means-tested basis. By end-June 2011, in consultation with the World Bank and other partners, we will conduct a thorough review ofall social expenditure in the education budget (scholarships, dormitory assistance, schoolmeals, etc.) to explore options for better targeting of such assistance to the most vulnerablegroups.In consultation with the World Bank, the Government will develop and, by end-March 2011, adopt a detailed action plan to implement this reform.20.We will reform the civil service in a way that increases efficiency without destabilizing the fiscal position. To this end, we have developed descriptions of new job functions and responsibilities for staff in central government administration along with a merit- and performance-based wage system for civil servants. Implementation of this reform will start in October 2011, and will ensure that the reform does not affect the aggregate public sector wage bill as a ratio to GDP. 21.As regards the energy sector, we will strive to achieve a stable framework for payments of current bills, pending a comprehensive sector restructuring strategy to be finalized and implemented in cooperation with the World Bank and other partners. To ensure a stablefunctioning of the sector, the Ministry of Economy, the Chişinău municipality authorities, and the key participants in the energy sector will seek to negotiate in good faith a MoU with the following key elements: (i) a monthly schedule of payments to energy suppliers that is consistent with typical collection lags in Termocom’s receivables during the heating season, (ii) full repayment of current arrears by Termocom before the following heating season; (iii) a mechanism for covering the cash gap arising from collection lags in Termocom or a bank guarantee from the Chişinău municipality backing Termocom’s adherence to the agreed payment schedule; (iv) creditors’ commitment to abstain from blocking bank accounts as long as the MoU is observed. In this context, the Chişinău municipality will budget for and pay in full its remaining debt to Termocom of MDL 64 million by end-March 2011.22.Meanwhile, we will adopt a number of legal and regulatory amendments which would help ensure cost recovery in the heating sector. By end-August 2011, we will adopt the necessary legal and/or regulatory amendments to raise the heating fee for apartments disconnected from central heating from 5 percent to 20 percent of the average heating bill. This increase is in line with regional practices and would mostly affect consumers with relatively high incomes. At the same time, the Ministry of Regional Development and Construction, the Chişinău municipality, Termocom, and the water distributor Apă Canal will seek to put an end to persistent losses caused by under-billing for hot and cold water delivery; other municipalities will seek to resolve this issue as well. And to facilitate timely collection of heating bills, by end-August 2011, we will adopt the necessary legal and/or regulatory amendments introducing a minimum payment of 40 percent of the monthly bill and setting August 1 as the deadline for settling all heating bills for the past heating season.23.With the international investment climate gradually improving, the government will accelerate the efforts to divest its noncore assets. In the first half of 2011 the government, with assistance from IFC, will put in place an advisor to review various options for private sector participation in Moldtelecom. At the same time, by mid-2011, the government will expand the list of state assets subject to privatization. This will pave the way for privatization of other large public companies. By end-September 2011, the government will approach various international financial institutions, seeking an advisor to explore options to divest Air Moldova as soon as possible. Also by end-September 2011, we shall develop a roadmap for the privatization of Banca de Economii, and, if need be, resume the engagement of the privatization advisor.Improving the Business Environment and Removing Barriers for Trade24.The wheat export ban introduced in response to dwindling grain stocks in early 2011 will be abolished as soon as possible, and we will not introduce any new barriers to trade. We plan to abolish this ban by end-April 2011, provided that domestic and regional grain shortages are alleviated. Moreover, we shall refrain from introducing any new tariff or non-tariff barriers to exports. In addition, by end-May 2011 we will conduct an assessment of the existing tariff and non-tariff barriers to trade and their consistency with Moldova’s WTO commitments with regard to market access, and will develop roadmap for their gradual elimination.。