数学一高等数学习题集大集合

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第一章函数·极限·连续

一. 填空题

a?1?x?1．设lim???tetdt, 则a = _2._______. ???x???x?ax2. =___

112n??_____.lim?2?2???2?

n??2n?n?n??n?n?1n?n?23. 已知函数f(x)??

?1?0|x|?1|x|?1, 则f[f(x)] ___1____.

4. lim(n?3n?n?n)=___limn??n??n?3n?n?nn?3n?n?n?2____.

5. limcotx?x?01??1??=______.

?sinxx?解. limcosxx?sinxx?sinx1?cosxsinx1??lim?lim?lim?

x?0sinxx?0x?0x?06xxsinxx33x26n19906. 已知limk?A(? 0 ? ?), 则A = ______, k = _______.

n??n?(n?1)k所以 k－1=1990, k = 1991;

111?A，A?? kk1991

二. 选择题

1. 设f(x)和?(x)在(－?, +?)内有定义, f(x)为连续函数, 且f(x) ? 0, ?(x)有间断点, 则 (a) ?[f(x)]必有间断点 (b) [ ?(x)]2必有间断点 (c) f [?(x)]必有间断点 (d) 所以(d)是答案.

2. 设函数f(x)?x?tanx?esinx?(x)必有间断点 f(x), 则f(x)是

(a) 偶函数 (b) 无界函数 (c) 周期函数 (d) 单调函数解. (b)是答案. 3. 极限lim?352n?1?的值是 ????2222?n??12?222?3n?(n?1)???(a) 0 (b) 1 (c) 2 (d) 不存在

解., 所以(b)为答案.

(x?1)95(ax?1)54. 设lim?8, 则a的值为 250x??(x?1) 1

(a) 1 (b) 2 (c) 解.所以(c)为答案. 5. 设lim58 (d) 均不对

(x?1)(x?2)(x?3)(x?4)(x?5)??, 则?, ?的数值为

x??(3x?2)?111 (b) ? = 5, ? = (c) ? = 5, ? = 5 (d) 均不对 333(a) ? = 1, ? = 解. (c)为答案.

6. 设f(x)?2x?3x?2, 则当x?0时

(a) f(x)是x的等价无穷小 (b) f(x)是x的同阶但非等价无穷小

x?0x(a) －1 (b) 1 (c) 2 (d) 3 解.所以(a)为答案. 8. 设limx?0atanx?b(1?cosx)cln(1?2x)?d(1?e?x)2?2，其中a2?c2?0, 则必有

(a) b = 4d (b) b =－4d (c) a = 4c (d) a =－4c 解.所以(d)为答案.

三. 计算题 1. 求下列极限 (1) lim(x?e)

x???x1x解. lim(x?e)?limex???x???x1xln(x?ex)x?eln(x?ex)x???xlim?ex???x?e?e1?e

xlim1?ex21?cos)x

x??xx1解. 令y?

x(2) lim(sinlim21xy?0ylim(sin?cos)?lim(sin2y?cosy)=ex??y?0xx1ln(sin2y?cosy)y?e2cos2y?sinyy?0sin2y?cosylim?e2

?1?tanx?(3) lim??x?01?sinx???1?tanx?解. lim??x?01?sinx??

1x3

11x3tanx?sinx???lim?1??x?01?sinx??2

taxn?sixn?tanx?sinx?? ?lim??1??x?0?1?sinx???limsinx(1?cosx)x31?sixntaxn?sixnn)x3?(1?sixtanx?sinxlim3?=ex?0x ??sinx?2sin2 =ex?0=ex?0limx3x2?e.

122. 求下列极限 (1) limx?1ln(1?3x?1)arcsin2x?132

解. 当x?1时, ln(1?3x?1)~3x?1, arcsin23x2?1~23x2?1. 按照等价无穷小代换 limx?1ln(1?3x?1)arcsin23x2?1?x? ?3?limx?1x?123x2?1?lim11 ?3x?123x?122(2) lim??12?cotx?0x2?解. 方法1：

?1lim?2?cot2x?0x??sin2x?x2cos2?1cos2x??x?=lim?=lim???22??x?0x2sin2xsinx?x?0????xx??? ??1?(x2?1)cos2 =lim?x?0?x4???2xcos2x?2(x2?1)cosxsinx?x? ???3?=lim?x?0?4x????2xcos2x?sin2x2x2cosxsinx?lim =lim 33x?0x?04x4x?2cos2x?4xcosxsinx?2cos2x1? =limx?012x22?2cos2x?2cos2x114cosxsinx?4sin2x11???lim?? =lim2x?0x?012x3224x32 =lim?2sin2x111112???????

x?024x3263233. 求下列极限 (1) limnn(n?1)

n??lnnnxnnn?1?1 解. lim(n?1)?limn 令nn?1?x limx?0ln(n??lnnn??lnn1?x) 3

1?e?nx(2) lim

n??1?e?nx1?en??1?e?nx?nx解. limx?0?1???0 x?0 ??1x?0?n?na?nb??, 其中a > 0, b > 0 (3) lim??n???2??解. lim??a?b??1?c? x?1/n,c?b/a alim????n???x?02?2??nnnx????ae?1xln(1?cx)?ln2xx?0?lim

=aeln(1?cx)?ln2xx?0?lim?aex?0?1?c?ac?alimcxlncxb?ab a

4. 求下列函数的间断点并判别类型

(1) f(x)?2?12?11x1x

解. f(0)?lim?x?0?2?12?11x1x?1, f(0?)?lim?x?02?12?11x1x??1

所以x = 0为第一类间断点.

?x(2x??)x?0??2cosx( 2 ) f(x)??

?sin1x?0??x2?1解. f(+0) =－sin1, f(－0) = 0. 所以x = 0为第一类跳跃间断点;

sin limf(x)?limx?1x?11不存在. 所以x = 1为第二类间断点; x2?1x?? f(??2)不存在, 而lim?2x(2x??)??,所以x = 0为第一类可去间断点;

2cosx2

x??k??lim?2?x(2x??)??, (k = 1, 2, …) 所以x =?k??为第二类无穷间断点.

22cosx1??xsinx?0?5. 讨论函数f(x)?? 在x = 0处的连续性. x

x?0x?e???

1)不存在, 所以x = 0为第二类间断点;

x?0x1?(xsin)?0, 所以当??0, limx?0?x(x?sin解. 当??0时lim? ???1时,在 x = 0连续, ???1时, x = 0为第一类跳跃间断点.

6. 设f(x)在[a, b]上连续, 且a < x1 < x2 < … < xn < b, ci (I = 1, 2, 3, …, n)为任意正数, 则在(a, b)内至少存在一个?, 使 f(?)?c1f(x1)?c2f(x2)???cn.

c1?c2???cn证明: 令M =max{f(xi)}, m =min{f(xi)}

1?i?n1?i?n所以 m ?

c1f(x1)?c2f(x2)???cn? M

c1?c2???cnc1f(x1)?c2f(x2)???cn

c1?c2???cn所以存在?( a < x1 ? ? ? xn < b), 使得f(?)?7. 设f(x)在[a, b]上连续, 且f(a) < a, f(b) > b, 试证在(a, b)内至少存在一个?, 使f(?) = ?. 证明: 假设F(x) = f(x)－x, 则F(a) = f(a)－a < 0, F(b) = f(b)－b > 0 于是由介值定理在(a, b)内至少存在一个?, 使f(?) = ?.

8. 设f(x)在[0, 1]上连续, 且0 ? f(x) ? 1, 试证在[0, 1]内至少存在一个?, 使f(?) = ?. 证明: (反证法) 反设?x?[0,1],?(x)?f(x)?x?0. 所以?(x)?f(x)?x恒大于0或恒小于0. 不妨设?x?[0,1],?(x)?f(x)?x?0. 令m?min?(x), 则m?0.

0?x?1因此?x?[0,1],?(x)?f(x)?x?m. 于是f(1)?1?m?0, 矛盾. 所以在[0, 1]内至少存在一个?, 使f(?) = ?.

9. 设f(x), g(x)在[a, b]上连续, 且f(a) < g(a), f(b) > g(b), 试证在(a, b)内至少存在一个?, 使 f(?) = g(?).

证明: 假设F(x) = f(x)－g(x), 则F(a) = f(a)－g(a) < 0, F(b) = f(b)－g(b) > 0 于是由介值定理在(a, b)内至少存在一个?, 使f(?) = ?. 10. 证明方程x5－3x－2 = 0在(1, 2)内至少有一个实根. 证明: 令F(x) = x5－3x－2, 则F(1) =－4 < 0, F(2) = 24 > 0 所以在(1, 2)内至少有一个?, 满足F(?) = 0.

?2?x2(1?cosx)?11. 设f(x)??1?1x??cost2dt?x0x?0x?0 x?0试讨论f(x)在x?0处的连续性与可导性.

x1x22costdt?1costdt?x??0f(x)?f(0)0x解. f?'(0)?lim ?lim?lim2??x?0?x?0x?0xxx1?x22cosx?12 ?lim?lim?0 x?0?x?0?2x2x2(1?cosx)?12f(x)?f(0)2(1?cosx)?x2x f?'(0)?lim ?lim?lim3???x?0x?0x?0xxx2sinx?2x2(cosx?1)?lim?0 ?lim2?x?0?x?06x3x 所以 f'(0)?0, f(x)在x?0处连续可导.

12. 设f(x)在x = 0的某领域内二阶可导, 且lim??sin3xf(x)??2??0, 求3x?0x??xf(0),f'(0),f''(0)及limx?0f(x)?3. 2xsin3x?f(x)sin3x?xf(x)?sin3xf(x)?x?2??lim?lim?0. 所以解. lim?332x?0x?0x?0xxxx?? lim?3x?sin??f(x)??0. f(x)在x = 0的某领域内二阶可导, 所以f(x),f'(x)在x = 0

x?0?x?连续. 所以f(0) = －3. 因为

sin3xsin3x?f(x)?3?f(x)?3 limx2?0, 所以limx?0, 所以 2x?0x?0xxsin3x3?f(x)?3sxx?lim3x?sin3x?lim3?3co3 lim ?limx?0x?0x?0x?0x2x2x33x23sin3x9? =limx?02x2f(x)?f(0)f(x)?3f(x)?39?lim?limx??0??0 f'(0)?lim2x?0x?0x?0x?0xx2f(x)?39?, 将f(x)台劳展开, 得由lim2x?0x21f(0)?f'(0)x?f''(0)x2?0(x2)?31992!f''(0)? lim, 所以, 于是 ?x?022x22f''(0)?9.

第二章导数与微分

一. 填空题 1. f(x)? f(k?1)1?x, 则f(n)(x)= _______. 1?x(?1)k?12?(k?1)!(?1)n2?n!(n), 所以f ??k?1?1n?1(1?x)(1?x)?x?1?t2d2y

2. 设? , 则2?______.

dx?y?costd2y?sint?dt2tcost?2sint1sint?tcostdy?sint??????解. , ??223dx4t2t4tdx2t?2t?tdx3. 设函数y = y(x)由方程ex?y'?cos(xy)?0确定, 则

dy?______. dx解. ex?y(1?y')?(y?xy')sinxy?0, 所以

ysinxy?ex?y y'?x?y

e?xsinxy4. 已知f(－x) =－f(x), 且f'(?x0)?k, 则f'(x0)?______. 解. 所以 f'(x0)?f'(?x0)?k

f(x0?m?x)?f(x0?n?x)?____(m?n)f'(x0)___.

?x?0?xf(x0?k?x)?f(x0)11?f'(x0), 则____k?____. 6. 设lim?x?0?x335. 设f(x)可导, 则lim7. 已知

d??1??1?1?, 则f'f????_______. ??2??dx??x??x?2?x2?1?21?1??1?解. ?f'?2??3?, 所以f'?2???. 令x2 = 2, 所以f'?2???1

x2?x?x?x??x?8. 设f为可导函数, y?sin{f[sinf(x)]}, 则解.

dy?_______. dxdy?f'(x)cosf(x)f'[sinf(x)]cos{f[sinf(x)]} dx2x?y9. 设y = f(x)由方程e为_______. y?1?

二. 单项选择题

?cos(xy)?e?1所确定, 则曲线y = f(x)在点(0, 1)处的法线方程

1x, 即 x－2y + 2 = 0. 2 7

1. 已知函数f(x)具有任意阶导数, 且f'(x)?[f(x)]2, 则当n为大于2的正整数时, f(x)的n阶导数是

(a) n![f(x)]n?1 (b) n[f(x)]n?1 (c) [f(x)]2n (d) n![f(x)]2n 解.. (a)是答案.

2. 设函数对任意x均满足f(1 + x) = af(x), 且f'(0)?b, 其中a, b为非零常数, 则 (a) f(x)在x = 1处不可导 (b) f(x)在x = 1处可导, 且f'(1)?a (c) f(x)在x = 1处可导, 且f'(1)?b (d) f(x)在x = 1处可导, 且f'(1)?ab 解., 所以. (d)是答案

注: 因为没有假设f(x)可导, 不能对于f(1?x)?af(x)二边求导. 3. 设f(x)?3x3?x2|x|, 则使f(n)(0)存在的最高阶导数n为

(a) 0 (b) 1 (c) 2 (d) 3 所以n = 2, (c)是答案.

4. 设函数y = f(x)在点x0处可导, 当自变量x由x0增加到x0 + ?x时, 记?y为f(x)的增量, dy为f(x)的微分, lim?y?dy等于

?x?0?x?y?dyo(?x)?lim?0. (b)是答案.

?x?0x?0?x?x(a) －1 (b) 0 (c) 1 (d) ? 解. 由微分定义?y = dy + o(?x), 所以lim1?2x?0?xsin5. 设f(x)?? 在x = 0处可导, 则 x

x?0??ax?b(a) a = 1, b = 0 (b) a = 0, b为任意常数 (c) a = 0, b = 0 (d) a = 1, b为任意常数

解. 所以 0 = a. (c)是答案. 三. 计算题

1. y?ln[cos(10?3x)]，求y'

2?sin(10?3x2)?6x解. y'???6xtan(10?3x2) 2cos(10?3x)2. 已知f(u)可导, y?f[ln(x?a?x2)]，求y'

2解. y'?f'[ln(x?a?x)]??2x?1??x?a?x2?2a?x21?? ?? =

f'[ln(x?a?x2)]a?x2

3. 设y为x的函数是由方程lnx2?y2?arctany确定的, 求y'. x解.

y'x?y22x?2yy'x ?22222yx?y2x?y1?2xx?y x?y x?yy'?y'x?y, 所以y'??x?etsintd2y4. 已知?, 求. 2tdx?y?ecostdyetcost?etsintcost?sint?t?解. , tdxecost?esintcost?sintd2yd?cost?sint?dt?(cost?sint)2?(cost?sint)21???? ??dxdx2dt?cost?sint?dx(cost?sint)2dtd2y2 ??dx2et(cots?sint)35. 设x?y?y，u?(x?x)解. dx?(2y?1)dy, du?

223/2, 求

dy du(2y?1)dy ?32dux?x(2x?1)dx2dy2 ?2du3(2y?1)x?x(2x?1)132(x?x)2(2x?1)dx 2dx

?x?f(t)??dyd2y6. 设函数f(x)二阶可导, f'(0)?0, 且?, 求, . 23tt?0t?0dxdx?y?f(e?1)dyf'(e3t?1)3e3tdy解. , 所以=3. ?dxf'(t)dxt?0d2y[f''(e3t?1)3(e3t)2?3e3tf'(e3t?1)]f'(t)?e3tf'(e3t?1)f''(t) ?3dx2[f'(t)]3所以

d2y[3f''(0)?3f'(0)]f'(0)?f'(0)f''(0)9f'(0)?6f''(0) ?3?232dxt?0[f'(0)][f'(0)]

?x?tet7. 设曲线x = x(t), y = y(t)由方程组?t确定. 求该曲线在t = 1处的曲率. y?e?e?2eetetetdyyt'et?2e1解. yt'??y?t. 所以 ??t?ttee?2edxxt'e?te(1?t)(e?2e) 所以

dy1??.

dxt?12e?dtd2yd?12et?2e?tet ?????2t3t2t??dxdt?(1?t)(e?2e)?dx(1?t)(e?2e)ed2y1 所以 . 在t = 1的曲率为 ??22dxt?18e|y''|(1?y'2)t?132 k??18e21???1?2??4e?x?0 x?032?e(1?4e)

?2?32四. 已知当x ? 0时, f(x)有定义且二阶可导, 问a, b, c为何值时 F(x)??二阶可导.

?f(x)2?ax?bx?c

F(x)?limF(x), 所以c = f(－0) = f(0); 解. F(x)连续, 所以lim??x?0x?0因为F(x)二阶可导, 所以F'(x)连续, 所以b = f?'(0)?f'(0), 且 F'(x)??x?0?f'(x)

2ax?f'(0)x?0??F''(0)存在, 所以F?''(0)?F?''(0), 所以

2ax?f?'(0)?f'(0)f'(x)?f'(0)?lim?2a, 所以 ?x?0x?0xx1 a?f''(0)

2 lim?x2(n)，求f(0). 五. 已知f(x)?21?x解. f(x)??1?1111??? 21?x21?x10

f(n)1n!1(?1)n (x)????2(1?x)n?12(1?x)n?1 f(2k?1)(0)?0, k = 0, 1, 2, … f2k(0)?n!, k = 0, 1, 2, … 六. 设y?xlnx, 求f(n)(1).

(n?1)!n?2(n?2)!?n(?1) nn?1xx解. 使用莱布尼兹高阶导数公式 f(n)(x)?x?(lnx)(n)?n(lnx)(n?1)?x(?1)n?1n?2 =(?1)n?1??(n?1)n?2 (n?2)!???(?1)(n?2)!n?1n?1?n?1xxx??所以 f(n)(1)?(?1)n?2(n?2)! 七. 已知

?y0edt??costdt?siny2,求y'.

0y222t2x2解. 两边对x求导, ey'?2xcosx?2yy'cosy,y'?2xcosx2ey2?2ycosy2

第三章一元函数积分学(不定积分)

一. 求下列不定积分: 1.

11?xln?1?x21?xdx

211?x11?x1?x1?1?x?lndx?解. ?lndln??ln??c ?1?x21?x21?x1?x4?1?x?11?x1?x1?x1?1?x?2. ?arctandx?arctandarctan?arctan???c 2?1?x1?x1?x2?1?x?1?x3.

2cosx?sinx?11?sinx?(1?cosx)2?1?cosxdx

2cosx?sinx?11?sinx1?sinx1?sinx1?1?sinx?解. ??dx???c ?1?cosxd1?cosx?2?(1?cosx)21?cosx1?cosx??4.

?dx 8x(x?1) 11

1解. 方法一: 令x?,

t?dxt7dt1?dt????ln(1?t8)?c 88??1?1x(x?1)t?18??8?1?t?t?1?1?ln?1?8??c 8?x??1t2 = ?

111(1?sinx?cosx)?(sinx?cosx)?1?sinx22dx 5．?dx??21?sinx?cosx1?sinx?cosx11cosx?sinx11dx??dx ??dx??221?sinx?cosx21?sinx?cosx11d(1?sinx?cosx)11 ?x????dx

xxx221?sinx?cosx22sincos?2cos22221111x ?x?ln|1?sinx?cosx|??dtan

x2222tan?12111x ?x?ln|1?sinx?cosx|?ln|tan?1|?c

2222二. 求下列不定积分: 1.

?(x?1)dx2x?2x?22

解.

?(x?1)2dt2dxd(x?1)cost 令x?1?tant ???tan2tsectx2?2x?2(x?1)2(x?1)2?1costdt1x2?2x?2 =????c???c

sin2tsintx?12.

?xdx41?x2

解. 令x = tan t,

?x4dt2dxcos3tdsintdsint11cost???dt??????c 44423???2tantsectsintsintsint3sintsint1?x 12

321?1?x =??3?x?2?1?x???c

?x?3.

?(2xdx2?1)1?x2

解. 令x?tant

sec2tcostdsint??dt?dt? ? 2222??22(2tant?1)sect2sint?cost1?sint(2x?1)1?xdx =arctansint?c?arctanx1?x2?c

??x2dxa?xx2dx22 (a > 0)

解. 令x?asint

a2sin2t?acostdt1?cos2t11???a2?dt?a2t?a2sin2t?c

acost224a2?x2?a2?x2??c

?a2?xx =?arcsin?22?aa5.

??(1?x2)3dx

解. 令x?sint

(1?cos2t)21?2cos2t?cos22t(1?x)dx??costdt??dt??dt

44234111311t?sin2t??(1?cos4t)dt?t?sin2t?sin4t?c 4488432311 =arcsinx?sin2t(1?cos2t)?c

844 =

314?1?2sin2t)?c =arcsinx?2sintcost(844 =arcsinx?381x1?x2(5?2x2)?c 86.

?x2?1dx x41 t解. 令x? 13

?x2?1dx??4x1?t2t21t4?1?22??2?dt???t1?tdt 令t?sinu??sinucosudu ?t?(x2?1)313 =cosu?c??c

33x37.

?xx?12x?12dx

dx?secttantdt

解. 令 x?sect,

?xx?12x2?1dx??sect?1secttantdt??(1?cost)dt?t?sint?c 2secttantx2?1?c x1 ?arccos?x三. 求下列不定积分:

e3x?exdx 1. ?4xe?e2x?1e3x?exex?e?xd(ex?e?x)x?x解. ?4xdx?dx??arctan(e?e)?c 2x2x?2xx?x2??e?e?1e?1?e(e?e)?12.

dx?2x(1?4x)

dt tln2x解. 令t?2, dx?

dxdt1?11?1arctant???dt????c ???2x(1?4x)?t2(1?t2)ln2ln2??t21?t2?tln2ln21(2?x?arctan2x)?c ln2 =?

四. 求下列不定积分:

x51. ?dx

(x?2)100x51x555?994?99解. ?dx??xd(x?2)???x(x?2)dx 10099??(x?2)9999(x?2)99 14

x55x45?4 =???x3(x?2)?98dx 9998?99(x?2)99?98(x?2)99?98x55x45?4x35?4?3x2 =? ???9998979699(x?2)99?98(x?2)99?98?97(x?2)99?98?97?96(x?2) ?5?4?3?2x5?4?3?2???c 959499?98?97?96?95(x?2)99?98?97?96?95(x?2)

?x?dx1?x4解.

1dt2dxtdt1dt2t令x?1/t???????

444222x1?x1t?11?t1?(t)tt4?1sec2u111?x4令t?tanu??du??ln|tanu?secu|?c??ln?c

2secu22x22

五. 求下列不定积分: 1.

2xcosxdx ?2xcosxdx??1121x(1?cos2x)dx?x??xdsin2x 2?441211 ?x?xsin2x??sin2xdx

4441211 ?x?xsin2x?cos2x?c

448解. 2.

3sec?xdx

解.

?sec3xdx??secxdtanx?secxtanx??tanxsecxtanxdx

=secxtanx?(secx?1)secxdx?secxtanx?ln|secx?tanx|?secxdx

3sec?xdx??2?311secxtanx?ln|secx?tanx|?c 22(lnx)3dx 3. ?2x(lnx)3113(lnx)233dx???(lnx)d??(lnx)??dx 解. ?x2xxx2 15

(lnx)33(lnx)26lnx(lnx)33(lnx)26lnx6???2dx??????2dx ??xxxxxxx(lnx)33(lnx)26lnx6????c ??xxxx4.

?cos(lnx)dx

解.

?cos(lnx)dx?xcos(lnx)??sin(lnx)dx?x[cos(lnx)?sin(lnx)]??cos(lnx)dx ?x[cos(lnx)?sin(lnx)]?c 2 ?cos(lnx)dx?5.

xcos4xxcos32211xx1x1x?2x??sin?2d??xsin?2?cot?c ??xsin824228242sin3六. 求下列不定积分: 1.

?x2dx?18?sin3xxcos4x2dx??111?2x?2x?2xxdsin??xsin?sindx ??828282?xln(x?1?x2)dx 22(1?x)解.

?xln(x?1?x2)112 dx?ln(x?1?x)d(1?x2)22?1?x211111ln(x?1?x2)??dx 22?221?x21?x1?x =

ln(x?1?x2)111 令x?tant ???sec2tdt 22?2(1?x)21?tantsectln(x?1?x2)1cost =?dt 22?2(1?x)21?2sintln(x?1?x2)1d2sint = ?2?2(1?x2)1?2sint22ln(x?1?x2)11?2sint =?ln?c 22(1?x)421?2sintln(x?1?x2)11?x2?2x?ln?c =222(1?x)421?x?2x

?xarctanx1?x2dx

1?x2dx 21?x解.

?xarctanx1?x22dx??arctanxd1?x?1?xarctanx??22 =1?xarctanx??11?x2dx?1?x2arctanx?ln(x?1?x2)?c

arctanexdx 3. ?e2xarctanex11?2x1?2xexx?2xxdx???arctanede??earctane??edx 解. ?2x2xe2221?e1?2x111?2x1e?xxxdx??earctane?dx ??earctane??2xx2x?221?e22e(1?e)1?2x11ex1?2xxx?x)dx??(earctane?e?arctanx)?c ??earctane??(x?2x22e1?e2?xln(1?x2)?3x?0七. 设f(x)??2 , 求?f(x)dx. ?xx?0?(x?2x?3)e解.

??(xln(1?x2)?3)dx??f(x)dx??

2?x(x?2x?3)edx???12?122xln(1?x)?[x?ln(1?x2)]?3x?cx?0? ??2 2x?02?x??(x?4x?1)e?c?1考虑连续性, 所以

c =－1+ c1, c1 = 1 + c

?12?122xln(1?x)?[x?ln(1?x2)]?3x?cx?0?f(x)dx??2 2x?02?x??(x?4x?1)e?1?c?x

八. 设f'(e)?asinx?bcosx, (a, b为不同时为零的常数), 求f(x). 解. 令t?e，x?lnt, f'(t)?asin(lnt)?bcos(lnt), 所以

x 17

f(x)?[asin(lnx)?bcos(lnx)]dx =

?x[(a?b)sin(lnx)?(b?a)cos(lnx)]?c 2九. 求下列不定积分: 1.

32x4?xdx ?解. 令x?2sint

323222x4?xdx?32sintcostdt??32(1?cost)costdcost ???3233214 =?cost?cos5t?c?(4?x2)2?(4?x2)2?c

35532.

53??x2?a2dx(a?0) x解. 令x?asect

x2?a2atantdx(a?0)??asecttant?a?tan2tdt?atant?at?c xasect22 =x?a?aarccosa?c x3.

?ex(1?ex)1?e2xdx

ex1?e2x解.

?ex(1?ex)1?e2xd??dx+?e2x1?e2xdx

?dex1?e2x1d(1?e2x)－?dx=arcsinex?1?e2x?c 21?e2x4.

?xxdx (a > 0)

2a?xxu4解. ?xdx 令u?x 2?du 令u?2asint 8a2?sin4tdt

2a?x2a?u2(1?cos2t)2dt?2a2?(1?2cos2t?cos22t)dt =8a?421?cos4ta422dt?3at?2asin2t?sin4t?c =2at?2asin2t?2a?24222 =3at?4asintcost?asintcost(1?2sint)?c

2222

=3at?3asintcost?2asintcost?c =3aarcsin22223xx2a?xx?3a2?2a22a2a2a2ax3a?x?x(2a?x)?c 2a2x2a?x?c

2a2a =3aarcsin2十. 求下列不定积分:

2?sinx?2?cosxdx 2?sinx1d(2?cosx)dx?2?dx??解. ?

2?cosx2?cosx2?cosx2dt2x2dt1?t 令tan?t 2??ln|2?cosx|?2?ln|2?cosx| 22?21?t3?t2?1?t21. =

4t41xarctan?ln|2?cosx|?c?arctan(tan)?ln|2?cosx|?c

23333sinxcosx?sinx?cosxdx

sinxcosx11?2sinxcosx?1dx??dx 解. ?sinx?cosx2sinx?cosx2.

1(sinx?cos)2?1111dx??(sinx?cosx)dx??dx =?2sinx?cosx22sinx?cosx)124 =(sinx?cosx)??24sin(x??)4 =

d(x??12x?(sinx?cosx)?ln|tan(?)|?c 24282十一. 求下列不定积分: 1.

x3??3x(2x?3)dx

223x?3xx?3xx?3x2(2x?3)dx??3d(x?3)??c 解. ?3ln322.

?(3x2?2x?5)(3x?1)dx

232332122解. ?(3x?2x?5)(3x?1)dx??(3x?2x?5)2d(3x?2x?5)

12 ?(3x?2x?5)2?c

53.

5?ln(x?1?x2)1?x2dx

12ln(x?x2?1)?c 2解.

?ln(x?1?x2)1?x2dx??ln(x?x2?1)dln(x?x2?1)?

?(1?xxdx2?x?1)ln(1?x?1)xdx222解.

?(1?x?x2?1)ln(1?x2?1)??dln(1?x2?1)ln(1?x2?1)?ln|ln(1?x2?1)|?c

十二. 求下列不定积分: 1.

xarctanx?(1?x2)dx

xarctanx1arctanx122?1dx?d(1?x)??arctanxd(1?x) 22?(1?x2)2??2(1?x)21arctanx111arctanx11?darctanx???dx

21?x22?1?x221?x22?(1?x2)2解.

??1arctanx11arctanx11?cos2t2?costdt????dt

21?x22?21?x2221arctanx111aextanx11?t?sin2t?c???arctanx?sintcost?c ??21?x24821?x2441aextanx11x?arctanx??c ??2221?x441?x 令x?tant?2.

?arcsinxdx 1?xx?t,1?x则x?tan2t

解. 令arcsin

?arcsinxdx??tdtan2t?ttan2t??tan2tdt?ttan2t?tant?t?c 1?xxxx?x?arcsin?c?(1?x)arcsin?x?c 1?x1?x1?x ?xarcsin 20

arcsinx1?x23. ??dx 22x1?xarcsinx1?x2解. ??dx令x?sint22x1?xt1?sin2t2?costdt?t(csct?1)dt 2?sintcost? ??tcottdt?tdt??tcott?cottdt? ??tcott?ln|sint|????12t?c 212t?c 21?x21 ??arcsinx?ln|x|?(arcsinx)2?c

x24.

arctanx?x2(1?x2)dx

arctanx?x2(1?x2)dx令x?tantt22?tan2tsec2tsectdt??t(csct?1)dt

解.

?tcsctdt?tdt??tdcott? ??tcott?ln|sint|?t?c???2??121t??tcott??cotdt?t2 22122arctanxx1?ln||?(arctanx)2?c x21?x2arctanx1x212?ln?(arctanx)?c ??2x21?x2十三. 求下列不定积分: 1.

?x34?x2dx

解.

32332x4?xdx令x?2sint8sint2cost2costdt?32sintcostdt ??? ?32(1?cost)costdtdcost???223232cos3t?cos5t?c 354122 ??(4?x)2?(4?x)2?c

352.

35??x2?a2 xx2?a2令x?asectxatant1?cos2t?asectasecttantdt?a?cos2tdt

解.

?atant?at?c?x2?a2?aarccosa?c x3.

?ex(1?ex)1?e2xdx

t(1?t)dt1?t1?sinu?dt令t?sinu?1?t2t?1?t2?cosucosudu

解.

?ex(1?ex)1?e2xdx令ex?t ?u?cosu?c?arcsinex?1?e2x?c 4.

?xxdx (a > 0)

2a?xxu4解. ?xdx 令u?x 2?du 令u?2asint 8a2?sin4tdt

2a?x2a?u2(1?cos2t)2dt?2a2?(1?2cos2t?cos22t)dt =8a?421?cos4ta422dt?3at?2asin2t?sin4t?c =2at?2asin2t?2a?24222 =3a2t?4a2sintcost?a2sintcost(1?2sin2t)?c =3at?3asintcost?2asintcost?c =3aarcsin22223xx2a?xx?3a2?2a22a2a2a2ax3a?x?x(2a?x)?c 2a2x2a?x?c

2a2a =3aarcsin2十四. 求下列不定积分: 1.

?sinx?sinxdx1?cosxdx

解.

1?cosx??sinxdxsin2x1?cosx???d(1?cosx)sin2x1?cosx??2?d1?cosx

1?cos2x 令1?cosx?u?2dudu??2?1?(u2?1)2?u2(2?u2)

??(

?11112?u?)du??ln||?c 22u22u2?u2?u22

?11?cosx?122ln|2?1?cosx2?1?cosx|?c

2?sinx?2?cosxdx 2?sinx1d(2?cosx)dx?2?dx??解. ?

2?cosx2?cosx2?cosx2dt2x2dt1?t 令tan?t 2??ln|2?cosx|?2?3?t2?ln|2?cosx| 21?t22?1?t22. =

4t41xarctan?ln|2?cosx|?c?arctan(tan)?ln|2?cosx|?c

23333sinxcosx?sinx?cosxdx

sinxcosx11?2sinxcosx?1dx??dx 解. ?sinx?cosx2sinx?cosx3.

1(sinx?cos)2?1111dx??(sinx?cosx)dx??dx =?2sinx?cosx22sinx?cosx)124 =(sinx?cosx)??24sin(x??)4 =

d(x??12x?(sinx?cosx)?ln|tan(?)|?c 2428十五. 求下列不定积分: 1.

??x1?xxxdx

2d(1?t3)4dt?????1?t3?c

331?t31?t32t2解.

1?xxdx令x?t3?4 ??1?x2?c

32.

??ex?1dx ex?1ex?1ex?1xdx?dx令e?sectx?2xe?1e?1sect?1?tanttantdt??(sect?1)dt

解.

?ln|sect?tant|?t?c?ln(e?ex2x?1)?arccos1?c ex3.

?x?1arctanx?1dx

x解. 令t?arctanx?1,tant?x?1,x?sec2t,dx?2sec2ttant

?ttantx?1arctanx?11?cos2t22dx??2secttantdt?2?ttantdt?2?tdt 22xsectcostt22dt?2tdt?2tdtant?t?2ttant?2tantdt?t ?cos2t??? ?2 ?2ttant?2ln|cost|?t2?c

?2x?1arctanx?1?ln|x|?(arctanx?1)2?c

第三章一元函数积分学(定积分)

一．若f(x)在[a，b]上连续, 证明: 对于任意选定的连续函数?(x), 均有

?baf(x)?(x)dx?0,

则f(x) ? 0.

证明: 假设f(?)? 0, a < ? < b, 不妨假设f(?) > 0. 因为f(x)在[a，b]上连续, 所以存在? > 0, 使得在[?－?, ? + ?]上f(x) > 0. 令m =

????x????minf(x). 按以下方法定义[a，b]上?(x): 在[?－?,

22? + ?]上?(x) =??(x??), 其它地方?(x) = 0. 所以

?baf(x)?(x)dx????????f(x)?(x)dx?m??22?0.

和

?baf(x)?(x)dx?0矛盾. 所以f(x) ? 0.

?二. 设?为任意实数, 证明: I???2011?2=. dxdx??01?(cotx)?1?(tanx)?4??证明: 先证:

?20f(sinx)?f(cosx)dx?=?2dx

0f(sinx)?f(cosx)4f(sinx)?f(cosx)令 t =

?2?x, 所以

?20

?f(sinx)dx?f(sinx)?f(coxs)?0?2f(cots)d(?t)

f(cots)?f(sitn) 24

? = 于是

??20f(cost)dt?f(cost)?f(sint)??20f(cosx)dx

f(cosx)?f(sinx)2?20f(sinx)f(sinx)dx??2dx?0f(sinx)?f(cosx)f(sinx)?f(cosx)f(sinx)?f(cosx)?dx??2dx?

0f(sinx)?f(cosx)2????20f(cosx)dx

f(cosx)?f(sinx)??=

?20所以

?20f(sinx)?f(cosx)dx?=?2dx.

0f(sinx)?f(cosx)4f(sinx)?f(cosx)???(cosx)?2 dx????0(cosx)??(sinx)?4?sinx?1????cosx??1所以 I??2dx??2?01?(tanx)01??同理 I??201?. dx??1?(cotx)4

三．已知f(x)在[0，1]上连续, 对任意x, y都有|f(x)－f(y)| < M|x－y|, 证明

1n?k?M ?f(x)dx??f???

0nk?1?n?2n1证明:

?10f(x)dx???knk?1k?1nn1nkf(x)dx, ?f()?nk?1nk??knk?1k?1nnkf()dx nn1nkk?n?f(x)dx?f()?|f(x)?f()?dx|??k?1??0?nk?1nn?k?1n?1 ???knk?1k?1nnnnkkf(x)?f()dx???kn?1M(x?)dxnnk?1nk

?M??

knk?1k?1nM?k??xdx???2?n?1M??22nk?1nn?四. 设In??40tannxdx, n为大于1的正整数, 证明:

?11?In?.

2(n?1)2(n?1)ntdt 证明: 令t =tanx, 则In??4tanxdx??001?t2n1 25

t111?t2?t???因为 ?> 0, (0 < t < 1). 所以 ?222?221?t1?12?1?t?(1?t)n1t11n11n?1dt?tdt 于是 ?tdt??2?00021?t2'立即得到

111. ?In??2(n?1)2n2(n?1)

五. 设f(x)在[0, 1]连续, 且单调减少, f(x) > 0, 证明: 对于满足0 < ? < ? < 1的任何 ?, ?, 有

??f(x)dx???f(x)dx

0???x证明: 令F(x)?x???0f(t)dt???f(t)dt (x ? ?), F(?)???f(t)dt?0.

?0?F'(x)??f(t)dt??f(x)?0??0[f(t)?f(x)]dt?0, (这是因为t ? ?, x ? ?, 且f(x)单减).

所以 F(?)?F(?)?0, 立即得到?

??0f(x)dx???f(x)dx

??六. 设f(x)在[a, b]上二阶可导, 且f''(x)< 0, 证明:

?ba?a?b?f(x)dx?(b?a)f??

?2?f''(?)(x?t)2?f(t)?f'(t)(x?t) 2!证明: ?x, t?[a, b], f(x)?f(t)?f'(t)(x?t)?令t?a?b , 所以f(x)?2a?b??a?b??a?b??f???f'???x??

2??2??2??bba?b??a?b??a?b??f??dx??af'???x??dx

222??????二边积分

?baf(x)dx??a =(b?a)f??a?b??.

?2?

七. 设f(x)在[0, 1]上连续, 且单调不增, 证明: 任给? ? (0, 1), 有

???0f(x)dx???f(x)dx

01证明: 方法一: 令F(x)??(或令F(x)?x??x0f(?t)dt???f(t)dt

0x0x0f(t)dt???f(t)dt)

F'(x)??f(?x)??f(x)?0, 所以F(x)单增; 又因为F(0) = 0, 所以F(1) ? F(0) = 0. 即

??f(?t)dt???f(t)dt?0, 即

0011

??0f(x)dx???f(x)dx

01八. 设f(x)在[a, b]上连续, f'(x)在[a, b]内存在而且可积, f(a) = f(b) = 0, 试证: |f(x)|?1b|f'(x)|dx, (a < x < b) ?a2证明: ?|f'(x)|?f'(x)?|f'(x)|, 所以 ?即 ??xax|f'(t)|dt?f(x)?f(a)??|f'(t)|dt,

axax?|f'(t)|dt?f(x)??|f'(t)|dt;

abbxx??|f'(t)|dt?f(b)?f(x)??|f'(t)|dt

即 ??bx|f'(t)|dt?f(x)??|f'(t)|dt

xbbaab所以 ??|f'(t)|dt?2f(x)??|f'(t)|dt

1b|f'(x)|dx, (a < x < b) ?a2即 |f(x)|?

九. 设f(x)在[0, 1]上具有二阶连续导数f''(x), 且f(0)?f(1)?0，f(x)?0, 试证:

?0f''(x)dx?4 f(x)证明: 因为（0，1）上f(x) ? 0, 可设 f(x) > 0

因为f(0) = f(1) = 0

?x0 ? (0,1)使 f(x0) =

1max (f(x))

0?x?1所以

?011f''(x)f''(x)dx （1） dx>?0f(x0)f(x)在（0，x0）上用拉格朗日定理

f'(?)?f(x0） ??(0,x0) x0在（x0, 1）上用拉格朗日定理

f'(?)??所以

f(x0) ??(x0,1) 1?x0?10f''(x)dx??f''(x)dx?????f''(x)dx??f'(?)?f'(?)

f(x0)??4f(x0)x0(1?x0)（因为(所以

a?b2)?ab） 211f''(x)dx?4 ?0f(x0)由（1）得

10f''(x)dx?4 f(x)bb11f(x)dx]?lnf(x)dx. 十. 设f(x)在[a,b]上连续, 且f(x)?0,则ln[b?a?ab?a?a解. 将lnt在点x0展开, 得 lnt?lnx0?11(t?x0)?2(t?x0)2 x02?1t(t?x0)?lnx0??1 x0x0t?f(x), 得

所以 lnt?lnx0? 令 x0?b1f(x)dx,?ab?ab1f(x)dx? lnf(x)?lnb?a?af(x)1bf(x)dxb?a?a?1

二边做定积分, 得

b1f(x)dx?(b?a)?(b?a) ?a?ab?abb1f(x)dx ?lnf(x)dx?(b?a)lnab?a?abb11f(x)dx]?lnf(x)dx 所以 ln[b?a?ab?a?a

blnf(x)dx?(b?a)ln

十一. 设f(x)在[0, 1]上有一阶连续导数, 且f(1)－f(0) = 1, 试证:

证明:

12?[f'(x)]dx?1

012?[f'(x)]dx??[f'(x)]dx?1dx?(?012121000f'(x)?1dx)2?(f(1)?f(0))2?1

2十二. 设函数f(x)在[0, 2]上连续, 且

?20f(x)dx= 0,

?xf(x)dx= a > 0. 证明: ? ? ? [0, 2], 使

0|f(?)| ? a.

解. 因为f(x)在[0, 2]上连续, 所以|f(x)|在[0, 2]上连续, 所以? ? ? [0, 2], 取?使|f(?)| = max |f(x)| (0 ? x ? 2)使|f(?)| ? |f(x)|. 所以

a?|(x?1)f(x)dx|?|x?1||f(x)|dx?|f(?)||x?1|dx?|f(?)|

000?2?2?2第四章微分中值定理与泰勒公式

一. 设函数f(x)在闭区间[0, 1]上可微, 对于[0, 1]上每一个x, 函数f(x)的值都在开区间(0, 1)内, 且f'(x)?1, 证明: 在(0, 1)内有且仅有一个x, 使f(x) = x.

证明: 由条件知0 < f(x) < 1. 令F(x) = f (x)－x, 于是F(0) > 0, F(1) < 0,

所以存在? ? (0, 1), 使F(?) = 0. 假设存在?1, ?2 ? (0, 1), 不妨假设?2 < ?1, 满足f(?1) = ?1, f(?2) = ?2. 于是 ?1－?2 = f(?1)－f(?2) = f'(?)(?1??2). (?2 < ? < ?1). 所以f'(?)?1, 矛盾.

二. 设函数f(x)在[0, 1]上连续, (0, 1)内可导, 且3?123f(x)dx?f(0). 证明: 在(0, 1)内存在一

个?, 使f'(?)?0. 证明: f(0)?3?12322f(x)dx?3f(?1)(1?)?f(?1), 其中?1满足??1?1.

33由罗尔定理, 存在?, 满足0 < ? < ?1, 且 f'(?)?0.

三．设函数f(x)在[1, 2]上有二阶导数, 且f(1) = f(2) = 0, 又F(x) =(x－1)2f(x), 证明: 在(1, 2)内至少存在一个?, 使 F''(?)?0.

证明: 由于F(1) = F(2) = 0, 所以存在?1, 1 < ?1 < 2, 满足F'(?1)?0. 所以

F'(1)?F'(?1)?0.所以存在?, 满足1 < ? < ?1, 且 F''(?)?0.

四. 设f(x)在[0, x](x > 0)上连续, 在(0, x)内可导, 且f(0) = 0, 试证: 在(0, x)内存在一个?, 使

f(x)?(1??)ln(1?x)f'(?).

证明: 令F(t) = f(t), G(t) = ln(1+t), 在[0, x]上使用柯西定理

F(x)?F(0)F'(?), ? ? (0, x) ?G(x)?G(0)G'(?)f(x)?(1??)f'(?), 即f(x)?(1??)ln(1?x)f'(?).

ln1(?x)所以

五. 设f(x)在[a, b]上可导, 且ab > 0, 试证: 存在一个? ? (a, b), 使

bn1

b?af(a)anf(b)?[nf(?)??f'(?)]?n?1

证明: 不妨假设a > 0, b > 0. 令F(x)?xnf(x). 在[a, b]上使用拉格朗日定理 bnf(b)?anf(a)?[n?n?1f(?)??nf'(?)]b(?a)

六. 设函数f(x), g(x), h(x)在[a, b]上连续, 在(a, b)内可导, 证明:存在一个? ? (a, b), 使

f(a) f(b)g(a)h(a)g(b)h(b)?0

f'(?)g'(?)h'(?)f(a)g(a)h(a)证明: 令F(x)?f(b)g(b)h(b), 则F(a) = F(b) = 0, 所以存在一个? ? (a, b), 使

f(x)g(x)h(x)f(a)g(a)h(a) F'(?)?f(b)g(b)h(b)?0

f'(?)g'(?)h'(?)

七. 设函数f(x)在[0, 1]上二阶可导, 且f(0) = f(1) = 0, 试证: 至少存在一个? ? (0, 1), 使 f''(?)?2f'(?)

1??证明: (f''(x)?2f'(x)f''(x)22?, 二边积分可得lnf'(x)(x?1)?c, 所以

1?xf'(x)1?xf'(x)(x?1)2?ec)

2令 F(x)?f'(x)(x?1). 由f(0) = f(1) = 0知存在? ? (0, 1), f'(?)?0. 所以F(?) = F(1)

= 0, 所以存在 ? ? (?, 1), F'(?)?0. 立即可得f''(?)?2f'(?)

1?? 30

八. 设f(x)在[x1, x2]上二阶可导, 且0 < x1 < x2, 证明:在(x1, x2)内至少存在一个?, 使

e11 x1e?ex2f(x1)xex2f(x2)?f(?)?f'(?)

证明: 令F(x)?e?xf(x)，G(x)?e?x, 在[x1, x2]上使用柯西定理. 在(x1, x2)内至少存在一个?, 满足

ex11F(x2)?F(x1) ?x1G(x2)?G(x1)e?ex2f(x1)ex2f(x2)?f(?)?f'(?).

九. 若x1x2 > 0, 证明: 存在一个? ? (x1, x2)或(x2, x1), 使 x1ex2?x2ex1?(1??)e?(x1?x2)

ex1，G(x)?, 在[x1, x2]上使用柯西定理. 在(x1, x2)证明: 不妨假设0 < x1 < x2. 令F(x)?xx内至少存在一个?, 满足

ex2ex1e???e??F(x2)?F(x1)x2x1?2 ??111G(x2)?G(x1)??2x2x1?立即可得 x1ex2?x2ex1?(1??)e?(x1?x2).

十. 设f(x), g(x)在[a, b]上连续, 在(a, b)内可导, 且f(a) = f(b) = 0, g(x) ? 0, 试证: 至少存在一个? ? (a, b), 使 f'(?)g(?)?g'(?)f(?)

证明: 令F(x)?f(x), 所以F(a) = F(b) = 0. 由罗尔定理至少存在一个? ? (a, b), 使 g(x) F'(?)?0,

于是 f'(?)g(?)?g'(?)f(?).

第五章常微分方程

一. 求解下列微分方程：

xx???x?y1. ?1?e?dx?ey?1?dy?0 ?????y?????x?e??y?1??dx??.

解. ?xdy1?ey令

xyx?u,x?yu.(将y看成自变量) ydxdudueu(u?1)?u?y, 所以 u?y ?udydydy1?eduueu?euu?eu y??u??dy1?eu1?eu?u?eu?1?eudyd(u?eu)dy1, , ??du????ln??lny?lnuu??u?eyu?eyy?c?cc1u?eu, y?, ??xu?euycx?eyy?y2?2xy?x2?y'?22. ?y?2xy?x2 ?y(1)??1?解. 令

x???x?yey??c. ????y?u,y?xu. xdyduduu2?2u?1?u?x?, 所以 u?x dxdxdxu2?2u?1duu2?2u?1?u3?u2?u?1x?2?u? 2dxu?2u?1u?2u?1u2?2u?1dxdu??

u3?u2?u?1x2u?dx??1?2??du??

x?u?1u?1?lnu?1u?1?lncx?cx. 由y(1)??1,得u(1)??1 , 22u?1u?1 32

所以 c = 0.

u?1yu?1?0?0?1?0, 即y??x. , 得到,

u2?1x

二. 求解下列微分方程： 1.

1?x2y'sin2y?2xsin2y?e21?x2

解. 令u?sin2y,则u'?y'sin2y. 得到

1?x2u'?2xu?e2解得 u?e21?x21?x2, u'?2x1?x2u?e21?x221?x为一阶线性方程

1?x2(c?ln|x?1?x2|). 即 sin2y?e2(c?ln|x?1?x2|).

2. (x?2xy?y2)dy?y2dx?0 解. 原方程可化为

dx2xx?1??2. dyyy即

dx?12???2???x?1, 为一阶线性方程(y为自变量, x为因变量). dy?yy??1y解得: x?y2?cy2e.

3. xy'lnxsiny?cosy(1?xcosy)?0 解. 令cosy?u, 则 u'??y'siny. 原方程化为

?u'xlnx?u(1?xu)?0

uu2?u'??, 为贝奴利方程.

xlnxlnx?u'111???. u2xlnxulnx1?u'11z?令z?, 则z'?2. 方程化为 z'?, 为一阶线性方程.

uuxlnxlnx解得 z?(x?c)1x?c?. 即 , (x?c)cosy?lnx.

lnxcosylnx

三. 求解下列微分方程： 1. edx?(xe?2y)dy?0 解. edx?xedy?2ydy?0.

yyyy

于是d(xey)?dy2?0. 所以方程解为 xey?y2?c.

?2. ?x?????x?dx??1??y2?x2?yy2?x2??1??dy?0 ??2解. xdx?dy?1y?x22dx?xyy?x1y?x222dy?0

xyy?x22设函数u(x,y)满足du(x,y)=

dx?dy.

x??(y) y所以

?u??x1y2?x2?xy2xy22, u(x,y)??1y2?x2xdx??(y)?arcsin所以

?u??y??'(y)??1?yy?x22. 于是?'(y)?0,?(y)?c

所以原方程的解为

12xx?y?arcsin?c 2y3. (x2?y2?2x)dx?2ydy?0

解. 由原方程可得 (x?y)dx?d(x?y)?0

2222d(x2?y2)得到 dx??0.

x2?y2于是原方程解为 x?lnx(?y)?c.

四. 求解下列微分方程:

22y2?x1. y'?

2y(x?1)解. 2yy'(x?1)?y?x

令y?u, 得到u'(x?1)?u?x

22u'?1xu??为一阶线性方程. 解得 x?1x?1?x?u?(x?1)??ln(x?1)?c?.

?x?1?

即 y2?c(x?1)?x?(x?1)lnx(?1) 2. xy'?y?x3y6 解. 该方程为贝奴利方程.

xy?6y'?y?5?y3.

xu'?u?x3 555u'?u??5x2. 解得 u?x5(c?x?2)

x253?55于是 y?cx?x

2令y?5?u, ?5y?6y'?u', ?

五. 设?(x)在实轴上连续, ?'(0)存在, 且具有性质?(x?y)??(x)?(y), 试求出?(x). 解. ?(0?0)??(0)?(0), ?(0)??2(0), ?(0)?0, ?(0)?1. i) ?(0)?0. 对于任何x有?(x??x)??(x)?(?x)

所以 ?(x)?lim?(x??x)??(x)lim?(?x)??(x)?(0)?0.

?x?0?x?0所以 ?(x)?0. ii) ?(0)?1

?(x??x)??(x)?x上式令?x?0, 得到

??(x)?(?x)??(x)?x??(x)(?(?x)?1)?x??(x)(?(x)??(0))?x

??'(x)??(x)?'(0) ???(0)?1解得 ?(x)?e?'(0)x.

六. 求解下列方程: 1. ??ydx?(y?x)dy?0

?y(0)?1?dxx????1解. 可得?dyy. 这是以y为自变量的一阶线性方程.

?x(1)?0?解得 x?y(c?lny).

x(1)?0, c?0. 所以得解 x??ylny.

?x(y'?1)?sin(x?y)?0?2. ??

y()?0??2?xu'?sinu?0?解. 令x?y?u. 可得?? ?u()???22dxducc??cscu?cotu. , ln?ln(cscu?cotu), xsinuxx???c??u()?, ?csc?cot?1, c?.

?222222?解为

?2x?cscx(?y)?cotx(?y).

七. 求解下列方程:

1. (1?x2)y''?(y')2?1?0 解. 令y'?p,则y''?所以 (1?x)2dp. dxdpdpdx?p2?1?0, 2?? dxp?11?x2arctanp??arctanx?c

所以

p?x?tanc?c1, p?x?c1?c1px, p(1?c1x)?c1?x

1?pxc12?1dyc1?x1于是 ????dx1?c1xc1c1(1?c1x)?1c12?1?dy????c?c(1?cx)??dx

11?1?c12?11解为 y??x?ln|1?c1x|?c2. 2c1c1?xy''?x(y')2?y'?02. ?

y(2)?2,y'(2)?1?解. 令y'?p,则y''?dp dx36

xdpdpp1dp11?xp2?p?0, ???p2, 2???1 dxdxxpdxxp令

11dp?u，则u'??2，u(2)?1 ppdx11u??1, u'?u?1 为u对于x的一阶线性方程 xx1c1解得 u?x?, u(2)?1, 得c = 0. u?x

22x于是得到 ?u'?11dx1?x, ?x, y?2lnx?c, y(2)?2,解得c?2?2ln2 p2dy2所以 y?2lnx?2?2ln2?ln()?2

x22?2y''?(y')2?y3. ? ?y(0)?2,y'(0)?1解. 令y'?p，则y''?pdp dy得到 2pdp?p2?y dydu?u?y为关于y的一阶线性方程. 且u?p2(0)?[y'(0)]2?1

|x?0dy?y令p?u, 得到

2解得 u?y?1?ce所以 1?u

|x?0?y(0)?1?ce?y(0)?2?1?ce?2, c?0.

于是 u?y?1, p??y?1

dy??dx, 2y?1??x?c1, y?1y(0)?2, 得到

c1?1, 得解 2y?1??xc1? 22y?1??x?1 2

八. 求解下列微分方程: 1. y(5)?y(4)?2y'''?2y''?y'?y?0

5432解. 特征方程 ????2??2????1?0

(??1)(?2?1)2?0

?1??1,?2,3?i,?4,5??i

于是得解 y?c1e?x?(c2?c3x)sinx?(c4?c5x)cosx

?y(4)?5y''?10y'?6y?02. ?

y(0)?1,y'(0)?0,y''(0)?6,y'''(0)??14?解. 特征方程 ??5??10??6?0, (??1)(??3)(?2?2??2)?0

42?1?1, ?2??3, ?3,4?1?i

得通解为 y?c1ex?c2e?3x?ex(c3cosx?c4sinx) 由 y(0)?1,y'(0)?0,y''(0)?6,y'''(0)??14

11, c2?, c3?1, c4?1 221x1?3x?ex(cosx?sinx) 得特解 y??e?e22得到 c1??

九. 求解下列微分方程:

1. y''?y?x?3sin2x?2cosx

2解. 特征方程 ??1?0, ???i

齐次方程通解 y?c1cosx?c2sinx

1x?x 2D?1113*3sin2x?32sin2x?sin2x??sin2x y2?2D?1D?1?4?11*2cosx y3?2D?1非齐次方程特解: y1?*考察

1111ixixixix12e?2e1?2e1?2e1 D2?1(D?i)2?1D2?2iDD2i?Dix11D111ix(?)1?2eix?xe?(cosx?isinx)(?ix) D2i4D2ii =xsinx?ixcosx

1*2cosx?xsinx 所以 y3?2D?1 =2ex?c2sinx?x?sin2x?xsinx 所以通解为 y?c1cos

?y''?y?2xex?4sinx2. ? ?y(0)?y'(0)?02解. 特征方程 ??1?0, ???i

齐次方程通解 y?c1cosx?c2sinx 非齐次方程特解 y1?*111xxx2xe?2ex?2ex 222D?1(D?1)?1D?2D?2 =2ex? y2?*?11??D?x?ex(x?1) ?22?114sinx??2xcosxeix的虚部) (计算方法同上题, 取22D?1D?1所以 y?c1cosx?c2sinx?ex(x?1)?2xcosx 由 y(0)?y'(0)?0可得c1?1,c2?2 得解 y?cosx?2sinx?ex(x?1)?2xcosx 3. y''?4y'?4y?eax

解. 特征方程??4??4?0,

2?1,2??2

y?(c1?c2x)e?2x

i) a??2

y*?ii)

*1112?2x?2x?2xe?e1?xe

(D?2)2(D?2?2)22a??2

1eaxax y?e?(D?2)2(a?2)21??2xax(c?cx)e?e122?a??2(a?2)?所以 y??

a??2?(c?cx)e?2x?1x2e?2x12??2

十. 求解下列微分方程: 1. xy''?xy'?y?2sin(lnx)

解. 令x?e,则t?lnx

tdydt得 2dydyx2y''?2?dtdtxy'?t. 解得 y?c1cot得到方程 y''?y?2sins?c2sint?tcots

所以得解 y?c1coslnx?c2sinlnx?lnxcoslnx 2. (x?1)2y''?(x?1)y'?y?6(x?1)ln(x?1) 解. 令x?1?et,则t?ln(x?1)

dydt得 2dydy(x?1)2y''?2?dtdt(x?1)y'?得到方程 y''?2y'?y?6tet. 解得 y?(c1?c2t)et?t3et 所以得解 y?(c1?c2ln(x?1))(x?1)?(x?1)ln3(x?1)

十一. 一质量为m的物体, 在粘性液体中由静止自由下落, 假如液体阻力与运动速度成正比, 试求物体运动的规律.

解. 取物体的初始位置为坐标原点, x坐标向下为正向. 并以x(t)表示在时刻t时的物体位置.物体所受的重力为mg, 阻力为kdx(k为比例系数). 由牛顿定律得到: dtk?dxd2x??m2?mg?k?x''?x'?g mdtdt. 即?????x(0)?x'(0)?0?x(0)?x'(0)?0解得 x?c1?c2ek?tm?mgt k?tkmg于是 x'??c2em?

mkkx(0)?0, 得到c1??c2

0?x'(0)??kmgc2? mkm2gm2g所以 c2?2, c1??c2??2

kkm2gm2g?mtmg所求解为 x??2?2e?t.

kkk

k第六章一元微积分的应用

一. 选择题

1. 设f(x)在(－?, +?)内可导, 且对任意x1, x2, x1 > x2时, 都有f(x1) > f(x2), 则 (a) 对任意x, f'(x)?0 (b) 对任意x, f'(?x)?0 (c) 函数f(－x)单调增加 (d) 函数－f(－x)单调增加

解. (a) 反例:f(x)?x3, 有f'(0)?0; (b) 反例:f(x)?x3; (c) 反例:f(x)?x3,f(?x)??x3 单调减少; 排除(a), (b), (c)后, (d)为答案. 具体证明如下: 令F(x) = －f(－x), x1 > x2, －x1 < －x2. 所以F(x1) =－f(－x1) > －f(－x2) = F(x2). 2. 设f(x)在[－?, +?]上连续, 当a为何值时, F(a)?值. (a) (c)

??[f(x)?acosnx]dx的值为极小

??2????f(x)cosnxdx (b)

1?????f(x)cosnxdx

??2?????1f(x)cosnxdx (d)

2???f(x)cosnxdx

解. F(a)???[f(x)?acosnx]dx

??2 ?a2????cos2nxdx?2a?f(x)cosnxdx??f2(x)dx

?????? ??a2?2a所以当a =

????f(x)cosnxdx??f2(x)dx 为a的二次式.

???1?????f(x)cosnxdx, F(a)有极小值.

3. 函数y = f(x)具有下列特征:

f(0) = 1; f'(0)?0, 当x ? 0时, f'(x)?0; f''(x)?x?0??0 , 则其图形 ?0x?0?(a) (b) (c) (d)

1 1 1 1

解. (b)为答案.

4. 设三次函数y?f(x)?ax3?bx2?cx?d, 若两个极值点及其对应的两个极值均为相反数, 则这个函数的图形是

(a) 关于y轴对称 (b) 关于原点对称 (c) 关于直线y = x轴对称 (d) 以上均错解. 假设两个极值点为x = t及 x = －t (t ? 0), 于是f(t) =－f(－t). 所以 at?bt?ct?d?at?bt?ct?d, 所以b + d = 0

3232f'(x)?3ax2?2bx?c?0的根为 x = ? t, 所以 b = 0. 于是d = 0. 所以

f(x)?ax3?cx 为奇函数, 原点对称. (b)为答案.

5. 曲线y?x(x?1)(2?x)与x轴所围图形面积可表示为 (a) ?x(x?1)(2?x)dx (b)

0?2?x(x?1)(2?x)dx??x(x?1)(2?x)dx

012012(c) ?x(x?1)(2?x)dx?0?1?x(x?1)(2?x)dx (d) ?x(x?1)(2?x)dx

12解.

0 1 2 由图知(c)为答案.

二. 填空题 1. 函数F(x)??x11???2??dt (x > 0)的单调减少区间______.

t??11?0, 所以0 < x < .

4x解. F'(x)?2?32. 曲线y?x?x与其在x?1处的切线所围成的部分被y轴分成两部分, 这两部分面积3之比是________.

解. y'?3x?1, 所以切线的斜率为k =3?切线方程: y??212?1?? 93222x?, 曲线和切线的交点为x??. (解曲线和切线的联立方程得

332742

x3?x21122??0, x?为其解, 所以可得(x?)2(x?)?0, 解得x??.)

33327333?2(x?x?0222x?)dx?3278?27? 比值为 31112233(x?x??)dx?0108327x2y2x2y23. 二椭圆2?2?1, 2?2?1( a > b > 0)之间的图形的面积______.

abba解.

二椭圆的第一象限交点的x坐标为 x?aba?baba2?b222. 所以所求面积为

? s??ab?4??0?aba?b22ba2?x2dx??0a?ab2?x2dx? b?abab??222222a?ba?b????baxxabxx2222? =?ab?4?????arcsin?a?x?arcsin?b?x???a??2a2b?2b2????00?????abarcsin=?ab?4??2???ab?2ab?arcsin?a2b2ab??arcsin22222(a?b)2a?bbba?b22a2b2?? 22?222(a?b)?a?ba?arcsin?

22?a?b?a令??arcsinaa2?b2 ?ab?2ab?????????= 4?ab? ?2?=4?abarctan??a?? b??4. x2 + y2 = a2绕x =－b(b > a > 0)旋转所成旋转体体积_______.

解.

－b a

由图知

V?2??[(a2?y2?b)2?(?a2?y2?b)2]dy

0a 43

?? =2??a024ba?ydy?8?ab?costdt?8?ab?0222222201?cos2tdt 2 =8?ab??4?2ba2?2

(5) 求心脏线? = 4(1+cos?)和直线? = 0, ? =

?围成图形绕极轴旋转所成旋转体体积_____. 2解. 极坐标图形绕极旋转所成旋转体体积公式

2? V?3所以

?????3(?)sin?d?

?2? V?32??(?)sin?d????33?2064(1?cos?)3sin?d?

?128?1 =??(1?cos?)434三. 计算题

2??032?32???16?160? 331. 在直线x－y + 1=0与抛物线y?x2?4x?5的交点上引抛物线的法线, 试求由两法线及

连接两交点的弦所围成的三角形的面积. 解. 由联立方程??x?y?1?0?y?x?4x?52解得交点坐标(x3,y3)?(1,2), (x2,y2)?(4,5)

由y'?2x?4求得二条法线的斜率分别为kx?1?11, kx?4??. 相应的法线为 24119y?2?(x?1), y?5??(x?4). 解得法线的交点为(x1,y1)?(6,).

242已知三点求面积公式为 S??所以

1x1?x32x2?x3y1?y3y2?y3

1x1?x3 S??2x2?x35y1?y31515. ???2y2?y323342. 求通过点(1, 1)的直线y = f(x)中, 使得解. 过点(1, 1)的直线为

y = kx + 1－k 所以 F(k) = =

?[x022?f(x)]2dx为最小的直线方程.

?[x022?kx?(1?k)]2dx

?[x024?2kx3?(k2?2k?2)x2?2k(1?k)x?(1?k)2]dx

?x52k4k2?2k?23??x?x?k(1?k)x2?(1?k)2x? =?43?5?0 =

2328?8k?(k2?2k?2)?4k(1?k)?2(1?k)2 53848F'(k)??8?(2k?2)?(4?8k)?4(1?k)?k??0

333 k = 2

所求直线方程为 y = 2x－1 3. 求函数f(x)?2?x20(2?t)e?tdt的最大值与最小值.

2?x2解. f'(x)?2x(2?x)e x = 0, x =??0, 解得

2?t?2??00f(0)?0, f(?2)??(2?t)edt?1?e, f(??)??(2?t)e?tdt=1

所以, 最大值f(?2)?1?e?2, 最小值f(0)?0.

4. 已知圆(x－b)2 + y2 = a2, 其中b > a > 0, 求此圆绕y轴旋转所构成的旋转体体积和表面积. 解. 体积 V?2?2? =8?ba2??b?ab?axa?(x?b)dx令x?b?asint4????2(b?asint)a2co2std t222??20cos2tdt?8?ba2??4?2?2ba2

表面积: y = f(x)绕x轴旋转所得旋转体的表面积为 S=

?ba2?f(x)1?f'2(x)dx

(x－b)2 + y2 = a2绕y轴旋转相当于(y－b)2 + x2 = a2绕x轴旋转. 该曲线应分成二枝: y?b?所以旋转体的表面积 S?2?a2?x2

aa?x?22?a?a(b?a2?x2)adx?2??(b?a2?x2)?aaaa?x22dx

=4?ab?dxa2?x2?a?8?ab?2dt?4?2ab.

05. 设有一薄板其边缘为一抛物线(如图), 铅直沉入水中,

i. 若顶点恰在水平面上, 试求薄板所受的静压力. 将薄板下沉多深压力加倍?

ii. 若将薄板倒置使弦恰在水平面上, 试求薄板所受的静压力. 将薄板下沉多深压力加倍? 解. 0 y

(20,6) x i. 由图知抛物线方程为y?3x. 于是 5x32 dp?x?2y?dx?6355dx

202 p??2006x25dx?62??x55?192 00假设将薄板沉到水中, 深为h处, 此时薄板的曲线方程为 y?3由题设知 6x?hx?h, dp?x?2y?dx?6xdx 55?h?20hxh?20x?hx?hdx?2?1920, 即?xdx?640

h55h?2053?1?2222 (x?h)?(x?h)??35?5?hh ?4, h = 12

3?640

ii. 由图知抛物线方程为y?320?x. 于是 520?xdx 5 dp?x?2y?dx?6x p?5620622x20?xdx???(20?x)?505520?031202?(20?x)25320?128 00假设将薄板沉到水中, 深为h处, 此时薄板的曲线方程为 y?3由题设知

20?h?x20?h?x, dp?x?2y?dx?6xdx 55 46

6?h?20h20?h?x6h?20xdx?2?1280, 即x(20?h?x)2dx?2560 ?55h521

62?(20?h?x)55h?20?h62?(20?h)(20?h?x)5332h?20

h －12 + 20 + h = 16, h = 18

四. 证明题

?1. 设f(x)为连续正值函数, 证明当x ? 0时函数?(x)??x0x0tf(t)dtf(t)dt单调增加.

xxx?f(x)x?f(t)dt??tf(t)dt?f(x)(x?t)f(t)dt??0?0?0??证明. ?'(x)??0 22xx?f(t)dt??f(t)dt???????0???0?上述不等式成立是因为

f(x) > 0, t < x.

2. 设f(x)在[a, b]上连续, 在(a, b)内f''(x)?0, 证明?(x)?证明. 假设a < x1 < x2 < b,

f(x)?f(a)在(a, b)内单增.

x?a?(x1)?f(x1)?f(a)?f'(?1) (a < ?1

x1?a

?(x2)?f(x2)?f(a)f(x2)?f(x1)?f(x1)?f(a) ?x2?ax2?af'(?2)(x2?x1)?f'(?1)(x1?a)

x2?af'(?1)(x2?x1?x1?a)?f'(?1)??(x1)

x2?a ? ?不等式成立是因为?1

x?a?a 在(a, b)内也F'(x)?0.

证明: 因为f'(x)?0, 所以f(x)单减.

F'(x)??1(x?a)2??axaf(t)dt?1?1f(x)=x?a(x?a)2?xaf(t)dt+

1(x?a)2?xaf(x)dt

1 =

(x?a)2x[f(x)?f(t)]dt?0

4. 证明方程tanx?1?x在(0, 1)内有唯一实根.

证明. 令F(x)?tanx?1?x. F(0) =－1 <0, F(1) = tan1 >0, 所以在(0, 1)中存在?, 使F(?) = 0.

1?1?0 (0 < x < 1), 所以F(x)单增, 所以实根唯一. 2cosxanan?1?0. 证明: 方程 5. 设a1, a2, …, an为n个实数, 并满足a1?2???(?1)32n?1又因为F'(x)? a1cosx?a2cos3x??ancos(2n?1)x?0 在(0, ?2)内至少有一实根.

证明. 令F(x)?a1sinx?a2则 F(0) = 0, F?sin3xsin(2n?1)x??an 32n?1ana2???n?1a????(?1)?0. 所以由罗尔定理存在? (0 < ? < ??132n?1?2?? ), 使F'(?)?0. 即acos??acos3???acos(2n?1)??0 12n2

五. 作图

2x2?2x?21. y? 2. y?(1?x2)e?x

x?1解. 1. 2.

第十二章函数方程与不等式证明

an?1an?a一. 证明不等式?2(n?1)lna证明: 令f(x)?ax, 在1111n?1an?2. (a > 1, n ? 1) n1?n?1,1n?上使用拉格朗日定理

f(1)?f(1)?f'(?)(1?1)nn?1nn?1 1 11?nn?1a?a?alnan(n?1) 即

a1n?alna1n?1a? ?n(n?1)111 所以

an?1an?a ?2(n?1)lna

二. 若a ? 0, b ? 0, 0 < p < 1, 证明 (a?b)p?ap?bp

1n?1an?2. (a > 1, n ? 1) n证明: 令f(x)?(x?b)p?xp?bp 显然f(0) = 0. 当x ? 0 时, 因为0 < p < 1

f'(x)?p(x?b)p?1?pxp?1?0

所以当x ? 0时, f(x)单减, 所以f(a) ? f(0) = 0. 所以 (a?b)?a?b?0 即得 (a?b)?a?b