高等数学A上册习题解答 至微分中值定理

华中师范大学专用

S)‰

M’u¥“‰ Æ2011c12 24F

gÿK3 ©¥ ½n A^

ÀK:

1. 3«m[0,1]þf (x)>0§Kf (0),f (1),f(1) f(0)½f(0) f(1)A ê ^S (B(A)f (1)>f (0)>f(1) f(0);(C)f(1) (0)>ff 2.¼êy=

2x(B)f (1)>f(1) f(0)>f (0);(D)f (1)>f(0) f(1)>f (0).

3(C)

(A)( ∞+∞)üNO\¶(B)( ∞,+∞)üN~ ;

(C)( 1,1)üNO,{«müN ¶(D)( 1,1)üN~ ,Ù{«müNO\.

3.¼êy=xarctanx ã/§3(B)(A)( ∞,+∞)??´à ¶

(B)( ∞,+∞)??´] ¶

(D)( ∞,0) à ,( ∞,0) ] .

f (x)x→0(C)( ∞,0) à ,( ∞,0) ] ¶

4. ¼êf(x)k ëY ꧅f (0)=0,lim(A)f(0)´f(x) 4 ¶(B)f(0)´f(x) 4 ¶

(C)(0,f(0))´- y=f(x) $:¶(D)f(0)Ø´f(x) 4 , !W K:

=1,K(B)

(0,f(0)) Ø´- y=f(x) $:.

5.éu¼êf(x)=x3§3«m[ 1,2]þ÷v. JF¥ ½n :ξ´1

1

华中师范大学专用

6. ~êk>0,¼êf(x)=lnx 7.® ¼ê

y=

1x+k3(0,+∞)S": ê´2

x3xx+1

x>0x≤0

3

(0.1)

x=

§y=e 4 ¶ x=0 §y=1 4 .

8.¼êf(x)=x4 2x2+53[ 2,2] 4.9. Ô y=4x x23Ùº:? -Ç 2.10.lim

√

√

+ 2

= 1.

x→0

n!)‰K:

11. f(x)3[0,1]þ §…0<f(x)<1§éu?Ûx∈(0,1)§Ñkf (x)=1§Áyµ3(0,1)S§k…=k êx§¦f(x)=x.

y²µ 35µ-g(x)=f(x) x,x∈[0,1],Kg(x)3[0,1]þ .Ï 0<f(x)<1§¤±g(0)×g(1)=f(0)×(f(1) 1)<0 â":½n, – k êξ∈(0,1)§¦g(ξ)=0.=f(ξ)=ξ.

5µ y{. 3ü Ø ξ1,ξ2∈(0,1)¦ f(ξ1)=ξ1Úf(ξ2)=ξ2.u´g(ξ1)=g(ξ2)=0.l dÛ ½n 3 êξ3∈(0,1)§¦g (ξ3)=0. ´db §éu?Ûx∈(0,1)§Ñkg (x)=f (x) 1=0§ù g (ξ3)=0gñ"

½öµdb §éu?Ûx∈(0,1)§Ñkg (x)=f (x) 1=0§ù`²g(x)3[0,1]üN~ " g(x)=0–õk .=–õk êx∈(0,1)§¦g(ξ)=0.=f(x)=x.12. x>0,y²µx

x2

<ln(1+x)<x.

y²µ {1(üN¼ê{)µ-x2

f(x)=ln(1+x) x+,

2g(x)=ln(1+x) x,

2

华中师范大学专用

x≥0.K

x21

f(x)= 1+x=>0,

1+x1+x

g (x)=

Ù¥x>0.

1 x

1=<01+x1+x

âüN5 O{§ f(x)3[0,+∞)þüNO\§ g(x)3[0,+∞)þüN~ " x>0 §

x2

>f(0)=0,f(x)=ln(1+x) x+2g(x)=ln(1+x) x<g(0)=0.

= x>0 §

x2x <ln(1+x)<x.

2

{2( Vúª)µr¼êln(1+x)3x=0? ‘. JF.{‘ ! VÐm§

ln(1+x)=x

1

x2,(1+θ1x)

x21

ln(1+x)=x +x3

323(1+θ2x)

Ù¥0<θ1,θ2<1.u´§ x>0 §k

ln(1+x)=x

Ú

1

x2<x,2(1+θ1x)

x21x23

ln(1+x)=x +x>x .

23(1+θ2x)32

x

x2

<ln(1+x)<x.2

= x>0 §

13. f(x),g(x)Ñ´ ¼ê§…|f (x)|<g (x),y²: x>a §|f(x) f(a)|<g(x) g(a).

3

华中师范大学专用

y²µ {1(üN¼ê{)µ-φ(x)=f(x) f(a) [g(x) g(a)], (x)=f(x) f(a)+[g(x) g(a)]

¦ §

φ (x)=f (x) g (x), (x)=f,(x)+g (x).

Ï |f (x)|<g (x), g (x)<f (x)<g (x),=

φ (x)<0, (x)>0

φ(x)´üN~ ¼ê§ (x)´üNO\¼ê"u´§ x>a §k

φ(x)<φ(a)=0 (x)> (a)=0

=

[g(x) g(a)]<f(x) f(a)<g(x) g(a)

= x>a ,

|f(x) f(a)|<g(x) g(a).

{2(…Ü¥ ½n)µÏ g (x)>|f (x)|≥0¤± x>a ,g(x)>g(a)d…Ü¥ ½n 3 ξ∈(a,x)¦

f (ξ)f(x) f(a)

=g(x) g(a)g(ξ)l §

|f(x) f(a)||f (ξ)|

= <1

g(x) g(a)g(ξ)= x>a ,

|f(x) f(a)|<g(x) g(a).

(0.3)(0.2)

4

华中师范大学专用

14. 0<a<b§¼êf(x)3[a,b]þëY§3(a,b)þ §Á|^…Ü¥

b

½ny²§ 3 :ξ∈(a,b)§¦f(b) f(a)=ξf (ξ)lny²µ-F(x)=lnx,x∈[a,b].K§F (x)=1=0,x∈[a,b].¦^…Ü¥

½n § 3 :ξ∈(a,b),¦

f(b) f(a)f (F(b) F(a)=ξ)

F

(ξ)

=

f(b) f(a)=ξf (ξ)ln

b

a

15.|^¼ê Vúª¦e 4 µ

x2

(1)xlimcosx e →∞

(2)xlim

sinx xcosx

→0

(1)):

cosx e x2

xlim

→∞x[x+ln(1 x)]x24x2x4=lim

[1

+

x+o(x4)] [1

+

+o(x4)]

x→∞

x2[x x

x+o(x2)]

x4

4

=lim

x+o(x4)

x→∞

x4

+o(x4)

=16

(2)):

lim

sinx xcosx

x→0tanx

x2=[x x3

+o(x3

)] x[1 +o(x2)]

xlim→0x3

=xlim x3

+x3+o(x3

)]→0x3=15

(0.4)

(0.5)

华中师范大学专用

µ

1

=

cosxdx1

2du1 u 1+u2

=2

du

=

1 u11

1 u+

1+udu= ln|1 u|+ln|1+u|+C

=ln 1+u 1 u

+C

1+u2=ln +2u 1 u +C

=ln 1+u2

2u 1 u2+1 u2

+C

=ln|secx+tanx|+C.

ؽȩúªµ(1)

11x

(x2+1)2dx=2arctanx+2(x2+1)

+C.

(2)

x2

(x2+1)2dx=12arctanx x2(x2+1)+C.(3) x(x+1)dx=1

12(x+1)d(x2

+1)= 12(x+1)+C.

(4)

x3

(x2+dx

= 1 1)2

12x2dx2

+1

=

x22(x+1)+

x

x+1dx= x2+1ln(x22(x+1)2+1)+C.(1) y²:

6

(0.6)

(0.7)

(0.8)(0.9)

(0.10)

华中师范大学专用

{1:(n {)x=tant

1

(x2+1)2dx

sec2=tsec4tdt

=

cos2tdt= 12[1+cos2t]dt

=1

2[t+12sin2t]=1x2arctanx+2(x2+1)+C. {2:(©ÜÈ©{)

1

(x2+1)dx

=

2 12xd1x+1

=

112xx+1

1 1

x+1d2x

= 1

12x(x2+1) 2(x2+1)x2

dx=

11

2x(x2+1) 12x2 1

x2+1

dx= 11 12x(x2+1) 2[x arctanx]+C=12arctanx+x2(x+1)+C.(2) y²:

7

(0.11)

(0.12)

华中师范大学专用

{1:(n {)x=tant

x2

(x2+1)2dx

tan2tsec2=t=

sec4tdt

sin2tdt= 12[1 cos2t]dt

=12[t 1

2

sin2t]=1

2arctanx

x2(x2+1)+C. {2:(©ÜÈ©{)

x2

=

(x2+1)2dx

x2d1x+1

= x1

12x+1 x+1d x2

=

x11

2x2+1+1

2x2+1dx=12arctanx x2(x+1)+C.8

(0.13)

(0.14)

华中师范大学专用

SK4.1ؽȩ Vg 5

¦e ؽȩ:

(1) (√+1)(√x3 √

+1)dx=

[x2

x+

√

x+1]dx=x3x23 2+25x5+x+C.(2)

3x4+3x2+1=

dx

(3x2+1

x+1)dx=x3+arctanx+C.(3)

2ex

+3 dx=2ex+3ln|x|+C.

(4)

3

1+x 2 dx=3arctanx 2arcsinx+C.

(5) bxebxdx=

(beb)xdx

=

(beb)x

b+lnb

+C.

9

(0.15)

(0.16)

(0.17)

(0.18)

(0.19)

1.

华中师范大学专用

2x+1 5x 1

(6)

dx=

10x

2(1x11x

5) 5(2)dx

=2 (1)xdx 1

15(2

)x5dx

=2(1)x

1(1)xln

5ln+C= 2ln55 x+15ln22 x+C.(7) secx(secx tanx)dx=

(secx secxtanx)dx

=tanx secx+C.(8) cos2x

cosx sinxdx

=

(sinx+cosx)dx

=sinx cosx+C.2.y=ln|x|+1.3.(1)27m,

(2)2√3(0.20)

(0.21)

(0.22)

10

华中师范大学专用

SK4.2 È©{

1.3e ª Òmà x?W\· Xꧦ ª¤á:(1)dx 1

=

d(3

5ln|x|);

(2)dx1=d(arctan3x);(3)dx

= d(1 arcsinx);(4)xdx= d(√

)2.¦e ؽȩ:

(1)

sin√dt2

sin√√d= 2cos√

+C.(2)

sinx+cosx3

=

dx13

d(sinx cosx)=3

2

(sinx cosx)2+C.(3)

1 x

=x==3

dx

1 3==sin==t=sint33cost2costdt=1 32(1 2sint)dt=12t+3

3cost+C=12arcsin2x1 3+4

9 4x2+C.(4)

dx

(x+1)(=

x+2)11

dx

=ln|x+1| ln|x+2|+C.11

(0.23)

(0.24)

(0.25)

(0.26)

华中师范大学专用

(5)=

sin2xcos3xdx

(0.27)

1

(sin5x sinx)dx211=cosx cos5x+C.2101+lnx

dx

(xlnx)

1

=d(xlnx)

(xlnx)2 1=+C.xlnx

(6)

(0.28)

(7)

1

dx

xππ,(0.29)

<t<π.K§dx=secttantdt,

)µ {1(n {)-x=sect,0<t<√

=|tant|.

1

dx

x

1

=secttantdt

sect|tant|

tant=dt

|tant|

t+C,0<t<π,=

t+C,π<t<π

1

+C,x>1,arccos=1

arccos+C,x< 1.

1

=|arccos|+C.

x

(0.30)

12

华中师范大学专用

{2.-u=

√

K,du=

xdx1

dx

x

xdx=

x2

1

=du

u+1

=arctanu+C

=arctanx2 1+C.

1

{3.-u=,K,dx=

1

du

(0.31)

1

dx

x

|u| 1

=udu

u2

1

du,0<u<1, =

1du, 1<u<0.

arccosu+C,0<u<1,=

arccosu+C, 1<u<0.

1

=|arccos|+C.

x √

(0.32)

(8)

dxx

ππ,(0.33)

<t<π.K§dx=

)µ {1(n {)-x=3sect,0<t<

13

华中师范大学专用

3secttantdt,

√

=|3tant|.

√

dx=

3|tant|3sect3secttantdt

=3

|tant|tantdt =3

tan2tdt,0<t<π,

3

tan2tdt,π<t<π.=

3(tant t)+C,0<t<π, 3(tant t)+C,π<t<π.√=√ 3arctan√

+C, 3arctan√x>3,

+C,x< 3.

= √

x2 9 3arctan

3

+C. {2.-u=√

K,du=xdx √

=

xdxx2 9=

x2xdxu2

du

=

u2+9[1 9

u2+9

]du

=u 3arctanu

3+C

= √

x2 9 3arctan3+C.

(9)

dx

1+dx=u==√

====

udu1+u

=u ln(1+u)=√

+ ln(1+√C

+C.

14

(0.34)

(0.35)

(0.36)

华中师范大学专用

(10)

1

x(x+a)

dx,

(a=0)

-u=xn+a,K,du=nxn 1dx,

dx=

du

1

x(x+adx

=

)du

nu(u a)=1 na[1u a 1u]du=1u analn|u

|+C=1

ln|xn|+C. 1

x(xn+adx

=

)11

xn 1a[x xn+a]dx

=1aln|x| 1naln|xn+a|+C.(11)

sinxcos3x

1+cos2xdx

cos3= x

dcosx

==u=cos====x

=

u31du

=

+u[u

1+u u]du

=1

[ln(1+u2) u22]+C=1

[ln(1+cos2x) cos2x]+C. 2=1

2

[ln(2 sin2x)+sin2

x]+C15

(0.37)

(0.38)

(0.39)

(0.40)

):

华中师范大学专用

(12)

1

sinxcos4dx

=

xsin2x+cos2xdx

=

sinxcos4xsinx

1 +dxsinx

sin2x+cos2=dx+xdx

= dcosx sinx1

cosx+[cosx+sinx

]dx

=

1

3cos3x+1cosx+ln|tanx2

|+C.(12)

1

sindx

==u=cos====x

==

xcosx 1(1 u2)udu

=

4

1

11 u2 1 u2 u4dx=11 u112ln1+u+u+3u

=112ln

1 cosx1+cosx+3cos3x+1cosx

+C.16

(0.41)

(0.42)

华中师范大学专用

SK4.3©ÜÈ©{

1.¦e ؽȩ:

(1) xln(x 1)dx

ln(x 1)d

x2

=2

=x22ln(x 1) x2

dln(x 1)x2

2

x2=2ln(x 1) 2(x 1)

dx=x2

1 2ln(x 1)

2[x+1+1x 1]dxx2=2ln(x 1) x24 x2 12ln(x 1)+C.(2)

(x2 1)sin2xdx

= 1

2

(x2 1)dcos2x= 12

2

(x 1)cos2x+cos2xxdx

= 12

1 2(x 1)cos2x+2

xdsin2x

= 1

2(x2 1)cos2x+12xsin2x 12

sin2xdx

= 1

2(x2 1)cos2x+112xsin2x+4cos2x+C.

(3)

coslnxdx

=xcoslnx+

sinlnxdx

=xcoslnx+xsinlnx

coslnxdx

l §

coslnxdx=

1

2

x[coslnx+sinlnx]+C.17

(0.43)

(0.44)

(0.45)

(0.46)

华中师范大学专用

(4)

(arcsinx)2dx

1

=x(arcsinx)2 x2arcsinxdx=x(arcsinx)2+ 2arcsinxd

1 x2

=x(arcsinx)2

+2 1 x2arcsinx 1 x22dx

=x(arcsinx)2

+2 1 x2arcsinx 2x+C.(5)

exsin2xdx=1 ex(1 cos2x)dx=1x1

2e 2

excos2xdxÏ §

excos2xdx=

cos2xdex

=ex

cos2x+2 exsin2xdx

=ex

cos2x+2

sin2xde

x

=excos2x+2[exsin2x

2excos2xdx]=excos2x+2exsin2x 4

excos2xdx]

¤±

excos2xdx

=1

5

[excos2x+2exsin2x]+C18

(0.47)

(0.48)

(0.49)

(0.50)

华中师范大学专用

(5)

exsin2xdx

=12ex 1

2

excos2xdx=1

11ex excos2x exsin2x+C2. f(x)=sinx

,¦xf (x)dx):

xf (x)dx=

xdf (x)

=xf (x)

f (x)dx=xf (x) f(x)+C

=cosx 2sinx

x

+C.3. f(x)=xex§¦

f (x)lnxdx. f (x)lnxdx

=

lnxdf(x)

=f(x)lnx

f(x)

xdx=f(x)lnx

exdx=xexlnx ex+C.

19

(0.51)

(0.52)

(0.53)

华中师范大学专用

SK4.4kn¼ê È©

e ؽȩ:

1.

x5+x4 8

=

xdx

xx2

+x+1+x2+x 8 x dx=x3x2

x 3+4 38

2+x+x+1+x 1+

xdxx3x2=3+2+x 4ln|x+1| 3ln|x 1|+8ln|x|+C.2.

1

(x2+1)(xdx

= 2

+x)x+12(x+1)+ 12(x+1)+1

xdx

= 12arctanx 14ln(x2+1) 1

2ln|x+1|+ln|x|+C.

3. 1

x(1+x)(1dx

= +x+x )111 x x+1 x2+x+dx

=ln|x| ln|x+1|

1

1

(x+dx)2

+=ln|x| ln|x+1| 22x+1

arctan+C

20

(0.54)

(0.55)

(0.56)

1.¦

华中师范大学专用

1

dxx+1

1

dx=

x4+2x2+1 2x2

1

dx=

(x2+1)2 2x2

1

=dx(x2++1)(x2 x+1)

√ √11 x+x = dxx2++1x2 +1

√√√ √11 + (2x+(2x = dxx2++1x2 x+1√√

√√√√22

[ln(x+x+1) ln(x x+1)]+[arctan(+1)+arctan(x 1)]+C=84√√√

√√x2++1=ln+[arctan(x+1)+arctan( 1)]+C

84x2 +1

(0.57)4.

1

dx3+sinx

2

=dx

7 cos2x

21u=tanx

=======du u1+u2

7 1

du

=

4u+3122u

=arctan+C412tanx=arctan+C.25.

(0.58)

21

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