11-12第一学期高数(1)练习

计算题 ；

121xsin2x1xsin1， 计算（1） = limxx?limx?xsin?1?0=0 xlimxx?0e?1x?0e?1xx?0e?1x2?3?xx?4xx2?3?xx?4x（2）求lim[?()]＝lim?lim()

x???x???x???x?2xx?2xxxx?43444x?41?2?1(1?)(1?)()xxx＝2+limx＝lim ＝2+lim ?limxx???x???x???x????22x?221(1?)x()x(1?)2xxxe42＝2+2＝2+e

e

2. 计算 limx?0tanx?sinx 3xsinx(＝limx?01x2x2?1)2sinx(sin)2sinx(sin)sinx(1?cosx)12cosx2lim ＝＝＝＝ limlimx?0x?0x?0xxx3cosx2x3cosxx34x???cosx22

1111exex?1?xex?1)=limlimlim3. 计算 lim(?x＝＝＝＝ limxxxxxxx?0xx?0x?0x?0x?0e?1x?22e?xe?ex(e?1)(e?1)?xe

1?t?t4. 计算 1. limcosx2 ?? ?cos2x .?cos2xedtedt,???e?(cosx)?sinx?e,x?0cosx1xdxdx

12

e?tdt?cos2x1sinx?ecosx ?.lim?lim2x?02ex?0x2x

?e?tdt2d?12d?cosx2?.2.limx?0?x0(arctant)2dtx3x

?limx?00(arctant)2dtx31(arctanx)2x2lim＝lim＝ ＝

x?0x?03x233x2

5. 设y?ln1?x2?x1?x?x2，求y''

(1?x2?x)2 解： y?ln?ln?2ln(1?x2?x)

11?x2?x

1

1?x2?xxy'?2[ln(1?x2?x)]'＝2?2 y''?(x?1?x)'x2?1?x2＝22x?1＝ 22x?1x?1?x2?1xx2?1＝x2?1?2x(x2?1)32

ydy226. 设arctan?lnx?y,求

xdx

?1y1??解： 2 2方程两边对x求导得???(x?y)???222 x?y?y??x?1??? ?x?

?y?x?y?x?yy?

dyx?y??

dxx?y

(x?1)3x?17.

设y?,求y?.2x(x?4)e

解：等式两边取对数得

1lny?ln(x?1)?ln(x?1)?2ln(x?4)?x

3

上式两边对x求导得

y?112????1

yx?13(x?1)x?4 (x?1)3x?1112?y??[???1]

(x?4)2exx?13(x?1)x?4

8. 设exy?axby,求dy,

解：两边取对数得

xy?xlna?ylnb两边对x 求导，有

?

y?xy??lna?y?lnbdylna?ylna?y??dy??dxdxx?lnbx?lnb

arccosx1?1?x29.. 设y?(x>0)，求dy. ?lnxx

2

解：y?

arccosx?ln(1?1?x2)?lnx x?y'?

x1?x2?arccosxx2?x211?x??

1?1?x2xx2arccosx1?1?x211?x＝???? 222xxxx1?x ＝? dy??

10．计算1。e xsinxdx ?sinxdex ?exsinx?exd(sinx) ?exsinx?excosxdx ex(sinx?cosx)?exsinxdxexsinx?(excosx?exdcosx)??exsinx?cosxdex?

exx ?esinxdx?(sinx?cosx)?C.2

arccosx x2arccosxdx 2x????????11. 计算

1?1?3x?2dx

32解：令3x?2?t，x?t?2,dx?3tdt

3t21?1?3x?2dx＝?1?tdt

t2?1?1dt ＝3?1?t＝3(t?1??1)dt 1?tt3＝3(?t?ln1?t)?C

2＝

33(x?2)2?33x?2?3ln1?3x?2?C 2dx12．计算 7x(x?2)

3

?1

解：

令

1x?t?dx??1dt,t261t1?t??2?dt???dt?x(x7?2)dx???1?7??7t1?2t?????2?t?

??111ln|1?2t7|?C??ln|2?x7|?ln|x|?C.1414213．计算

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

解：?11111dx?dx?dx ???2x?1x?22x?3?11111dx?dx??x?22?x?3dx 2?x?1?11ln|x?1|?ln|x?2|?ln|x?3| ?C 22?ln

(x?1)?(x?3) ?C

x?23e4dx.14. 计算 ? e xlnx(1?lnx)

?3?. e46?2arcsin(lnx)

??3e4ed(lnx)??lnx(1?lnx)3e4ed(lnx)?2?lnx(1?lnx)3e4dlnx1?(lnx)2e??e?15. 计算解：

??20x2sinxdx

?20?20xsinxdx＝??2x2d(cosx)

??＝[?xsinx]?202?202xcosxdx

?＝2?20xdsinx

??2?2＝2[xsinx]0?20sinxdx

?2??2＝??2[cos]0=

16. 计算

?10xarctanxdx

4

解：

?10xarctanxdx

112＝?arctanxd(x) 201211x21dx ＝xarctanx0??2201?x2111)dx ＝??(1?8201?x2?1111＝?x0?arctanx0 822＝

??4?1 2

1x2sint)?dt,求 17. 设 f(x xf(x)dx.10t

11解：

xf(x)dx?1 f(x)d(x2)0 20

112112

?xf(x)0?xf?(x)dx 220

x2sint1sint?f(x)?dt,f(1)?dt?0,

11tt 22sinx2sinx f?(x)??2x?,2???????????10xx1?f(1)?1xf(x)dx22?10x2f?(x)dx

11???2xsinx2dx??11sinx2dx2202?0111?cosx20?(cos1?1).22??

18. 试确定 a , b 的值使

x 解：

I ?lim

?lim(b?cosx)a?x?0? (b?1)a?0x?0a = 0 或 b =1 将 a = 0 代入知不合题意 故b =1

x22

x2?1x?0(b?cosx)a?xt2dt?a?tlim0?1x?0bx?sinx?limx2?0x?0?limx?0(1?cosx)a?x?a?1,?a?45