高等数学第5章定积分

第五章 定积分

习题5.1

1.填空

n（1）lim?f??i??xi

??0i?1(2)介于x轴，函数f?x?的图像及两条直线x?a,x?b之间的各部分面积的代数和。 2．利用定积分的定义计算 （1）?xdx

01解：?f?x?在区间?0,1?上连续

?将?0,1?分成n等分，不妨设分点为xi??n,?i?1,2,3,?,n?

小区间?xi,xi?1?的长度为?xi?取?i?xi,?i?1,2,3,?,n? 则由定积分定义得

nniini1n,?i?1,2,3,?,n?

?f????xi?1???i?1??xi??i?1i11??2nnnn?i?11n?n?1? i?2?n2当???时n?? ??xdx?lim01n??0??i?xi?limi?1n1n?n?1?n2n??2?12

（2）

?10niiedx?limxn???f????xi?11?limn???i?12n?11?1?1?i?1nnnf???lim?e?e???e?en??n?nn????nn1????n?ne1??e??11?????1enen??lim??1?e?lim??1?e?lim11n??n??n??n???1?nnn?1?en?1?e?????n???

?e?13. 画出被积函数的图像，由定积分的几何意义就可以得到要证得等式成立。

4.根据定积分的几何意义、题中积分区间关于原点对称和奇函数、偶函数的性质就可以

得到。

5.由定积分的几何意义:介于x轴，函数f?x?的图像及两条直线x?a,x?b之间的各

部分面积的代数和。就可以说明两式相等

习题5.2

1.填空

（1）区间长度，函数表达式；自变量 （2）?f?x?dx

ab2.填空：

?1??fac?x?dx??bcfa?x?dx;f?2?m?b?a???baf?x?dx?M?b?a?,?a?b??3??ab

f?x?dx???b?x?dx;（4）曲边梯形各部分面积代数和等于以f???与b?a为邻边的矩形面积； （5）下列每组积分的大小关系是：

?102xdx?2?10xdx

23??21xdx?x2?1021xdx

3??010xdx??10ln(x?1)dx

??1lnxdx??1(lnx)dx

21?0edx??(x?1)dx

2sinxdx??20xdx

3、估计下列各定积分的值： （1）?(x?2)dx

132结果：6??31(x?2)dx?22

x?[1,3]

2过程：令 f(x)?x2?2,' ?f(x)?0 其中x?[1,3] ? fmin?f(1)?3,fmax?f(3)?11

由定积分的估值定理得： 3??3?1???6??3131(x?2)dx?11??3?1?

2?3(x?2)dx?22

2（2）

?3xarctanxdx

??3333结果:

19?xarctanxdxdx?23?

过程：令 f(x)?xarctanx,x?[33,3]

'?f(x)?0 其中x?[33,3]

? fmin?f(33)?318?,fmax?f(3)?33?

由定积分的估值定理得：

?19???333xarctanxdxdx?23?

5（3）

?414?(1?sin2?x)dx

5结果：???414??(1?sinx)dx?2?

15x?[?,?]

442过程：令 f(x)?1?sin? fmin?f(?)?1,2x,fmax?f(?2)?2

?1?f(x)?2

由定积分的估值定理得：

5?????(1?sin414?2x)dx?2?

（4）?e20x?x2dx

结果：?2e?2?02ex?x2dx??2e?14

过程：令 f(x)?ex2?x,x?[0,2]

由复合函数的性质得

fmin?f()?e21?14,fmax?f(2)?e

2由定积分的估值定理得：

??2e2??02ex?x2dx??2e?14

4、证明不等式： （1）?21x?1dx?2

x?1?2x?[1,2]

证明：令 f(x)?'?f(x)?0 其中x?[1,2]

?fmin?f(1)?0

?f(x)?0

??21x?1?x?1?x?1?12?2dx?0

????212?212dx

12

22（2）???2sinxxdx?sinxx

?44证明：令 f(x)?'x?[,?2]

?f(x)?0 其中x?[?4,?2]

? f(2?2)?f(x)?f(?4)

??12?f(x)??22?dx? 22????2sinxx2

4（3）?25??x1?x21dx?12

证明：令 f(x)?x1?x2x?[1,2]

?f(x)?0 其中x?[1,2]

'? f(2)?f(x)?f(1) ??2525?f(x)??12

12??221f(x)dx??

（4）??20(1?sinx)dx???2???20(1?sinx)dx??

证明：令 f(x)?(1?sinx),x?[0,'?f(x)?1?cosx 其中x?[0,?2]

?2]

?f(x)?0,x?[0,'?2]

?f(0)?f(x)?f(?1?f(x)?2?)2

??2???20(1?sinx)dx??

?5.利用积分中值公式求极限：lim令f(x)?cosx,x?[0,因为f(x)在[0,?4nn???40cosxdxn

?4]

]上为减函数，且连续

所以由积分中值定理得：???(0,??4)，使得

?40cosxdx?cos??(nn?4?0)??4?cos?

n因为0?cos??1

??limn???40cosxdx?lim(n??n?4?cos?)?0

n6.设f?x?及g?x?在?a,b?上连续，证明：

（1）若在?a,b?上，f?x??0，且?f?x??0则在?a,b?上f?x??0

ab（2）若在?a,b?上，f?x??0，且f?x?不恒为零，则?f?x??0

ab（3）若在?a,b?上，ff?x??g?x?

x,?x??g??且

?baf?x?dx??bag?x?dx，则在?a,b?上

（1）证明：反证法

假设f?x?不恒为零不妨设存在x0??a,b?,f?x0??0 ①若x0??a,b?,?f?x?连续，在

?x0??,x0???,内f?x??0.在

闭区间

??????x0?上f2,x0??x??m?0，2??fxx?dx?m??0

???ba??dx??x0?2x0??f?2②若x??a?0?a,同1，在闭区间?a,a?上，

?2??f?x??mb?aa?0??f?x?dx??a?2aaf?x?dx?ma??0

③

x0?b，

在闭区

间

??b??b,b?上f???x?2??b?0m??bf?x?dx??bab??bf?x?dx?mb??0

2综上分析与??bf?x?dx?0矛盾，所以假设不成立，即在?a,b?上f?x??0

a（2）、（3）的证明同（1），略

习题5.3

1.填空题（以下均设f(x)可导） (1)0 （2）f(x)?f(a) （3）?3xln(x2?1)

(4) 0 (5) 0 (6) f(x) (7) f(t) (8) (x?a)f?(x)?f(x)

2. 求导数

y（1）?t0edt??x0costdt?0

解：等式两边同时对x求导，得：

ey?y??cosx?0y???cosxey

dy?cosxdx?ey?yet0dt??x又因为，0costdt?0

，

?[e]0?[sint]0?0?(e?1)?sinx?0?e?1?sinx?dydx??cosxeyyytyx

?cosxsinx?1t?2dy?x??1ulnudu,?t?1? 求2 （2）设?1dx2?y?ulnudu2?t?2?dydtdy??2tlnt,5dxdt?2tlnt53 dydt?2tlnt2??????t3dxdtdx2tlntdydx22d?ddxdy1dx?dt??2t?? 32dtdx2tlnt2tlnt（3）

?cosxsinxcos(?t)dt

2利用变限求导公式

??cosx?cos??sinx??sinx?cos??cosx?22?(sinx?cosx)?cos??sinx?2

（4）利用变上限求导公式

g??x??2x1?x662?g???x??2?10x?1?x6?

?g???1???23.计算下列各定积分：

?x1?17???（1）结果： 原式? ?3x66??117321（2）结果：

?3 原式??arcsinx??42?12??3

（3）结果：1?0?1

原式??3xdx??3a2???11?x2dx???x??1??arctanx??1?1?4

30100?（4）结果：

1a

1?x?d?arctanx2aa?a???0(1?)a1x3a原式??3a0??3a

（5）结果：

10?6 1x1原式??x???d??arcsin? ?2226??0?x?1????2?也可以采用换元的方法（令x?2sint）

?（6）结果:1?

4原式???sec40?2??1?d???tan????04??1??4

（7）结果：4 原式???0sinxdx?174??2???sinx?dx???cosx?0??cosx???4

?2?（8）结果：

10

23原式??xdx??1xdx?174

（9）结果：1 原式???1?x?dx???x?1?dx?1

0112（10）结果：

1083

2原式???x?1?dx??x212dx?83

4.求下列极限

（1）原式?limcosx?2x2xex24x?0?limcosx?1

x?04（2）原式?limx?11x2ex2?e

（3）原式?lim?x0e2t2dt?limx???2?e0xt2dt?limx???2eex2x2x???e2xex2?0

?2x（4）原式?lim(1?cosx)?25231x?lim1?cosx5x2x??0x??0?limsinx10xx??0?110

x25.设I(x)??x0te?t2dt,讨论当x为何值时，I(x)有极值

2?x解：I??x??xe

令I??x??0 得

x?0

?当x?0时，I?x?取得极值

6.解：1）求??x?的表达式

当x??0,??时

??x???x0f(t)dt??x012sintdt?1?cosx2

当x????,0???0,???时

??x???x00?dt?0

0,x????,0????,???

?????x??????1?cosx2,x??0,??2）讨论??x?的连续性

在x?0点处

?x??0lim?(x)?lim0?0x??0x??0lim?(x)?lim1?cosx2?1?12x??0?0 ?0?(0)?1?cos02????0?????0????0???x?在x?0处连续

在x??点处

?x??lim??(x)?1x??lim??(x)?01?cos?2

?1?(?)???????????????x?在x??处不连续

???x?在???,?????,???上是连续的，在x??处不连续

习题5.4

1. 填空

43?1??5??9?0;0;?2??6???;In?2;?3??7??10??2;1?2e?4?;?3324n?2n?8?14?e?1;

2??13?13???ln;???4?92??2?4??122.计算下列定积分：

?（1）原式=??20?cos??21cos?dcos??????

44??024（2）换元方法

令x?tant ?dx?sectdt

?x?3,????x?1,??t?t?2?3 4??3dxx2?21?1?x??3sectdttan22?4t?sect?3???3costsin2?4tdt???31sin24tdsint

2?2331????????sint??4（3）换元的方法

令1?x?t

?dx??2tdt?x?1,??3x?,??4t?0

1t?2??134dx1?x?1??012?2tdtt?11??012?2dt?1?0122t?111dt??202dt??202t?1dt

2?2?ln?1?t??2?1?2ln2??2t?0??0（4）换元的方法

令x?sect

?dx?sect?tantdt??x?2,??x?1,?22t??3

t?0???x?1x1dx??30tantsect??sect?tantdt??30tantdt?2??sec302t?1?dt?3??3（5）换元的方法

令x?sect

?dx?sect?tantdt??x?2,??x?1,?t??3

t?02??2x?1x2?1dx??30tantsect2??30?sect?tantdt????sect?cost?tdt??30sectdt??30cosdt???lnsect?tant??

（6）原式=

?30??sint?03?ln(2??3)?32??02cosxdx?2??02cosxdx??202cosxdx????2cosxdx?22 2?（7）原式=

???2??24cos?d???4??2cos??d????1?cos2??d???2??222??22?2?2??2??1d??2?2?cos2?d???2?2??21?cos4?2d?3?2（8）换元的方法

令

x?t

?dx?2tdt?x?4,??x?1,?t?2 t?12?4lnxx1dx??lntt2112tdt?4?lntdt?4?t?lnt?1?4?t?dt?8ln2?4

11t222（9）采用分部积分的方法

?????20cosxedx?2x?20e2x?dsinx???e?2x2x?sinx??0?2?20sinx?2edx??2x02x?e?2?2edcosx?e?2??e0???2x22cosx???4?cosxedx0

?5?2cosxedx?e?202x????2cosxedx?02xe?25?（10）原式=

?0?xsinx??2dx???0xsinxdx???022??0x21?cos2x2dx????0x22?0dx?2??0x2cos2x2dx?x3?1?????6?04?????x3?12xcosxd2x?????6?04?xcos2xd2x?0???363?1?4?0xdsin2x?2?36?1?1?x2sin2x???04?4?1sin2x?2xdx

?063?0??1418??0xdcos2x???336??14?x?cos2x?0??4?cos2xdx?46??sin2x?0???46（11）采用分部积分的方法

?e1sin(lnx)dx?

?x?sin(lnx)?1??1eex?cos(lnx)dx?esin1?x1?e1cos(lnx)dxe?esin1??x?cos(lnx)?1?e?e1x?(?sin(lnx))1xdx?esin1?ecos1?1??sin(lnx)dx1

?2?sin(lnx)dx?esin1?ecos1?11e??sin(lnx)dx?1eesin1?ecos1?12（12）原式=

?11e?lnxdx??e1lnxdx???x?lnx?x?1??x?lnx?x?1?1?e1e2e?1?2?2e

（13）原式=

?x??arcsinx?2????0?1?10x?2arcsinx?211?x2dx1102??242??10arcsinxd21?x1??2???21?x?arcsinx????04?2?21?x?11?x2dx?4?0??2x?0??24?23.

解：?f(x?1)dt

02令x?1?t

?dx?dt?x?2,??x?0,t?1 t??1??20f(x?1)dx?1?1?1?1f(t)dt??1?1f(x)dx

求定积分?f(x)dx即可

?1?1f(x)dx?0?1?0?111?exdx??1011?xdx??0?111?exdx??ln(1?x)?0?1?0?111?exdx?ln2

其中?11?exdx

令ex?t

?dx?1tdtt?1t?e1?1?x?0,??x??1,0

1??dx???e?11?ttdt??11?ex1111?1??e?1??t1?t11??dt???lnt?e?1???ln?1?t???e?1?1??ln2?ln(1?e)??1??f(x?1)dt?02?1?1f(x)dx??0?111?exdx??1011?xdx?1?ln2?ln(1?e)?ln2?1?ln(1?e)?1?1

4.证明

等式右端为??a?b?x?dx a令a?b?x?t

?dx??dt?x?b,??x?a,t?a t?bb???bab?a?b?x?dx????abbaf(t)(?dt)?f(x)dx?baf(t)dt??baf(x)dx

??a?b?x?dxa5.证明

等式左端为?x01m?1?x?ndx

令1?x?t

?dx??dt?x?1,??x?0,t?0 t?1n??x01m?1?x??1?x?2dx?dx??1?1?t??100mt???dt??nm?0?1?t?1mtdt?n?0?1?x?1mxdx?n?10xn?1?x?mdx ??x01mnxn?1?x?dx

6.解：

?f(x)?tanx2?f?(x)?2tanx?secx222422?f??(x)?2secx?secx?2tanx?2secxtanx?2secx?4secxtanx

????40f?(x)f??(x)dx??40?4sec66x?tanx?8secx?tanx?dx43????4secx?tanxdx?04?408secx?tanxdx?43???44secx?tanxdtanx?04?408secx?tanxdtanx?23?

3??44(1?tanx)?tanxdtanx?022?408(1?tanx)?tanxdtanx?2?44???4?44664??4tanx?2tanx?tanx???tanx?2tanx?6??0?3?0?87.证明：

??0??f?x??f???x???sinxdx??????????0??0f?x?sinxdx??0?f???x?sinxdxffff?x?sinxdx???0sinxdf??x???0?x?sinxdx???sinxf??x???0??x?sinxdx?0??0???0f??x?cosxdx

?0cosxdf??x???0?x?sinxdx???cosxf?x???0??0f?x?sinxdx?f????f?0??2?1?3习题5.5

1.填空题

（1）p?1,p?1 （2）q?11,q?1 （3）k?1,k?1 （4）发散 （5）1

（6）过点x平行于y轴的直线左边，曲线y?f?x?和x所围成的面积 2.判别下列各广义积分的收敛性，如果收敛，则计算广义积分 （1）???0e?axdx,?a?0?

xe2?limxf?x??limxe22x???x????ax?limx???ax?0

????0e?axdx,?a?0?收敛

b?ax???0e?axdx?limdx

b???0?e?1?ax??11?ab?1dx?lim??e??lim??e?? b???b????a?0?aa?ab（2）?????2x?2x?2

?limxf?x??lim2x???x22x???x?2x?2?1

??????2dxx?2x?2收敛

ba2??????2dxx?2x?2??n?x?lima???b????dxx?2x?2?lim?arctan(x?1)?a??

a???b???b（3）?0xedx（n为自然数）

?limxf?x??limx2x???x???n?2e?x?lim?n?2?!exx????0

????0nxedx收敛

?xn?x???0xedx?lim20a????a0xedx?n!

n?x（4）???20dx?1?x?22

10dx?1?x???dx?1?x?122??2dx1?1?x?2

?limxf(x)?xx?1?1?x??????20dx?1?x?6232发散

（5）?dx?4?x?132 1?f?x???4?x?22?32?4?x?2 ??623dx收敛

?4?x?dx?623?4?x?xdxx?12??423dx?4?x?2??643dx?4?x?2???????3?4?x?3????3?4?x?3??632??2??41416（6）?21

?f?x??xx?1?2x?1,x?(1,2]

??2xdxx?1收敛

1?2xdxx?111??10t?1t28?23?2tdt??t?2t??

?3?031（7）?lnnxdx

011?lim?xfx?02?x??x?0lim?x2lnx?0

n??lnxdx收敛

01n?10lnxdx???1?n! （采用分部积分的方法）

nen（8）?edxx1??lnx?dx?21

?1x1??lnx???2?edlnx1??lnx?21???arcsin?lnx???1?arcsin1?arcsin0?e?2

（9）?0epxsinxdx,?p?0?

aa???0epxpxpxsinxdx?lim??e?cosx??lim?pecosxdx??0a???a???0?1?lim?pea?????1?plim??p?1??2??02a???pxpxsinx??plimesinxdx??0a???0a2a?a0epxsinxdx

epxsinxdx?11p?1??????0epxsinxdx?22. 利用定义判定广义积分?发散。

lnxxp1dx的敛散性，p为何值时该积分收敛，p为何值时该积分

???lnxxp1dx?limb???blnx1??lnxdx?lim??pp?1b???1?pxx?????p?1p?1?11????2p?1?x???1?p??1b???1???1?p2???p?1时发散；p?1时收敛

?0,???x?0?x?14已知f?x???x,0?x?2，试用分段函数表示?f?t?dt

???22?x??1,解： 当x?0时

x??x????f?t?dt???0?dt?0

当0?x?2时

x??f?t?dt?0??f?t?dt??x0f?t?dt?0??x0t2dt?x24

当2?x时

?x??f?t?dt??0??f?t?dt?2?20f?t?dt??x2f?t?dt??0??0?dt??20t2dt??x21?dt?t2?x?0?????t?2?x?1?4?0?0,??12f?t?dt??x,?4x?1,??ba

????0??x?0?x?2?

???x???2?x?q5.讨论广义积分?（1）当p?1时 lim??x?a?pdx?x?a??b?x?p,?p?0,q?0?的敛散性

x?af?x??lim??x?a?x?ap1?x?a??b?x?pq?1?b?a?q

所以原广义积分在x?a时是收敛的 当p?1时

x?alim??x?a?pf?x??lim??x?a?x?ap1?x?a??b?x?pq?1?b?a?q?0

所以原广义积分在x?a?时是发散的 （2）当q?1时

lim??b?x?f?x??lim??b?x?x?aqq1x?b?x?a??b?x?pq?1?b?a?p

所以原广义积分在x?b?时是收敛的 当q?1时

lim??b?x?f?x??lim??b?x?x?aqq1x?b?x?a??b?x?pq?1?b?a?p?0

所以原广义积分在x?b?时是发散的

综上所述：当且仅当p?1,q?1时，原广义积分才是收敛的，否则就是发散的。

习题5.6

1.（闵可夫斯基不等式）设f?x?、g?x?在

11b21b?a,b?上连续，证明

??f???a?b?x??g?x???2dx????f????a2?x?dx??22???g?x?dx? ????a?2且等号成立当且仅当f?x??Cg?x?或g?x??Cf?x?，（C为常数） 证明：提示（利用柯西-许瓦兹不等式）

??ba??f?x??g?x???2dx???babaf22?x?dx??abbg22?x?dx?2?a?bff1?x??g?x?dx11b2?f?x?dx??a2g?x?dx?2???ab?x?dx??22???g?x?dx?????a?2?b????f??a??1?b222????x?dx????ag?x?dx???????1b2121b????f??a?b?x??g?x???22dx????f????a?x?dx??22???g?x?dx?????a?2由柯西-许瓦兹不等式得 ：等号成立的条件当且仅当f?x??Cg?x?或g?x??Cf?x?，（C为常数）

2.设f?x?在?a,b?上有二阶导数，且f证明： f''''证明?f?x?dx??x??0，

abb?a2 ??f?a??f?b???。

?x??0得f?x?为凸函数

f?a??f?b??f?a?b?ab?a?f??x??b?x?a??af?x?dx??f?a??b?a2f?b??f?a???x?a?dx

ab??f?a??f?b???3.设f?x?、g?x?都是?a,b?上的连续函数，且f为凸函数，证明

1b?aba??1f?uxdx?f??????b?a?1b?a?ba?u?x?dx?

?b分析：设c??bau?x?dx，问题转而证明?f?u?x??dx??b?a?f?c?

a由f?x?为可微凸函数，?x?E,有f?x??f?c??f'?c??x?c?

以x?u?x?代入，并在?a,b?上求积分，便可征得结论 证明：由以上分析，则有f?u?x???f?c??f'?c??u?x??c?，

'?baf?u?x??dx??b?a?f?c??f1b?ab'?c??u?x?dx?fab?c??c??b?a???b?a??f?c?

把c??u?x?dx代入上式即得要证明的结论

ab4.设f?x?在?a,b?上单调增加的连续函数，证明?xf?x?dx?aa?b2?baf?x?dx

2证明：提示（作辅助函数（把b改为变量t）F?t??F'?tax?f?x?dx?增

a?t?taf?x?dx,

，

且

?t??0，所

ta以

a?t2F?t?单调加

F?a??0?F?t???F?b???b

a?x?f?x?dx?a?b2f?taf?x?dx?0

?x?f?x?dx?a?b2?baf?x?dx?0

?bax?f?x?dx??ba?x?dx综合练习题5

1.求下列极限：

（1）

???11??lim??22n?n?n???n?????1?2??n??1?????n???lim?n????1n?122?1n?222???1?1?1????n?2?1?2?1????n?2?????n?1?lim???n???ni?1?????1?2??i??1?????n??11?x21?i?1????n?2可以看做函数f?x??在

in处的函数值，如果用n?1个分点将区间

?0,1?分为n个长度相等的子区间，每一个子区间的长度都是

坐标为

in1n,，第i个子区间的右端点的

10?i?1,2,?,n?。f?x?在?0,1?上连续，所以

?11?x2dx存在，且

?10?n?dx=lim?xi??2??0?i?11?x?111???i?2??。其中极限与积分区间的分割方式无关，与?的取

i??in法无关，如果把积分区间?0,1?分为n个长度相等的子区间，?i取做每个子区间的右端点就

??1?n????11?x2。

得???i??n??1?????n????n?1lim??n????ni?1???到

??1?m=2??i?1??????n??n???l?1?1????n?2??2?1????n????1=

12?10dx

（2）

?11?1?111?1lim??????lim??????n???01nn?n?n???n??nn?11?1?1?nnn?n???????1011?xdx?ln2

（3）limn????nk?11sink?n??10sin?xdx?2?

（4）

11??1??2??n??nlim??1???1????1????limen???n???n??n??n?????en???（5）

lim1n1??2??n??ln??1???1????1?n??n??n????n????1n?limen???1???ln?nn??i?1?i?

lim1ni???ln?1??ni?1?n??e?0ln?1?x?dx1?e?121n?n?1???2n?1??limennn?n?1???2n?1??en??lim1nnn?n?1???2n?1??en??lim1n?1?i??ln?1??ni?o?n??e?0ln?1?x?dx1?4n??nn??2.求下列积分

1）?2?1xsgn?x?1?dx

012=??01?2xdx??x2??x2??x2?1?1xdx??0?xdx?1??2?????????

?1?2?0?2?122）?baxxdx

当b?0时 ?b??bax?xdx???b?a?ttd??t????bttdt???25t2?2?55??a???a2?b2?5??

?a5??当b?0 ??0axxdx??b0xxdx??0ax?xdx??b0xxdx??0?a?ttd??t???b0xxdx0b

???25t2??5??55a2?b2??5????2x2??2????a?505????2?55?a2?b2??,b?0??baxxdx??5???

?2?55??a2?b2?,b?0?5???3）?bxe?x?eadx

当b?e时 ?bx?edx??x?e?b?e?e??x?eb??axe?xe?a??baexdx?beb?e?aea?e??a

?beb?e?aea?e??eb?e?ea?e???b?1?eb?e??a?1?ea?e当b?e

e??eaxex?edx??bexee?xdx??e?1??a?1?ee?ba?ee?x??xe?????eb?beee?xdx??e?1??a?1??e?2e??a?1??e??xeab?x?ea?e???bee?x??e???e??eb

a?e??b?1?ee?bb?ea?e???a?1?e,b?e??b?1?e dx??a?ee?b2e?a?1?e?b?1e,b?e??????4）?sinxdx

ab当b?0时 ???ba?sinxdx??cosx?a?cosb?cosa

b当b?0 ?0asinxdx?0?b0sinxdx?b?0a?sinxdx??b0sinxdx

??cosx?a??cosx?0?2?cosa?cosb??ba?cosb?cosa,b?0sinxdx??

2?cosa?cosb,b?0??3?45）?dxxx?42

采用换元的方法 x?2sect,??x??3,???x??4,???dx?2secttantdt?2?t?arccos????3?t?2?3

???3?4dxxx?42???2?arccos????3?2?312sect???2tant?2secttantdt??12??2?arccos????3?2?3dt??3?1?2?arccos???2?3?6）?20dxasinx?bcosx2222,?a,b?0?

??2022dxasinx?bcosx22???202secxdxatanx?b1?222???202dtanxatanx?b22dabtanx2?ab?20?a?tanx???1b???

1??a??2??arctan?tanx??ab??b??0???2ab7）?20cosxdxasinx?bcosxcosxdx2222,?a,b?0?

???202asinx?bcosx222???2022022dsinxasinx?b(1?sinx)dsinx2222????ab4?b2?sin2x?b22

a?barctana?bb228）此题积分区间由错误，把区间改了之后做的

?10x31?xdx

令1?x?t，则t??0,1?

?10x31?xdx=??1?t012?3?t???2t?dt??2?t?3t?3t?tdt?2246801?32315

?9）?2sinxcosxdx

02????20sinxcosxdx?12?20?sinx?21sinxdsinx????

3?3?02310）?lnx?1?x2dx

0???10lnx??1?x2?dx??xlnx????ln(1??ln(1??1?x2?????011?10xdx1?x222)??1?x???0

2)?2?111）?x?arctanx?dx

012?10x?arctanx?dx?2?0?arctanx???212?x2d??2?2xx22102?arctanx?2????01?10x222arctanx11?x2dx??322????101arctanx1?xdxarctanxx1?x12320arctanxdx???11?x2dx????01??232??xarctanx?0?110??arctanx?2dx??2??2??12?????ln(1?x)??32432?2?0???2??216?4?12ln2

12）?xsinnxdx

0?2当n为偶数时

??0xsinnxdx????????21n?2?0?12??xdcosnx???xcosnx??0nn21??02x?cosnxdx????n2??2nn2???0xdsinnx??sinnxdnx????2n2??2n2?xsinnx?0??2n2??0sinnxdx23?02n

n2n?cosnx?03n当n为奇数时

??0xsinnxdx??

????x212?n????0xdcosnx??221?1?x2cosnx???0nn????02x?cosnxdx2n2???n2nnn2???0xdsinnx?sinnxdnx???2n2??2n2?xsinnx?0????0sinnxdx23?02n

n2n?cosnx?0343n13）?ecosnxdx

??当n为偶数时

??e??x?cosnxdx???ecosnx?????e?e?e?e????x???ne?x?xsinnxdxx??nesinnx???????????necosnxdx2x???????necosnxdx??2

??n?1??2???ecosnxdx?e?ee?e2x?????????ecosnxdx?xn?1当n为奇数时

????xx?ecosnxdx??e?cosnx????????x?nesinnxdx?x??e?e?e?e??n?1??2???x??nesinnx??????????necosnxdx2x????????necosnxdx??2

???ecosnxdx?e?ee??2x??????ecosnxdx?x?e?n?1综上所述

??e??xcosnxdx???1?10ne?e2???n?1

14）?x?dx?2?x?22 （采用换元的方法）

2sint,t?0t?dx?2costdt?x?0,???x?1,?

?4?10dx?2?2?x?2??4014cost24121212????42costdt?2424??40sectdt??3?????2????2410?40sectdtant???secttant?04?24?40tant?secttantdt24242412?28???40sect?sect?1?dt2?40sectdt?sectdt?2ln332424??40sectdt??404??lnsect?tant??024???440sectdt?33?28ln4ln?2?1??0sectdt??1414??2?1?

dx?2?x?22?2?1?15）?xlnxdx

122?21xlnxdx?2?21?x3?lnxd??lnx??3?3?1x832?22x13x31dx?x3??ln2???3?9?1?12x

83ln2?7916）?xedx（采用换元的方法0

0x?t,?x?1,??x?0,dx?2tdtt?1t?0

?10xe2xdx??10te?2tdt?14t?11012tedt5t5t??10?t4etdt?2?te??004t3t??2e?10??te?0?40?0tedt1??120?t2etdt?2e?10e?40?te??002t??240?tetdt?2e?10e?40e?120?te??00t??240?etdt?2e?10e?40e?120e?240?te??00t??2e?10e?40e?120e?240e?240?e??0111113t11

?240?88e3.利用积分中值定理计算极限 （1）limn???2n?1nxe2?x2dx

由积分中值定理得：存在点?，使得 limn???n?1nxe?x2dx?limf??n??22f???,???n,n?1????n?1??n??limn???limf????limn???e?n??2?0

?limn???n?1nxe102?xdx?0xx（2）limn???xen1?edx

由积分中值定理得，存在存在点?，使得

10limn???xenxx1?edx?limf??n??nf???,???0,1???1?0??limn??limf????limn???e??n??xx1?e?0

?limn???10xen1?edx?04.计算下列导数 （1）f?x???2xe3dt1?t?e3x

f??x??3xx3x1?x9

1?ed?12（2）sinxtdt? ?????0?dx?d?1d?11?cos2xt?2?sin?xt?dt?dt???0??0?dx?dx?2????d?1sin2x??dx?24x???4sin2x?8xcos2x16x2

5.计算下列极限 （1）采用洛比达法则 ?lim?x?0tan?sinx??cosxsin?tanx??secx22?lim?x?0tan?sinx?sin?tanx??1

（2）采用洛比达法则

?lim?x?0ln?1?sinx??2sinxcosx2(1?x)4?12?lim?x?0sinx?2sinxcosx2(1?x)4?122?14

?4x3?4x36.设f?x?为连续函数，证明?f?t??x?t?dt?0x?x0?tfudu?dt ??????0?证明：对?x0?tfudu?dt进行分部积分 ??????0?x?x0?tfudu?dt??t?tfudu????????????0???0?0?x0t?d?f?u?du??x?f?t?dt?00tx?x0t?f?t?dt

???x?t??f?t?dt0x所以结论成立

7.设f?x?在?a,b?上连续，g?x?在?a,b?上连续且不变号，证明存在???a,b?使得

?baf?x?g?x?dx?f????g?x?dx

ab证明：不妨设g?x??0

fmin?m,fmax?M

则m?g?x??f?x??g?x??M?g?x?

所以由积分中值定理得：???a,b?使得?f?x?g?x?dx?f????g?x?dx

aabb8.设函数f?x??x??f?x?dx,其中a不等于-1的常数，证明?f?x?dx?200aaa33?a?1?

证明：令?f?x?dx?b，则由题意得f?x??x?b

20a则?f?x?dx?0a?a0x?bdx?213a?ba,

3又??f?x?dx?b

0a所以得?f?x?dx?0aa33?a?1?

9.设函数f?x??0在?a,b?上连续，令F?x??'?xaf?t?dt??xb1f?t?dt，证明

（1）F?x??2；

（2）方程F?x??0在?a,b?内有且仅有一个根。 证明：（1）F'?x??f?x??1f?x??2

（2）F'?x??2?0

所以F?x?为单调函数， ?F?a??F?b???ab1f?t?dt?0

?baf?t?dt?0所以由零点定理得：方程F?x??0在?a,b?内有且仅有一个根。 10．设函数g?x?连续，f?x??解：f?x???f'1??x?t?g?t?dt,求f?x? 2'0x1?2xx?g?t?dt?2x?t?g?t?dt??002?

x212x?x0?x?t?g?t?dt,=

2?x0tg?t?dt?

??2?x??2x?tgtdt?xg?x? ?????0111.已知函数f?x?在?a,b?内可导，在?a,b?上连续，且f?0??3?2f?x?dx，证明在

3?a,b?内存在一点?，使得

1f?????0

证明：?f?0??3?2f?x?dx

3 ?f?0?是区间?,1?上的积分平均值 3???2??2??必存在一点x0??,1?，满足f?3??f?x0??f?0?

?x?是连续函数

?在?0,x0?之间，至少存在一点?满足f????f?x?,???0,x0?

?f?????0

12.设函数f?x?连续，试证明?xf?x30a2?dx?2?1a02xf?x?dx

证明：提示（换元令x2?t）就可以直接得出要证明的结论

13.设函数f?x?连续且为奇函数，证明其原函数必为偶函数，反之成立吗？如果不成立，请举出反例

证明：f?x?的一切原函数可写作F?x??经验证其原函数必为偶函数 ?x?c?14.求函数c，使得lim???x????x?c?x?x0f?t?dt?C

?c??2ttedt

解：

2c??x?c??lim??lim1????x???x???x?c??x?c??xx?c2cx?2cx?c2c?e?c??tedt?2c2t1?1?2cc????e2?2?

?e?521?1?2cc????e2?2??c?15.判断下列广义积分的收敛性，若收敛，计算其值。 （1）?发散 （2）?收敛

??02??0dxxlnx

dx?1?x??1?x??,???0?

???dx0?1?x2??1?x????1dx0?1?x2??1?x??????dx1?1?x2??1?x????1dx?2dt10?1?x2??1?x????0?t1?1?t?2??1?t?????1?110?1?x2???1?x???1?x???dx ???arctanx?1?0?4（3）???arctanx1x2dx

收敛

p???arctanxdx?p11x2dx?plim????aarctanx1x2plim?????1arctanx?1??x??lim?1p????1x?1?x2dx??1p112?plim???2?1x2?21?x2dx??1p?12?plim???2???1??x21?x2?dx21?p???22?12plimx?????ln?1?x2??1??2?12ln24）?2ex01dx

?ex?1?3收敛

2ex?32201dx?d?ex?1??2?e?1?3

x?210e?1?3?1ex?1?316. 设

f?x?在

?a,b?上单调增加，且f''?x??0，证?b??a?f???baa??f??x?df??x?a??b?f ab2证明：因为f?x?在?a,b?上单调增加，?b?a?f?a???baf?x?dx显然成立

又因为f''?x??0，所以f?x?为凸函数

?f?x??f?a??f?b??f?a?b?a?x?a???bf?b??f?a?af?x?dx?f?a??b?a?ba?x?a?dx

?b?a2??f?a??f?b???

（明

17.已知u?x?在?0,1?上连续可微，u?0??0，证明：?u2?x?dx?011?210u'2?x?dx。

?u?x???x0u?t?dt'?u2?x???0u?t?dt'?x?

'22利用柯西-许瓦兹不等式得

???u1x0ut?dt'?x2??x01dt?2??x0u'2?t?dt?x?u01?x?dx

1'22?x??2??0u?t?dt'1?12?'2x01dt?2?x0u1'2?t?dt?x?u01'20?x?dx

??u0?x?dx??0?x?u0?x?dx?dx?2?u?x?dx所以结论成立

18.已知f?x?在?0,b?上连续，?0,b?内连续可微且fb0'?x??M,f?0??0.证明：

?x?f?x?dx?b33?M

证明： f?x????x0f'?t?dt?Mxb0?b0x?f?x?dx??b?13?2Mxdx?M??x??M3?3?0b3

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