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(完整word版)微积分(数学分析)练习题及答案doc

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(i)???af(x,y)dx关于y在任何闭区间?c,d?上一致收敛,???cf(x,y)dy

关于x在任何区间?a,b?上一致收敛; (ii)积分

???adx???cf(x,y)dy与?dy?c????af(x,y)dx (18)

中有一个收敛,

则(18)中另一个积分也收敛,且

?

31. 解: 因为

??adx???cf(x,y)dy??dy?c????af(x,y)dx.

?ba1bxb?xaxdy?,所以I??dx?xydy. 由于函数xy在R??0,1???a,b?上

0alnxy满足定理19.6的条件,所以交换积分顺序得到

I??dy?xydx??a0b111?bdy?ln.

a1?y1?ab32. 解: 因为

ba?1?x?xlimsin?ln??0, x?0?xlnx??ba?1?x?xlimsin?ln??0 x?0??x?lnx所以该积分是正常积分. 交换积分次序, 得

ba1?1?x?x?1?I??sin?ln?dx??sin?ln?00?x?lnx?x?1??bab?1?1??xydydx????xysin?ln?dx?dy.

a?x???0?在上面的内层积分中作变换ln1y1?u,有 x??1?1??(y?1)uxsinlndx?esinudu?, ??2?0?0x1?(y?1)??于是

b?1b1?1??I????xysin?ln?dx?dy??dy?arctan(b?1)?arctan(a?1). 2a0a1?(y?1)?x???解法二: 取b为参量, 利用积分号下求导数的方法,有

11?1?I'(b)??sin?ln?xbdx? 201?(1?b)?x?积分上式,可得

16

I(b)?arctan(b?1)?c

由于I(a)?0,即有c??arctan(a?1),于是有

I?I(b)?arctan(b?1)?arctan(a?1).

bsinbx?sinax33. 解: 因为??cosxydy,所以

ax????bsinbx?sinaxI??e?pxdx??e?px?cosxydy???dx ?00a??x ?由于e?pxcosxy?e?px及反常积分反常积分

???0dx?e?pxcosxydy (21)

ab???0e?pxdx收敛,根据魏尔斯特拉斯M判别法,含参量

?在?a,b?上一致收敛.由于e?px??0e?pxcosxydx

cosxy在?0,??)??a,b?上连续,根据定理19.11交换积分(21)

的顺序,积分I的值不变.于是

I??dy?e?pxcosxydx??a0b??bapdy 22p?y ?arctanba?arctan. pp在上述证明中,令b?0,则有

F(p)??e?px0??sinaxadx?arctan(p?0), (22) xp由阿贝耳判别法可得上述含参量反常积分在p?0上一致收敛.于是由定理19.9,F(p)在

p?0上连续,且

F(0)??又由(22)式

??0sinaxdx. xa??sgna. p2I(a)?F(0)?lim?F(p)?lim?arctanp?0p?0在上式中,令a?1,则有I?34. 解: 由于e?x22?2.

??2cosrx?e?x对任一实数r成立及反常积分?e?x收敛①,所以原积分在

017

r????,???上收敛.

考察含参量反常积分

???由于?xe?x202?e?x2cosrxdx???xer0?'???x2sinrxdx, (24)

??2sinrx?xe?x对一切x?0,???r???成立及反常积分?xe?xdx收敛,

0根据魏尔斯特拉斯M判别法,含参量积分(24)在???,???上一致收敛. 综合上述结果由定理19.10即得

?'(r)???xe?xsinrxdx?lim0??2A???0?A?xe?xsinrxdx

2?1?x2?A1A?x2? ?limesinrx??recosrxdx?? 0A????202?? ??于是有

r???x2recosrxdx????r?. ?022r2ln??r????lnc,

4??r??ce从而??0??c,又由原积分,??0???r24.

???0e?xdx?2?2?r24,所以c??2,因此得到

??r??35. 解: 把含参数a的反常积分

?2e.

Ik(a)??e?kx0??sinaxdx(k?0,a?0). x中的被积函数关于a求偏导数, 可得

?当k?0时, 有

??0e?kxcosaxdx,

e?kxcosxy?e?kx,

因此,由M判别法, 号下求导,即

??dkIk(a)??e?kxcosaxdx?2.

0daa?k2???0e?kxcosaxdx关于参量a?0是一致收敛的,因此对Ik(a)可以在积分

因为Ik(0)?0,所以

18

Ik(a)??于是

a0kada?arctan. 1a12?k2kI(a)??令a?1,有I?36.解:

??0??sinaxa??kxsinaxdx?limedx?limarctan?. ??0?k?0k?0xxk2???0sinx?dx?. x22?22?|y|ds??L0|sin?|sin??cos?d??2?sin?d??4.

0?37.解: 直线段的参数方程是:

?x?t??y?2t0?t?1, ?z?3t?于是,

?(x?y)dx?(y?z)dy?(z?x)dz??[(t?2t)?2(2t?3t)?3(3t?t)]dt

L01??7tdt?0138.解:原式?7. 2?Pdx?Qdy??C?B,A??ABbAB??Pdx?Qdy y

????4dxdy????2?0?dx C?A,B? D0 ??4S?2b

A?0,0? B?b,0? x 39.解:

13??2323xy?ydx?x?3xydy ??????3?C?D2222????3x?3y?4?3x?3y???????dx???4dx?4ab?. D40.解: 由于

d(x2y?y2z?z2x)?(2xy?z2)dx?(2yz?x2)dy?(2zx?y2)dz,因此,全微分(2xy?z2)dx?(2yz?x2)dy?(2zx?y2)dz的原函数是x2y?y2z?z2x?C.

41.解:(Ⅰ).画出积分区域

y 19

y?x y?x

o x (Ⅱ).

???x?y?dxdy??D 1 0dx? x x?3132?32x?ydy??x?x?xdx?????0?22?20.

142.解:

????x?y?z?dxdydz??222V 2? 2? 0d??4d??r2?r2sin?dr

0 0 ? R?? 0d???4sin?d???rdr?2????cos?? 044 0 0? R?R51??2?2?R5. 55??43.解: y v

v?u v?u

o x o u

21221xyxy?xy?(Ⅰ). 由?????,得2x?xy?2y2???0.

aabbab?ab?ab244xy?xy?2于是??B?4AC?22?22?0,故?????是抛物线.令y?0,得

abab?ab?ab2xy?xy?x?0,x?a.故?????与x轴相交于?0,0?,?a,0?.

?ab?ab?xa?xy???u,x?u?v,??????x,y??u?ab?2?(Ⅱ).令? ,则?,故

??u,v??y?x?y?v.?y?b?u?v?.???u?2?ab(Ⅲ).

2?xa?v?2?yb?v2a2??ab.

b2?2 uababab 1ab 1ab2S???dxdy???dudv?dudv. ?du?dv?u?udu???????? 0 u2 0222212DD'D' 2? 1?x2y2?2244.解:??sin?2?2?dxdy ??d??sinr?rdr

0 0?ab?D21?cos2r221?1?????sinrdr???dr???r2?sin2r2??.

0 024?2?04 1222 1145.解:

???zdxdydz??V 2? 0d?? 3 0dr?r2 3 4?r2z?rdz

20

(完整word版)微积分(数学分析)练习题及答案doc

(i)???af(x,y)dx关于y在任何闭区间?c,d?上一致收敛,???cf(x,y)dy关于x在任何区间?a,b?上一致收敛;(ii)积分???adx???cf(x,y)dy与?dy?c????af(x,y)dx(18)中有一个收敛,则(18)中另一个积分也收敛,且
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