《微积分》习题6(3)

2019-01-07 11:28

11?(?cosx?x)26?(x?cosx)2?026?3?1???

?12

?x(14)

?x01?sin2xdx??x0sinx?cosxdx??40(cosx?sinx)dx??x4(sinx?cosx)dx

??(sinx?cosx)4?(?cosx?sinx)x?22x04

?(15)

??2xcosx?sinx(xsinx)2??1(x sinx)2d(xsinx)??12?42?2x sinx??4dx??244?

(16)

?3?1max{1 , x2}dx??1?1ldx??31x2dx?x1?1?x232?10313

11.设f(x)? 解:设

?t(l?t)e0x?2tdt,问x取何值时,f(x)取极大值或极小值.

?t(t?l)e-2tde?g(t)

f(x)?g(x)?g(e)

所以 因为 所以

f ' (x)?g ' (x)?x(x?1)e?2x??x(x?1)e?2x

f ' (x)在(?? ,0)f(x)在

,(?1 , ?? ) , (-1 , ?)上大于0,在(0 , 1)内小于0

(?? , 0) , (1 ,??)上单调递增,在(0 , 1)内单调递增.

所以当x?0时,f(x)取极大值,x?1时,f(x)取极小值。 12.设

?I1???2?sinx1?x2cos2xdx2?I2???2?(sinx?cosx)dx

2?I3???2(sin5x?cosx)dx?2比较I1 ,I2 ,I3的大小. 解:

?I1??2sinx??21?x2cos2xdx

?0

?I2?

?2??(sinx?cosx)dx2

?2

?I3?

?2??(sin5x?cosx)dx2

??2

????cosxdx2

???2??cosxdx?02

???2???cosxdx?02

?I3?I1?I2

13.用换元积分法计算下列各定积分 (1)

??sinx1?cos2xd0x (2)

?ln3dx 01?ex(4)

?1dx (5)

0(1?x2)3?2x2?11xdx 3)

?e2dx1?ln

1x(6)?ax2a2?x2dx0( (7)

?3dxx1?x21 (8)

?1xexe?eee??x0dx (9)?40tan(lncosx)dx

(10)(13)

??e6e223lnx?2dx (11)xdxx2?dxx(1?lnx)lnx (12)

?2?sin9xdx

0x?12 (14)

?1x?3?2x?5?1x2dx

解:(1)??x0?11?cos2xdcosx

x??arctan(cosx)

0?? 2(2)令1?ex?t

则x?ln(t2?1)

x?ln3,时t?2;x?0,时t?2

?2?221t2?1dt

?lnx?12 x?12??[ln3?2ln(2?1)]

(3)=

?e2dxx1?lnx1

2? ???0xsinxdx???xsinxdx?21?lnxe2 1?23?2

(4)?x31?322x12?03?dx(1?322x)

?x3(1?x2)1302?12dx

3(1?x2)102 ?2

2(5)??21x2?1dx x212 x1 ?x2?1?arccos1 ?3?x8?3

aa4xaarcsin 8a0(6)?(2x2?a2)a2?x2?0 ?0??16a4??16a4

(7)令x?tant,则积分区域为

3??到. 43?dx1?x2?1????34(sect)2?dttant?sect

???34?dt?dcost??3sint1?cos2t?

41d(?cost)13dcost???3???21?cost21?cost???4?4???1313ln(1?cost)?ln(1?cost)?2?24411?ln(2?2)?ln3(2?2)221?ln(2?2)?ln62

(8)令ex?t

??1exex?e?xe0dx??e11dt1tt?tt?1??edt??lnx?x2?1?

??1t2?11e?e2?11?2?ln(9)

(10)令t?lnx则x?et

?e6e3lnx?2dx?x3?613t?2dt62?(3t?2)219?14

(11)令lnx?t,则积分上下限变为与1.

12?eedxx(1?lnx)lnx??112dt(1?t)t

令t?sin2x积分上下限为:,

42?????22sinxdsinx1?sin2x?sinx???244?2dsinx1?sin2x??ln(sinx?sin2x?12?

4?ln1?22?2322??ln2?21?3(12)

??2??0sin9xdx

?????(cosx?4cos?6cos?4cos02(1?cos2x)4dcosx86420x?1)dcosx

?0(13)令x2?sec3a则dx?seca?tana

?22dxx2x2?1????34?seca?tana?dasec2?tana?

3?23???3cosada??sina??244?(14)令x?1?a


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