a2xfxdx
0
20
2令
x

1t
则dx

1t2
dt

1dxx1x2
11
1t
1

1t2



1t3
dt


11
1t
1

t
2
dt

1t
1
dt
11t2
1x
1
dx
11x2
即
1dx
x1x2

1x
dx
11x2

x
4若ft是连续函数且为奇函数，证明ftdt是偶函数；若ft是连续函数且为偶函数，0
x
证明ftdt是奇函数0
证令Fxxftdt0若ft为奇函数则ftft从而
FxxftdttuxfuduxfuduFx
0
0
0
所以Fxxftdt是偶函数0若ft为偶函数则ftft从而
FxxftdttuxfuduxfuduFx
0
0
0
所以Fxxftdt是奇函数0
5※设fx在∞，∞内连续，且Fx
x
x2tftdt试证：若fx单调不减，则Fx
0
单调不增
证
F

x


x
xftdt
0
x0
2tf
tdx

xftdtxfx2xfx
0
8
fxftdtxfxfxxfxxffx0
其中在x与0之间当x0时x由fx单调不减有ffx0即Fx0当x0时x由fx单调不减有ffx0即Fx0综上所述知Fx单调不增
1计算下列定积分：
11xexdx0
习题64
e
2xl
xdx1
4l
x
3
dx
1x

52e2xcosxdx0
7πxsi
x2dx0
4

3
4
xsi
2
dxx

2
61xlog2xdx
e
8si
l
xdx；1
9l
2x3ex2dx0
10
12
x
l

1

xdx

01x
解
1
1xexdx
0
1xdex
0
xex
10

1exdx
0
e1ex
10
e1e1e0
12e
2
exl
xdx1
1
2
e
l
1
xdx2

12
x2
l

x
e1

12
exdx1e21x2
1
24
e1

14
e2
1
3
4l
xdx2
4
l
xd
1x
1
x2
x
l

x
41
2
41
1dx8l
24x
x
41
8l
24
4
3
x
4
si
2
dxx

3
xdcotx


x
cot
x
4

3
4


3
cot
xdx
4
π4
9
π

l

si

x

3


1

44
9

π

12
l

32



52e2xcosxdx
2e2xdsi
xe2xsi
x
2
2
2e2xsi
xdx
0
0
0
0


eπ2
2e2xdcosxeπ2e2xcosx

24
2e2xcosxdx
0
0
0

eπ242e2xcosxdx0
故
2
e2x
cos
xdx

1
eπ

2

0
5
9
f6
21
xlog2
xdx

12l

2
2
l

xdx2

1
1
2l
2
x2
l

x
21

2
xdx
1
14l
2323
2l
2
2
4l
2
7
πxsi
x2dx1
0
2
πx21cos2xdx11x3
0
23
π0

12
πx2dsi
2x
0

π36

1x24
si
2x
π0
2
πxsi
2xdxπ31
0
64
π
xdcos2x
0
π31xcos2x
64
π0

π
cosxdx
0

π36

π4
1si
2x8
π0

π36

π4
e
e
8
si
l
xdxxsi
l
x
1
e1

1
cosl
xdx
esi
1xcosl
x
e1

e
si
l
xdx
1
e
esi
1ecos111si
l
xdx
故
esi
l
xdx1esi
1ecos11
1
2
9l
2x3ex2dx1l
2x2dex21x2ex2l
2l
2xex2dx
0
20
2
0
0
l
21l
2dex2l
21ex2l
2l
21
20
20
2
10
12
x
l

r