习 题 五
习 题 五
1. 由定积分的几何意义计算下列定积分 (1) 2π 0sin d x x ?
;
(2
) R
x -?
;
(3) 0
13d x x -?;
(4) π
cos d x x ?.
1. 解:由定积分的几何意义 (1) 2π 0sin d x x ?
π
2π
sin d sin d A (A)0x x x x =+=+-=??π
(2
) R
x -
?
2 1
2
R
x R -==?
π
(3) 0
13d x x -?32
=
(4) π
cos d x x ?π
2π 0
2cos d cos d A (A)0x x x x =+=+-=??π
2. 用定积分的定义,计算由曲线21y x =+与直线1,4x x ==及x 轴所围成的曲边梯形的面积.
解:因为被积函数2()1f x x =+在[14],上是连续的,故可积,从而积分值与区间[14],的分割及点i ξ的取法无关. 为了便于计算,把区间[14],分成n 等份,每个小区间的长度都等于3
n
,分点仍记为
012114n n x x x x x -=<<<
<<=
并取(12)i i x i n ξ==,,,,得积分和
22
21
1
1
1
33
()(1)(1)11)n
n
n
n
i
i
i
i i i i i i i i f x x x x n n
====?=+?=+?=+∑∑∑∑ξξ
((
+)
23211
27186n n
i i i i n n ===++∑∑ 32
19181
(1)(21)(1)622n n n n n n n =
+++++ 9111
(1)(2)9(1)62n n n
=+++++ 令n →∞(此时各小区间的长度都趋于零,故0λ→),对上式取极限,由定积分的定义,
得
4
2
21
01
9111
(+1)d lim (1)lim[(1)(2)9(1)6]242n
i i n i x x x n n n →→+∞==+?=+++++=∑?
λξ
3. 判断下列式子是否一定正确 (1) ()d 0b
a f x x ?≥(其中()0f x ≥);
(2) ()d ()d ()b
b
a
a
f x x f x x
a b
?≥.
3. 解:
(1)不一定正确,这是因为题中未指明a 与b 的大小关系. 当a b ≤时,有 ()d 0b
a
f x x ?≥;当a b ≥时,有 ()d 0b
a
f x x ≤?.
(2)一定正确.
由定积分的性质,已知a b <,()()f x f x ≥,则 ()d ()d .b b
a
a
f x x f x x ??≥
4. 试比较下列各组积分值的大小,并说明理由 (1) 1
1
1
23 0
d d d x x x x x x ???,,;
(2) 4 4 4
2 3
3
31
ln d (ln )d d ln x x x x x x
???
,,; (3) 1
1
1
d ln(1)d
e d x x x x x x +???,,.
4. 解:
(1)当[0,1]x ∈时,有23x x x ≥≥,因此 1
1
1
23 0
d d d x x x x x x ≥≥???.
(2)当[3,4]x ∈时,有ln 1x ≥,21(ln )ln ln x x x
≥≥, 因此 4
4
4
2 3
3
3
1
(ln )d ln d d ln x x x x x x
≥≥???
5. 计算
(1) 3 0
(1cos )d lim
sin x
x t t x x
→--?;
(2) 3 0
(1cos )d lim
tan x
x t t x x
→--?.
解:(1)根据洛必达法则和积分上限函数导数的性质
3 0
(1cos )d lim
sin x
x t t x x
→--?
301cos lim
1cos x x
x
→-=- 20
lim(1cos cos )3x x x →=++=
(2)同理
3 0
(1cos )d lim
tan x
x t t x x
→--?
3201cos lim
sec 1
x x
x →-=- 24003cos sin 3cos sin 3
lim lim 2sec tan sec 2tan 2x x x x x x x x x x →→=== 6.
求21 d x y t t =?(0)x >的导函数().y x '
解: 0
2
2
1 0
()[cos d d ]x
y x t t t t ''=+??
1
220 0
[cos d d ]x t t t t '=-+??
2
2
211cos ()x x =-?-+2211cos x x x = 7. 计算下列定积分 (1) 3
22 11
()d x x x
+
?; 解: (1) 3
22 11()d x x x +
?3311()13x x =-=193
(2) 4
1d x ? 9
4)d x x =?3229211
()454326
x x =+=
(3) 4
1
?
22
,,d 42
t t
t x dx t -==
4
2
1
1
2)d 8
t t =-?
3
1[283t t =-=
(4) 1
x ?
2,1,2d t x t dx t t ==+=
2221
021t dx t =+?
?22[-arctan ]0t t =2201
2[1]1dx t =-+?42arctan 2=-
(5) 1
1d x x x -? 0
1
22 1 0
d d x x x x -=-+??=0
(6)π
2π 2
sin d x x -?
π02π0
2
sin d sin d x x x x -
=-
+?
?=2
(7)e
1 e
ln d x x ?1
e
1 1e
ln d ln d x x x x =-
+??11
111
1[ln ]ln e
e e
e
x x
dx x x dx =--+-??1
2-e =
(8) x ?
d x x =
2d x x =
2cos x =-3
244cos 233
x π
=-= (9) ln 2
2 0e (1e )d x x x +?
ln 2
23 0
(e 2e +e )d x x x x =+?23ln 21(e e +e )03x x x =+1
63
=
(10) 1
x -?
1
1
x -=?
1
1
x -=-+??
02
20
11))22x x -=
---?? 33
2021
2210112(1)|(1)| 333
x x -=---= (11) e
1
2ln d x
x x
+? e 12ln dln x x =+?2e 152ln ln 122x x =+=
(12
) ln 0
x ?
22
2,ln(1),d d 1
t
t x t x t t =+=
+
ln 0
x ?
2
l
2 02d 1t t t =+?12(arctan )0t t =-π22
=-
(13
) 0
a
x x ?
sin ,d cos d x a t x a t t ==令
0 a
x x ?
222220
=sin cos d a t a t t π?? 2
20
=[1cos 4]d 8
a t t π
-?
2220
=sin 2d 4
a t t π
?
2sin 4 =[t ]2820
a t
π
-2 =
16a π (14
)x ?
2
1
02d 1t t t
=+?
21
0112d 1t t t -+=+?
1012[1]d 1t t t
=-++?
2
12[t ln(1)]02
t t =-++2ln 2-1=
(15
) 4
?
,d 2d t x t t =
4
?
2
0d 21t t
t =+?
2 0(11)2d 1t t t
+-=+? 2 012(1)d 1t t =-
+?2=2(ln(1))0t t -+=42ln3-
(16
)3
e 1
?3
e 1
2=
?
3=2=
(17
) 1x 22
24
cos t d sin t t π
π
=?22241sin t d sin t t ππ-=?2241[1]d sin t
t π
π=-? 24
[cot ]
14
t t ππ
π=--=-
(18)π
5
2 0cos sin 2d x x x ?6
20
cos 2sin d x x x π
=?
6202
2cos dcos 7
x x π
=-=?
(19)
π
cos d x x x ?
π
dsin x x =? π
sin sin d x x
x x π=-?0
cos x
π=2=-
(20) e
1ln d x x x ? e 2 11ln d 2x x =? e 2e
1 111ln d 22x x x x =-? e
2 111e d 22x x =-?21(e 1)4
=+ (21) 1 0e d x
x x -? 1 0d(e )x x -=-? 110 0e e d x x x x --=-+? 11 0e e d x x --=-+?21e
=- (22)
220
sin d x
e x x π
?
22
d cos x
e x π
=-?222
cos cos d 20
x
x
e x x e π
π
=-+?2220
cos 2cos d 20x
x e x e x x
π
π
=-+?2220
cos 2d sin 20
x x
e x e x π
π
=-+?22220cos 2sin 4sin d 2200x x x e x e x e x x π
ππ=-+-?
22
sin d x e x x π
?
221[cos 2sin ]225
00
x x e x e x ππ=-+2155
e π=+ (23)
1
arctan d x x ?
1201arctan d 01x x x x x
=-+? 12
20111arctan d(1)021x x x x
=-++? 2111arctan ln(1)002x x x =-+1
ln 242
π=-
(24
)
3
x ?
2,1,2d t x t dx t t ==-=
3
x ?
22112ln 2d 4ln d t t t t t t ==??2124[ln d ]1t t t =-?224[ln ]11t t t =-8ln 24=- 8. 求函数 2 031
()d 1
x
t I x t t t +=-+?
在区间[01],上的最大值与最小值.
解:被积函数231
()1
t f t t t +=
-+在[01],上连续,因此 2 031()d 1x t I x t t t +=-+?可导.
231
()01
x I x x x +'=
>-+,因此 2 031()d 1x t I x t t t +=-+?在[01],上为增函数. 将0,1x =代入求得最小值为(0)0I =
,最大值为 12 031(1)d 1t I t t t +==-+?.
9. 试证
(1) 1
1
(1)d (1)d m n n m
x x x x x x -=-??
证明:令1x t =-,则
1
(1)d m n x x x -? 0 1
(1)d n m t t t =--? 1 0
(1)d n m t t t =-? 1
(1)d n m x x x =-? (2)1
1
2
2 111d d 11x x t t t t =++?? 证明:令1
t u
=
,则 1
2 1d 1x t t +? 1 12 2
11()d 11x u u u
=?-+? 112 1d 1x u u =-+?1 2 11d 1x u u =+?1 2 11d 1x u u =+?1
2 11d 1x t t =+? (3)ππ
22 0 0sin d cos d n
n x x x x =??.
证明:令2
x t π
=
-,则
π2 0
sin d n
x x ?
0π
2
cos d n
x x =-?π
2 0cos d n x x =?
10. 判断下列广义积分的收敛性,若收敛,则算出广义积分的值 (1) 4
1
d x x +∞?
解:收敛.
4 1d x x +∞
?
31
1133
x -+∞=-= (2
) 1
+∞?解:发散. (3) 2
e
d (ln )
x
x x +∞?
解:收敛.
2 e
d (ln )x x x +∞
?
2 e dln (ln )x x +∞=?e
1ln x
+∞
=-
1=
(4)发散 (4
) e
x +∞
?
解:发散 (5) 2
1
arctan d x
x x
+∞?
解:收敛.
2
1
arctan d x
x x +∞
?
11
arctan d()x x +∞
=-?
11arctan d()x x
+∞=-? 12
1111arctan 1x dx x x x +∞+∞
=-+-?+?
2 11()41x
dx x x π+∞=---+?
4π=--π1ln 242=+ (6) 2
d 22
x
x x ∞
∞
++?
+- 解:收敛.
2 d 22x
x x ∞
∞
++?
+- 2 d(1)(1)1x x ∞∞+=++?+-arctan(1)x ∞
∞
=++-π=
(7
) 0
a
?
解:收敛.
a
?
d()
a
x =?
arcsin a x a =2
π
=
(8
) 2 1
?
解:收敛. 令sec x t =
,则 2
1
? arcsec2
arcsec1
d t =?
3
π=
(9) 1
1d (2)
x
x x --?
解:发散 (10
) e
1
?
解:收敛.
e
1
?
e
1
=?
1
arcsin(ln )
e x =2
π
=
11.
用抛物线线法计算x ?
的近似值(取10n =,计算到小数点后三位)
. 解:简要步骤如下:
(1)用分点01291001i x x x x x x ==,,,,,,,把区间[01],10等分,每个小区间的