(完整版)高中数学选修2-2金版教程1-6
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1.6
一、选择题
1.已知F ′(x )=f (x )则下列等式正确的是( )
A.⎠⎛a
b f (x )dx =F (b )-F (a ) B.⎠⎛a
b f (x )dx =F (a )-F (b ) C.⎠⎛a b f (x )dx = b -a n
F (ξi ) D.⎠⎛a b f (x )dx =
b -a n f (ξi )Δx 解析:B 中F (a )-F (b )应是F (b )-F (a );
C 中F (ξi )就是f (ξi );
D 中Δx 应去掉,因为b -a n
就是小区间长度Δx .故选A. 答案:A
2.若F (x )满足F ′(x )=sin x ,则F (x )的解析式一定是( )
A .F (x )=cos x
B .F (x )=-cos x
C .F (x )=1-cos x
D .F (x )=-cos x +c (c ∈R )
解析:因为(-cos x +c )′=-(cos x )′+c ′=sin x +0=sin x ,所以F (x )=-cos x +c (c ∈R ).故选D.
答案:D
3.设f (x )=⎩⎪⎨⎪⎧
x 2,x ∈[0,1],2-x ,x ∈[1,2],则⎠⎛02f (x )dx =( ) A.34 B.45
C.56 D .不存在
解析:⎠⎛02f (x )dx =⎠⎛01x 2dx +⎠⎛1
2(2-x )dx =x 33|10+(2x -x 22
)|21 =13+[(2×2-222)-(2-12)]=56
.故选C. 答案:C
4.(2009·福建)
(1+cos x )dx 等于( ) A .π
B .2
C .π-2
D .π+2
解析:因为原式=(x +sin x )⎪⎪ π2-π2=(π2+sin π2)-[-π2+sin(-π2
)]=π+2.故选D. 答案:D
5.⎠⎛a
b f ′(3x )dx =( ) A .f (b )-f (a ) B .f (3b )-f (3a )
C.13
[f (3b )-f (3a )] D .3[f (3b )-f (3a )] 解析:∵[13
f (3x )]′=f ′(3x ), ∴⎠⎛a
b f ′(3x )dx =13f (3x )|b a =13
[f (3b )-f (3a )]. 答案:C
6.f (x )是一次函数,且⎠⎛01f (x )dx =5,⎠⎛0
1xf (x )dx =176,那么f (x )的解析式是( ) A .4x +3
B .3x +4
C .-4x +2
D .-3x +4
解析:设f (x )=ax +b (a ≠0),
则⎠⎛01(ax +b )dx =⎠⎛01axdx +⎠⎛0
1bdx =12ax 2|10+bx |10=12
a +
b =5;
⎠⎛01x (ax +b )dx =⎠⎛01(ax 2
+bx )dx =13ax 3|10+12bx 2|10=13a +12b =176
, 所以由⎩⎨⎧ 12a +b =5,
13a +12b =176. 解得a =4,b =3,故f (x )=4x +3.
答案:A
二、填空题 7. =________.
答案:2(e 2-1)
8.已知函数f (x )=3x 2+2x +1,若f (x )dx =2f (a ),则a =________. 解析:∵f (x )dx = (3x 2+2x +1)dx
=(x 3+x 2+x )|1-1=(1+1+1)-(-1+1-1)=4,
∴2f (a )=4,即f (a )=2,
即3a 2+2a +1=2,3a 2+2a -1=0.
解得a =-1或a =13
. 答案:-1或13
9.(2008·山东高考)设函数f (x )=ax 2+c (a ≠0).若
⎠⎛01
f (x )dx =f (x 0),0≤x 0≤1,则x 0的值为________. 解析:由⎠⎛01f (x )dx =⎠⎛0
1(ax 2+c )dx
=(a 3x 3+cx )|10=a 3+c , ∵⎠⎛01f (x )dx =f (x 0), ∴a 3+c =ax 20+c ,∴a (x 20-13
)=0. ∵a ≠0,∴x 20-13=0,∴x 0=±33
. 又∵0≤x 0≤1,∴x 0=
33
. 答案:33 三、解答题
10.求下列定积分的值.
(1)⎠⎛231-x x
2dx ; (2)⎠⎛1
2(x -x 2+1x )dx . 解:(1)∵(-1x -ln x )′=1x 2-1x =1-x x
2, ∴⎠⎛2
31-x x 2dx =(-1x -ln x )|32 =(-13-ln 3)-(-12
-ln 2) =16+ln 23
. (2)⎠⎛12(x -x 2+1x )dx =⎠⎛12xdx -⎠⎛12x 2dx +⎠⎛1
21x dx =12x 2|21-13x 3|21
+ln x |21 =32-73+ln2=-56
+ln2. 11.计算下列定积分.
(1) (|2x +3|+|3-2x |)dx ;
(2) 1-sin2xdx .