(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 .

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