高二数学 双基限时练12

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双基限时练(十二)

1.下列各式中,正确的是( ) A .⎠⎛a

b F ′(x )d x =F ′(b )-F ′(a )

B.⎠⎛a b F ′(x)d x =F ′(a)-F ′(b)

C .⎠⎛a

b F ′(x )d x =F (b )-F (a ) D.⎠⎛a

b F ′(x)d x =F(a)-F(b)

答案 C

2.∫π

20( sin x -cos x)d x =( ) A .0 B .1 C .2 D .π

2

解析 ∫π

20(sin x -cos x)d x =∫π20sin x d x -∫π

20cos x d x =(-cos x)

⎪⎪⎪ π20-(sin x)

⎪⎪⎪ π20 =1-1=0. 答案 A

3.若∫a 1(2x +1

x )d x =3+ln 2(a>1),则a 的值为( ) A .6 B .4 C .3 D .2 解析 ∵⎠⎛1

a (2x +1

x )d x

=(x 2

+ln x)

⎪⎪ a 1=a 2+ln a -1, 又⎠⎛

1

a (2x +

1

x )d x =3+ln 2,

∴a =2. 答案 D

4.⎠

⎛π-πcos x d x 等于( )

A .2π

B .π

C .0

D .1

解析 ⎠

⎛π-πcos x d x =sin x

⎪⎪ π-π=sinπ-sin (-π)=0. 答案 C

5.设f(x)=⎩⎪⎨⎪⎧

x 2 (0≤x<1),

2-x (1

则⎠⎛02f(x)d x 等于( )

A .34

B .4

5 C .5

6 D .不存在

解析 ⎠⎛0

2f(x)d x =⎠⎛0

1x 2d x +⎠⎛1

2(2-x)d x

=13x 3 ⎪⎪⎪ 10+(2x -12x 2)

⎪⎪⎪ 2

1 =13+2-32=5

6. 答案 C

6.由曲线y =x 2-1,直线x =0,x =2和x 轴围成的封闭图形的面积(如图阴影部分)是( )

A .⎠⎛0

2(x 2-1)d x

B .|⎠⎛0

2(x 2-1)d x |

C.⎠⎛0

2|x 2-1|d x

D .⎠⎛0

1(x 2-1)d x +⎠⎛1

2(x 2-1)d x

答案 C

7.若a =⎠⎛0

2x 2d x ,b =⎠⎛0

2x 3d x ,c =⎠⎛0

2 sin x d x ,则a ,b ,c 的大小关

系是________.

解析 ∵a =⎠⎛

2x 2d x =

13x 3

⎪⎪ 20=8

3, b =⎠⎛

2x 3d x =

14x 4

⎪⎪ 2

0=4, c =⎠⎛0

2

sin x d x =(-cos x )

⎪⎪ 2

0=-cos2+1<2. ∴b >a >c . 答案 b >a >c

8.计算⎠

⎛2-2( sin x +2)d x =________.

解析 ⎠⎛2-2(sin x +2)d x =⎠⎛2-2sin x d x +⎠

⎛2-22d x

=(-cos x )

⎪⎪ 2-2+2x

⎪⎪ 2

-2 =-cos2+cos(-2)+2×2-2×(-2) =8. 答案 8

9.设函数f (x )=ax 2+c (a ≠0),若0≤x 0≤1.且

⎠⎛0

1f (x )d x =f (x 0),则x 0=________.

解析 ∵⎠⎛

1f (x )d x =⎠⎛0

1(ax 2+c )d x =⎝

⎛⎭

⎪⎫a 3x 3+cx ⎪⎪

10=a

3+c , 又⎠⎛0

1f (x )d x =f (x 0),∴ax 20

+c =a

3+c .

∵a ≠0,∴x 2

0=13, 又0≤x 0≤1,∴x 0=33. 答案 3

3

10.计算下列定积分:

(1)⎠⎛14x -x 2x +x

d x ; (2)⎠⎛0

2(2-|1-x |)d x ;

(3)∫π2-π

2(sin x -cos x )d x .

解 (1)⎠⎛14x -x 2x +x d x =⎠⎛14(x +x )(x -x )x +x d x = ⎠⎛

1

4(

x -x )d x =⎝ ⎛⎭⎪⎫23x 32-12x 2⎪

4

1=

⎝ ⎛⎭⎪⎫23×432-12×42-⎝ ⎛⎭⎪⎫23-12=163

-8-23+12=-176.

(2)∵y =2-|1-x |=⎩

⎪⎨⎪⎧

1+x ,0≤x ≤1,3-x ,1

∴⎠⎛

2 (2-|1-x |)d x =

⎠⎛0

1 (1+x )d x +

⎠⎛

1

2 (3-x )d x =⎝

⎭⎪⎫x +

12x 2⎪

⎪⎪

10+

⎭⎪⎫3x -12x 2⎪⎪

2

1=

32+4-52=3.

(3)∫π2-π

2 (sin x -cos x )d x =(-cos x -sin x )⎪

π2-π

2=-1-1=-

2.

11.f (x )是一次函数,且⎠⎛0

1f (x )d x =5,⎠⎛0

1xf (x )d x =

17

6,求f (x )的解析

式.

解 设f (x )=ax +b (a ≠0), 由⎠⎛0

1f (x )d x =5,⎠⎛0

1xf (x )d x =17

6,

得⎠⎛

1(ax +b )d x =(

12ax 2

+bx )⎪

10=

1

2a +b ,

⎠⎛0

1x (ax +b )d x =(

13ax 3+12bx 2)

⎪⎪ 10=13a +1

2b ,

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