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 ,