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人教新课标版数学高二-2-2限时练 1.6 微积分基本定理

1.6 微积分基本定理
周;使用时间17年 月 日 ;使用班级 ;姓名
一、选择题
1.若F ′(x )=x 2,则F (x )的解析式不正确的是( )
A .F (x )=13x 3
B .F (x )=x 3
C .F (x )=13x 3+1
D .F (x )=13x 3+c (c 为常数)
2.ʃ0-4|x +2|d x 等于( )
A .ʃ0-4(x +2)d x
B .ʃ0-4(-x -2)d x
C .ʃ-
2-4(x +2)d x +ʃ0-2(-x -2)d x
D .ʃ-
2-4(-x -2)d x +ʃ0-2(x +2)d x
3.若S 1=ʃ21x 2d x ,S 2=ʃ211x d x ,S 3=ʃ21e x d x ,则S 1,S 2,S 3的大小关系为(
) A .S 1<S 2<S 3 B .S 2<S 1<S 3
C .S 2<S 3<S 1
D .S 3<S 2<S 1
4.已知函数f (a )=ʃa 0sin x d x ,则f (f (π2))等于( )
A .1
B .1-cos 1
C .0
D .cos 1-1
5.已知f (x )=⎩⎪⎨⎪⎧ x 2,-1≤x ≤0,1,0<x ≤1,则ʃ1-1f (x )d x 的值为( )
A.32
B.43
C.23 D .-23
6.已知f (a )=ʃ10(2ax 2-a 2x )d x ,则函数f (a )的最大值为( )
A.19
B.29 C .-19 D .-29
7.若f (x )=x 2+2ʃ10f (x )d x ,则ʃ10f (x )d x 等于( )
A .-1
B .-13
C.13 D .1
二、填空题
8.ʃa -a (x cos x -5sin x +2)d x =________.
9.f (x )=sin x +cos x ,则⎠⎜
⎜⎛-π2π
2f (x )d x =________. 10.设f (x )=⎩⎪⎨⎪⎧
lg x ,x >0,x +f a 03t 2d t ,x ≤0,若f [f (1)]=1,则a =____________. 三、解答题
11.已知二次函数y =f (x )的图象如图所示,则它与x 轴所围图形的面积为多少?
12.已知f (x )=ʃx -a (12t +4a )d t ,F (a )=ʃ10[f (x )+3a 2]d x ,求函数F (a )的最小值.
一、选择题
1.答案 B
2.答案 D
解析 ∵|x +2|=⎩
⎪⎨⎪⎧
x +2,-2≤x ≤0,-x -2,-4≤x <-2, ∴ʃ0-4|x +2|d x =ʃ-2-4(-x -2)d x +ʃ0-2(x +2)d x . 故选D.
3.答案 B
解析 S 1=ʃ21x 2d x =
⎪⎪13x 321=13×23-13=73
, S 2=ʃ211x
d x =ln x |21=ln 2, S 3=ʃ21
e x d x =e x |21=e 2-e =e(e -1). ln 2<ln e =1,且73
<2.5<e(e -1), 所以ln 2<73
<e(e -1),即S 2<S 1<S 3. 4.答案 B
解析 f (π2)=⎠⎜⎛0
π2sin x d x =-cos x 20π=1,
f (f (π2
))=f (1)=ʃ10sin x d x =-cos x |10=1-cos 1. 5.答案 B
解析 ʃ1-1f (x )d x =ʃ0-1x 2d x +ʃ101d x = ⎪⎪x 330-1+1=13+1=4
3
,故选B.
6.答案 B
解析 f (a )=ʃ10(2ax 2-a 2x )d x =(23ax 3-12a 2x 2)|10=-12a 2+23
a , 由二次函数的性质可得f (a )max =-(23)24×(-12)=29. 7.答案 B
解析 ∵f (x )=x 2+2ʃ10f (x )d x ,
∴ʃ10f (x )d x =(13
x 3+2x ʃ10f (x )d x )|10 =13
+2ʃ10f (x )d x , ∴ʃ10f (x )d x =-13
. 二、填空题
8.答案 4a
9.答案 2
解析 ʃπ2-π2f (x )d x =⎠⎜⎜⎛-π2π2(sin x +cos x )d x =(-cos x +sin x ) 2

π
-
=(-cos π2+sin π2)-[-cos(-π2)+sin(-π2
)] =1+1=2.
10.答案 1
解析 因为x =1>0,所以f (1)=lg 1=0.
又x ≤0时,f (x )=x +ʃa 03t 2d t =x +t 3|a 0=x +a 3,
所以f (0)=a 3.
因为f [f (1)]=1,所以a 3=1,
解得a =1.
三、解答题
11.解 由图可得f (x )=1-x 2与x 轴所围图形的面积为ʃ1-1(1-x 2)d x =
⎪⎪⎝⎛⎭⎫x -13x 31-1
=(1-13)-[-1-13(-1)3]=43
. 12.已知f (x )=ʃx -a (12t +4a )d t ,F (a )=ʃ10[f (x )+3a 2]d x ,求函数F (a )的最小值.
解因为f(x)=ʃx-a(12t+4a)d t=(6t2+4at)|x-a =6x2+4ax-(6a2-4a2)=6x2+4ax-2a2,F(a)=ʃ10[f(x)+3a2]d x=ʃ10(6x2+4ax+a2)d x =(2x3+2ax2+a2x)|10=2+2a+a2
=a2+2a+2=(a+1)2+1≥1.
所以当a=-1时,F(a)的最小值为1.。

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