4.C [解析] ⎠⎛1-3(3-x 2
-2x)d x =⎝ ⎛⎭
⎪⎫3x -13x 3-x 2⎪⎪⎪
1-3=32
3. 【能力提升】
5.C [解析] ⎠⎛0
1f(x)d x =⎠
⎛0
1(ax 2+1)d x =ax 3
3+x ⎪⎪⎪
10=a 3+1=2,解
得a =3.
6.D [解析] 根据定积分的相关知识可得到:由直线x =-π
3,x =π
3,y =0与曲线y =cos x 所围成的封闭图形的面积为: )
⎪⎪⎪S =∫π3-π3cos x d x =sin x π3-π3=sin π3-sin ⎝ ⎛⎭
⎪⎫-π3=3,
故选D .
7.D [解析] ⎠⎛4
8
+d t =+⎪⎪⎪
8
4=×64+×8-×16-×4=+52--26=.
8.C [解析] ⎠⎛0
k
(2x -3x 2
)d x =⎠⎛0
k
2x d x -⎠⎛0
k
3x
2
d x =x 2⎪⎪⎪⎪⎪⎪k 0-x 3k
0=k 2-
k 3=0,∴k =0或k =1.
9.D [解析] 由F(x)=kx ,得k =100,F(x)=100x ,错误!100x d x =(J ).
10.2ln 2+1 [解析] 由题设f 1
(x)=⎩⎨⎧
1,1
x ≤1,
1x ,1
x >1,
于是定积分⎠
⎛21
4
f 1(x )d x =⎠⎛1141x d x +⎠
⎛1
21d x =ln x ⎪⎪⎪ 114+x
⎪⎪⎪ 21=2ln 2+1. [解析] ⎠⎛0
1
(x -x 2
)d x =
⎪⎪⎪⎝ ⎛⎭⎪⎫23x 32-13x 310=13. 12.1 [解析] ∫π
20(sin x +a cos x)d x =(a sin x -cos x)错误!=错误!-