1、课时分层作业(十)微积分基本定理(建议用时：40分钟)一、选择题1.(ex2x)dx等于()A1Be1CeDe1C(ex2x)dxe11e，故选C.2已知积分(kx1)dxk，则实数k()A2B2C1D1A(kx1)dxk1k，k2.3设f (x)则f (x)dx()ABCDDf (x)dxx2dx(2x)dxx3.4若函数f (x)xmnx的导函数是f (x)2x1，则f (x)dx()ABCDAf (x)xmnx的导函数是f (x)2x1，f (x)x2x，f (x)dx(x2x)dx.5设axdx，bx2dx，cx3dx，则a，b，c的大小关系是()AabcBcabCacbDcbaAax
2、dx，bx2dx，cx3dx，abc.二、填空题6 d_. d cosdsin .7. (2|x|)dx_.因为f (x)2|x|所以 f (x)dx (2x)dx(2x)dx2.8已知x(0,1，f (x)(12x2t)dt，则f (x)的值域是_0,2)f (x)(12x2t)dt(t2xtt2)2x2(x(0,1)f (x)的值域为0,2)三、解答题9计算定积分：(|2x3|32x|)dx.解设f (x)|2x3|32x|，x3,3，则f (x)所以 (|2x3|32x|)dx (4x)dx6 dx4x dx2x26x2x226245.10设函数f (x)ax2c(a0)，若f (x)d
3、xf (x0)，0x01，求x0的值解因为f (x)ax2c(a0)，且ax2c，所以f (x)dx(ax2c)dxcaxc，解得x0或x0(舍去)即x0的值为.1若y(sin tcos tsin t)dt，则y的最大值是()A1B2C1D0By(sin tcos tsin t)dtsin tdtdtcos tcos 2tcos x1(cos 2x1)cos 2xcos xcos2xcos x(cos x1)222.2若f (x)x22f (x)dx，则f (x)dx等于()A1BCD1Bf (x)dx是常数，所以可设f (x)x2c(c为常数)，所以c2f (x)dx2(x2c)dx2，解得
4、c，f (x)dx(x2c)dxdx.3设抛物线C：yx2与直线l：y1围成的封闭图形为P，则图形P的面积S等于_ .由得x1.如图，由对称性可知，S22.4已知f (x)若f (f (1)1，则a_.1因为f (1)lg 10，且3t2dtt3a303a3，所以f (0)0a31，所以a1.5已知f (x)(12t4a)dt，F(a)f (x)3a2dx，求函数F(a)的最小值解因为f (x)(12t4a)dt(6t24at)6x24ax(6a24a2)6x24ax2a2，因为F(a)f (x)3a2dx(6x24axa2)dx(2x32ax2a2x)2132a12a21(a1)211.所以当a1时，F(a)的最小值为1.
