2024-2025学年高中数学第一章导数及其应用1.6微积分基本定理课时素养评价含解析新人教A版选修2-2

PAGE课时素养评价十一微积分基本定理(15分钟30分)1.计算QUOTEdx= ()A.0B.1C.2D.-2【解析】选B.QUOTEdx=QUOTE(1-x)dx+QUOTE(x-1)dx=QUOTE+QUOTE=QUOTE+QUOTE-QUOTE=1.2.已知定积分QUOTE(kx+1)dx=k,则实数k= ()A.2 B.-2 C.1 D.-1【解析】选A.因为QUOTE(kx+1)dx=k,所以QUOTE=k,所以QUOTEk+1=k,所以k=2.【补偿训练】若QUOTE(2x-3x2)dx=0,则k= ()A.0B.1C.0或1D.不确定【解析】选B.QUOTE(2x-3x2)dx=(x2-x3)QUOTE=k2-k3=0,解得k=0(舍去)或k=1.3.若S1=QUOTEx2dx,S2=QUOTEdx,S3=QUOTEexdx,则S1,S2,S3的大小关系为 ()A.S1<S2<S3 B.S2<S1<S3C.S2<S3<S1 D.S3<S2<S1【解析】选B.S1=QUOTEx2dx=QUOTEx3QUOTE=QUOTE,S2=QUOTEdx=lnxQUOTE=ln2<1,S3=QUOTEexdx=exQUOTE=e2-e=e(e-1)>QUOTE,所以S2<S1<S3.4.曲线y=2x2与直线x=1,x=2及y=0所围成的平面图形的面积为________.

【解析】依题意,所求面积为S=QUOTE2x2dx=QUOTEx3QUOTE=QUOTE-QUOTE=QUOTE.答案:QUOTE5.计算下列定积分.(1)QUOTEdx.(2)QUOTE(cosx+2x)dx.【解析】(1)因为QUOTEdx=QUOTEdx=[lnx-ln(x+1)]QUOTE=lnQUOTE.(2)QUOTE(cosx+2x)dx=QUOTEQUOTE=2+QUOTE·QUOTE.(30分钟60分)一、选择题(每小题5分,共25分)1.QUOTEdx= ()A.e2-2 B.e-1C.e2 D.e+1【解析】选C.QUOTEdx=(x2+lnx)QUOTE=e2.2.曲线y=cosx(0≤x≤2π)与直线y=1所围成的图形面积是 ()A.2π B.3π C.QUOTE D.π【解析】选A.如图所示,S=QUOTE(1-cosx)dx=(x-sinx)QUOTE=2π.【补偿训练】设f(x)=ax2+c(a≠0),若QUOTEf(x)dx=f(x0),0≤x0≤1,则x0的值为()A.-QUOTEB.QUOTEC.-QUOTED.QUOTE【解析】选B.因为QUOTEf(x)dx=QUOTE=QUOTEa+c,所以aQUOTE+c=QUOTEa+c,所以QUOTE=QUOTE,因为0≤x0≤1,所以x0=QUOTE.3.计算QUOTEdx等于 ()A.7 B.QUOTE C.QUOTE D.QUOTE【解析】选C.QUOTE|x2-4|dx=QUOTE(4-x2)dx+QUOTE(x2-4)dx=QUOTE+QUOTE=QUOTE.4.QUOTEsin2QUOTEdx等于 ()A.QUOTE B.QUOTE-1 C.2 D.QUOTE【解析】选D.QUOTEsin2QUOTEdx=QUOTEdx=QUOTE(x-sinx)QUOTE=QUOTE.5.如图,设D是图中所示的矩形区域,E是D内函数y=cosx图象上方的点构成的区域(阴影部分),向D中随机投一点,则该点落入E中的概率为 ()A.QUOTE B.QUOTE C.QUOTE D.QUOTE【解析】选D.因为QUOTEcosxdx=sinxQUOTE=1,故所求概率为QUOTE=QUOTE.二、填空题(每小题5分,共15分)6.已知t>1,若QUOTE(2x+1)dx=t2,则t=________.

【解析】QUOTE=t2+t-2,从而得方程t2+t-2=t2,解得t=2.答案:27.设f(x)=QUOTE若f(f(1))=1,则a=________.

【解析】因为x=1>0,所以f(1)=lg1=0.又因为f(x)=x+QUOTE3t2dt=x+a3(x≤0),所以f(0)=a3,所以a3=1,所以a=1.答案:18.等比数列{an}中,a3=9,前3项和为S3=3QUOTEx2dx,则公比q的值是________.

【解析】S3=3QUOTEx2dx=x3QUOTE=27.由题意得QUOTE即QUOTE两式相除整理得2q2-q-1=0,解得q=1或q=-QUOTE.答案:1或-QUOTE三、解答题(每小题10分,共20分)9.若QUOTEdx=QUOTEdx,求a的值.【解析】因为QUOTE′=QUOTE,QUOTE′=-QUOTE,由QUOTEdx=QUOTEdx,得QUOTEQUOTE=QUOTEQUOTE,所以QUOTE-QUOTE=QUOTE-1,所以QUOTE=QUOTE,QUOTE=QUOTE,所以a=QUOTE=QUOTE.故a=QUOTE.10.计算定积分QUOTE(|2x+3|+|3-2x|)dx.【解析】方法一:令2x+3=0,解得x=-QUOTE;令3-2x=0,解得x=QUOTE.QUOTE(|2x+3|+|3-2x|)dx=QUOTE(-2x-3+3-2x)dx+QUOTE(2x+3+3-2x)dx+QUOTE(2x+3-3+2x)dx=QUOTE(-4x)dx+QUOTE6dx+QUOTE4xdx=-4·QUOTE+6xQUOTE+4·QUOTE=45.方法二:设f(x)=|2x+3|+|3-2x|=QUOTE如图,所求积分等于阴影部分面积,即QUOTE(|2x+3|+|3-2x|)dx=S=2×QUOTE×(6+12)×QUOTE+3×6=45.1.已知2≤QUOTE(kx+1)dx≤4,则实数k的取值范围为________.

【解析】QUOTE(kx+1)dx=QUOTE=(2k+2)-QUOTE=QUOTEk+1,所以2≤QUOTEk+1≤4,解得QUOTE≤k≤2.答案:QUOTE2.已知f(x)是二次函数,其图象过点(1,0),且f′(0)=2,QUOTEf(x)dx=0,求f(x)的解析式.【解析】设f(x)=ax2+bx+c(a