PAGE課時素養評價十一微積分基本定理(15分鐘30分)1.計算QUOTEdx= ()A.0B.1C.2D.-2【解析】選B.QUOTEdx=QUOTE(1-x)dx+QUOTE(x-1)dx=QUOTE+QUOTE=QUOTE+QUOTE-QUOTE=1.2.已知定積分QUOTE(kx+1)dx=k,則實數k= ()A.2 B.-2 C.1 D.-1【解析】選A.因為QUOTE(kx+1)dx=k,所以QUOTE=k,所以QUOTEk+1=k,所以k=2.【補償訓練】若QUOTE(2x-3x2)dx=0,則k= ()A.0B.1C.0或1D.不確定【解析】選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,則S1,S2,S3的大小關系為 ()A.S1<S2<S3 B.S2<S1<S3C.S2<S3<S1 D.S3<S2<S1【解析】選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.曲線y=2x2與直線x=1,x=2及y=0所圍成的平面圖形的面積為________.

【解析】依題意,所求面積為S=QUOTE2x2dx=QUOTEx3QUOTE=QUOTE-QUOTE=QUOTE.答案:QUOTE5.計算下列定積分.(1)QUOTEdx.(2)QUOTE(cosx+2x)dx.【解析】(1)因為QUOTEdx=QUOTEdx=[lnx-ln(x+1)]QUOTE=lnQUOTE.(2)QUOTE(cosx+2x)dx=QUOTEQUOTE=2+QUOTE·QUOTE.(30分鐘60分)一、選擇題(每小題5分,共25分)1.QUOTEdx= ()A.e2-2 B.e-1C.e2 D.e+1【解析】選C.QUOTEdx=(x2+lnx)QUOTE=e2.2.曲線y=cosx(0≤x≤2π)與直線y=1所圍成的圖形面積是 ()A.2π B.3π C.QUOTE D.π【解析】選A.如圖所示,S=QUOTE(1-cosx)dx=(x-sinx)QUOTE=2π.【補償訓練】設f(x)=ax2+c(a≠0),若QUOTEf(x)dx=f(x0),0≤x0≤1,則x0的值為()A.-QUOTEB.QUOTEC.-QUOTED.QUOTE【解析】選B.因為QUOTEf(x)dx=QUOTE=QUOTEa+c,所以aQUOTE+c=QUOTEa+c,所以QUOTE=QUOTE,因為0≤x0≤1,所以x0=QUOTE.3.計算QUOTEdx等于 ()A.7 B.QUOTE C.QUOTE D.QUOTE【解析】選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】選D.QUOTEsin2QUOTEdx=QUOTEdx=QUOTE(x-sinx)QUOTE=QUOTE.5.如圖,設D是圖中所示的矩形區域,E是D內函數y=cosx圖象上方的點構成的區域(陰影部分),向D中隨機投一點,則該點落入E中的概率為 ()A.QUOTE B.QUOTE C.QUOTE D.QUOTE【解析】選D.因為QUOTEcosxdx=sinxQUOTE=1,故所求概率為QUOTE=QUOTE.二、填空題(每小題5分,共15分)6.已知t>1,若QUOTE(2x+1)dx=t2,則t=________.

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

【解析】因為x=1>0,所以f(1)=lg1=0.又因為f(x)=x+QUOTE3t2dt=x+a3(x≤0),所以f(0)=a3,所以a3=1,所以a=1.答案:18.等比數列{an}中,a3=9,前3項和為S3=3QUOTEx2dx,則公比q的值是________.

【解析】S3=3QUOTEx2dx=x3QUOTE=27.由題意得QUOTE即QUOTE兩式相除整理得2q2-q-1=0,解得q=1或q=-QUOTE.答案:1或-QUOTE三、解答題(每小題10分,共20分)9.若QUOTEdx=QUOTEdx,求a的值.【解析】因為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.計算定積分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:設f(x)=|2x+3|+|3-2x|=QUOTE如圖,所求積分等于陰影部分面積,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,則實數k的取值范圍為________.

【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)是二次函數,其圖象過點(1,0),且f′(0)=2,QUOTEf(x)dx=0,求f(x)的解析式.【解析】設f(x)=ax2+bx+c(a