2020版高中数学第一章导数及其应用1.6微积分基本定理练习（含解析）新人教A版.docx

1.6微积分基本定理课时过关能力提升基础巩固1.下列定积分的值等于1的是()A.01xdxB.01(x+1)dxC.011dxD.0112dx解析:011dx=x|01=1-0=1,故选C.答案:C2.241xdx的值为()A.-2ln 2B.2ln 2C.-ln 2D.ln 2解析:241xdx=lnx|24=ln 2.答案:D3.24(x3+x2-30)dx的值为()A.56B.28C.14D.563解析:24(x3+x2-30)dx=14x4+13x3-30x|24=14(44-24)+13(43-23)-30(4-2)=563.故选D.答案:D4.若0Tx2dx=9,则常数T的值为.解析:0Tx2dx=x33|0T=T33=9,T=3.答案:35.若01(2xk+1)dx=2,则k=.解析:01(2xk+1)dx=2k+1xk+1+x|01=2k+1+1=2,解得k=1.答案:16.若f(x)=x2,0x1,2-x,1x2,则02f(x)dx=.解析:02f(x)dx=01x2dx+12(2-x)dx=x33|01+2x-x22|12=13+22-222-2-12=56.答案:567.如图,曲线y=cos x与直线x=-,x=34,y=0所围成图形的面积S=.解析:S=-34|cosx|dx=-2(-cosx)dx+-22cosxdx+234(-cosx)dx=1+2+1-22=4-22.答案:4-228.已知函数f(x)=ax2+c(a0),若01f(x)dx=f(x0),0x01,则x0的值为.解析:01f(x)dx=01(ax2+c)dx=13ax3+cx|01=a3+c=ax02+c.0x01,x0=33.答案:339.计算下列定积分:(1)02(2x+3)dx;(2)-13(4x-x2)dx;(3)12(x-1)5dx.解:(1)因为(x2+3x)=2x+3,所以02(2x+3)dx=(x2+3x)|02=22+32-(02+30)=10.(2)因为2x2-x33=4x-x2,所以-13(4x-x2)dx=2x2-x33|-13=232-333-2(-1)2-(-1)33=203.(3)因为16(x-1)6=(x-1)5,所以12(x-1)5dx=16(x-1)6|12=16(2-1)6-16(1-1)6=16.能力提升1.若S1=12x2dx,S2=121xdx,S3=12exdx,则S1,S2,S3的大小关系为()A.S1S2S3B.S2S1S3C.S2S3S1D.S3S2S1解析:S1=12x2dx=13x3|12=1323-1313=73,S2=121xdx=lnx|12=ln 2-ln1=ln2e73,所以S2S11,|x|,x1(e为自然对数的底数),则02f(x)dx=()A.12+e2-eB.12+eC.12+e-e2D.-12+e-e2解析:02f(x)dx=01|x|dx+12(-ex)dx=01xdx-12exdx=12x2|01-ex|12=12+e-e2.答案:C4.若f(x)是一次函数,且01f(x)dx=5,01xf(x)dx=176,则f(x)为()A.4x+3B.3x+4C.-4x+2D.-3x+4解析:设f(x)=ax+b(a0),则01f(x)dx=01(ax+b)dx=01axdx+01bdx=12a+b=5,01xf(x)dx=01x(ax+b)dx=01(ax2)dx+01bxdx=13a+12b=176.由12a+b=5,13a+12b=176,解得a=4,b=3,故f(x)=4x+3.答案:A5.若f(x)=x2+201f(x)dx,则01f(x)dx=()A.-1B.-13C.13D.1解析:设01f(x)dx=t,则f(x)=x2+2t,因此01f(x)dx=01(x2+2t)dx=13x3+2tx|01=13+2t,即t=13+2t,解得t=-13,即01f(x)dx=-13.答案:B6.若01(2ax2-a2x)dx=16,则实数a的值是.解析:01(2ax2-a2x)dx=2a3x3-a22x2|01=2a3-12a2=16,解得a=1或a=13.答案:13或17.计算定积分:(1)3173(2sin x-3cos x)dx;(2)-aax2dx(a0);(3)121x(x+1)dx;(4)-23(|2x-4|+|x+1|)dx.解:(1)3173(2sinx-3cosx)dx=23173sinxdx-33173cosxdx=2(-cosx)|3173-3sin x|3173=2-cos173+cos3-3sin173-sin3=2-12+12-3-32-32=33.(2)由x2=x,x0,-x,x0,得-aax2dx=0axdx+-a0(-x)dx=12x2|0a-12x2|-a0=a2.(3)f(x)=1x(x+1)=1x-1x+1,取F(x)=lnx-ln(x+1)=lnxx+1,则F(x)=1x-1x+1.所以121x(x+1)dx=121x-1x+1dx=lnxx+1|12=ln43.(4)令f(x)=|2x-4|+|x+1|=3-3x,x-1,5-x,-1x2,3x-3,x2.则-23(|2x-4|+|x+1|)dx=-2-1(3-3x)dx+-12(5-x)dx+23(3x-3)dx=3x-32x2|-2-1+5x-12x2|-12+32x2-3x|23=512.8.已知f(x)=2x+1,x-2,2,1+x2,x(2,4,求k的值使k3f(x)dx=403.解:分2k3和-2k2两种情况讨论:当2k0,k=-1.又2k3,k=-1(舍去).当-2k2时,k3f(x)dx=k2(2x+1