2019-2020学年高二数学人教A版选修2-2训练：1.6 微积分基本定理 Word版含解析.doc
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.01 xdxB.01 (x+1)dxC.01 1dxD.01 12dx解析:01 1dx=x|01=1-0=1,故选C.答案:C2.24 1xdx的值为()A.-2ln 2B.2ln 2C.-ln 2D.ln 2解析:24 1xdx=ln x|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0T x2dx=9,则常数T的值为. 解析:0T x2dx=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,1<x2,则02 f(x)dx=. 解析:02 f(x)dx=01 x2dx+12 (2-x)dx=x33|01+2x-x22|12=13+2×2-222-2-12=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 |cos x|dx=-2 (-cos x)dx+-22 cos xdx+234 (-cos x)dx=1+2+1-22=4-22.答案:4-228.已知函数f(x)=ax2+c(a0),若01 f(x)dx=f(x0),0x01,则x0的值为. 解析:01 f(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+3×2-(02+3×0)=10.(2)因为2x2-x33'=4x-x2,所以-13 (4x-x2)dx=2x2-x33|-13=2×32-333-2×(-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=12 x2dx,S2=12 1xdx,S3=12 exdx,则S1,S2,S3的大小关系为()A.S1<S2<S3B.S2<S1<S3C.S2<S3<S1D.S3<S2<S1解析:S1=12 x2dx=13x3|12=13×23-13×13=73,S2=12 1xdx=ln x|12=ln 2-ln 1=ln 2<ln e=1,S3=12 exdx=ex|12=e2-e=e(e-1)>e>73,所以S2<S1<S3,故选B.答案:B2.05 (ex-sin x)dx的值为()A.e5-1B.e5-2C.e5-3D.e5-4解析:05 (ex-sin x)dx=05 exdx-05 sin xdx=ex|05+cos x|05=e5-e0+cos 5-cos 0=e5-1-1-1=e5-3.答案:C3.若f(x)=-ex,x>1,|x|,x1(e为自然对数的底数),则02 f(x)dx=()A.12+e2-eB.12+eC.12+e-e2D.-12+e-e2解析:02 f(x)dx=01 |x|dx+12 (-ex)dx=01 xdx-12 exdx=12x2|01-ex|12=12+e-e2.答案:C4.若f(x)是一次函数,且01 f(x)dx=5,01 xf(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01 f(x)dx=01 (ax+b)dx=01 axdx+01 bdx=12a+b=5,01 xf(x)dx=01 x(ax+b)dx=01 (ax2)dx+01 bxdx=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+201 f(x)dx,则01 f(x)dx=()A.-1B.-13C.13D.1解析:设01 f(x)dx=t,则f(x)=x2+2t,因此01 f(x)dx=01 (x2+2t)dx=13x3+2tx|01=13+2t,即t=13+2t,解得t=-13,即01 f(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)-aa x2dx(a>0);(3)12 1x(x+1)dx;(4)-23 (|2x-4|+|x+1|)dx.解:(1)3173 (2sin x-3cos x)dx=23173 sin xdx-33173 cos xdx=2(-cos x)|3173-3sin x|3173=2-cos 173+cos 3-3sin 173-sin 3=2-12+12-3-32-32=33.(2)由x2=x,x0,-x,x<0,得-aa x2dx=0a xdx+-a0 (-x)dx=12x2|0a-12x2|-a0=a2.(3)f(x)=1x(x+1)=1x-1x+1,取F(x)=ln x-ln(x+1)=ln xx+1,则F'(x)=1x-1x+1.所以12 1x(x+1)dx=12 1x-1x+1dx=ln xx+1|12=ln 43.(4)令f(x)=|2x-4|+|x+1|=3-3x,x-1,5-x,-1<x<2,3x-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k3 f(x)dx=403.解:分2<k<3和-2k2两种情况讨论:当2<k<3时,k3 f(x)dx=k3 (1+x2)dx=x+x33|k3=(3+9)-k+k33=403.整理,得k3+3k+4=0,即k3+k2-k2+3k+4=0.(k+1)(k2-k+4)=0.k2-k+4>0,k=-1.又2<k<3,k=-1(舍去).当-2k2时,k3 f(x)dx=k2 (2x+1)dx+23 (1+x2)dx=(x2+x)|k2+x+x33|23=(4+2)-(k2+k)+(3+9)-2+83=403-(k2+k)=403,k2+k=0,解得k=0或k=-1,满足条件.综上所述,当k=0或k=-1时,k3 f(x)dx=403.
