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.-2ln2B.2ln2C.-ln2D.ln2解析:∫24❑1xdx=lnx|24=ln2.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:315.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¿={x2,0≤x≤1,2-x,1e>73,所以S21,|x|,x≤1¿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(a≠0),4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+2∫01❑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:B56.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¿∫π317π3❑¿2sinx-3cosx)dx;(2¿∫-aa❑√x2dx(a>0);(3¿∫12❑1x(x+1)dx;(4¿∫-23❑¿|2x-4|+|x+1|)dx.解:(1¿∫π317π3❑¿2sinx-3cosx)dx=2∫π317π3❑sinxdx-3∫π317π3❑cosxdx=2(-cosx¿|π317π3−3sinx|π317π3=2(-cos17π3+cosπ3)−3(sin17π3-sinπ3)=2(-12+12)−3(-√32-√32)=3√3.(2)由√x2={x,x≥0,-x,x<0,得∫-aa❑√x2dx¿∫0a❑xdx+∫-a0❑¿-x)dx6¿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∫12❑1x(x+1)dx¿∫12❑(1x-1x+1)dx=lnxx+1|12=ln43.(4)令f(x)=|2x-4|+|x+1|¿{3-3x,x≤-1,5-x,-10,∴k=-1.又2
