PAGEPAGE61.6微積分基本定理課時過關·實力提升基礎鞏固1.下列定積分的值等于1的是()A.01xdx BC.011dx解析:011dx=x|01答案:C2.241xdxA.-2ln2 B.2ln2C.-ln2 D.ln2解析:241xdx=答案:D3.24(x3+x2-30)dx的值為(A.56 B.28 C.14 D.解析:24(x3+x2-30)dx=14x4+13x3-30x|24=14(答案:D4.若0Tx2dx=9,則常數T解析:∵0Tx2d答案:35.若01(2xk+1)dx=2,則k=解析:01(2xk+1)dx=2k+1答案:16.若f(x)=x2,0≤x≤1解析:02f(x)dx=01x2答案:57.如圖,曲線y=cosx與直線x=-π,x=3π4,y=0所圍成圖形的面積S=解析:S=-π3π4|cosx|dx=-π-π2(-cosx)dx+-答案:48.★已知函數f(x)=ax2+c(a≠0),若01f(x)dx=f(x0),0≤x0≤1,則x0的值為解析:01f(x)dx=01=∵0≤x0≤1,∴x0答案:39.計算下列定積分:(1)02(2x+(2)-13(4x-x(3)12(x-1)5解:(1)因為(x2+3x)'=2x+3,所以02(2x+3)dx=(x2+3x)|02=22+3×2-(02+(2)因為2x2所以-13(4x-x=(3)因為16(x所以12(x-1)5dx==16×(2-1)實力提升1.若S1=12x2dx,S2=121xdx,S3=A.S1<S2<S3 B.S2<S1<S3C.S2<S3<S1 D.S3<S2<S1解析:S1=12x2dx=13x3|12=13×23-13×1所以S2<S1<S3,故選B.答案:B2.05π(ex-sinx)dx的值為(A.e5π-1 B.e5π-2 C.e5π-3 D.e5π-4解析:05π(ex-sinx)dx=05πe=ex|05π+cosx|=e5π-1-1-1=e5π-3.答案:C3.若f(x)=-ex,x>1,A.12+eC.12+e-e2 D解析:02f(x)dx=01|x|d=01xdx-12ex答案:C4.若f(x)是一次函數,且01f(x)dx=5,01xf(x)dxA.4x+3 B.3x+4C.-4x+2 D.-3x+4解析:設f(x)=ax+b(a≠0),則01f(x)dx=01axd01xf(x)dx==01(ax2)dx由解得故f(x)=4x+3.答案:A5.若f(x)=x2+201f(x)dx,則01fA.-1 B.C.D解析:設01f(x)dx=t,則f(x)=x2因此01f(x)dx=01(x2+2t)dx=13x3+2tx|答案:B6.若01(2ax2-a2x)dx=16,則實數解析:01(2ax2-a2x)dx=2a答案:137.計算定積分:(1)π317π3(2sin(2)-aax(3)12(4)-23(|2x-4|+|x+1解:(1)π317π3(=2π317π3sinx=2(-cosx)=2=2(2)由得-aax2dx=1(3)f(x)取F(x)=lnx-ln(x+1)=lnx則F'(x)所以1=1=lnx(4)令f(x)=|2x-4|+|x+1|=則-23(|2x-4|+|x+1|)dx=-2-1(3-3x)dx+=8.★已知f(x)=2x+1,x∈[-解:分2<k<3和-2≤k≤2兩種狀況探討:當2<k<3時,k3f(x)dx=k3(1+x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).當-2≤k≤2時,k3f(x)dx=k2