Question

求定积分题

1、∫(1,0) (x-1)^2/(√x) dx
2、∫(e,1) lnxdx

求解，需详细过程

Answer 1

1、∫(1,0) (x-1)²/(√x) dx
解：原式=[0,1]∫[(x²-2x+1)/(√x)]dx=[0,1]∫[x^(3/2)-2x^(1/2)+x^(-1/2)]dx
=[(2/5)x^(5/2)-(4/3)x^(3/2)+2√x]︱[0，1]=2/5-4/3+2=16/15。
2、∫(e,1) lnxdx
解：原式=[1，e]∫lnxdx=[1，e][xlnx-∫dx]=(xlnx-x)︱[1，e]=(elne-e)-(1ln1-1)=1

Answer 2

1，设√x=t,则x=t^2代入得
∫(1,0) 2*(t^2-1)^2 dt
=2[1/5t^5-2/3t^3+t](1,0)
=2[1/5-2/3+1]-0
=16/15

2,∫(e,1) lnxdx
=x*lnx - ∫(e,1) xdlnx
=x*lnx - ∫(e,1) dx
=[x*lnx - x](e,1)
=[e-e]-[0-1]
=1

Answer 3

∫(1,0) (x-1)²/(√x) dx
=∫(1,0) (x²-2x+1)/(√x) dx
=∫(1,0) (x√x-2√x+1/√x)dx
=∫(1,0) [x^(3/2)-2x^(1/2)+x^(-1/2)]dx
=[(2/5)x^(5/2)-2*(2/3)x^(3/2)+2x^(1/2)]|(1,0)
=2/5-4/3+2
=16/15

∫ lnxdx
=xlnx-∫ xd(lnx)
=xlnx-∫ dx
=xlnx-x+C
∴∫(e,1) lnxdx
=(xlnx-x)|(e,1)
=(e-e)-(0-1)
=1