Question

定积分问题

Answer 1

追答

Answer 2

(3)
∫(0->ln2) √(e^x -1) dx
let
e^x = (secu)^2
e^x dx = 2(secu)^2 . tanu du
dx = 2tanu du
x=0, u=0
x=ln2 , u= π/4

∫(0->ln2) √(e^x -1) dx
=2∫(0->π/4) (tanu)^2 du
=2[secu]|(0->π/4)
=2( 2-1)
=2

(4)
∫(0->1) xe^(2x) dx
=(1/2)∫(0->1) xde^(2x)
=(1/2) [x.e^(2x)]|(0->1) -(1/2)∫(0->1) e^(2x) dx
=(1/2)e^2 - (1/4)[e^(2x)]|(0->1)
=(1/2)e^2 - (1/4)( e^2 - 1)
=1/4 + (1/4)e^2

Answer 3

3、令t=√(e^x-1)，则x=ln(t^2+1)，dx=2t/(t^2+1)dt
原式=∫(0,1)2t^2/(t^2+1)dt
=∫(0,1)[2-2/(t^2+1)]dt
=(2t-2arctant)|(0,1)
=2-π/2
4、原式=(1/2)*∫(0,1)xd(e^2x)
=(1/2)*xe^2x|(0,1)-(1/2)*∫(0,1)e^2xdx
=(1/2)*e^2-(1/4)*e^2x|(0,1)
=(1/2)*e^2-(1/4)*e^2+1/4
=(1/4)*(e^2+1)