解:(1)原式=1/2∫(1,2)d(2x-1)/(2x-1)
=(1/2ln|2x-1|)|(1,2)
=1/2(ln3-0)
=1/2ln3
(2)∵∫(0,1)(2x+k)dx=(x²+kx)|(0,1)=1+k
又∫(0,1)(2x+k)dx=2
∴1+k=2
故k=1
(3)原式=∫(0,1)2t²dt/(t²+1)
(用t=√(e^x-1)代换)
=2∫(0,1)[1-1/(t²+1)]dt
=2(t-arctant)|(0,1)
=2(1-π/4)
=2-π/2
(4)原式=∫(a,x)f(t+a)d(t+a)
=∫(2a,x+a)f(u)du
(用u=t+a代换)
=∫(2a,x+a)F'(u)du
(∵F'(x)=f(x))
=[F(u)]|(2a,x+a)
=F(x+a)-F(2a)
(5)原式=1/2∫(0,1)e^(x²)d(x²)
=1/2[e^(x²)]|(0,1)
=1/2(e-1)
(6)原式=∫(1,e)lnxdx
=(xlnx)|(1,e)-∫(1,e)dx
(应用分部积分法)
=(e-0)-(e-1)
=1
(7)原式=(-xcosx)|(0,π/2)+∫(0,π/2)cosxdx
(应用分部积分法)
=(0-0)+(sinx)|(0,π/2)
=1-0
=1
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