Question

计算定积分

谢了 大家了

Answer 1

Answer 2

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

Answer 3

∫(1,2)1/(2x-1)dx=1/2ln(2x-1)|x=2-1/2ln(2x-1)|x=1 =1/2ln3

∫(0,1)(2x+k)dx=x^2+kx|x=1-x^2+kx|x=0 =1+k=2 k=1

∫(0,ln2)根号(e^x-1)dx
令根号(e^x-1)=t,则x=ln(t^2+1),dx=2t/(t^2+1)
则∫(0,ln2)根号(e^x-1)dx=∫(0,1)2t^2/(t^2+1)dt
=∫(0,1)[2-2/(t^2+1)]dt
=2t-2arctant|t=1 - 2t-2arctant|t=0
=2-π/2

F'(X)=f(x)
则F'(x+a)=f(x+a)
∫(a,x)F'(t+a)dt
=F(t+a)|t=x-F(t+a)|t=a
=F(x+a)-F(2a)

∫(0,1)xe^x^2dx
=1/2xe^x^2|x=1 -1/2xe^x^2|x=0
=1/2

∫(1,e)|lnx|dx
=xlnx-x|x=e -xlnx-x|x=1
=0+1
1