Question

高数定积分5个题，

高数定积分5个题，最好有电子版的，，，在线等

Answer 1

(1)
∫(0->π/2) xsin2x dx
=-(1/2)∫(0->π/2) x dcos2x
=-(1/2)[ xcos2x]|(0->π/2) +(1/2)∫(0->π/2) cos2x dx
=-π/2+ (1/2)∫(0->π/2) cos2x dx
=-π/2 +(1/4)[sin2x]|(0->π/2)
=-π/2

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

(3)
∫(0->2π) x.(cosx)^2 dx
=(1/2)∫(0->2π) x.(1+cos2x) dx
=(1/4)[x^2]|(0->2π) +(1/2)∫(0->2π) x.cos2x dx
=π^2 +(1/4)∫(0->2π) x dsin2x
=π^2 +(1/4)[x.sin2x]|(0->2π) -(1/4)∫(0->2π) sin2x dx
=π^2 +(1/8)[cos2x]|(0->2π)
=π^2

(4)
∫(1->e) coslnx dx
let
u=lnx
du = (1/x) dx
dx = e^u du
x=1, u=0
x=e, u=1

∫(1->e) coslnx dx
=∫(0->1) e^u. cosu du
=∫(0->1) cosu de^u
=[cosu.e^u]|(0->1) +∫(0->1) sinu.e^u du
=e.cos1- 1 +∫(0->1) sinu de^u
=e.cos1- 1 + [sinu.e^u]|(0->1) - ∫(0->1) cosu.e^u du
=e.cos1- 1 + sin1.e - ∫(0->1) cosu.e^u du
2∫(0->1) e^u. cosu du = e(sin1+cos1) -1
∫(0->1) e^u. cosu du = (1/2)[e(sin1+cos1) -1]

=> ∫(1->e) coslnx dx =(1/2)[e(sin1+cos1) -1]

(5)
∫(-∞->0) e^(ax) dx
=(1/a) [e^ax]|(-∞->0)
=1/a

Answer 2

1、分部积分
2、x*e^-(x*x) dx = 0.5*e^-(x*x) d(x*x)
3、先降次，再分部积分
4、变量代换 t = lnx => x = e^t
5、dx = (1/a)dax

追问

写一下