求定积分。高等数学,大一题目

2024-11-08 09:04:37
推荐回答(2个)
回答1:

回答2:

(3)
∫(0->2π/ω) tsinωt dt
=-(1/ω)∫(0->2π/ω) tdcosωt
=-(1/ω)[t.cosωt]|(0->2π/ω) + (1/ω)∫(0->2π/ω) cosωt dt
=2π +(1/ω^2)[sinωt]|(0->2π/ω)
=2π

(5)
∫(1->4) lnx/√x dx
=2∫(1->4) lnx d√x
=2[lnx.√x]|(1->4) -2∫(1->4) dx/√x
=8ln2 - 4[√x]|(1->4)
=8ln2 - 4(2-1)
=8ln2-4