Question

微积分0到正无穷 0.02te^-0.02t dt 帮个忙

Answer 1

作变量代换令 u = 0.02t,则 dt =50du

[0,+∞] ∫0.02te^(-0.02t) dt

= [0,+∞] ∫u*e^(-u) *50du

= 50* [-u*e^(-u) - e^(-u)] |[0,+∞]

=50