How to solve the limit of #int_0^x t^2e^{-t}dt# when # x -> #infinite?

Question

598 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-17T18:54:58

#lim_(x->oo) int_0^x t^2e^(-t)dt = 2#

Integrate by parts:

#int_0^x t^2e^(-t)dt = int_0^x t^2d(-e^(-t))#

#int_0^x t^2e^(-t)dt =[-t^2e^(-t)]_0^x + 2 int_0^x te^(-t)dt#

#int_0^x t^2e^(-t)dt =-x^2e^(-x) + 2 int_0^x td(-e^(-t))#

#int_0^x t^2e^(-t)dt =-x^2e^(-x) -[2te^(-t)]_0^x + 2 int_0^x e^(-t)dt#

#int_0^x t^2e^(-t)dt =-x^2e^(-x) -2xe^(-x) - 2 [e^(-t)]_0^x#

#int_0^x t^2e^(-t)dt =-x^2e^(-x) -2xe^(-x) - 2 e^(-x) +2#

Now we now that:

#lim_(x->oo) x^n e^(-x) = 0# for any #n>=0#

So:

#lim_(x->oo) int_0^x t^2e^(-t)dt = 2#