How do you integrate #int xe^(-x/2)# by parts from #[0,4]#?

Question

2323 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-22T01:38:13

#-4(3e^-2-1)#

Let's select our values of #u, dv# and integrate and differentiate for #du, v#:

#u=x#
#du=dx#

#dv=e^(-x/2)dx#
#v=-2e^(-x/2)#

Note that our bounds of integration will not change -- we're not performing any substitutions.

So, for definite integrals, we integrate by parts in the following way:

#int_a^budv=uv|_a^b-int_a^bvdu#

Thus,

#int_0^4xe^(-x/2)dx=-2xe^(-x/2)|_0^4 +2int_0^4e^-x/2dx=(-8e^-2)-4e^-(x/2)|0^4=-8e^-2-4e^-2+4=-12e^-2+4=-4(3e^-2-1)#