How do you find the integral from 0 to 2 of #xe^(2x) dx#?

1 Answer
Jun 6, 2015

#\int_0^2xe^{2x}dx=3/4e^4+1/4#

Use integration by parts

#\int u dv=uv-int vdu#

Let #u=x, \implies du=dx#

Let #dv=e^{2x}dx, \implies v=1/2e^{2x}#

Substitute #v# and #u# into the top expression

#\int_0^2xe^{2x}dx=[x/2e^{2x}]_0^2 -int_0^2 1/2e^{2x}dx#

#\int_0^2xe^{2x}dx=(e^4-0)-[1/4e^{2x}]_0^2#

#\int_0^2xe^{2x}dx=3/4e^4+1/4#