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

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Answer 1 · 2015-06-06T05:14:12

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

#\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#