Question

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

Answer 1

`\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`