# Use integration by parts to find the given integral: `int _0^1x/e^(2x) dx`Can you show it in the simplest form possible please.

### 1 Answer | Add Yours

`int_0^1 x/e^(2x) dx`

To solve using integration by parts, we need to use the formula `intudv=uv-intvdu ` .

So let,

`u=x` and `dv = 1/e^(2x)dx`

`du=1dx` `v = int dx/e^(2x) = int e^(-2x)dx`

`du= dx` `v = e^(-2x)/(-2) = -e^(-2x)/2`

Substitute these to the formula.

`int x/e^(2x)dx = x(-e^(-2x)/2)-int-e^(-2x)/2dx = -(xe^(-2x))/2+1/2inte^(-2x)dx`

`= -(xe^(-2x))/2-1/4e^(-2x) = -x/(2e^(2x)) - 1/(4e^(2x)) =-(2x+1)/(4e^(2x))`

Then, evaluate the limits of the integral.

`int_0^1 x/e^(2x)=-(2x+1)/(4e^(2x))` `|_0^1` `= [-(2*1+1)/(4e^(2*1))]- [-(2*0+1)/(4e^(2*0))]=-3/(4e^2)+1/4= 1/4 - 3/(4e^2)`

**Hence, `int_0^1 x/e^(2x)dx = 1/4 - 3/(4e^2)` .**