Question

What is `F(x) = int x-xe^(-x) dx` if `F(0) = 2`?

Answer 1

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

Explanation:

`F(x) = intx - xe^-xdx = 1/2x^2 - intxe^-xdx`
Finding `intxe^-xdx` is tricky, we'll have to use integration of parts.
`intxe^-xdx = -xe^-x - e^-x + C`
Therefore, `F(x) = 1/2x^2 +xe^-x + e^-x + C`.
Now we have to solve for `C`.
`F(0) = 1/2 * 0^2 + 0e^0 + e^0 + C`
`=1+C = 2` (since `F(0) = 2`)
Therefore, `C = 1`.
`F(x) = 1/2x^2 +xe^-x + e^-x + 1`.