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

1 Answer
Nov 29, 2017

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.