Guided Integration Question

$y' = e^{-x} - xe^{-x}$

That is part I. How can we use that information to to find $xe^{-x}$

Integrate both sides. Let's isolate $xe^{-x}$ on one side of the equation before integrating.

$xe^{-x} = e^{-x} - y'\ int xe^{-x} dx = int e^{-x} dx - int y' dx\ int xe^{-x} dx = -e^{-x} - y+C\ int xe^{-x} dx = -e^{-x} - xe^{-x}+C$

Remember that one way to show that $$int xe^{-x}dx=-xe^{-x}-e^{-x}+c$$ is to show that $$frac{d}{dx}left(-xe^{-x}-e^{-x}+cright) = xe^{-x}.$$ The first step in doing that is knowing how to differentiate $xe^{-x}$, which you presumably already did in Part 1.