Q. Determine the value of f(1).

$\begin{aligned} f(x) + f'(x) & = xe^{-x} & \small \color{#3D99F6} \text{Multiply }e^x \text{ on both sides.} \\ e^xf(x) + e^xf'(x) & = x \\ \frac {d}{dx}e^xf(x) & = x \\ e^x f(x) & = \frac {x^2}2 + \color{#3D99F6} C & \small \color{#3D99F6} \text{where }C \text{ is the constant of integration.} \\ f(x) & = \frac {x^2}{e^x} + \frac C{e^x} & \small \color{#3D99F6} \text{Given that }f(0) = 0 \\ f(0) & = 0 + C = 0 & \small \color{#3D99F6} \implies C = 0 \\ \implies f(x) & = \frac {x^2}{2e^x} \\ f(1) & = \boxed{\dfrac 1{2e}} \end{aligned}$