We can find a simple for f ( x ) {\displaystyle f(x)} by integration by parts. Let u = t , d u = d t {\displaystyle u=t,\;du=dt} and d v = e − t d t , v = − e − t . {\displaystyle dv=e^{-t}dt,\;v=-e^{-t}.} Then we have
f ( x ) = ∫ 0 x t e − t d t = [ t ( − e − t ) ] 0 x − ∫ 0 x ( − e − t ) d t = − x e − x + [ − e − t ] 0 x = − x e − x − [ e − x − 1 ] = 1 − ( x + 1 ) e − x {\displaystyle {\begin{aligned}f(x)&=\int _{0}^{x}te^{-t}dt\\&=\left[t(-e^{-t})\right]_{0}^{x}-\int _{0}^{x}(-e^{-t})dt\\&=-xe^{-x}+\left[-e^{-t}\right]_{0}^{x}\\&=-xe^{-x}-\left[e^{-x}-1\right]\\&=1-(x+1)e^{-x}\end{aligned}}}
![{\displaystyle {\begin{aligned}f(x)&=\int _{0}^{x}te^{-t}dt\\&=\left[t(-e^{-t})\right]_{0}^{x}-\int _{0}^{x}(-e^{-t})dt\\&=-xe^{-x}+\left[-e^{-t}\right]_{0}^{x}\\&=-xe^{-x}-\left[e^{-x}-1\right]\\&=1-(x+1)e^{-x}\end{aligned}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/34a27639133f9c51b76110f40f0629d1acc61b61)