Chapter 8 - Techniques of Integration - 8.1 Integration by Parts - Exercises - Page 395: 51

$$-\frac{2}{e}+1$$

Given $$\int_{0}^{1} x e^{-x} d x$$ Let \begin{align*} u&= x\ \ \ \ \ \ \ \ dv= e^{-x}dx\\ du&= dx\ \ \ \ \ \ \ \ v=-e^{-x} \end{align*} Then \begin{align*} \int_{0}^{1} x e^{-x} d x&= -xe^{-x} \bigg|_{0}^{1} +\int_{0}^{1} e^{-x} d x \\ &= -xe^{-x} -e^{-x}\bigg|_{0}^{1} \\ &= -\frac{2}{e}+1 \end{align*}