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Let $I=\int_{0}^{1} x e^{x} d x$

Integrating by parts, we obtain

$\begin{aligned} I &=x \int_{0}^{1} e^{x} d x-\int_{0}^{1}\left\{\left(\frac{d}{d x}(x)\right) \int e^{x} d x\right\} d x \\ &=\left[x e^{x}\right]_{0}^{1}-\int_{0}^{1} e^{x} d x \\ &=\left[x e^{x}\right]_{0}^{1}-\left[e^{x}\right]_{0}^{1} \\ &=e-e+1 \\ &=1 \end{aligned}$

Hence, the given result is proved.